The Physics

THE PHYSICS COMPANION

A C Fischer-Cripps

Institute of Physics PublishingBristol and Philadelphia

© IOP Publishing Ltd 2003

c© IOP Publishing Ltd 2003

All rights reserved. No part of this publication may be reproduced, storedin a retrieval system or transmitted in any form or by any means, electronic,mechanical, photocopying, recording or otherwise, without the prior permissionof the publisher. Multiple copying is permitted in accordance with the termsof licences issued by the Copyright Licensing Agency under the terms of itsagreement with Universities UK (UUK).

British Library Cataloguing-in-Publication Data

A catalogue record for this book is available from the British Library.

ISBN 0 7503 0953 9

Library of Congress Cataloging-in-Publication Data are available

Commissioning Editor: Tom SpicerProduction Editor: Simon LaurensonProduction Control: Sarah PlentyCover Design: Frederique SwistMarketing: Nicola Newey and Verity Cooke

Published by Institute of Physics Publishing, wholly owned by The Institute ofPhysics, London

Institute of Physics Publishing, Dirac House, Temple Back, Bristol BS1 6BE,UK

US Office: Institute of Physics Publishing, The Public Ledger Building, Suite929, 150 South Independence Mall West, Philadelphia, PA 19106, USA

Printed in the UK by MPG Books Ltd, Bodmin, Cornwall


This book is dedicated to Rod Cameron, OSA, myformer high school science teacher, who, when I asked

him “Why do we study these things?” replied “To knowthe truth”, thus beginning my career as a physicist.


Preface1. Thermal Physics1.1 Temperature 21.2 Heat and solids 71.3 Heat transfer 131.4 Gases 271.5 Work and thermodynamics 331.6 Gas processes 381.7 Kinetic theory of gases 431.8 Heat engines 511.9 Entropy 592. Waves and Optics2.1 Periodic motion 702.2 Waves 752.3 Superposition 862.4 Light 1042.5 Mirrors 1112.6 Lenses 1162.7 Optical instruments 1292.8 Interference 1352.9 Diffraction 1422.10 Polarisation 1513. Electricity3.1 Electricity 1583.2 Magnetism 1773.3 Induction 1913.4 Magnetic circuits 2003.5 R-C & R-L circuits 2063.6 AC circuits 2153.7 Electromagnetic waves 2314. Mechanics4.1Scalars and vectors 2444.2 Statics 2524.3 Moment of inertia 2594.4 Linear motion 2654.5 Forces 2714.6 Rotational motion 2814.7 Rotation 2864.8 Work and energy 2964.9 Impulse and momentum 3055. Properties of Matter5.1 Fluids 3155.2 Solids 3275.3 Matter 3385.4 Universe 352Materials data 372


When I was a physics student, I found things hard going. Myphysics text book and my professor assumed that I knew fartoo much, and as a result, I did not do very well inexaminations despite my personal interest in the subject.Later, when it was my turn to be in front of the class, Idecided to write up my lecture notes in such a way so as toprovide clear and succinct accounts of the wide variety oftopics to be found in first year university physics. This book isthe result, and I hope that you will find it helpful for yourunderstanding of what is an extraordinary subject.

In writing this book, I was assisted and encouraged by JimFranklin, Hillary Goldsmith, Suzanne Hogg, Walter Kalceff,Les Kirkup, John O’Connor, Joe Wolfe, Tom von Foerster,my colleagues at the University of Technology, Sydney, andall my former students.

Special thanks are due to Dr. Robert Cheary and Prof.Richard Collins. I hope that through this book you will in turnbenefit from whatever I have been able to transmit of theirprofessional and enthusiastic approach to physics that was myprivilege to experience as their student. My sincere thanks tomy wife and family for their unending encouragement andsupport. Finally, I thank Tom Spicer and the editorial andproduction team at Institute of Physics Publishing for theirvery professional and helpful approach to the wholepublication process.

Tony Fischer-Cripps,Killarney Heights, Australia, 2003

Preface


ThermalPhysics

Part 1


1.1 Temperature

Summary

Zeroth law of thermodynamics:

Two systems are in thermal equilibrium when they are atthe same temperature.

Absolute zero: 0K −273.15 °CIce point: 273.15K 0 °CTriple point: 273.16K 0.01 °CBoiling point 373.15K 100 °C

Temperature scales:

When two systems are in thermal equilibrium with a third,they are also in thermal equilibrium with each other.

Thermal equilibrium:

The Physics Companion2

( )18010032FC oo

−=


BoundarySurroundings

Ener

gy &

Mat

ter t

rans

fers

Everythinginside thedotted lineis included

in the“system”

Describing a thermodynamic system:

an isolated portion of space orquantity of matter.

separated from the surroundingsby a boundary.

analysed by consideringtransfers of energy and matteracross the boundary between thesystem and the surroundings.

Volume Pressure Temperature Mass

A thermodynamic system is:

macroscopic quantities microscopic quantities

Kinetic energy ofmolecules.

1.1.1 Thermodynamic systems

Heat flows from the system at hightemperature to the system at lowtemperature.

SystemB

SystemA

At this condition, System A isnow in thermal equilibriumwith System B.

System A SystemB

Both systems eventually reachthe same temperature.

Consider two separate thermodynamic systems A and B which are initiallyat different temperatures and are now brought into contact.

3Thermal Physics

T2T1 T

T

When two systems are separately in thermal equilibrium with a third, theyare also in thermal equilibrium with each other. This statement is known asthe zeroth law of thermodynamics.


What is temperature? The microscopic answer is that temperature is ameasure of the total kinetic energy of the molecules in a system.How can it be measured?• Our senses (touch, sight)• With a thermometer

1.1.2 Temperature

Celsius temperature scale:defined such that0 oC = ice point of water100 oC = boiling point of water.

Fahrenheit temperature scale:defined such that32 oF = ice point of water212 oF = boiling point of water

18010032FC oo

−=

The measurement of temperature is naturally associated with thedefinition of a temperature scale.

The International Temperature Scale is based on the definition ofa number of basic fixed points. The fixed points cover the range oftemperatures to be normally found in industrial processes. The mostcommonly used fixed points are:

Note: Standard atmospheric pressure(1 atm) is defined as 760 mm Hg(ρ = 13.5951 g cm−3) at g = 9.80665 ms−2.

1. Temperature of equilibrium between ice and air-saturatedwater at normal atmospheric pressure (ice point) is: 0.000 oC

2. Temperature of equilibrium between liquid water and its vapourat a pressure of 1 atm pressure (steam point) is: 100.000 oC

The ice and steam points are convenient fixed points. A morereproducible fixed point is the triple point of water.

Vapour

Ice

Water

The state of pure water existing asan equilibrium mixture of ice, liquidand vapour. Let the temperature ofwater at its triple point be equal to273.16K. This assignmentcorresponds to a an ice point of273.15K or 0oC. Triple point is usedas the standard fixed pointbecause it is reproducible.

4 The Physics Companion


Gas in chamber is at low pressure(so gas is “ideal”)

Gas must not condense into a liquid(Helium is a good gas to use forlow temperatures)

Pressure of the gas is an indicationof temperature

Must maintain constant gas volume

1.1.3 Kelvin temperature scale

kPa

1. Pressure readings for ice point and boiling point of water are recorded.2. Divide these readings into 100 divisions and call the ice point 0 and the

boiling point 100 degrees. This is the Celsius scale.

273.15 K

Pressure

0K

Boilingpoint ofwater

Icepoint ofwater

373.15 K

100divisions

K

oC

3. But, it probably makes more sense to let zero on the temperature scalebe that at zero pressure. So, we can extrapolate back to zero P and callthis zero temperature. As shown later, (see page 27) this works if thevolume is kept constant.

A constant volume gas thermometer (CVGT) is a specialthermometer that gives a reading of pressure as an indicationof temperature.

• Molecules occupy avery small volume

• Molecules are atrelatively largedistances from eachother

• Collisions betweenmolecules are elastic

The ice point of water then turnsout to be 273.15 divisions fromzero and the boiling point 373.15divisions from zero. This newtemperature scale is calledKelvin and is the official SIunit of temperature.

0 K is 273.16divisions belowtriple point. Thus,273.16 K = triplepoint of waterwhich is 0.01 oC.

Thus, 0K = −273.15 oCwhich is sometimescalled absolute zero.

The CVGT can be used to define a temperature scale:

5Thermal Physics

Ideal gas


1.1.4 Examples

1. Give a short description of the following states ofequilibrium:(a) Chemical equilibrium.(b) Mechanical equilibrium.(c) Thermal equilibrium.

(a) Chemical equilibriumSolution:

(b) Mechanical equilibrium

(c) Thermal equilibrium

No chemical reactions are taking place inside the system boundary.

Any external forces on the system boundary are in balance. Thesystem is not accelerating or decelerating.

No heat flow is taking place within the system or across the systemboundary.

2. Can a system be in a state of thermal equilibrium if the temperature ofthe system is different to that of the surroundings?

Yes, but only if the system boundary is perfectly insulating. The conditionfor thermal equilibrium is that there is no heat flow within the system andacross the system boundary. For the most part, this means no temperaturegradients anywhere within the system and surroundings. If the systemboundary is perfectly insulating, then the system and the surroundings canbe at different temperatures and the system still be in a state of thermalequilibrium even though the system and the surroundings are not in thermalequilibrium with each other.

Solution:



1.2 Heat and solids

Summary

TLL o∆α=∆

TA2A o∆α=∆

TVV o∆β=∆

Thermal expansion

( )12 TTmcQ −=∆ Specific heat

( )12 TTnCQ −=∆

cMC m=

fmLQ =

vmLQ =

Latent heat

Energy exists in many forms. There is a flow, or transfer, of energy when a change of

form takes place. Heat and work are words which refer to the amount of

energy in transit from one place to another. Heat and work cannot be stored. Heat and work refer to

energy in transit.

7Thermal Physics


When a solid is heated, there is an increase inthe linear dimensions as the temperature rises.

Coefficient oflinear expansion

Volume

∆LLoLength

Area

Both hole and objectchange size. The holebehaves as if filled withobject’s material.

Coefficient ofvolume expansionfor solids, β ≈ 3α

VoVo+∆V

Ao

To explain the phenomenon ofthermal expansion and contraction,it is necessary to examine the natureof the bonding between atoms ormolecules within the solid.

1.2.1 Thermal expansion

TLL o∆α=∆

( )( )

TA2ATA2AAAA

TLL2AATLL

AL;0L

LLL2L

LLLLA

o

ooo

ooo

o

o2

o2

2o

2o

oo

∆α=∆

∆+≈−=∆

∆α+≈

∆α=∆

=≈∆

∆+∆+=

∆+∆+=

TVV o∆β=∆


Lo

Lo

AAA o ∆+=


Consider the forces acting between two atoms:

Note that near the equilibrium position,the force required to move one atomaway from another is very nearly directlyproportional to the displacement x fromthe equilibrium position: Hooke’s law orlinear elasticity.If one atom is held fixed, and the otheris moved away by the application of anopposing force, then the area under theforce-distance curve gives the increasein potential energy of the system.As an atom is bought closer to another,the potential energy decreases until itreaches a minimum at the equilibriumposition where the long range attractionis balanced by the short-range repulsion.Since zero potential energy isassigned at infinity, then the minimumat the equilibrium position correspondsto a negative potential energy.

xEquilibriumposition

xo

U

Energy becomesmore positive withincreasing xcommensurate withan increasing sweptarea under F–x curve

1.2.2 Atomic bonding in solids

F+

F−

xArea under curvegives increase inpotential energy

dx

xo“Strength” ofthe bond Fmax

Attra

ctio

nR

epul

sion

Equilibriumposition

The shape of the potential energyvariation is not symmetric about theminimum potential energy.The temperature, or internal energy isreflected by the amplitude of theoscillations and hence by the width of thetrough in the energy distribution. Raisingthe temperature makes the averageenergy more positive. As well, due to theasymmetric nature of this distribution,the average or mean distance increaseswith increasing temperature leading tothermal expansion

Averageenergyincreaseswithincreasingtemperature

Mean distances

U

x

9Thermal Physics


1.2.3 Specific heat

2. The amount of energy (∆Q) required tochange the temperature of a mass ofmaterial is found to be dependent on:

Specific heat orheat capacity

Material c (kJ kg−1 K−1)Water 4.186Steel 0.45Cast iron 0.54Aluminium 0.92

1. When heat is transferred into a system, the temperature of thesystem (usually) rises.

The specific heat is the amount of heatrequired to change the temperature of1 kg of material by 1 oC

3. Putting these threethings together in aformula, we have:

• the mass of the body (m) kg• the temperature increase (∆T) oC, K• the nature of the material (c) J kg−1 K−1

( )12 TTmcQ −=∆

Heat is energy in transitfrom one system to

another

Mass

Syst

em B

SystemA

∆Q

( )12 TTT −=∆

Molar specific heat (C) is the amount of heat required to raise thetemperature of one mole of the substance by 1 oC

6.02 × 1023 molecules

The mass (in kg) of one mole of substance iscalled the molar mass Mm. The molecularweight m.w. is the molar mass in grams. Inall these formulas, molar mass Mm in kg isused. Hence, for water, where m.w. for H2 =2 g and m.w. for O2 = 16 g, the molar massis: Mm = (16+2)/1000 = 0.018 kg mol−1.

Number of moles Molar specific heat (or molar heat capacity)

Heat capacity formula: ( )12 TTnCQ −=∆

Note: We use small c forspecific heat, large C formolar specific heat.

cMC m=



∆Q heat in to raisetemperature

Latent heat is that associated with a phase changee.g. Consider the heating of ice to water and then to steam:

Water:Lv (liquid to gas) LV= 2257 kJ kg−1

Lf (solid to liquid) LF= 335 kJ kg−1

Formation orbreakage ofchemical bondsrequires energy

Solid

LatentHeat in

Ice @0 oC

T1

Water@ 0oC

Cold water & ice

LatentHeat in

Steam @100oC

T2

Water @100oC

Hot water & steam

1.2.4 Latent heat

( )12 TTmcQ −=∆

Temperaturerise

fmLQ =

vmLQ =

11Thermal Physics


1. An aluminium pot of mass 0.6 kg contains 1.5 litres of water. Bothare initially at 15 oC.

( )( )

( )( )

( )

MJ97.3109.4610385.3107.533Q

MJ385.3257.25.1QkJ6.580

9.467.533QkJ9.46

15100920.06.0QkJ7.533

15100186.45.1Q

363Total

vw

wAl

Al

w

10015

10015

=

×+×+×=

=

=∆

=

+=

=

−=∆

=

−=∆

−

+

−

−

Data: cwater = 4186 J kg−1 K−1

cAl = 920 J kg−1 K−1

Lv = 22.57 × 105 J kg−1

Lf = 3.34 × 105 J kg−1

ρwater = 1000 kg m−3

cice = 2110 J kg−1 K−1

(a) Calculate how many joules are required to raise both the saucepanand the water to 100 oC.

(b) Calculate how many joules are required to change the water fromliquid at 100 oC to steam at 100 oC.

(c) What is the total energy needed to boil all the water away ?

Water from 15 oC to 100 oC

Boil water (liquid to vapour)

Aluminium from 15 oC to 100 oC

Total energy required.

Note: 1.5 L water = 1.5 kg

1.2.5 Examples

2. 0.5 kg of ice at −5 oC is dropped into 3 kg of water at 20 oC. Calculatethe equilibrium temperature (neglecting external heat exchanges).

( )( )

( )( ) ( ) ( )( )

C4.5TT14665172275251400

251400T12570T209516700052750QQ0

0T41865.01034.35.05021105.0Q

20T41863Q

2

2

22

waterice

25

ice

2water

°=

+=

−+++=

+=

−+×++=

−=

With these types of problems, it is easiest toequate all the terms involving a transfer of heatto zero. In this problem, there are no externalheat exchanges, thus, the heat lost by the wateris equal to the heat gained by the ice.


(a)

(b)

(c)

Solution:

Solution:


1.3 Heat transfer

Summary

( )coldhot TTL

kAQ −=D

dxdTkA

dtdQ

−=

Convection

Rate of cooling

Total

12R

TTQ −

=D

++=

cbatotal C

1C1

C1R

( )

1

2

12.

rrln

TTkL2Q −π−

= Conduction - radial pipe

Conduction - composite wall

Heat flow by conduction

General heat conduction

)TT(hAQ 12 −=D

( )So TTmcLkA

dtdT

−−=

4e ATeQ σ=

D

41i 1

AQT

σ=

D

Equilibrium temperature

Stefan–Boltzmann law

( )41

422 TTAeQ −σ=∆ Radiative heat transfer

13Thermal Physics


1. Conduction is microscopic transfer of heat. More energetic moleculestransfer internal energy to less energetic molecules through collisions.Energy is transferred, not the molecules, from one point to another.

Thot

Tcold

L

A

∆Q∆t

( )coldhot TTL

kAQ −=D

Note: This equationapplies to a uniformcross-section ofmaterial where k, Aand L are constants

2. The rate of heat flow (J s−1) is found to bedependent on:

3. Putting these four things together in aformula, we have:

Heat conduction formula

Units• the surface area of the body (A) m2 • the length of the body (L) m• the temperature difference (∆T) oC, K • the nature of the material (k) W m−1 K−1

1.3.1 Conduction

dxdTkA

dtdQ

−=

For the formulas to work, we must keep the signs straight and be aware ofwhat direction we mean as being positive.

xTkA

tQ

∆

∆−=

∆

∆ If the differences ∆are made very small,then we have:

Note: This equation ismore general and mayapply to non-uniformcross-sections (e.g.where A = f(x)).Note: the thickness Lof the body is nowexpressed in terms ofan interval “∆x” on thex axis.

T1

T2

x

This is a“negative”slope sinceT2 − T1 isnegative

ColdHot

T1

T2

x

This is a“positive”slope sinceT2 − T1 ispositive

HotColdIf ∆T=T2 − T1 ispositive then dQ/dtcomes out negativeindicating thedirection of heat flowin the negative xdirection.

If ∆T=T2 − T1 isnegative then dQ/dtcomes out positiveindicating the directionof heat flow is in thepositive x direction.



Thermal conductivity (k)

Material k (W m−−−−1 K−−−−1)Aluminium 220Steel 54Glass 0.79Water 0.65Fibreglass 0.037Air 0.034

• a material property• units: W m−−−−1 K−−−−1

• high value indicates goodthermal conductor

• low value indicates goodthermal insulator

Heat ElectricalTemperature (T) Voltage (V)Heat flow (Q) Current flow (I)Thermal conductivity (k) Conductivity (1/ρ)Thermal conductance (C) Electrical resistance (R)

TR1Q

TCQR1C

LkAC

∆=

∆=

=

=

ThermalconductanceW K−1

Thermalconductivity

Thermalresistance

VR1I

ALR

=

ρ=

Electrical resistivity

Note: some definitions of thermal conductance omit the area term A inwhich case the units of C are W m−−−−2 K−−−−1 instead of W K−−−−1

thus

W K−1

Compare

1.3.2 Thermal conductivity

Mathematically, temperature gradient is very much like potential gradientin electricity. Heat flow is similar to current flow.

15Thermal Physics


T1

T1'T2'

T2

La Lb Lc

Hot

Cold

a

bc

a

T1

T2

positive x direction

For slab (a)

( )

a1

'1

1'

1a

CQTT

TTCQ

=−

−=

(b) (c)

Thermal resistances inseries add just likeelectrical resistors in series.

++=−

cba12 C

1C1

C1QTT

Add (a)+(b)+(c) together:

Total

12R

TTQ −

=

Composite wallformula

1.3.3 Composite wall

cbaTotalTotal

RRRRC

1++==

b

'1

'2 C

QTT

=−

c

'22 C

QTT

=−

The radial pipe problem is different to the case ofa block because the heat flows from the inside ofthe pipe to the outside across an ever expandingcross-sectional area A. In the case of the block, thearea A through which the heat flowed was aconstant (did not change with x). Here the area A isa function of r.

T2

T1

r1r2

rdr

( )drdTrL2kQ

.π−=

dxdTkAQ

.−=

( )

( )

1

2

12.

1

2.

12

r

r

T

T.

.

rrln

TTkL2dtdQQ

rrln

Lk21

Q

TT

drr1

Lk21dT

Q

1

drrLk21dT

Q

1

2

1

2

1

−π−

==

π

−

=

−

π

−

=

π

−

=

∫∫

Note: If T2-T1 is positivethen dQ/dt isnegative (i.e. oppositeto direction of positive r)

For heat flow Qin terms of T1,T2 and r1 and r2,we mustintegrate theeffect of theincreasing valueof A withrespect to r.

Radialconductionformula


rL2A π=

T1'T2'


( )

( )

( )

( )

( )

( ) St

mcLkA

So

So

S

So

o

S

S

S

S

TeTTT

tmcLkA

TTTT

ln

TTlnC0t@TT

tmcLkACTTln

dtmcLkAdT

TT1

dtmcLkAdT

TT1

TTL

kAdtdTmc

+−=

−=

−

−

−=∴

==

−=+−

−=−

−=−

−−=

−

∫∫

dtdTmc

dtdQ

=

Rate oftemperaturechange(degrees persecond)

Differentiate w.r.t. time to get(assuming no change of state)

Question: If heat flows from a body (by conduction) and the body cools,what is the rate at which the temperature of the body changes (dT/dt)?

Answer: Start with heat capacity formula:

( )STTL

kAdtdQ

−−=

But, for conduction, the rate of extraction of heat is:

TmcQ ∆=∆

Negative indicates atemperature decreasew.r.t. time

Thus:

T

t

To

TS

Initial rateof cooling

At t = 0, T = To so,inserting into Eqn. 1, weget an expression for theinitial rate of cooling:

( )So0t

TTmcLkA

dtdT

−−=

=

Eqn. 1

This formula applies for coolingvia conduction only.

1.3.4 Rate of cooling

)TT(KdtdT

S0t

−−=

=

)TT(1

dtdTK

)TT(KdtdT

So

So

−

−=

−−=

A similar analysis can be done for a generalised heat transfer coefficient Kthat takes into account conduction, convection and radiation. Thisempirical relationship is known as Newton’s law of cooling.

If the initial rate of cooling is measured, then theheat transfer coefficient K can be calculated:

Surroundings

Initialtemperature

17Thermal Physics


Fluid travels from oneplace to another:• Due to density change -

Natural or freeconvection

• Due to mechanicalassistance (pump or fan)- forced convection

)TT(hAQ 12 −=C

Convective heat transfer coefficient depends on:• Fluid properties: thermal conductivity, density,

viscosity, specific heat, expansion coefficient

Fluid (e.g. air)absorbs heat

(increase in temperature)

Fluid (in this case, air) losesheat as it cools

(decrease in temperature)

Con

vect

ion

curr

ent

1.3.5 Convection

• Flow characteristics: whether natural or forced, surfacegeometry, flow regime (laminar or turbulent)

kc

Pr pη=

Viscosity (kg m−1 s−1)

Thermal conductivity

Specific heatPrandtlnumber

The effectiveness or measure of heat flow by convection is given by theconvective heat transfer coefficient. This is not entirely a materialproperty, but depends also upon the circumstances of the heat flow.

The fluid takes heat withit, thus the flow of the fluidrepresents a heat flow.

η

ρ=

vLReReynoldsnumber

Velocity of fluid

Viscosity

Density

The Prandtl number and Reynolds number are dimensionless quantitiesthat combine to form the convective heat transfer coefficient h. The way inwhich they are combined depends on the circumstances of the flow. Thefinal value of convective heat transfer coefficient is usually obtained byexperiment under controlled conditions. From the equations above, we cansee that convection is assisted with a large velocity (v), and high density,high specific heat and low thermal conductivity.



The transfer of energy by electromagnetic waves in the thermal range.Let us call energy transferred by electromagnetic radiation: radiantenergy.

Thermal radiation is in therange 0.1 to 100 µ m.

XraysRadio waves

0.1 µm100 µ m

0.76 µ m visible 0.38 µ minfrared ultra violet

Thermal range

1.3.6 Thermal radiation

Absorptivity + Reflectivity + Transmittivity = 1

Incident

Fractionabsorbed

Fractionreflected

Fractiontransmitted

When radiant energy coming from a source falls on body, part of it isabsorbed, part of it may be reflected, and part transmitted through it.

Reflected

Absorbed

Source

Transmitted

Incident radiation Q.

19Thermal Physics


4e ATeQ σ=

D

Stefan–Boltzmannconstant5.67 × 10-8 W m−2 K−4

Absolutetemperature

Surface area

e is the emissivity of surface (variesbetween 0 and 1) and depends on:

• nature of surface• temperature of surface• wavelength of the radiation being

emitted or absorbed

Material ePolished aluminium 0.095Oxidised aluminium 0.20Water 0.96Black Body 1.0

e indicates how well a body emits or absorbs radiant energy. A goodemitter is a good absorber. Emissivity is sometimes called relativeemittance.

Radiant energy(electromagneticradiation)emission rate

Any body whose temperature is above absolute zero emits radiant energyaccording to the Stefan–Boltzmann law.

1.3.7 Emission

Now, the emissivity of a surfacedepends on the wavelength of theradiation being emitted (orabsorbed). A surface whoseemissivity is 0.8 at say 10 µmmay have an emissivity of only0.1 at 100 µm. This is called aselective surface.

Quantifies the rate of emission orabsorption for different surfaces.

Despite these variations in emissivity, for any particular wavelength, it sohappens that the absorptivity α of a surface is equal to its emissivity e fora surface in equilibrium. This is known as Kirchhoff’s radiation law.

e

λ

1

0

Characteristics of aselective surface

Goodemitter andabsorber atshortwavelengths

Poor emitterand absorberat longwavelengths



1. The fraction ofenergy absorbed(αQ) isconverted tointernal energy(U) and thus isequivalent toheat flow havingtaken place. Thetemperature ofthe body thusrises (if nochange of state).

2. But, all bodiesabove absolutezero emitradiant energyand the rate ofemissionincreases as thetemperature ofthe body rises(Stefan -Boltzmann law).

If this didn't happen, thenthe body would continue toabsorb radiation and thetemperature would riseindefinitely till melt-down!

3. Thus, if noconduction orconvection orany other energyinput, then thetemperature ofthe body risesuntil the rate ofemissionbecomes equalto the rate ofabsorption -radiative(dynamic)equilibrium.

1.3.8 Absorption and emission

Absorbed

Emitted

We might well ask the question“What happens to the energy that isabsorbed?”

21Thermal Physics


1. Suppose the surroundings (1) are maintained at a low temperature T1and the body (2) maintained at a higher temperature T2. e.g. by electricity.

2. All bodies simultaneously emitand absorb radiant energy.

to surroundings

from surroundings

411i ATeQ σ=

D

3. The rate of emission from thebody is given by:

since α = e

4. The rate of absorption by thebody is given byand depends on:• the temperature of the surroundings• the emissivity (or absorptivity)

of the body

ia QQ DDα=

T1

QeQi

T2

Pin

Qa

(1)

(2)...

1.3.9 Radiative heat transfer

422e ATeQ σ=

D

Temperatureof body

Pin

Electricalenergy in

422e ATeQ σ=

D

4112a ATeeQ σ=

D

Radiationabsorbed

Temperature ofsurroundings

Radiationemitted

4112

422 ATeeATeQ σ−σ=∆ D

This quantity represents the net heat flow ∆Qfrom the body and is equal to the rate ofelectrical energy that must be supplied tomaintain the body at T2.

At radiative dynamic equilibrium,ENERGY OUT = ENERGY IN

Radiantenergyemitted

Electricalenergy in

Radiantenergyabsorbed

=+

Radiantenergyemitted

Electricalenergy in

Radiantenergyabsorbed

= -

Consider the energy flow into and out from the body:

If the surroundings have anemissivity of 1, then e1 =1then we have :

( )41

422 TTAeQ −σ=∆ D

Temperatureof body (K)

Temperature ofsurroundings (K)Net Heat flow (positive in this

case indicates net radiant heat flowfrom the body and transfer of internalenergy from body to surroundings)

Emissivityof body

4112

411ia ATeeATeQQ σ=σα=α=

DD


where


For a black-body, e =1no matter what the wavelength.The emission spectrum for ablack body is the maximumpossible for any given temperatureand has the shape which can becalculated using quantum theory. For an actual body, e < 1 and variesdepending on the wavelength being emitted. This leads to deviations fromthe black-body emission curve.

Radiant energy is emitted orabsorbed in the thermal range -0.1 to 100 µm.

Within this range, the amountof energy emitted is notuniformly distributed.

Black-body spectrum. e = 1(maximum possible emission)over all wavelengths

Actual body

As temperature increases:• total energy (area under curve)

increases• peak in emission spectrum shifts to

shorter wavelengths• can only be explained using

quantum theory

1.3.10 Radiation emission

If the source of radiation may beregarded as a point source (due to thesource being small or distance to objectbeing large) then the energy reaching aperpendicular surface varies inverselyas the square of the distance.If distance d is doubled, then energy reaching thesurface in W m−2 is reduced by a factor of four.

θ

If the surface is not perpendicular, then theradiation intensity (W m−2) is reduced by cos θ− i.e. the radiation is spread over a larger areaand thus the intensity is reduced.

Variation of emission with angle

Variation of emission with distance

d

Pointsource

A

A2

23Thermal Physics

Wavelength

Ener

gy

4000 K

3000 K

5000 K


Consider an insulated plate with incidentradiation. If the temperature of the plate isneither increasing or decreasing, then, inthe absence of any other energy input, rateof absorption = rate of emission.

4i

ea

ATeQ

QQ

σ=α

=

C

CC

then

assuming

Radiative equilibrium

This does not necessarily meantemperature of plate is equal totemperature of source. If the platereceives all the energy emittedfrom the source and the source isa black-body, then temperaturesare equal at radiative equilibrium.

The final steady-state temperature depends on radiation intensity (W m−2)falling on the body and not on the emissivity of the surface! i.e., a blacksurface and a white surface reach the same equilibrium temperature.

At equilibrium:4

e ATeQ σ=

ia QQ α=

1.3.11 Equilibrium temperature

at this condition

41i

4i

ae

1AQ

T

ATQ

e

σ=

σ=

α= λλ

Now, α = e is true for any particularwavelength of radiation we are talkingabout. But, the sun, being very hot,transmits a lot of energy at shortwavelengths. Objects placed in the sunreceive a portion of this energy with acertain value of α.As an object heats up, it begins to radiate more and more energy and therate of emission depends upon T4. Thus, the temperature reaches a pointwhere the rate of emission = rate of absorption. BUT!, the rate of emissionalso depends on the value of e as well as T4. Generally, e and α both varywith wavelength, so if a body has a high value of α for short wavelengths,and a low value of e at long wavelengths (e.g. a black-painted car) then thebody would have to reach a higher steady state temperature to keep the rateof absorption = rate of emission compared to the case where α = e.

People with dark skin absorb and emit radiation at a greater rate than peoplewith light skin. But for skin, unlike paint, the absorptivity at short wavelengths isless than the emissivity at long wavelengths. Hence the balance is for a lowerequilibrium temperature compared to the case where α = e for both short andlong wavelengths.



1.3.12 Examples

1. Calculate the heat flow through a glass sliding door of a house if itis 25 °C inside and 5 °C outside and the door has an area of 1.9 m2

and thickness 3 mm where kglass = 0.79 W m−1 K−1.Solution:

( )

( )( )

kW0.10003.0

5259.179.0L

TTkAQ 12

=

−

=

−

=

3. The oil sump on a motor vehicle is made from an alloy casting with finsto assist in cooling. If the fins have a total surface area of 600 cm2 inaddition to the flat surface area of the casting also of 600 cm2, and theoil temperature is 85 °C, determine the rate of heat dissipation from thesump casting when the vehicle is stopped and h = 10 W m−2 K−1 andthe ambient air temperature is 30 °C.

Solution:

( )

( )( )W66

308510120010

TThAQ4

12

=

−×=

−=

−

2. Calculate the temperature of a steel block 60 seconds after being heatedto 500 °C and then allowed to cool by being placed on a steel table topmaintained at 20 °C. The contact surface 20 × 20 mm, the thickness is20 mm, the specific heat is 0.45 kJ kg−1 K−1, the mass is 0.125 kg andthe coefficient of thermal conductivity is 54 W m−1 K−1. Ignore anycooling by convection or radiation.

Solution:

( )

( )( )( )

C6.173

206002.0450125.0

0004.054exp20500

TeTTT St

mcLkA

So

=

+

−−=

+−=−

25Thermal Physics


4. A bright chrome seat belt buckle rests on the seat of a parked car andreceives radiation of intensity 0.75 kW m−2 from the midday sun.

Stefan-Boltzmann constant: σ = 5.67 × 10−8 W m−2 K−4

(a) Calculate the rate of energy absorption per m2 of surface areaof the buckle per second (emissivity of chrome: 0.07).

(b) Determine an expression for the rate of energy emission perm2 of surface area and hence calculate the equilibriumtemperature of the buckle (neglect conduction and convectionand any variations of emissivity with wavelength).

(c) Calculate the equilibrium temperature of the buckle if it werepainted with black paint (emissivity of black painted steel:0.85 ) - justify your answer.

( )( )

( )

( )

( )

C66

K339T1097.35.52

1067.507.075007.0T

QQ

T1067.507.0Q

ATQ

W5.5275007.0

IQ

07.0kWm75.0I

9

84

ae

48e

4e

a

2

DD

D

D

D

=

=

×

=

×

=

=

×=

εσ=

=

=

α=

α=

=ε

=

−

−

−

−

Solution:

Equilibrium temperature would be the same.Emissivity e = 0.07 cancels out in expression for T


Consider 1 m2 of area:


1.4 Gases

Summary

atmg ppp +=

2

22

1

11TVp

TVp

=

2211 VpVp =

2

2

1

1TV

TV

=

nRTpV =

Combined gas law

Boyle's law

Charles’ law

Equation of state

BA ppp +=

Absolute pressure

Partial pressures

27Thermal Physics


1.4.1 Solids, liquids and gases

The molecules of the gas occupy a very small volumecompared to the volume of the container

The molecules are very distant from one another relativeto their size and only interact during collisions

Collisions between molecules are elastic

The thermal properties of a perfect or ideal gas are the most convenient tostudy. The properties of an ideal gas are:

Forces betweenmolecules:• long-range

attractive• short-range

repulsive

Intermolecular forces are negligibleand molecules move with rapid,random motion filling the space

available to it

Intermolecular attractive forces arestrong enough to bind molecules

loosely

Intermolecular forces are stronglyattractive and bind molecules

together

GAS

LIQUID

SOLID

Application ofpressure

Application ofpressure

gets stronger asmolecules getcloser together

very strong forcebut only acts overa very shortdistance

Real gases often behave like ideal gases at conditions in which no phasechanges occur.



%Vol Partial pressure (mbar)

N2 78.08 791.1O2 20.95 212.2Ar 0.93 9.4CO2 0.03 0.3

Impact of a moleculeon the wall of thecontainer exerts aforce on the wall.There are many suchimpacts per second.The total force per unitarea is calledpressure. Pressure isthe average effect ofthe many impactsresulting frommolecule to wallcollisions.

Newtons

metres2Nm-2 = Pa

One of the most important macroscopic properties of a gas is its pressure.

1.4.2 Pressure

1 atmosphere:= 760 Torr= 1013 millibars= 101.3 kPa= 760 mm Hg

A pressure gaugeusually measuresthe pressure aboveor belowatmosphericpressure

Absolute zero ofpressure is nopressure at all.Pressures abovezero pressure arecalled absolutepressures.

It is important to realise that pressuregauges measure the pressure aboveor below atmospheric pressure. Thusin thermodynamic formulae, absolutepressure must always be used.

In a mixture of gases, the total pressure is the sum ofthe pressures of the component gases if those gasesoccupied the volume of the container on its own.These pressures are called partial pressures.

+

=

pa

pb

ptotal

AreaForcePressure =

atmg ppp +=

Total: 1013 mbar

Dry air

29Thermal Physics


• Pressure• Temperature• Volume• Mass

Macroscopic properties of a gas

These quantities specify the state of a gas

Consider a mass (m) of gas:

2

22

1

11TVp

TVp

=

If temperature T is a constant:

2211 VpVp =

If pressure p is a constant:

2

2

1

1TV

TV

=

Combined gas law

Boyle's law

Charles’ law

Notes: These lawscannot be applied whenthe mass of gaschanges during theprocess. Pressures andtemperatures areabsolute.

Can be obtained from kinetictheory or from experiment.

If p,V and T all vary then:

1.4.3 Gas laws

Let us express the mass of a gasindirectly by specifying the numberof moles:

By using moles, we get the ideal gasequation with the universal gas constantR (units J mol−1K−1). otherwise, value for Rdepends on the nature of the gas ( i.e. nolonger universal) and has units J kg−1K−1 .

mMmn =

6.02 × 1023

molecules

Mass in kg

Molar massNo. moles

Experiment shows that Boyle'slaw and Charles law leads to:

no. moles

Absolutetemperature

Absolutepressure

Volume

Universal gasconstant8.3145 J mol−1 K−1

This equation links all the macroscopic quantities needed to describe the(steady) state of an ideal gas and is thus called an equation of state. Thereare other equations of state, mostly used to describe the state of real gases.

nRTpV =

Example: Calculate the volume occupiedby one mole of an ideal gas at 273 K atatmospheric pressure.

( ) ( )( )L406.22V

273314.81V3.101nRTpV

=

=

=



Liquid

1.4.4 Phases of matter

V

p Ideal gas

Gas

Liquid+vapour

V1p ≈

T1

T3 = TC

T2

(1)

(2)(3)

(no phasechanges)

Vapour pressureThe partial pressureexerted by thevapour when it is inequilibrium with itsliquid. It depends ontemperature andnature of thesubstance. Thetemperature at whichthe vapour pressureequals the prevailingatmosphericpressure is calledthe boiling point.

Critical Temperature TcThere is, for each gas, atemperature above whichthe attractive forcesbetween molecules arenot strong enough toproduce liquefaction nomatter how high apressure is applied.Tc H2O = 647 K at 218 atmTc He = 5.2 K at 2.3 atm

Gas starts condensing intoliquid, no change in pressureas volume decreases

All gas condensed intoliquid, attempts to furtherreduce volume producelarge increase in pressureas liquid is compressed.

In a p–V diagram, the temperature iskept constant, volume decreased andpressure recorded.

p

T

Liquid

Solid

Vapour

Constant pressureheating exampleBoiling

MeltingTriple point

Critical point(pc,Tc)

At each point (p,T) only a single phase can exist except on the lines wherethere is phase equilibrium. At the triple point, solid, liquid and vapour existtogether in equilibrium.

In a phase diagram, we keep the volume V a constant and plot pressure vstemperature.

Fusion

Vapourisation

Sublimation

31Thermal Physics


1.4.5 Examples

1. A motorist checks the pressure of his tyres after driving at high speedand measures 300 kPa. He notices that the temperature of the tyre is50 oC. What would be the pressure when the tyre is at roomtemperature (assume the volume of the tyre has remained constant) ?

kPa272p323300

293p

Tp

Tp

1

1

2

2

1

1

=

=

=

Solution:

3. An air compressor has a tank of volume 0.2 m3. When it is filledfrom atmospheric pressure, the pressure gauge attached to the tankreads 500 kPa after the tank has returned to room temperature (20oC). Calculate the mass of air in the tank.

Solution:

2. Calculate the mass of air in a room of volume 200 m3 at 20 oC and101.3 kPa given that the molecular mass of air is 28.92 g mol−1

.Solution:

( ) ( )( )

( )kg5.24002892.8312m

moles8312n27320314.8n200101300

nRTpV

=

=

=

+=

=

( )( )( ) ( )( )

( )kg43.1

02892.4.49mmoles4.49n

27320314.8n2.010005003.101nRTpV

=

=

=

+=+

=



1.5 Work and thermodynamics

Summary

( )12 TTmcQ −=

( )12 TTnCQ −=

( )12 UUWQ −=−

vp CCR −=

( )∫=

2

1

V

V

dVVpW

Specific heat

Molar specific heat

First law of thermodynamics

Universal gas constant

Work done on or by a gas

33Thermal Physics


Experiments show that when a gas is heated at constant volume, thespecific heat cv is always less than that if the gas is heated at constantpressure cp.

cp = 1.005 kJ kg−1 K−1

cv = 0.718 kJ kg−1 K−1

Air

1.5.1 Gas

Constant volumeheating

p1V1T1

p2V1T2

Constant pressureheating

Qv

Heat added, finaltemperature of bothsystems = T2

p1V2T2

Qp

dV

W

W

For a given temperature rise ∆T, there will alwaysbe a volume change ∆V with the constant pressureprocess meaning that the energy into the systemhas to both raise the temperature and do work, thuscp is always greater than cv.

For a constantpressure process,a volume changeinvolves a forceacting through adistance andhence work isdone on or by thesystem. Volumechange in solids &liquids very smallhence distinctionbetween cp and cvnot usually made.

( ) ( )

( )12p

1212vp

TTmc

VVpTTmcQ

−=

−+−=( )12vv TTmcQ −=

Qp is a very importantquantity in industrialprocesses and is giventhe name enthalpy.


JmN

mmNpV 3

2

=

⋅=

=


( ) ( )

( )

( ) ( ) ( )12v1212p

12p

1212vp

TTmcVVpTTmc

TTmc

VVpTTmcQ

−=−−−

−=

−+−=

For the constant pressure process:

Heat in or out of gas

Work done on or by gas

Change in internal energy

( )12 UUWQ −=−

Heat in orout ofsystem

Work doneon or by thesystem

Changes in internalenergy may be readilycalculated from:

( )12v12 TTmcUU −=−

Heat transferred in constantvolume heating (even if it’s nota constant volume process)

For ALL processes:

When V2>V1 then W ispositive and work is doneby the gas. When V2<V1,W is negative and work isdone on the gas.

1.5.2 1st law of thermodynamics

( )( ) ( ) ( )

( ) ( ) ( )

vp

vp

12v1212p

22

11

12v1212p

12

CCR

CRC

TTnCTTnRTTnCnRTpVnRTpV

but

TTnCVVpTTnCUUWQ

−=

=−

−=−−−

=

=

−=−−−

−=−

Consider a constant pressure process p1 = p2 and the 1st law.

The universal gas constant R isthe difference between the molarspecific heats.

Change ininternal energyfor all processes

Work done on or by thegas at constant pressure

Heat flow intoor out fromsystem atconstantpressureIdeal gasequation ofstateSubstituteinto 1st lawequation

1st law ofthermodynamics

Internal energy isthe kinetic energyof vibration of themolecules thatmake up the gas.

Thermal Physics 35


For a system at equilibrium, the properties are the same throughout thesystem. For a gas, only two independent properties are required (e.g. pand V). The equation of state gives the connection between these and theothers (e.g. T and n).

If the change of state from P1 to P2 occurs as aseries of small steps, each of which represents anequilibrium condition, then a line joining the twopoints represents an equilibrium or quasi-staticprocess. If the system undergoes a process, thestate of the gas may change from the initial stateto the final state. In a quasi-static process, the outof balance condition that drives the change ofstate is very small so that there are no “dynamic”effects influencing the process.

Initial state

Final state

P1

P2

1.5.3 p–V diagram

On a p–V diagram, the area under the curve between P1 and P2 is

• If gas expands V2 − V1 > 0 then work is done by the gas• If gas contracts from V2 − V1 < 0 then work is done on the gas

Pressure is somefunction of Vp

V

P1

P2

p = p(V)

( )

W

dVVpArea2

1

V

V

=

= ∫

Work done onor by the gas

p

V


JmN

mmNpV 3

2

=

⋅=

=


1. A milkshake is prepared by mixing the 500 gof milk and ice cream in an electric blender. Ifthe blender has a power rating of 100 W, andthe initial temperature of the milk and icecream were 2 oC, calculate the temperature ofthe mixture after being mixed for 3 minutes(ignore heat flow from the surroundings andassume the mixture has the specific heat thesame as water).

( )( )( ) ( )( )

C6.11T

3T41865.06031000UUWQ

o2

2

12

=

−=−−

−=−

Note, in this example, it is seen that the first law applies to liquids (andsolids) as well as gases. However, here, work is done on the liquid, thusW is actually negative (which is consistent with V2-V1 < 0 when work isdone on a gas).

Heat flow into(+ve) or out from(-ve) system

Work done by(+ve) or on (-ve)the system

Change ininternalenergy ofsystem

1.5.4 Example

Solution:

Thermal Physics 37


1.6 Gas processes

Summary

( )12 VVpW −=

)TT(nCQ 12p −=

)TT(nCU 12v −=∆

2

2

1

1TV

TV

=

0W =

2

2

1

1Tp

Tp

=

)TT(nCQ 12V −=

)TT(nCU 12v −=∆

2211 VpVp =

0U =∆

1

211 V

VlnVpW =

WQ =

nn2211 VpVp =

)TT(nCU 12v −=∆

1CC

C

)TT(nCQ

vp

12

−

−

=

−=

nn

n

n

1VpVpW 2211

−

−

=

n

v

p

CC

=γ=n

Constant pressure

Constant volume

Constant temperature

Polytropic

Adiabatic



( )12 VVp −=

Work done on or by thesystem (i.e. the gas)

Heat flow intoor out of system

Specific heat at constant pressure(different to that at constant volume)


Specific heat atconstant volume

( )12 UUWQ −=−

)TT(nCQ 12p −=

)TT(nC 12v −=

The difference between the heat in and the work outis the change in internal energy of the system.

Naturalincrease involume is notrestrained. Gasis able to pusha frictionlesspiston upwards.

W

gas

If a thermodynamic system undergoes a process, the state of the gas maychange from the initial state to the final state. The process may occur at aconstant volume, temperature or pressure, or all of these may vary.

p

V

P1 P2

(p1V1) (p2V2)

2

2

1

1

21

TV

TV

pp

=

=

1.6.1 Gas processes

A gas process at constant volume is called isochoric.

0pdV2

1

V

V

== ∫

Heat flow is equal to change in internal energy and no work is done onor by the system.

p

V

P1

P2

(p1V1)

(p2V2)

2

2

1

1

21

Tp

Tp

VV

=

=

gasNaturalincrease involume isrestrained

Work done on or by thesystem (i.e. the gas)

Heat flow intoor out of system


Specific heat atconstant volume

( )12 UUWQ −=−

)TT(nCQ 12V −=

)TT(nC 12v −=

A gas process at constant pressure is called isobaric.

39Thermal Physics


A gas process at constant temperature iscalled isothermal.

2211

21VpVp

TT=

=

For a constant temperature process, there is no changein internal energy and heat in equals work done (orheat out equals work in).

( )12 UUWQ −=−

0)TT(nC 12v =−=

Heat flow into orout of system.(Cannot beeasily calculateddirectly. Need touse Q = W).

p

V

P1

P2

(p1V1)

(p2V2)[ ]

1

222

1

211

VV

V

V

VVlnVp

VV

lnVpW

pVK

VlnK

dVVKW

VKp

2

1

2

1

=

=

=

=

=

=

∫

but

Looks similar toisothermal but is ofdifferent slope whichdepends on n

nn

n

2211 VpVp

KpV

=

=

n is called thepolytropic indexwith values usuallybetween 1 and 1.6

Constant

( )12 UUWQ −=−

)TT(nC 12v −=

Heat flow into or out of system,where it can be shown that:

1CC

C

where)TT(nCQ

vp

12

−

−

=

−−=

nn

n

n Cn is called the polytropic molarspecific heat. When T2 > T1, Q isnegative, but it may be positive sinceCn can be negative depending on thevalue of n.

1VpVp

dVKVW

VKp

2211

V

V

2

1

−

−

=

=

=

∫−

n

n

n

Work done on orby the system(i.e. the gas)

p

V

P2

P1(p1V1)

(p2V2)

A gas process in which p, V and T all change is called polytropic.

2

22

1

11TVp

TVp

=


Constant

Constant


Adiabatic process is specialcase of the polytropic processwith n = γ

( )12 UUWQ −=−

)TT(nC 12v −=

Heat flow into orout of system = 0

1VpVpW 2211

−γ

−=

For adiabatic process, work done on or bythe gas is equal to the change in internalenergy.

This is a very important processfor heat engines since itrepresents the most efficienttransfer of internal energy intowork without loss tosurroundings. Adiabaticprocesses can be approximatedin systems that are insulated,have a small temperaturedifference to surroundings, orproceed very quickly such thatheat has no time to flow into orout from system.

( )

γ=

=

−

−=

v

p

vp

CC

1CC

0

n

nnn

Adiabaticindex

γγ

γ

=

=

2211 VpVp

KpV

p

V

P2

P1

(p1V1)

(p2V2)

( )

1CC

C

nC00

TTnCQ

vp

12

−

−

=

=∴

=

−−=

nn

n

n

n

2

22

1

11TVp

TVp

0Q

=

=

A gas process in which no heat flow occurs is called adiabatic.

p

V

n < 1

n = 1 (isothermal)

n = γ (adiabatic)

n > γ

n = 0 (constant pressure)

n =

(

cons

tant

vol

ume)

8

Constant volume, constant pressure,isothermal and adiabatic processes are allspecial cases of the polytropic process.

41Thermal Physics

T2 <> T1 becausethen the processwould be isothermal.


1. 1 mole of nitrogen is maintained at atmospheric pressure and isheated from room temperature 20 oC to 100 oC:(a) Draw a pV diagram(b) Calculate the heat flow into the gas.(c) Calculate change in internal energy of the gas.(d) Calculate the work done by the gas.

J6488.16656.2313W

UWQ

=

−=

∆=−

( ) ( )( )J6.2313

2010002892.010001

)TT(nCQ 12p

=

−=

−=

( ) ( )( )J8.1665

8002892.07201)TT(nCU 12v

=

=

−=∆

p

V

P

(pV1) (pV2)

Nitrogencp = 1000 J kg−1 K−1

cv = 720 J kg−1 K−1

R = 8.3145 J mol−1 K−1

Molecular weight = 28.92

Note: C = cM where Mis the molar mass (kg)

Note: positive W indicatesthat work is done by the gas.

1.6.2 Example

(a)

(b)

(c)

(d)

Solution:



1.7 Kinetic theory of gases

Summary

2vm21kT

23

=

2vmVN

31p =

mkT8vavπ

=

mkT3vrms =

mkT2vmp =

Pressure

Average kinetic energy

Average velocity

rms velocity

Most probable velocity

2Av kT

21KE =

67.1R25Cp ==

4.1R27Cp ==

Energy per degree of freedom

Molar specific heat for monatomic gas

Molar specific heat for diatomic gas

kT2mv

223 2

evkT2

m4)v(f−

ππ=

Maxwell velocity distribution

Thermal Physics 43


Consider N molecules of an ideal gas inside a container of volume V atan absolute temperature T.

• The molecules are in rapid motion and moverandomly around colliding with each other and thewalls of the container.

• The molecules exert forces on the walls of thecontainer during collisions

• During a collision, the velocity component vy ofthe molecule is unchanged but vx changes indirection but not in magnitude

Pressure is the result of the total average force acting on the walls of thecontainer. Consider the collision of one molecule with the contain wall:

1.7.1 Pressure

tv2m

tvmF x

∆=

∆

∆=

The change in velocity of a moleculeduring a time interval ∆t is:

vy

vx

vy

vx

∆v/∆t is acceleration

x

xxxv2vv2vv

=∆

=−−

Thus, the force imparted to thewall by the molecule is:

VtAvN

21 x∆=

Total No.molecules

Volume of container

Volume ∆V within which half themolecules hit the wall of area A

Half the molecules hit the wall,the other half are travelling inthe other direction

During a time ∆t, molecules a distance lessthan or equal to vx∆t away from the wallwill strike the wall. Thus, the number ofcollisions will be the number of moleculeswithin the volume element ∆V = Avx∆t.A

vx∆t

∆V

If there are N molecules in the totalvolume V, then the number within thevolume element ∆V is:

VVN ∆

No. of collisions



The total force on the wall atany instant during time ∆t is thus:

2x

xx

total

mvVN

AF

tAvVN

21

tv2m

F

=

∆

∆=

Total number ofcollisions

Force fromeach collision

But, so far we have assumed thatvx is the same for each molecule.The molecules in the containeractually have a range of speeds. The average value of vx

2 componentsleads to the average force (and hence pressure) on the wall:

2x

av

23x

22x

21x2

x

vmVN

AF

N...vvv

v

=

+++

=

= Pressure

But, it would be more convenient to have an expression which included thetotal velocity v rather than the x component vx, thus:

2

2x

av

22x

2z

2y

2x

2

vmVN

31p

vmVN

AF

v31v

vvvv

=

=

=

++=

Since random motion in alldirections thus velocitycomponents are all equal

Magnitude of average velocity2 given bysum of average components

The square root of the average of allthe velocity2 is called the root meansquare velocity

2rms vv =

=

=

=

=

=

2A

2

2

2

vm21nN

32

vm21N

32nRT

nRT

vm21N

32

vNm31pV

The average translational kineticenergy of a single molecule dependsonly on the temperature T.

2

Avm

21T

NR

23

=

But, R and NA are constants. Theratio of them is a new constant,Boltzmann’s constant k.

2vm21kT

23

=

AnNN =

Averagetranslationalkinetic energyof a singlemolecule

Since 2rmsmv

21kT

23

=or

Averagetranslational kineticenergy

1.38 × 10−23 J K−1

Thermal Physics 45

22x

2z

2y

2x

v31v

vvv

=

==


Gas molecules in a container do not all have the same speed.

Indicates the numberof molecules with aparticular velocity v

∫=

2

1

12

v

vv,v dv)v(NfN

Curve flattens outand peak shifts tohigher velocitiesas temperature isincreased.

Area under the curveindicates the number ofmolecules with a velocity inthe range v1 to v2

kT2mv

223 2

evkT2

m4)v(f−

ππ=

Maxwell's distribution ofmolecular speeds - from"statistical mechanics"

1.7.2 Velocity distributions

Averagevelocity:

mkT8

dv)v(vfv0

av

π

=

= ∫∞

rmsvelocity:

mkT3

dv)v(fvv0

2rms

=

= ∫∞

Mostprobablespeed:

mkT2v

0dv

)v(df

mp =

=

f(v)

v

High T

Low Tv1 v2

vmp

vav

vrms

We saw previously that the average kinetic energy of a single gas moleculein a container could be calculated from the temperature of the gas. It is oftendifficult to decide which velocity to use in a given situation.

To calculate the rms velocity, we square thevelocities first, then divide by the total number ofmolecules, and then take the square root. This isdifferent to finding the average velocity since theact of squaring the velocities first weights the finalanswer to those larger velocities in the velocitydistribution. Thus, vrms is a little large than vav.

mkT3

vv 2rms

=

=

We should use rms velocity when dealing with the kinetic energy of themolecules. We should use average velocity when the process underconsideration (e.g. flow through a pipe) is affected by the molecules’velocity.



Molecules in a gas are capable of independent motion. Consider adiatomic gas molecule:

(a) the molecule itself can travelas a whole from one place toanother

Translational motion

(b) the molecule can spinaround on its axis

(c) the atoms within the molecule canvibrate backwards and forwards

(only for solids)

Rotational motionVibrational motion

1.7.3 Molecular motion

Three translationalvelocity components arerequired to describe themotion of a monatomicgas molecule before andafter any collisions. Thesecomponents are calleddegrees of freedom.

Rotation about theaxis of the atom isnot counted sincethis does not changeduring collisions

vx

vy

vz1.

2.

3.

Three translational andtwo rotational velocitycomponents are requiredto describe the motionof a diatomic gasmolecule before andafter any collisions. Thisrepresents five degreesof freedom.

Rotations about thisaxis do not changeduring collisionstherefore notincluded as a degreeof freedom vx

vz

vy

rz

rx

1.

2.

3.

4.

5.

Thermal Physics 47


The average kinetic energy for a gas molecule is equally partitionedbetween the rotational and translation components. For each degree offreedom, the average kinetic energy is given by kT21

Type of gas Degrees of Total internalfreedom energy

Monatomic gas 3 3/2 kTDiatomic gas 5 5/2 kTPolyatomic gas 6* 6/2 kT

all translationalkinetic energy

Translational androtational kinetic energy

* Could be more or lessdepending on the gas

1.7.4 Specific heat and adiabatic index

( )12v TTnCU −=∆

( ) ( )12v12A TTnCTTk23nN −=−

Consider a temperature change ∆T = (T2-T1) for n moles of a monatomicgas. For a given temperature rise, the change in average translationalkinetic energy is equal to the change in internal energy.

Avogadro's number

No.molecules

R23C

kN23C

v

Av

=

= R25Cv =

Since

R25C

CCR

p

vp

=

−=

R27Cp =

and

Monatomic ideal gas Diatomic ideal gas

For M

onat

omic

gas

, we

only

nee

d to

con

side

rtra

nsla

tiona

l kin

etic

ene

rgy

then

67.135

R1

32R

25

=

=

⋅=γ

4.157

R1

52R

27

=

=

⋅=γ

33.168

R1

62R

28

=

=

⋅=γ

Monatomic Diatomic Polyatomic

The principle of equi-partition of energy allows us to calculate the adiabaticindex of different types of gases.

v

p

CC

=γ

For any ideal gas:



kT2mv

223 2

evkT2

m4)v(f−

ππ=

dv)v(NfdN =

Maxwell velocitydistribution function

Low value of kT means that themolecules tend to all have similar speeds(ordered motion). High value indicates alarge range in speeds (disorder).

kT = 0.025 eV at 300 K

Shape of distribution curve depends on relativesize of the translational kinetic energy and kT.

The number of particles N with a total energy E was computed byBoltzmann using the Maxwell velocity distribution function:

kTE

CeN−

=

and thus the ratio of numbers ofmolecules in a gas with energies E1 andE2 at a particular temperature T is givenby:

kTE

kTE

2

12

1

ee

NN

−

−

=

a constant

Maxwell-Boltzmann energy distribution

1 eV = 1.6 × 10−19 Joules

1.7.5 Energy distributions

f(v)

v

High T

Low T

Thermal Physics 49


1.7.6 Examples

1. Compute the average kinetic energy of a molecule of a gas atroom temperature (300K):

2. If the gas in the previous question was H2, calculate the rmsvelocity of a single molecule.

3. For one mole of gas at 300K, calculate the total translationalkinetic energy.

J102.6

3001038.123

kT23KE

21

23

av

−

−

×=

×=

=

( )

1

27

23

rms

sm1934

1032.33001038.13

mkT3v

−

−

−

=

×

×

=

=

J3750

3001038.1231002.6

kT23NKE

2323

ATotal

=

××=

=

−

mH = 3.32 x 10-27kg

k = 1.38 x 10-23 J K−1

NA = 6.02 x 1023Solution:

Solution:

Solution:



1.8 Heat engines

A heat engine is any device which is capable of continuousconversion of heat into work.

Summary

H

C

S

Ro

TT

1

QQ1

−=

−=η

CH

CTT

TCOP

−

=

Carnot efficiency

Coefficient ofperformance

51Thermal Physics


All these are thermodynamic, quasi-static, reversible processes.

1. A gas undergoes aconstant volumeprocess from P1 to P2

3. And finally a constantpressure compressionback to P1

p

V

P2

P1

V

pP2

P3

V

p

P1 P3

V

p

P1

P3

P2

The cyclic nature of this arrangement means that continuous work outputmay be obtained. The continuous conversion of heat into work is theoverriding characteristic of a heat engine. Other devices that convert heatinto work in a non-cycle manner (e.g. a rifle that converts heat into kineticenergy of a bullet) is not a heat engine.

1.8.1 Cyclic processes

2. It then undergoesan isothermalexpansion to P3

Let us now combine these processes on theone diagram. This combination, where thepressure, temperature and volume of theworking fluid have returned to their initialvalues, is a thermodynamic cycle.

Area enclosed bycycle denotes workdone on or by system

The word reversible means thatthe direction of the process maybe reversed by simply reversingthe temperature, pressure,volume, etc (or whatever iscausing the state to change).An isothermal expansion maybe reversed by reversing thedirection of the work W in whichcase work W will be convertedinto heat Q and transferredback to the heat source.

Quasi-static means that the pressure,temperature and volume are changingvery slowly, or slowly enough so thatdynamic effects (such as momentumchanges or viscosity) are insignificant.



0 - 1 Constant pressure intake(intake stroke).

1 - 2 Adiabatic compression with increase inpressure and temperature and decrease involume (compression stroke).

2 - 3 Constant volume addition of heat (ignitionof fuel by spark plug). Pressure andtemperature increase with no change involume. No work flow.

3 - 4 Adiabatic expansion (power stroke).Decrease in pressure and temperature andincrease in volume. Work output.

4 - 1 Constant volume heat rejection (exhaustvalve opens). Pressure and temperaturefall to their initial values.

1 - 0 Constant pressure exhaust(exhaust stroke).

Two cyclesjoinedtogether

p

V0 1

2

3

4

The Otto cycle starts off with the first half of a pumping cycle, which isinterrupted by the actual working cycle, after which the pumping cycleresumes. This cycle forms the basis of nearly all the world’s motorvehicles.

1st half ofPumping cycle

Working cycle

2nd half ofPumping cycle

1.8.2 Otto cycle

1

2

1i

VV

11−γ

−=η

The maximum theoretical (or indicated) thermal efficiency of an engineusing the Otto cycle depends on the compression ratio V1/V2 and isgiven by:

γ is the adiabatic index of the working gas.

ω

V1

V2

53Thermal Physics


1.8.3 Thermal efficiency

QD representsdissipativelosses (friction,turbulence, etc).

W is usefulmechanical workdone on thesurroundings

1) Heat source @ TH

3) Heat sink @ TC

QS

QR

W

QD

2) Workingcycle

A heat engine uses a thermodynamic cycle to perform work. After a heatengine completes a cycle, the system returns to its original state. The onlyobservable difference is that heat ∆QS has been taken from the source, aquantity of this heat has been converted to work ∆W, and the remainder∆QR has been rejected into the heat sink.

If all processes in thecycle are reversible,then to transfer ∆Qs backto the source, we requirethe work ∆W (to besupplied) plus the heat∆QR (required to betransferred from the sinkback to the source). Thepresence of anirreversible processsomewhere within thecycle ∆QD would meanthat extra work would berequired to supply thesame amount of heat ∆QSback to the source.

Heat energysupplied

(QS)

Work (W)

Heat (QR+QD)

Heatengine

useful output

non-usefuloutput

A heat engine always has two outputs:

S

Ro

S

RSo

RS

RS

S

o

QQ1

QQQQQWWQQ

QW

−=η

−=η

−=

+=

=

=ηinputEnergy

outputEnergy Thermalefficiency

but

hence

The maximum possible efficiency obtainableis called the Carnot efficiency and mustalways be less than 1.

WQQWQQ

UWQ

RS

RS+=

=−+

∆=−

Relevant signsalready included inthis formula (enterall quantities aspositive).

Heatdissipatedby friction

Heatrejected

If no change in internalenergy of working fluid∆U = 0



• Heat is supplied isothermally at TH• Heat is rejected isothermally at TC• No heat is transferred anywhere else in the cycle

Conditions for maximum efficiency:

1 - 2 Isothermal compressionHeat rejected into sink at TC aspressure increased

2 - 3 Adiabatic compressionPressure and temperature bothincrease without any heat flow

3 - 4 Isothermal expansionHeat supplied from source at TH andpressure decreases p1V1 = p2V2

4 - 1 Adiabatic expansionPressure and temperature bothdecrease to initial value, no heat flow(i.e. lost to surroundings)

1.8.4 Carnot cycle

12

3

4

p

V

QS, TH

QR, TC

T3=T4=THisothermal

T1=T2=TCisothermal

3

43343S V

VlnVpWQ ==−

2

11121R V

VlnVpWQ ==−

H

Cc

34H

21Cc

S

Rc

2

1CR

3

4H

3

43S

333

TT

1

VVlnTVVlnT

1

QQ1

VVlnnRTQ

VVlnnRT

VVlnnRTQ

nRTVp

−=η

−=η

−=η

=

=

=

=Now,

Also,

Then

Since

T3 = T4 = THisothermal

for adiabatic compression 2 - 3

3

4

2

1

C

H1

4

1

1

4

C

H1

3

2

2

3

VV

VV

TT

VV

TT

TT

VV

TT

=

=

=

=

=

−γ

−γ

for adiabatic expansion 4 - 1

thus

Maximum possible efficiencyattainable by any heat engineoperating between temperaturelimits TH and TC

1

2

1

1

2

2211

2

22

1

11

VV

TT

VpVp

TVp

TVp

−γ

γγ

=

=

=

For 3 – 4:

For 1 – 2:

55Thermal Physics


QS

F

Gas underpressure

Perfectvacuum

Perfectinsulator

Frictionlesssurfaces

Perfectconductor

Opposing forcealways maintained alittle lower thanpressure forceexerted by piston.Work is done bypressure force aspiston movesdownwards.

A quantity of heat is supplied from QS and work is done by piston againstopposing force. If there is no friction, and the cylinder is perfectlyinsulating, then all the heat supplied QS goes into work.

p

V

1

2

3

Isothermalexpansion

To obtain a cyclic orcontinuous conversion ofheat into work, it isnecessary to stop theexpansion at some point (2)and then reject heat (by saycooling gas at constantpressure) (3) and then(adiabatically) compressinggas back to p1V1 at (1). If wedidn’t do this, then wewouldn’t be able tocontinuously convert heatinto work. The area enclosedby the curve on the p–Vdiagram is the work done.

Continuous or cyclic conversion of heat to work requires heat to berejected at some point in the cycle.

usually to the surroundings

1.8.5 Heat sink in an engine



Heat source @ TH

Heat sink @ TC

QS

QR

WS

QD

Heatengine

WG

Heat source @ TH

Heat sink @ TC

QS

QR

WS+

QD

Heatpump

WG

DRSS QQQW −−=

DRSS

DSRS

DSG

GRS

QQQWQWQQ

QWWWQQ

+−=

−+=

−=

+=

+

+

+

compare

In a real heat engine, irreversible processes lead to dissipation of heat intothe surroundings. Schematically, an engine can be thus treated as a seriescombination of a reversible cycle, the output of which is the work WG doneby the working substance, being then connected to an irreversible process,where some of the energy WG is dissipated as heat QD, and the remainderavailable as useful work WS.

Driven in reverse, we maysupply mechanical work WSto be converted to heat flowto the source and at thesame time transfer heat QRfrom the sink to the source.But, in the presence ofdissipative processes, QDstill remains directed to thesink. Thus, compared to thefirst case, additional workWS+ is required to deliverthe same amount of heat QSto the source.

1.8.6 Reversibility

A refrigerator is a heat engine inreverse.

( )

oi

io

12

QWQWQQ

UUWQ

=+

−=+−

−=−

Heat“rejected” intohot reservoir

Heat takenfrom coldreservoir

Work inputto refrigerator

Assuming no netrise in temperatureof workingsubstance ∆U = 0

Relevant signshave alreadybeen included inthis formula

The best refrigeratortransfers heat from Tc,using the minimum workW. Thus the coefficientof performance is: CH

C

i

TTT

WQCOP

−

=

=

57Thermal Physics


1. A Carnot cycle is operated between two heat reservoirs at 800 Kand 300 K. If 600 J are withdrawn from the hot reservoir in eachcycle, calculate the amount of heat rejected to the cold reservoir.

J225375600Q

J375W

625.0600W

QWQ%5.62

8003001

TT

1

R

RS

H

Cc

=

−=

=

=

+=

=

−=

−=ηHeat source @ TH

Heat sink @ TC

QS

QR

WG

600J

800 K

300 K

1.8.7 Examples

Solution:

2. The output from a petrol engine is 50 kW. The thermal efficiency is15%. Calculate the heat supplied and the heat rejected in one minute.

Solution:

MJ17QWQQ

MJ20Q

15.0QW

J103W

5000060W

R

Rs

s

S

6

=

+=

=

==η

×=

=



1.9 Entropy

Summary

1. No heat engine can continuously convertall the heat it receives into work due tothe need to reject heat.

2. Heat will not spontaneously flow from alower temperature to a highertemperature

Two most popular statements of the 2nd Law

Planckstatement

Clausiusstatement

TQS = Entropy

Thermal Physics 59


The mixing of the water is an example of an irreversible process. Heatflows from the hot to the cold water, no work is done. There is a preferredor natural direction for all processes involving heat flow. The second lawis concerned with this preferred direction.

Heat flow

Preferred or natural direction

Hot Cold

1.9.1 Reversible and irreversible processes

A kilogram of water at 0 oC is mixed with 1 kilogram of water at 100 oC.What happens?

Question:

( )( ) ( )( )C50T

100T418610T41861TmcQ

2

22°=

−−=−

∆=∆

We get 2 kg water at 50 oC

ColdHotWarm

BUT WHY?Certainly we can calculate that the heat lost by the hot water is equal tothe heat gained by the cold water, but why do we not just get 1 kg of hotwater and 1 kg of cold water in the same container. Energy would still beconserved. Why does heat flow from the hotter to the colder body?Why doesn’t the water ever unmix itself into hot and cold regions?



The associated heat flow andinternal energy change aregiven by the first law:

( )12 UUWQ −=−

( )∫=

2

1

V

V

dVVpW

Consider a gas process: The area under the curve is thework done on or by the gasaccording to:

BUT! The first law says nothing about a very important experimentalobservation, and that is, what is the preferred or natural direction of theprocess? Is there a natural direction for this process anyway? Is it easierfor the gas to expand and do work, or for us to do the work and compressthe gas?

1.9.2 Entropy and reversibility

p

V

P1

P2

TQS =

S is a function of Q and T since there is a natural tendency for Q to travelfrom a hot body at T2 to a cold body at T1. That is, heat and temperatureare the important variables.

A measure of reversibilityis called entropy and canbe calculated using:

Note: Entropy is calculated from Q and T. Work does not comeinto it since there is no preferred or natural direction associatedwith work. There is a preferred or natural direction associatedwith heat and this is why entropy is calculated using Q.

Entropy

The word reversible means that the direction of the process may bereversed by simply reversing the temperature, pressure, volume, etc (orwhatever is causing the state to change). There is no preferred or naturaldirection associated with work, it may be done on or by the gas with equalfacility. But, there is a natural direction associated with heat flow. We shallsee that heat flow to the surroundings leads to a process being irreversible.Work flow done by a system is reversible.

Thermal Physics 61


Heat Sink@Ts

p

V

1

2

W QD = 0

Qs

Heatsource@TH

Hea

t and

wor

k flo

w a

cros

s th

e sy

stem

bou

ndar

y

A source supplies heat QSisothermally to the workingsubstance (eg. a gas).

2211 VpVp =

1

211 V

VlnVPW =

Answer:In calculating changesin entropy, we needonly consider thoseitems or componentsin the system whichaccept or reject heat.Work transferred tosurroundings does notaffect entropy (thereis no preferreddirection for work).

Question: A reversible gas process happens within the system boundary.What is the entropy change of the system due to this event?

Component Heat flow Working substance + QS @ TH

H

STQ

TQS

+=

∆=∆

Entropy change of the system:

1.9.3 Reversible process

The pressure on the piston acting through a distancecauses mechanical work W to be done.

Since, in thisexample, thereare nodissipativelosses to thesurroundingsthen QD = 0. Ifthe sameamount of workW is done onthe system,then heat Qs istransferredback to theHeat source –a reversibleprocess.

The gas accepts this heatand expands such that itstemperature remainsconstant. The expansioncauses an increase involume and a fall inpressure according to:

If this is a reversible process,then perhaps we might thinkthat the change in entropyshould be zero. We shall seewhy this is not so shortly.



The system and thesurroundings (heatsources and sinks)together may be referredto as the universe.

0

0TQ

TQ

TQS

H

S

H

S

=

+−

++

=

∆=∆

Entropy change of sink

Entropy change ofsystem

Component Heat flow Heat source − QS @ THWorking substance + QS @ THHeat sink + QD = 0 @ TS

Entropy change of source

1.9.4 Entropy change of the universe

So far, we have only considered the entropy change of the system. We nowneed to include the entropy changes of anything that absorbs or rejects hear inthe vicinity of the system. Thus, for a reversible process, the entropy changeof the system and things directly affected by the system is found from:

The entropy change of the universe for areversible process is zero.

Heat flow to heat sink atTS from dissipativemechanisms (e.g. friction)

Component Heat flowHeat source − QS @ THWorking substance + QS @ THHeat sink + QD @ TS

S

D

S

D

H

S

H

S

TQ

TQ

TQ

TQ

TQS

=

+−

++

=

∆=∆

Total entropy change:

Isothermal

Energy balance:

For an irreversible process, Qd is not zero, hence, the entropy changeof the system and things directly affected by the system is given by:

QD does no workand hence can onlybe recovered (sentback to source) ifadditional work issupplied.

WQQWQ

S

DS≠

+=

Entropychange of sink

Entropy change ofworking substance

Entropy change of source

This is greater than zero, hence, there is a net increase inthe entropy of the universe for an irreversible process.

Thermal Physics 63


Let us look at a heat engine utilising thermodynamic cycle and aworking substance which consists of a series of reversible processes,e.g. a Carnot cycle

c

R

H

s

s

R

H

c

H

c

s

Rc

TQ

TQ

QQ

TT

TT

1QQ1

=

=

−=−=η

Thus

Carnot efficiency

TC

QS

QR

W

TH

… (1)

Now, let us examine the change in entropy of the system by considering theheat gained and lost by the working substance. The system underconsideration is just the working substance undergoing a series of processes.

1-2 Isothermal2-3 Adiabatic3-4 Isothermal4-1 Adiabatic

0TQ

TQ

0TQ

TQ

H

s

c

R

=

+

=

−

0dTdQ

0TQ

0TQ

TQ

c

R

H

s

=

=

=

−

+

∫

∑

Thus, for the wholecycle

Negative indicatingheat out of workingsubstance

Positive indicatingheat into workingsubstance

from (1)

For any reversible orirreversible heat engine,there is no net change inentropy. This holds forirreversible processesbecause we are talkingabout a cycle.

1.9.5 Entropy in a cycle

1

2

3

4

p

V

or

A non-reversible process somewhere in the cycle represents a flow of heatto the surroundings QD at the expense of useful mechanical work availablefrom the system.If the heat engine itself is the system under consideration, then at the end ofany cycle, the pressure, temperature, etc have all resumed their initialvalues. The fact that some heat QD has been generated at the expense ofmechanical work does not affect the state of the working substance at theend of the cycle and there is no gain in entropy within the engine.



Note: It would appear that for a reversibleengine (QD= 0), the fact that QR goes to thesurroundings would lead to an overallincrease in entropy since heat is being“lost” to the surroundings. However, thisQR, even though it goes into thesurroundings, is recoverable since it canbe gotten back by supplying work W.

Heat source @ TH

Heat sink @ TC

QS

QR

WS

QD

Universe

Heatengine WG

For the system and the surroundings (sources and sinks), there is no netchange in entropy if all processes within the system are reversible. Thetotal entropy increases if any processes within the system are irreversible.• The heat source loses entropy −QS/TH• The heat sink gains entropy +QR/TC• The heat engine itself is returned to the same state at the end of each

cycle so heat flow into and out of the working substance within theengine causes no net change in entropy within the engine.

• Work is done by the engine and may leave the system, but this does notlead to any change in entropy (work is ordered energy).

• For a reversible heat engine, QS/TH = QR/TC• If there were a dissipative process, then less mechanical work would be

performed and the heat sink would gain additional entropy +QD/TC

1.9.6 Entropy

Consider the mixing of hot and cold water. The temperature differencebetween the two may have been used as TH and TL for a heat engine. That is,we have the opportunity to do mechanical work. Once mixed and at uniformtemperature, the opportunity to do mechanical work has been lost. Entropy isa measure of lost opportunity.Entropy is a measure of randomness or disorder. Thermal energy arises dueto random motion of molecules. Work flow, however, is ordered energy sinceit can be controlled. There is a natural tendency for things to become moredisordered. Thus, any processing involving heat involves some naturaltendency to disorder. This tendency towards disorder makes all real processesirreversible since the energy lost to the disordered state cannot be recoveredunless additional work is done which is at the expense of additional disordersomewhere else in the universe. Entropy is a quantitative measure of theamount of disorder or randomness associated with a real process.

Thermal Physics 65


The 2nd law of thermodynamics can be expressed in many ways. Thereis no single equation like the 1st law. There are three popular statementsof the 2nd law:

1. No heat engine cancontinuously convert heatinto work due to the needto reject heat.

2. Heat will notspontaneously flow froma lower temperature to ahigher temperature

3. The total entropy of anyisolated system cannotdecrease with time.

An engine which could convert allthe heat it receives into workwould spontaneously create orderout of disorder. This has neverbeen observed to happen.

There is a natural tendencytowards the disorderedstate.

All real processes are irreversiblewhich always leads to a totalincrease in entropy.

1.9.7 The 2nd Law of Thermodynamics

∆S > 0 irreversible process∆S = 0 reversible process∆S < 0 impossible process

All real processes are irreversible. The greater the irreversibility of aprocess, the greater the increase in entropy.

Another (but quantitative)statement of the 2nd Law

Note: Entropy is not energy.There is no “Law ofConservation of Entropy”.Indeed, irreversible processes“create” entropy.

When a system undergoes a process, the entropy change of the system,added to the entropy change of the surrounding heat sources and sinks, isthe total entropy change of the universe bought about by the process. Thesign of ∆S (the total entropy change) signifies the presence of thefollowing types of processes:

All real processes involvesome dissipative loss

Energy has not been lost or destroyed, but the opportunity to use it has. Theenergy has become unavailable. Entropy is a quantitative measure of lostopportunity.



1. A kilogram of ice at 0 oC ismelted and converted to waterat 0 oC. What is the change inentropy of this system?

Solution:Temperature is a constant at T = 273 KHeat added = +335 kJ/kg (latent heat of fusion)

1-

Q

Q

Q

Q

KJ1227273

335000

dQT1

dQT1S

2

1

2

1

=

=

=

=∆

∫

∫

T is aconstant

Is this a reversible or irreversible process? Wemust consider the entropy change of the universe.If heat is transferred isothermally from somesource, then the process is reversible since theentropy change of the source will be -1227 J K−1.If heat is transferred from a source at a highertemperature, then the process is irreversible sinceQs/Ts will now be less than 1227 J K−1.

1.9.8 Examples

Cold

This is the entropy change of thesystem, not the universe.

2. A kilogram of water at 0 oC isheated to 100 oC. What is thechange in entropy of this system?

Solution:Temperature is not a constant but Q may be expressed in terms of T.

1-

T

T

Q

Q

J K1306273373lnmc

dTmcT1

dQT1S

mcdTdQ

2

1

2

1

=

=

=

=∆

=

∫

∫

cp water = 4186 J kg− 1K−1

Here we areapproximating theprocess as an infiniteseries of isothermalprocesses taking in heatin infinitesimalquantities dQ.

Cold Hot

This is the entropy change of thesystem, not the universe.

Thermal Physics 67

Ice


3. A kilogram of water at 0 oC is mixed with 1 kilogram of water at 100 oC.What is the total change in entropy?

Is this an irreversible process? Yes it is.How can we tell? Well, in this process, theonly heat flow is from the hot water to thecold water. There are no heat flows to andfrom external sources and sinks. Hence,the entropy change of this system is alsothe entropy change of the universe. Sincethere is a net increase in entropy (100JK−1), then the process must be irreversible.

Solution:

Before mixing, total entropy is 1306 J K−1 + 0After mixing, we get 2 kg water at 50 oC (323 K).What is the entropy of the system after mixing?Then can work out total ∆S.

1beforeafter

1-

T

T

Q

Q

K J100=

1306-1406 =-SSKJ1406273323lnmc

dTmcT1

dQT1S

mcdTdQ

2

1

2

1

−

=

=

=

=∆

=

∫

∫

Entropy of system aftermixing (relative to 0 oC)

Net increase or totalchange in entropy

Take water at 0 oC tohave zero entropy

m = 2kg

Change in entropy aftermixing compared to thatbefore mixing.

ColdHot

This is the entropychange of the universe.



Waves &

Optics

Part 2


2.1 Periodic Motion

Displacement of particle fromequilibrium position for SimpleHarmonic Motion (SHM)

Summary

tsinAy ω=

( )φ+ω= tsinAy

( )φ+ωω= tcosAdtdy

( )φ+ωω−= tsinAdt

yd 22

2

kyF −=

km2T π=

22Am21E ω=

Lg

=ω

Displacement as a function of timewith initial phase angle

Velocity

Acceleration

Restoring forcecharacteristic of SHM

Period of SHM

Energy in SHM

Frequency of oscillation of apendulum



2.1.1 Periodic motion

time

θ= sinAy

Now, sin θ is a fraction which varies between 0 and ±1. If we multiply thisfraction by the amplitude A of the motion then we get the displacement ofthe mass as a function of θ.

Since the mass undergoes periodic motion there is a frequency ofoscillation. If the mass is moving up and down at a frequency ω, then theproduct ωt is the angle θ.

tsinAy ω=

Fraction between 0and 1 (or -1)Amplitude (largest

displacement fromequilibrium position)

Displacementfrom equilibriumposition

0

+y

1 cycle

Time for one cycle iscalled the period.

A

A

Periodic motion of themass consists ofdisplacements around anequilibrium position

Plot of displacement vs time

Consider the motion of a mass attached to a spring.

71Waves and Optics


ωt0

+y

π

Period

+A

-A

2π

Now, this expression: assumes that y = 0 when t = 0which is true if we start our time measurements from y = 0. Whathappens if time measurements are started when the object is at y =A?

0

tsinAy ω=

We have to add an initialoffset φ to the angle:

( )φ+ω= tsinAy

Now, at t = 0, we have: wherey0 is the initial displacement from theequilibrium position. Consider theseexamples of initial displacement y0 for somepossible values of initial phase angles φ:

φ= sinAy0

initial phase angle

The general expression for position (i.e. displacement from the equilibriumposition) for the mass is thus:

( )φ+ω= tsinAy

φ yo

π/2 Aπ 03/2π -A

2.1.2 Initial phase angle

( )φ+ω= tsinAy

( )φ+ωω= tcosAdtdy

( )φ+ωω−= tsinAdt

yd 22

2

Displacement

Velocity

Acceleration



( )( )

ymF

tsinAytsinAmmaF

2

2

ω−=

φ+ω=

φ+ωω−==Now,but

thusHowever, the product mω

2 is actually a constant “k” for a fixed frequency:

kyF −=Thus The minus sign indicates that the force acting on the mass is in adirection opposite to the displacement and acts so as to bring themass back to the equilibrium position. The magnitude of thisrestoring force is a function of displacement of the mass from theequilibrium position. k is often called the spring or force constant.

2.1.3 Simple harmonic motion

Consider the force applied to a body moving with an oscillatory motion.

The combination of periodic motion and a restoring force whose magnitudedepends on the displacement from the equilibrium position is called simpleharmonic motion or SHM.

Because the displacements, velocities andaccelerations of the mass involve smoothlyvarying sine and cosine functions.

km2T

2T

T2f2

mk

π=

ω

π

=

π

=π=

=ωNow,

also,

Note: the period of oscillation only dependson the mass and the spring (or force) constantand not on the amplitude.

In general, at any instant, the mass has kinetic and potential energy.

( )φ+ω=

=

−== ∫∫

tsinkA21

ky21.E.P

dykydyF.E.P

22

2

Potential Kinetic

( )φ+ωω=

=

tcosAm21

mv21.E.K

222

2

The total energy is P.E.+K.E., thus:

( ) ( )

22

22222

Am21E

tcosAm21tsinkA

21E

ω=

φ+ωω+φ+ω=

As the sin term increases, thecos term decreases. In otherwords, for a constant m, ω andA, the total energy is a constant.

In SHM, energy is constantly being transferredbetween potential and kinetic.

2mω=

73Waves and Optics


L

θ

mg

F

x

Equilibrium position

Let x, the arc length, be thedistance, or displacement, from theequilibrium position.

θ= Lx

Whenever the pendulum is displaced fromthe equilibrium position, there is arestoring force F acting along x towardsthe equilibrium position.

θ−= sinmgF

minus sign indicates force acts inopposite direction to positive x

For small angles of θ:θ≈θsin

kxFL

mgk

xL

mgLxmg

mgF

−=

=

−=

−=

θ−=

A pendulum does move insimple harmonic motion if thedisplacements from theequilibrium position are small.

Thus:

this equationcharacteristicof SHM

LgLmmgmk

=ω

=

=ω

In the case of a pendulum, themass term cancels out and theperiod or frequency of motiondepends only on the length of thependulum and g.

Period of motion:

2.1.4 Example

θcosmg

1. Determine an expression for theperiod of a pendulum.

Solution:



2.2 Waves

Summary

λ= fv

ρ=

Tv

( )φ+−ω= kxtsinAy

( )φ+−ωω= kxtcosAvy

( )φ+−ωω−= kxtsinAa 2y

2

2

22

2

ty

v1

xy

∂

∂=

∂

∂

( )ρλω= 22A21E

22AT21P ωρ=

Velocity of a wave

Stretched string

Displacement of a particle

Velocity of a particle

Acceleration of a particle

General wave equation

Energy transmitted by onewavelength of stretched string

Power in one wavelength onstretched string

75Waves and Optics


2.2.1 Waves

A wave is some kind of a disturbance.The disturbance travels from one place to another. Some examples are:

• mechanical waves• electromagnetic waves• matter waves (probability waves)

• Transverse (displacement ofparticles perpendicular to thedirection of travel of wave)

• Longitudinal (displacementof particles parallel todirection of travel)

Simple harmonicmotion ofparticles (y)

Direction ofwave travel (x)

There are two types ofmechanical waves:

Sideways motion ofparticles (y)

Wave travel (x)

Mechanical waves travelthrough a medium (eg.water). The particles whichmake up the medium aredisplaced from theirequilibrium position in aregular periodic manner(which may be SHM - itdepends on the system).

In a longitudinal wave, the displacements of theparticles (y) is in the same direction as the motionof the wave (x). However, it is convenient to labelthe sideways movements as y to distinguish themfrom the wave travel x.

The highest pointis called a crest.

The lowest point iscalled a trough.



This plot shows the displacement (upand down) of ONE particle in themedium as a function of time.

Now, in a medium, there are many such particles. These particles (whichmight be molecules or atoms in a solid) are connected together bychemical bonds which act like springs.Let’s look at a line ofparticles in the medium.If one particle isdisturbed downwards,then this particle, after ashort instant, will dragthe next particledownwards. Thisparticle, which of courseis connected to the nextparticle, will drag thatone down as well. Butby this time, the firstparticle may be on theway back up. Theparticles all follow oneanother as they aredragged up and down bythe previous particle. Inthis way, thedisturbance or wavetravels along throughthe medium.

Equilibriumposition

Distance “x” within themedium

t = 1

t = 2

t = 3

2.2.2 Wave motion

0

+y

T

A

• A mechanical waveis some kind ofdisturbance in amedium.

• The disturbancetravels from left toright (which is why itis called a travellingwave).

• The particles in themedium move upand down.

• The movement ofany one particle maybe just simpleharmonic motionabout zero.

Summary:

77Waves and Optics


There are two types ofplots which showsomething about waves.The first is a plot of theposition of ONE particleagainst time and showsthe period of the up anddown motion of theparticles.

λ

Wavetravel

The distance betweentwo identical points onthe wave in the secondtype of plot is called thewavelength λ.

Dis

tanc

e in

the

med

ium

indi

rect

ion

of m

otio

n of

par

ticle

s

At any particular instant, some particles in the medium are above theirequilibrium position, some are below and some are at the equilibriumposition. A snapshot of the particles in the medium looks like a wave:

Dis

tanc

e in

the

med

ium

indi

rect

ion

of m

otio

n of

par

ticle

s

These plots are different. The top one shows thedisplacement from equilibrium position against timefor one particle. The bottom one shows thedisplacement from equilibrium position for a lot ofparticles at a particular instant in time.

2.2.3 Wavelength

0

+y

T

A

t

0

+y

A

x

The second plot showsthe position of ALLparticles along a line inthe medium and showsthe displacement fromthe equilibrium positionas a function of thedistance that particle isfrom the first particle.



Consider a transverse wave in a medium whose particles undergo SHM:

This is a “snapshot” ofthe disturbance, orwave, at some time “t”.

• The shape of the wave is arepeating pattern.λ is called the wavelength and isthe length of one complete cycle

The disturbance travels with a velocity v. The time for one complete cycleis T. Thus, since:

λ=

λ=

=

fvT

v

tdv

then

since one complete wavelengthpasses a given point in a time T

and f in cycles per second

2.2.4 Velocity, frequency and wavelength

SHM ofparticles inmedium

λ

0

+y

A

x

I’m moving thestring just a littleand quite slowly.

I’m moving thestring a long waybut still quiteslowly.

Now I’m movingthe string a longway but nowquite quickly.

ActionLong wavelengthSmall amplitudeλ

λ

λ

A

A

A

Large amplitudeLong wavelength

Large amplitudeShort wavelength

Observation

T1f =since

79Waves and Optics


The velocity of a wave is the speed of the disturbance as it passes throughthe medium - not the velocity of the particles within the medium.

Wave Velocity

Stretched string(transverse)

Fluid(longitudinal)

Solid(longitudinal)

Gas(longitudinal)

ρ=

Tv

ρ=

Bv

ρ=

Ev

MRT

p v

γ=

ρ

γ=

T − tension in stringρ − mass per unit length

B − bulk modulusρ − density of fluid

E − elastic modulusρ − density of solid

γ − adiabatic indexp − pressureT − abs. tempM − molar mass

Note thatthe velocityof the waveis a functionof an elasticproperty/inertialproperty

It takes time for the motion of one atom or molecule to affect the next atomor molecule. This means that the original disturbance in the medium takesa finite amount of time to travel from one place to another. The velocity ofa wave in a particular medium depends upon its physical properties(usually the density and elastic modulus).

2.2.5 Wave velocity



Consider a transverse wave on a string. We wish to calculate thedisplacement y of any point P on the string as a function of time t. But, notall points on the string have the same displacement at any one time.

x

Points separated by awhole number ofwavelengths have thesame displacement y.

Thus, what we really want is a formula whichgives displacement y as a function of both xand t. If we had this formula, then we couldfind the displacement of a point P located at xat any time t.

This picture isa snapshot ofthe positionsof all pointson the stringat some giventime t.

y = f(x,t)Let’s consider the motion of the point located at x =0. If the points onthe string are moving up and down with SHM, then:

( )φ+ω= tsinAy if y = 0 att = 0, then φ = 0

The disturbance, or wave,travels from left to rightwith velocity v = x/t. Thus,the disturbance travels from0 to a point x in time x/v.

x

y

0

P@ t

P'@ (t-x/v) Shape ofwave at t

Shape of waveat (t − x/v)

x

P'@ t

Now, let us consider the motionof a point P located at positionx. The displacement of point Plocated at x at time t is thesame as that of point P' locatedat x = 0 at time (t − x/v).

Points with samedisplacement and same(vertical) velocity are said tobe in phase with each other.

2.2.6 Particle displacement

0

+y

Ax

P

81Waves and Optics


Thus, to get the displacement of the point P at (x,t) we use the sameformula for the motion of point P' located at x = 0 but put in the time t = (t− x/v):

φ+

−ω=

vxtsinA

Now, it is convenient to let:λ

π=

2k

and since:

k

k2f

fv

ω=

π=

λ=

then:

( )φ+−ω=

φ+

ω

−ω=

kxtsinAy

kxtsinAyThus:

( )φ+ω= tsinAy

Wave number

Displacement of particle in the medium fromequilibrium position as a function of x and t.

This applies to both transverseand longitudinal waves.

If t is held constant, andy plotted as a functionof x, then shape of wave(snapshot) is displayed.

If x is held constant,and y plotted as afunction of t, thenSHM motion of singlepoint is displayed.

x

y

0

t

y

0

eg. @t = 0

v

Note: as wave moves to right, pointP at x = 0 goes upwards (+ve y)

2.2.7 Wave equation



( )φ+−ω= kxtsinAy

( )φ+−ωω= kxtcosAvy

( )φ+−ωω−= kxtsinAa 2y

Displacement

Velocity = dy/dt

Acceleration = dv/dt

(holding x constant)

The velocity and acceleration of the particle in the medium are found bydifferentiating while holding x constant:

( )

( )

( )

( )

2

2

22

2

2

2

2

2

22

2

ty

v1

xy

kxtsinAvx

y

thusvk

but

kxtsinAkx

y

kxtcoskAxy

kxtsinAy

∂

∂=

∂

∂

−ωω

−=∂

∂

=ω

−ω−=

∂

∂

−ω−=∂

∂

−ω=

Let us now find dy/dx while holding t constant:

ay = acceleration of particle

This is called the wave equation and givesinformation about all aspects of the wave by tyingtogether the motion of the particles and the wave.

v is the velocity of the wave

2.2.8 General wave equation

(holding x constant)

The symbol ∂ is used to remind us that weare taking the derivative with one (or more)of the variables in the equation heldconstant (i.e. in this case, t). Derivatives ofthis type are called partial derivatives.

Velocity of wave

83Waves and Optics


A wave can be used to transfer energy between two locations.

Energyfromoscillatingsource

Waves in medium

Targetlocationreceivesenergy

2.2.9 Energy transfer by wave motion

1. The external source performs work on the first particle in the string.2. The particle moves with SHM. Energy of particle is converted from

P.E. to K.E. etc. Total energy of particle is unchanged.3. Particle loses energy to next particle at the same rate it receives

energy from the external source. Total energy of particle remainsunchanged but energy from source gets passed on from one particleto the next till it arrives at the target location.

4. Energy from the external source travels along the string withvelocity v.

5. The total energy of each particle is:6. The total energy for all oscillating particles in a segment of string

one wavelength long is:

22Am21E ω=

( )ρλω= 22A21E

mass per unit length

since ρλm

In one time period T, the energy contained in onewavelength of string will have moved on to the nextwavelength segment.

( )

22

22

22

22

AT21P

AT21P

vA21

T1A

21T

EP

ωρ=

ωρ

ρ=

ωρ=

ρλω=

= λ

Rate of energytransmission

Energy in onewavelength

λ=T1vsince

Mass perunit length

for a stretched string

Power transmittedby wave on astretched string

sinceThus:

222 fvA2P ρπ=or

T - tension (N)ρ - mass per unit lengthω - frequency in rads s−1

A - amplitude

ρ=

Tv

λ

t t+T



2.2.10 Example1. The following equation describes the displacement y (metres) of

particles in a medium as a function of x and t:( )π+−= 75.0x8t4.151sin30.0y

( ) 1-

1-

s m 258

283.31v

28

2k;fv

Hz1.242

4.151f

s rad 4.151

m30.0A

=

π=

λ

π=+

λ

π=λ=

=π

=

=ω

=

(a) Calculate the amplitude of this wave?(b) Calculate the frequency (in Hertz)?(c) Calculate the velocity of the wave?(d) What is the particle’s velocity at a position x = 200 mm and t = 6 secs?

( )

( )( ) ( ) ( )( )1-

y

m s9.14

75.020.0864.151cos3.04.151

kxtcosAv

−=

π+−=

φ+−ωω=

Solution:

All these angles are in radians, somake sure your calculator is in radiansmode before taking the cos!

Now, in this book, and in some other text books, the wave function is written:

( )φ+−ω= kxtsinAyBut, in some books you will see:

( )φ+ω−= tkxsinAyBoth are correct, but it depends on whichdirection is defined as being +ve for the verticaldisplacements of the particle. In this book, +ve yhas meant upwards. Hence, in the figure here,the wave shape is drawn initially going down at t= 0 because as the wave moves to the right, theparticle at x = 0 actually moves up (+ve).

x

+y

0

v

t = 0

However, some books show waves as initially going “up”. In this case, the directionof +ve y is actually downwards. The formulas give different answers to problems. Toget the same answer, φ = π may be used in the first equation. In this book, the +vedirection for y is upwards and hence formula (1) above will be used.

(1)

(2)

As wave moves to right, particleat x = 0 moves upwards.

Sign conventions:

85Waves and Optics


2.3 Superposition

Summary

ρ=

TL2nf

2121

22 ω−ω=

ω−ω

vfA2I 222ρπ=

o10 I

Ilog10=β

18

oo

ms103

1c

−

×=

εµ

=

( )φ+−ω= kxtsinA2yR

+ω

π+ω

π+ω

π= ...t5sin

54t3sin

34tsin4y

tsinx2cosA2yy 21 ω

λ

π=+

( )kxtcoskAvp 2−ωρ=∆

Superposition of two waves ofsame amplitude, phase anddirection

Superposition of two waves ofsame amplitude, phase butopposite direction

Superposition of two waves ofsame amplitude, phase butopposite direction

Beat frequency

Pressure amplitude

Intensity

Decibels

Speed of light

Fourier analysis



What happens when two or morewaves arrive at the same point at thesame time?

The resulting displacementof particles within themedium is the sum of thedisplacements that wouldoccur from each wave if itwere acting alone.

This is the PRINCIPLE OF SUPERPOSITION

If two wave functions y1(x,t) and y2(x,t) both satisfy the general waveequation, then so does the resulting combined function y1+y2.

2

2

22

2

ty

v1

xy

∂

∂=

∂

∂

The general wave equation ties together information about the shape of thewave, and velocity and accelerations of the particles within the mediumthrough which the wave is travelling.

This leads to a more formal definitionof superposition.

2.3.1 Superposition

87Waves and Optics

General waveequation


Consider two waves travelling in the same direction having the sameamplitude, frequency and wavelength.

Interference is the term used to describe the result of the superposition ofwaves.

Constructive interference

Destructive interference

Amplitude of resultant is largerthan either component

Amplitude of resultant is lessthan one of the components.

( )( )22

11

kxtsinAykxtsinAy

φ+−ω=

φ+−ω=

Case 1: Both have the same initial phase angle φ

Case 2: One wave has a phase angle φ =π and the other φ =0

( ) ( )( )φ+−ω=

φ+−ω+φ+−ω=

+=

kxtsinA2kxtsinAkxtsinA

yyy 21R

Resultant wave has twicethe amplitude but thesame frequency as thecomponent waves.

( )( )( ) ( )

( ) ( )( )

0yysinsinbut

sinAsinAkxt let

kxtsinAkxtsinAyykxtsinAykxtsinAy

21

21

2

1

=+

θ−=π+θ

π+θ+θ=

−ω=θ

π+−ω+−ω=+

π+−ω=

−ω=

Waves cancel out.

ωt

A

−A

yR

y1,y2

2.3.2 Interference

0


A

−AyR

y1

0

y2


Case 3. Waves have a phase difference of φ =π/4

Case 4: Waves have a phase difference of φ =π/3

( )( )( ) ( )

( ) ( )

( ) ( )

( )8kxtsin8

cosA2

24kxtkxtcos

.2

4kxtkxtsin2Ayy

2BAcos

2BAsin2BsinAsin

4kxtsinAkxtsinAyy4kxtsinAy

kxtsinAy

21

21

2

1

π+−ωπ

=

π+−ω−−ω

π+−ω+−ω=+

−

+=+

π+−ω+−ω=+

π+−ω=

−ω=

Amplitude ofresultant wave

( )( )

( ) ( )

( ) ( )

( )6kxtsin6

cosA2

23kxtkxtcos

.2

3kxtkxtsin2Ayy

3kxtsinAykxtsinAy

21

2

1

π+−ωπ

=

π+−ω−−ω

π+−ω+−ω=+

π+−ω=

−ω=

Resultantwave is out ofphase by π/6to wave 1

Resultant waveis out of phaseby π/8 to wave 1

Amplitudeof resultantwave

but

thus

89Waves and Optics

A

−A

yR

y1

0

y2

A

−A

yR

y1

0

y2


GENERAL CASE: frequency, amplitude and phase are all different.

Any periodic waveform, no matter how complicated, can be constructedby the superposition of sine waves of the appropriate frequency andamplitude. The component frequencies can be found by a mathematicaltechnique called Fourier analysis.

A Fourier transform of awave is a “frequency map”,it tells us what componentwaveforms are needed toproduce the overall wave.

This means that a square wave can be obtained by superimposing a seriesof sine functions each of a different amplitude and frequency.

Consider a square wave. The complete Fourier series for thisis:

+ω

π+ω

π+ω

π= ...t5sin

54t3sin

34tsin4y

2.3.3 Fourier analysis

Amplitude ofcomponent Frequency of

component

tsin4y ω

π

=

ω

π+ω

π= t3sin

34tsin4y

ω

π+ω

π+ω

π= t5sin

54t3sin

34tsin4y

ωt

y

π

1

−1

2π


We start off with a sinewave with a frequencyequal to that of the finalsquare wave to besynthesised. We then addto this another sine waveof a higher frequency andlower amplitude whichproduces a small blip inthe first sine wave. Wethen more and more sinewaves until eventually, theresultant wave begins tolook like a square wave.


Let the waves have the same amplitude, wavelength and frequency but betravelling in opposite directions.

Wave travelling to the right:

Wave travelling to the left:

( )kxtsinAy1 −ω=

( )kxtsinAy2 +ω=

( )( )

( ) ( )

( ) ( )

tsinx2cosA2yy

tsinkxcosA22

kxtkxtcos

.2

kxtkxtsin2Ayy

kxtsinAykxtsinAy

21

21

2

1

ω

λ

π=+

ω=

+ω−−ω

+ω+−ω=+

+ω=

−ω=

λ

π=

2ksince

Resultant wave:

Amplitude varies between 0 and 2Aas x varies. The motion is still simpleharmonic motion but the amplitudevaries according to the value of x.

y2

y1

2.3.4 Superposition for waves in opposite directions

2nx2 π

=λ

πThe resulting displacement is always zero when:

no matter what value of t.

n = 1,3,5….

These values of x are called nodes and occur when:

The resultant displacements are a maximum when:

These positions are called antinodes and occur when:

...4

5,4

3,4

x λλλ=

...2

3,,2

,0x λλ

λ=

π=λ

π mx2 m = 0,1,2,...

91Waves and Optics

−

+

=+

2BAcos

2BAsin2

BsinAsin;


All particles in the mediumundergo SHM. It just sohappens that the particles atthe nodes always have anamplitude of zero and henceare stationary. The otherparticles usually oscillatevery rapidly and hence theeye only sees the envelopeof the waveform, hence theterm standing wave.

θ

A

-A

0

y

23π

25π

2π

NodeAntinode2.3.5 Standing waves

A very common example of standing waves appears in a stretched string.If the string is fixed at both ends then there must be at least a node at eachend - because y = 0 at both ends. For a standing wave to be produced, the

length of the string must be equal to anintegral number of half-wavelengths.

2nL λ

=

n = 1,2,3,...

L = λ/2

n = 1 The frequency of this mode ofvibration is called the fundamentalfrequency.

L = λ

n = 2 2nd harmonic

1st harmonic

ρλ=

ρ=

λ=

λ=

T1f

Tv

vf

fv

ρ=

λ=

TL2nf

2nLBut

Thus

For astretchedstring

Allowablefrequencies forstanding waves (ornormal modes) for astring length L.

Now,

L

Because only then is thereat node at each end



Consider a travelling wave on a stretched string:

22AT21P ωρ=

Now consider a standing wave on a stretched string:

If there were no energylosses (eg. friction, sound)then waves would travelbackwards and forwardsbeing reflected from eachend indefinitely.

If we kept adding energy to the system, then since there is nodissipation, the amplitude of the waves would increase indefinitely.The amplitude increases because the frequency (via the wavelengthand velocity) is fixed by the length of the string. The only thing thatmay increase is A.This is called resonance and occurs when:

• there are no energy dissipative mechanisms• the oscillations are supplied at a frequency at

or near a normal mode frequency.If the oscillations are not provided at or near a normal mode (or resonantfrequency) then energy is not transferred into the system as effectively.

2.3.6 Resonance

Energyfromoscillatingsource.

Target locationreceives energyand converts itinto some otherform.

L = λ

n = 2

Power

93Waves and Optics


Consider two waves travelling in the same direction,same A, but with different frequencies.

( )( )kxtsinAy

kxtsinAy

22

11

−ω=

−ω=

( ) ( )

−

ω+ω

ω−ω=

−ω+−ω=

+=

kxt2

sint2

cosA2

kxtsinAkxtsinAyyy

2121

21

21R

Resultant:

Amplitude termoscillates with timeat a frequency of:

Frequency of resultantoscillation is the averageof ω1 and ω2

Displacementplotted as afunction of time(keeping x fixed).

The beat frequency is:

221 ω−ω

2121

22 ω−ω=

ω−ω

Amplitude of yR oscillates with a frequency (ω1-ω1)/2 but the ear hears twopulses or beats in this one cycle.

−

+

=+

2BAcos

2BAsin2

BsinAsin

using

yR

2.3.7 Beats

ωt

y1y2

−A

0

A



The wave equation is exactly the same as before except that y indicateslongitudinal displacements of particles from their equilibrium positions:

( )kxtsinAy −ω=

Consider some particles connected by series of springs as shown:

In a longitudinal wave, particles undergo oscillations in a direction parallelto the direction of the wave propagation. Longitudinal waves are really justlike transverse waves except that the particles in the medium move backand forth instead of up and down.

As the particle moves to the right, spring is compressed and this compressionacts on the next particle and so on. If the first particle moves with a regularperiod motion backwards and forwards about its equilibrium position, theneventually all the particles will move with this same oscillatory motion. If thefirst particle moves with SHM, then all all particles will move with SHM butwith a phase lag compared to the first particle.

Examples of longitudinal waves:

• Sound waves

• Waves on a slinky spring

• Earthquake waves (primary waves - thesecondary earthquake waves are transverse).

• Any waves in a liquid or a gas (liquids andgases cannot transmit transverse waves- except on the surface of liquids)

Condensation orcompression

Expansion orrarefaction

Wavetravel

Sideways motion ofparticles

λ

2.3.8 Longitudinal waves

95Waves and Optics


Longitudinal waves are alternate compressions and expansions betweenparticles in the medium. For sound waves, the compression corresponds topressure changes ≈1 Pa in the medium.

Hence, for soundwaves in a solid:

( )

( )kxtcoskAvp

vB

Bv

kxtcoskAxy

2

2

−ωρ=∆

ρ=

ρ=

−ω=∂

∂

xyB

xyB

VVBp

VVpB

o

o

∂

∂−=

∆

∆−=

∆−=∆

∆

∆−=

Pressure changes result in volume changeswithin the medium. The bulk modulus B is ameasure of the relationship between the two.For longitudinal waves, the volume changesoccur in a direction parallel to thedisplacements from equilibrium positions andthe fractional volume change is proportionalto ∆y/∆x or δy/δx.

Note: the displacementsy depend on sin(ωt − kx)where the pressurevariations ∆p depend oncos(ωt − kx) and are thusout of phase by π/2. Thatis, the pressure is agreatest when thedisplacements fromequilibrium position iszero.

y = 0

∆pmin ∆pmax

y = 0∆p = pmax

y =−A∆p = 0 y =+A

∆p = 0

y = 0∆p = pmin

2.3.9 Sound waves

Blah blah..You don’t say?

The velocity of sound wavesdepends on the elastic propertiesof the medium and the density ρ.

Medium VelocityAir (@20oC) 340 ms−1

Water 1400 ms−1

Polystyrene 1840 ms−1

Steel 5000 ms−1

Although the velocity of sound generally decreases with increasing densityof the medium, it also depends on the medium’s elastic properties. Thus, thevelocity of sound in solids is greater than that in gases.

MRT p v γ

=ρ

γ=

γ - adiabatic indexp - pressureT - abs. tempM - molar mass

Pressure and density effects cancel and thespeed of sound depends on the temperatureT and nature of the gas γ only.

Velocity of sound in air

Velocity ofwave



Two longitudinal waves interfere constructively when the density orpressure is enhanced by the superposition. Let us consider sounds waves ina pipe where the waves have identical amplitude and frequency and aretravelling in opposite directions.

Case 1. Pipe is closed atone end.

Now, at the closed end ofthe pipe, the displacementsof particles within the airinside the pipe must bezero since the closed end ofthe pipe prevents anydisplacements.The allowablefrequencies for standingwaves are thus fixed bythe length of the pipe.

4nL λ

= n = 1, 3, 5...

Odd harmonics ofthe fundamental.

y = 0

Displacement node(y = 0) or apressure antinode(∆P = max or min)

L

At the open end of thepipe, ∆P = 0, and thus,there occurs a pressurenode and adisplacement antinodey = ±A.

∆p =0

y = −A

Plot of displacement yof particles within the

air along the length ofthe pipe for resultant

wave. +vedisplacements of

particles vary between−A and +A.

L= λ/4

L= 3λ/4

L= 5λ/4

A

-A

y =0

2.3.10 Superposition of longitudinal waves

97Waves and Optics


Case 2. Pipe is open at both ends.

At the open ends of the pipe, ∆P = 0, and thus, there is a pressure nodeand a displacement anti-node y = ±A. The allowable frequencies forstanding waves are again fixed by the length of the pipe.

2nL λ

= n = 1, 2, 3... All harmonics of thefundamental.

L= λ/2

L= 2λ/2

L= 3λ/2

Plot of displacementy of particles withinthe air along thelength of the pipe forresultant wave. +vedisplacements ofparticles varybetween −A and +A.

A

-A

y =0

The diameter of the pipe makes a difference to the volume of the sound.



Sound waves are longitudinal waves characterised by pressure (or density)variations. The power for a 1 m2 square area of wave is given by:

kAvp 2max ρ=∆

W m−2

vfA2I 222ρπ=

f2π=ωsince

2.3.11 Sound intensity

o10 I

Ilog10=β

For a sound intensity I in W m−2,the sound intensity in db is givenby:

Io = 1 × 10−12 W m−2 and is approximatelythe threshold of hearing at 1kHz

Sound level or intensity I is the amount of power per square metre (inW m−2). This is a linear measure of sound intensity. Doubling the Watts persquare metre doubles the sound intensity. However, the ear can detectsounds which vary in intensity over a wide range. If we were to use W m−2

as a measure of sound intensity experienced by the ear, then the numberswould go from very small (1 × 10−12 W m−2) to very large (1 × 102 W m−2).This is very inconvenient, so we usually express sound intensity usingdecibels (db) which is a logarithmic scale.

Sound Typical levelwhisper 20 dbordinary conversation 65 dbmotor car 60 dbtrain 90 db

When something increases linearly, thereis a steady increase in the quantity. Whensomething increases logarithmically, thereis a rapid increase in the quantity:I

db

Intensity Intensity

ILinear scale

Log scale

Linear and log scales

AmplitudeIv

pArea

P

kvvA21

vA21

AreaP

2max

222

22

=

ρ

∆=

ρ=

ωρ=

since

ρ in this formula is the mass per unitvolume (or density) of the medium. Thisis similar to the expression given for thepower transmitted on a string, but here,we have a 3 dimensional situation.

orIntensity

The pressure variations insound are in the order of 1Pa. The maximum in thepressure difference ∆pmax iscalled the pressureamplitude.

BkAkAvp 2

max=

ρ=∆

99Waves and Optics


The human ear is subjective when it comes to interpreting pressurevariations as sound. Objective quantities such as intensity, and frequencyare measurable using scientific instruments. Subjective qualities such asloudness and pitch are not measurable using scientific instruments.

Loudness (volume)• interpreted as intensity by the ear but is

frequency dependent. For a givenintensity, sounds at low frequency or veryhigh frequency are not as loud as those atmoderate frequencies (a few kHz).

Intensity• measured in decibels and

is an indication of theenergy carried by thesound wave. The intensitydepends on the amplitudeA of the sound waves.

Frequency• measured in cycles

per second.

Pitch• interpreted as frequency by the ear but

depends on the intensity. The pitch of asingle frequency, or a “pure tone”becomes lower as the intensity increases.

Objective Subjective

Timbre• harmonic content

Tonal quality• depends on the component frequencies

of the sound. A pure tone has a“single” frequency. Bright sounds havemore power in the high frequencycomponents (to musicians, sharpsounds will mean higher pitched).

2.3.12 Hearing



When energy is lost from the sound wave as it passes through the mediumwe say that the sound has undergone absorption. The absorption of soundin most materials increases as with increasing frequency of the soundwaves and also increases with decreasing density of the material.

• Sound waves strike the wall.

• Some of the sound isreflected by the wall and itis called an echo.

• Whatever doesn’t getreflected goes through intothe wall and undergoessome degree of absorption.

• Anything that is notcompletely absorbed, getstransmitted through to theother side of the wall.

What happens when a loudnoise is produced inside aroom?

Incidentsoundwave

Absorbed inwall

Reflectedsound wave

(echo)

How do we know what is reflected and what goes on through the material?When sound goes from one medium to another, there is a large reflectioncomponent when there is a large difference in acoustic impedance of thetwo materials. The acoustic impedance is given by:For sound going from air and striking abrick wall, there is a very large acousticimpedance mismatch and most of thesound is reflected. For sound wavesstriking a wall covered with a cloth,there is less of an echo because there isless of an acoustic impedance mismatchand the sound is absorbed by the cloth.So with a brick wall, sound is easilyreflected, but any that does go into it isreadily absorbed!

cz ρ=

Material z kgm-1s-1

Air 415Water 1.5 × 106

Concrete 8 × 106

Steel 33 × 106

During absorption, the energy ofthe sound wave is converted toheat within the medium.For large open spaces, absorptionwithin the medium is important. Forenclosed spaces, reflection andabsorption at the walls is important.

2.3.13 Reflection and absorption of sound

Tran

smitt

edso

und

wav

e

DensityVelocity

101Waves and Optics


1. Plane waves

No divergence of rays.Wavefronts are parallelplanes. Same energypasses through sameare, thus no variation inintensity.

I = constant2. Cylindrical waves

A1

A2

r1

r2

h

θ Expanding wavefrontincreases surface areawith distance fromsource:

( )rfAAPI

API

hrAhrA

22

11

22

11

=

=

=

θ=

θ=

3. Spherical waves

Pointsource

function of r2

If the distance rfrom the sphericalwavefront to thesource is doubled,then the areaincreases by 4 times.Hence:

r1I ∝

2r1I ∝

A1

A2

Area of sphere = 4πr2

2.3.14 Intensity variationsWavefronts

RaysRays arealwaysperpendicularto wavefronts.

λ

λ

The direction in which waves radiate from the source is shown by straight lines calledrays. A wavefront is a surface on which all points have the same phase ofoscillation and is normal to the rays.



2.3.15 Examples

1. Calculate the amplitude of the superposition of the two waves whichare described by the equations below at a position x = 2 m:

Solution:

( )( )x5t8sin20y

x5t8sin20y

2

1+=

−=

( )

m45.33256.1

22cos202y

m256.1

25

21

−=

π=

=λ

λ

π=

+

2. For the superposition of the waves in the previous question,determine the position x of the first antinode (assuming x = 0 is anode).

m628.02256.1x

m256.125

=

=

=λ

π=

Solution:

3. Calculate the fundamental frequency of an air filled tube which isopen at both ends, has a diameter of 10 mm, and a length of 400mm and is at 0 oC.

Solution:

( )mm8004.02

2L

=

=λ

λ=

( )( )( )

Hz4218.0

336f

fvms336

028.0273314.84.1

MRTv

1

=

=

λ=

=

=

γ=

−

103Waves and Optics


2.4 Light

Summary

i

r

r

inn

sinsin

=θ

θSnell’s law

i

rC n

nsin =θ Critical angle

( )22oo

2e

m2Nq

1nω−ωε

+= Dispersion equation



2.4.1 Light rays

Consider a light source:

2. Rays of light emanate fromthe source and radiateoutwards in all directions.

3. Some rays enter an observer’seye and are focussed by thelens in the eye to form animage on the retina.

4. Other rays are reflectedand/or absorbed by thesurroundings

1. The directions in which light wavesradiate from the source are shown bystraight lines called rays.

Rays are imaginary lines drawnperpendicular to the wavefronts andindicate the direction of travel of thewaves

In this example, the light source can be considered a point sourceemitting spherical waves. Light rays thus radiate outwards. Far awayfrom the source, the radius of the wavefronts are very large and can beapproximated by plane waves in which the rays are parallel.

Geometrical optics: raysPhysical optics: waves

Planewaves

Spherical waves oflarge radius

Light raysapproximatelyparallel

105Waves and Optics


If the surface is smooth, the angle ofreflection = the angle of incidence.

The eye sees the light asif it were coming from apoint behind the surface.

If the surface is rough, the angle of reflection still equals the angle ofincidence for each ray, but these angles are now such that the light isscattered.

Many of the light rays do notenter the eye at all and an imagecannot be formed. Or, the onesthat do enter the eye appear notto come from a single point soagain, no image is formed.

This type of reflection is calleddiffuse reflection.

This type of reflection iscalled specular reflection.

2.4.2 Reflection



If a light wave strikes a medium (such as a block of glass) at an angle, thenthe waves which enter the medium slow down and the wavelengthbecomes shorter to compensate. This has the effect of altering the directionof travel of the wave.

Experiments show that the angle ofincidence = angle of reflection

Incidentlight Reflected

light

Refractedlight

θlθi

θr

Normal

All angles measuredw.r.t. the normal tothe surface

The speed of light in a material is always less than that of the speed of lightin a vacuum. The ratio of these two speeds is called the refractive index.

vcn =

Vacuum 1Air at STP 1.0003Glass 1.52

Medium n

Experiments show that formonochromatic light, the angles ofincidence and refraction are related by therefractive indices of the two materials: i

r

r

i

rrii

nn

sinsin

sinnsinn

=θ

θ

θ=θ

li θ=θ

Snell’slaw

The frequency of light does not changewhen the wave passes from one material toanother. The velocity does change, hence,since v = fλ, the wavelength also changes.

If light travels from a material with low value of n to one with high valueof n, then the velocity of the wave is reduced and the wavelength becomesshorter to compensate. The frequency remains the same. The path taken bylight rays is reversible. It doesn’t matter whether the light rays travel froman optically more dense to less dense material or vice versa.

Optically lessdense

Optically moredense

2.4.3 Refraction

λi

λr

Incidentlight ray

Refracted light ray

Incident lightwaves

Refracted lightwaves

107Waves and Optics


Let us examine the path of light rays when passing from anoptically more dense to a less dense medium (ni > nr).

Incidentlight

Refractedlightθr

θi ni

nr

What happens when the angle of incidence is made increasingly larger?

Incidentlight

Refractedlightθr

θi

ni

nr

Internallyreflectedlight

θC

90o

(a) (c) (d)(b)

As θi is made larger, there comes a point where the angle of refractionθr = 90o and the refracted ray grazes the surface of the boundary betweenthe two mediums. This angle of incidence is called the critical angle.Thus:

At angles of incidence greater than the critical angle, the refraction nolonger takes place (since by Snell’s law, sin θr >1 which is not possible).The ray is then totally internally reflected.

i

rC n

nsin =θ

ir

ir

i

r

r

i

sinnnsin

nn

sinsin

θ=θ

=θ

θ

In keeping with Snell’s law:

Here we have ni/nr >1 hence sin θr >sin θi and thus θr > θi and therefracted ray is bent away from thenormal.

2.4.4 Total internal reflection



Most refractive indicesare quoted withreference to yellowsodium light 589 nm.

The speed of light in a vacuum is the same for all wavelengths. The speedof light in a medium depends on the wavelength as well as the properties(the permittivity) of the medium.

Incidentlight

Refractedlight

θi

ni

nr

( )22oo

2e

m2Nq

1nω−ωε

+= Dispersion equation

Frequency ofincident light

Because of the interaction betweenelectrons in the medium and the action ofthe electric field of the light waves.

Since the refractive index is a measure of the relative speeds of light wavesin a vacuum and a medium, then the value of n for a medium thus dependson the wavelength or the frequency of the incoming light.Consider white light incidenton a transparent medium.When light is incident on themedium, the incoming electricfield E causes a distortion ofthe internal charge distributionof the molecules (polarisation).This causes molecules to try toalign themselves with the E field– thus altering the net field withinthe material. For a rapidly varyingE field, the molecules may not be able to move fast enough to keep up withthe changing field. The multitude of frequencies in the incoming whitelight above results in different responses, on an atomic scale, within thematerial. The result is that for higher frequency components, the velocityof the light is reduced and the refractive index is increased. This effect canbe quantified by the dispersion equation.

The dispersion equation gives n for variations in frequency of incidentlight. This frequency itself does not change for the refracted wave, butwavelength and velocity do. If everything except ω is held constant, thenfor an increase in ω, (decrease in λ) n also increases.

2.4.5 Dispersion

109Waves and Optics

Natural or resonant frequency of theoscillating charges in the material.

Magnitude of theoscillating charge

Number of oscillatingcharges per unit volume

Mass of theoscillating charges

Permittivity offree space


1. A light at the bottom of a 2m deep swimming pool is switched on.What is the diameter of the beam at the surface of the water ifnwater=1.33?

d

2m

Incidentlight

Refractedlightθr

θini

nr

Solution:

o

1C

i

rC

75.4833.11sin

nnsin

=

=θ

=θ

−

m6.4dm28.2r

2r75.48tan

=

=

=

48.75o

2 m

r

2.4.6 Example



2.5 Mirrors

2rf

r2

's1

s1

=

=+

s's

hh

m1

2

−=

−=

Summary

Mirror equation

Mirror magnification

111Waves and Optics


2.5.1 Mirrors

Mirrors have two sides:

Light rays coming from objects are reflected in mirrors to form an imageWhat we see in a mirror:

Principal axis

Concave mirror

Real Virtual

+s

+s'

Realobject

Realimage

The object distance “s” is positive and the object is real if the object is onthe real side. If the object appears to exist on the virtual side, the objectdistance is negative and the object is a virtual object.

The image distance “s’ “ is positive and the image is “real” if the image ison the real side. The image distance is negative and virtual if it lies on thevirtual side.

+s −s'

Realobject Virtual

image

Real Virtual

Rays of light appearto come from theimage but do notactually do so.

A real object is one where the rays of light actually come from the object.A virtual object is one where rays of light appear to come from the objectbut do not actually do so.

Real side Virtual side

the side fromwhich the incidentlight comes.

The back side wherethere are no light rayspresent.

Convex mirror



Radius of curvature: The radius of curvature “r” is positive if the centreof curvature “c” is on the real side, and negative if the centre of curvatureis on the virtual side.

r is negative(convex mirror)

r is positive(concave mirror)

r

The focal length (f) is positive if the centre of curvature lies on the real sideand negative if the centre of curvature lies on the virtual side. The focallength f has the same sign as r.

Realside

Virtualside

Convex mirror: f is negativeConcave mirror: f is positive

Focal length: Any incident rays which are parallel to the principal axis arereflected such that the reflected rays appear to come from a point half waybetween the surface of the mirror and the centre of curvature (this is for aspherical mirror only - not a lens).

Conversely, any rays passing through, or travelling towards, the focal pointwill be reflected parallel to the principal axis.

r

Real side Virtualside

Principalaxis

rr

c c

−f +f

c

c

Convex Concave

2.5.2 Sign conventions

113Waves and Optics


The path of two light rays through a mirror system are known no matterwhat kind of mirror or object is involved. The two rays are:1. Any incident rays which are parallel to the principal axis are reflected

such that the reflected rays appear to come from the focal point “f”.

2. All rays passing through (or travelling towards) the centre of curvature“c” of the mirror strike the mirror at right angles and hence arereflected back along their original path.

Realside

Virtualside

Object

Image

c f

1

2

Concave

The eye interprets the light rays emanating from a point in space ascoming from an object. The image formed by the mirror is the object forthe lens of the eye.

c

ImageActualobject

Realside

ObjectImage

cf

Convex

1

2

2.5.3 Ray diagrams

The eye “sees” the image of theactual object in the mirror. Theimage may be magnified ordiminished in size, inverted orupright in orientation, dependingon where the object is situatedwith respect to the mirror and thecurvature of the mirror.The object distance, image distance,focal length and radius of curvatureare related by:

2rf

r2

's1

s1

=

=+

The linear magnification “m” is theratio of the image size to the objectsize:

If m is +ve, thenimage is uprightif m is -ve, thenimage is inverteds

'shh

m1

2

−=

−=

This formula onlyapplies to rays whichmake a small angle tothe principal axis(paraxial rays).



2.5.4 Example

1. Professor Smith uses a concave shaving mirror with a focal length of450 mm. How far away should Prof. Smith’s face be from the mirrorfor him to see an image of his face which is upright, and twice itsactual size?

f = 0.450 mmm = +2

upright

m225.0s450.0's

450.01

's1

's2

2'ss

s'sm

f1

's1

s1

=

−=

=+

−

−=

−=

=+

Object Image

cf

s =?

Now

Thus

Solution:

115Waves and Optics


2.6 Lenses

f1

's1

s1

=+

s's

h'hm −=−=

f1P =

−

−=

211

2

r1

r11

nn

f1

( )

−+= ls

f11

ssM o

fsM o

=

1fs

M o+=

Angular magnification


Thin lens equation

Lens maker’s equation

Linear magnification

Lens power


Summary



Rays passing through a transparentmaterial with parallel sides aredisplaced, but not deviated indirection.

Rays passing through atransparent material withnonparallel sides are deviatedin direction.

The surfaces of a lens are shaped sothat all parallel rays are deviated sothat they meet at a single point.

Why is this curious propertyof a lens important? Itunderlies the operation of:• optical instruments• eyes• eye glasses and contact

lenses

A thin lens isone whosespherical surfaceshave a radiuslarge incomparison withthe thickness ofthe lens.Equations forthin lenses arerelatively simple.

R

t

Parallelrays meetat focalpoint

and thendivergefrom focalpoint

Para

llel r

ays

plane wave frontssphericalwavefronts

get changed into

Every lens has twofocal points, one oneach side.

2.6.1 Thin lens

117Waves and Optics


Object

Image

Light rays travel in straight lines. When an object is illuminated, or isself-luminous (e.g. a candle or light bulb), rays of light emanate from theobject.

Light raysemanating in alldirections fromobject

What happens to these light rays?Some of them go off and strike othersurfaces and are reflected, refractedor absorbed. Some go off into space,and some may enter someone’s eye.

Now, if it is desired to form an image of the object on a screen (say aprojector screen, or the retina of the eye), then a lens is necessary. Considerthe rays of light which emanate from the top of the arrow shown below:

These rays go off and strike othersurfaces

These rays strike thesurface of the lensand get refracted soas to meet at a singlepoint.

If a screen is placed where the rays meet, then an image of the object willappear on the screen. Only those rays which are intercepted by the lenscontribute to forming the image. A larger lens gathers more of the rays andso the image is brighter. Large telescopes have large lenses so as to gatheras many of the light rays coming from a faint star as possible.

2.6.2 Lens action



A lens consists of two refracting surfaces. Here we consider twospherical surfaces which are positioned close together - a thin lens.

For light rays travelling from left to right through a lens, an image is “real”if it is formed on the right hand side of the lens and “virtual” if formed onthe left hand side. These terms are just labels for identifying whichside of the lens the image or objects are. If the rays diverge from an object,the object is real and is on the left side of the lens. If the rays (going fromleft to right) converge to the object, then the object is virtual and lies on theright side of the lens.

The general convention,for objects or images, isthat “real” is positive.Note that unlike amirror, the virtual sideof a lens for images isthe real side for objects.

Virtualside

Realobject

Real side

+s

+s’

+r1c

Convexlens:

+f Realimage

Labels shown here refer to the image

The radius of curvature is +ve if itlies on the real side. In all formulas,r1 is the radius of curvature of thefirst surface upon which light isincident.

For a lens, the focal length is positive ifthe incident parallel light converges toform a real image (e.g. convex). Focallength is negative if incident parallellight diverges to form a virtual image(e.g. concave).

The image shown here is a virtualimage since it is formed on the lefthand side of the lens and thus theimage distance s’ is negative.

Virtualside

Realobject

Realside

-r1c

Concavelens:

-f

Virtualimage

−s'

2.6.3 Lenses

+s

119Waves and Optics


Convex lens:

The path of two light rays through a thin lens system are known no matterwhat kind of lens or object is involved. The two rays are:1. A ray that passes through the centre

of a lens continues on in a straightline.

2. A ray that travels parallel to theprincipal axis will emerge from thelens in a direction towards the focalpoint (and vice versa).

Note, compared to a mirror: a lens hastwo focal points, one on each side ofthe lens and two radii of curvature. r1 isthe radius of curvature of the firstsurface upon which light is incident.The focal length is not one half of theradius of curvature for a lens. The twofocal lengths for a lens are equal, evenif the radii of curvature for each sideare different (if the medium is the sameon each side of the lens e.g. air).

The eye interprets the light rays emanating from a point in space ascoming from an “object”. The image formed by the lens is the object forthe lens of the eye. Compared to the actual object, the image may bemagnified or diminished in size, inverted or upright in orientation,depending on where the object is situated with respect to the lens and thenature of the lens.

Concave lens:

Virtualside

object

image

Realside

What does theeye see?

object

f

image

f

2.6.4 Ray diagrams



For image to be in focussed on the retina, focal length is madesmaller (eye muscles contract and increase curvature of lens) sothat image distance s’ remains fixed (see thin lens eqn later)

The retina of the eye is a “screen” upon which a real image is focussed. Imagedistance s’ is fixed by the dimensions of the eye.

Eye muscles for lens encirclethe lens. Contraction causeslens to bulge more, r and fdecrease. When musclerelaxed, f is largest.

2. Examine what happens if the object isbrought closer to the eye:

1. Object is at a distance s from the lens of the eye.

s

s'

f

f

s

s'

f

f

Note that mostof the focussingis really doneby the cornea.Lens providesonly a “fine”adjustment.

3. If the object is now placed a great distancefrom the eye.

s = infinity

s' = f

f

f

The eye muscles relax, f is increased, parallel raysare brought to a focus on the retina where the focallength = image distance.

2.6.5 The eye

121Waves and Optics


For far-sighted people, the near point is much greater than 250 mm andobjects far away from the eye can only be focussed. Light from nearbyobjects is focussed beyond the retina. A correcting lens (convex) isrequired to allow nearby objects to be brought into focus.

The near point is the closest distance that an object may be placed in frontof the eye which may be brought into focus on the retina. Usually about250mm for a normal eye. This distance is limited by the elasticity of thecrystalline lens in the eye.

s

s'

f

f s

s'

f

f

Far-sightedeye

Normal eye

Blurredimage

The far point is the furthest distance that an object may be placed for itsimage to be focussed on the retina (usually infinity).

The lens in a near-sightedeye cannot relax enough toallow parallel incoming lightto be focussed on the retina.Light is focussed in front ofthe retina. A diverging orconcave lens is required forimage to be focussed onretina.

s = infinity

s'

f

f

Normal eye

s'

f

f

Near-sighted eye

Blurredimage

2.6.6 Near point and far point

s = infinity



The object distance, image distance, focal length and radius of curvatureare related by the thin lens and lens-makers equations:

f1

's1

s1

=+

The linear magnification “m” is the ratio of the image size to theobject size:

If m is positive, thenimage is upright.if m is negative,then image isinverted.s

'sh'hm

−=

−=

Medium

Lens material

Thin lens equation(paraxial rays)

Lens-makersequation

The power of a thin lens in air is given by the reciprocal of the focal lengthand has units dioptres.

f1P =

All distances aremeasured from thecentre of the lens

metres

Virtualside

Realobject

Realside

+s

+s'

+r1 c

+f Realimage h’

h

2.6.7 Lens equations

−

−=

211

2

r1

r11

nn

f1

123Waves and Optics


2. If the object to be examined is positioned a little closer than the focalpoint. This causes the rays to diverge from the lens thus forming avirtual image behind the actual object.

1. If the object is placed a little further away from the focalpoint, then the image is real, inverted and magnified.

Object

Imagef

3. The eyes interpret light rays as if they are emanating from a largerobject at distance s'.

Note, at s = 2f, then s’=2fand image is inverted butnot magnified.

mfs

fs's

fsfs's

f1

's1

s1

=

−

−

=−

−

=

=+

This formula shows thatthe magnification for afixed focal length lensdepends on the objectdistance. e.g., movingobject closer to lensincreases m.

−s'

s

f

4. How muchmagnification is there?

hh’

2.6.8 Magnifying glass



In an optical instrument, the important thing is the size of the image formedon the retina. This determines the apparent size of the object. It can beshown that:

o

iMθ

θ=

Objects can be madeto appear larger bybringing them closerto the eye. But theeye can onlyaccommodateobjects bought to adistance not lessthan the near point(usually about 250mm) .

( )

−+= ls

f11

ss

M o

where so is the distance from the object to the eye (usually taken to be thenear point) without lens, and f is the focal length of the lens. When the lensis placed so that the eye is at the focal point l = f, or the object is placed atthe focal s = f point, then:

fsM o

=

s'

so = 250 mm

θo

s'

s

θi

Actualobject

Imageformed bymagnifierbecomesobject forlens in eye.

Depends on where actual object is placedw.r.t. focal point of lens. Illustrated here is s <fso that virtual upright image is obtained.

Lens

The magnifying power ofan optical instrument isdefined as the angularmagnification:

Angle of objectat eye whenviewed withoutlens.

Angle of imageof the object ateye whenviewed withlens.

l = f

Magnifying power can be increased by decreasing the focal length. But,various aberrations limit the value of M to about 3X to 4X for a singleconvex lens. If more magnifying power is needed, then need to use acompound arrangement of lenses.

≈250mm

With the magnifying lens in place (hereshown at l = f), the object may be boughtcloser to the eye to increase M.

1fs

M o+=

If eye is bought up close to thelens (l = 0) and image at s = so,then:

2.6.9 Magnification

125Waves and Optics


For a simple magnifier, what do you see when the object is placed at thefocal point of the lens?

∞=

=

=+

's

0's

1f1

's1

f1Consider the thin

lens equationwith s = f.

the image isformed atinfinity!

Parallel rays enter the eye which then focuses these rays onto the retina. Thefocal length of the lens in the eye is thus the diameter of the eyeball in thisinstance and the eye muscle is relaxed - most comfortable viewing.

∞=

−=

s'sm

What about the linearmagnification?

the image isinfinitely large!

“At infinity” means that the object (for the eye) is sufficiently far away sothat light rays from a point on the object arrive at the eye are virtuallyparallel.The image becomes an “object” for the lens of the eye. The object ( as faras the eye is concerned) is now at a very large distance from the eye. But,no matter how far away, it still subtends some (small) angle.

Remember, it is the angular magnification which isperceived by the eye. For an image at infinity, theangular magnification depends only on f (i.e. withreference to a standard near point of 250 mm).

fsM o

=

Object

Imagef



We have seen that the refractive index of materials depends upon thewavelength of the light being refracted since the speed of light in a mediumdepends on the wavelength as well as the properties of the medium.

Chromatic aberration

Thus, when incident white light is refracted by the lens surfaces, smallwavelengths (e.g. blue) will be refracted at greater angles at each surfacethan long wavelengths (red light).

Rays ofdifferentwavelengths donot meet atsame point

The refractive index ofthe lens material dependson the wavelength of thelight being refracted. Foroptical instruments, bestto use yellow light tocalculate focal lengthsand radii of curvature.

Spherical aberrationFormulas and discussions so far have assumed that rays of light passingthrough the spherical lens are paraxial (i.e. make small angles with the axisof the lens). This is not always the case as shown below:

A point in the object is focussedas a diffused circle in the im ageplane. The circle has a minimumradius at what is called the circleof least confusion.

Non paraxial rays

This is because in the derivation of the thin lens equation, it is assumed thatsinθ = θ in Snell’s law so that n1θ1 = n2θ2.

θ

2.6.10 Lens aberrations

127Waves and Optics


1. George has trouble seeing the blackboard in Physics lectures. He getshis eyes tested and finds that he can only see objects clearly if they areless than 100 mm away from him.

10m1.0f

01.0

1f1s

m1.0's

−=

−=

+

−

=

∞=

−=(a) (b)

Note, objects at infinity need to appear as if they were 0.1 m fromGeorge’s eyes for him to see them sharply. s' is negative since imageformed by spectacles will be on virtual side of lens and will serve as anobject for the lens in the eye.

( )

( )

( )

mm110Rmm110R55.0100

1R2

R1

R1155.1

1001

R1

R11n

f1

2

1

11

21

=

−=

=

−−−=

−

−−=

Concave

Solution:

(a) What power of lens would George need forspectacles to enable him to see objects that arefar away?

(b) If glass of refractive index 1.55 is to be used tomake the lenses, what radius of curvature isrequired (assume R2=-R1)?

2.6.11 Example


Dioptres


2.7 Optical instruments

oe

'o

oeTotal

ffs250

MmM

=

= Microscope

Astronomicaltelescope

e

o

ff

M =

Summary

129Waves and Optics


1. Thin lenses in contact

2. Thin lenses not in contact

1fs

M o+=

In an optical instrument, the eye is usually bought very close to theeyepiece or ocular. In this case, the angular magnification is:

For a single lens magnifier, this is usually limited to about 4X due to lensaberrations. How then to increase the magnification of an object and retaina good quality image? We use lenses in combination. There are two waysof combining lenses:

When two thin lenses are placed very close together, we say they are “incontact”. The power of the combination of the lenses is the sum of thepowers of each lens.

21 f1

f1

f1

+=This formula applies to combinations of+ve (convex) and -ve (concave) lenses.

Combinations of lenses not in contact enables optical instruments ofvarying design to be constructed.

The general procedure is to treat the image made by the first lens as theobject for the second lens. One must be very careful about image andobject distances, since some objects must be classed as being “virtual” or“real” depending on which side of the lens they exist.

the +1 factor comes into itwhen the eye is bought veryclose to the lens

2.7.1 Optical instruments



Lens 1 forms a real,inverted image o f theobject O1. The imageI1 becomes the objectO2 for lens 2. In thisexample, the objectO2 is inside the focallength of lens 2. Thefinal image, I2 is avirtual image (on theleft of lens 2).

Each lens is treated separately. Rays are not drawn through lens 1 and thenthrough lens 2 in one operation. Rays can only be drawn using the twoknown directions of rays (through focus and centre) in a step-by-stepprocedure.

Virtualside

RealobjectO1

Realside

+so +so'

Realimage I1

Lens 1 (objective) Lens 2 (eyepiece)

For the largest magnification (i.e. retinal image),the final image should be placed at the near point(se' = 250mm). It can be shown that the totalangular magnification is: oe

'o

oeTotal

ffs250

MmM

=

=

Virtualside

Realside

+se

−se'

VirtualimageI2

For the purposes of rayconstruction, lenses canbe assumed to be ofinfinite transverse size.

Real image fromlens 1 becomesreal object forlens 2

2.7.2 Microscope

1.

2.

131Waves and Optics


In an astronomical telescope, the object O1 is usually at infinity hence theimage I1 is formed at the focal length of the objective lens (i.e. lens 1). Forrelaxed viewing, the final image should also be at infinity, hence, lens 2(the eyepiece) is placed so that the object O2 is at f2 ( = fe).

The angular magnification is the ratio of the angle subtended by the finalimage I2 formed by lens 2 at the eye θe to the angle θo subtended by theobject O1 at the objective lens (lens 1) (which is the same as that subtendedby the object at the unaided eye) .

e

o

ff

M =

If h' is the height of the image I1, then:

e

o

o

e

ee

e

oo

o

ff

M

f'htan

f'htan

=θ

θ=

θ==θ

θ==θ

objective(lens 1)

eyepiece (lens 2)

since bothangles are small

Virtualside

Parallelrays fromreal objectat infinity Real

side

+so = infinity

+so' = fo

Lens 1 objective

Lens 2eyepiece

θo

Parallel raysfrom final imageat s2' = infinity

se = fe

Virtualside

Realside

θe

I1

2.7.3 Astronomical telescope

We might well ask, if thefinal image is at infinity,how then do we seeanything? We will get animage on a screen if thescreen is placed at infinity.But, if the rays enter oureye, then the lens in theeye focuses the rays on tothe screen of our retina.



A particularly interesting case arises when the image formed by the first lens(the objective) is not between the two lenses in combination. For example,with the lens combination shown below, with lens 2 not in position, theimage I1 is to the right of the position of lens 2. I1 then becomes a virtualobject for lens 2.

Object

f1 f2

I1

I2

A

Virtualimage

Real image fromlens 1 becomesa virtual objectfor lens 2

1

2

2

Galileantelescope

Object

f1

f1 f2f2

I1

I2

2

Rays from tipof objectmeet here

Real image for lens1 becomes virtualobject for lens 2.Object distance s2is thus negative.

1

Note again thatlenses are“infinite” in sizefor thepurposes of rayconstruction.

1

2

2.7.4 Special case

f1

f2

133Waves and Optics


2.7.5 Example

Solution:

041.0mmm756.011588m

053.06000315m

mm88's's

1115

1501

1158.315200s

mm8.315's's

16000

1300

1

21T

2

1

2

==

−=

−

−

−=

−=

−=

−=

+

−

=−

−=

−=

=

+=

2. A convex lens (f = 300 mm) is placed 200 mm from a concave lens(f = −50mm). An object is placed 6 m away from the convex lens.Determine the position, magnification and nature of the final image.

88 mm virtual side of lens 2 with magnification 0.041

object

200 mm

#1 #2



2.8 Interference

Summary

λ=θ nsindn = 0,1,2,...

λ+

=θ2

1n2sindn = 0,1,2,...



θ

λ

π= sindcosII 2

max Fringe intensity

d2mx

=

λ=∆

d221mx

=

λ

+=∆

Thin film interference maxima

Thin film interference minima

135Waves and Optics


Interference arises from the superposition of waves.

phase difference (ifany) between themremains constant.

Constructive DestructiveAmplitude ofresultant is less thancomponents.

Amplitude ofresultant is greaterthan components.

Consider two point sources S1 and S2 of monochromatic, coherent lightseparated by a distance d emitting spherical wave fronts.

light of a “single”wavelength

Ray 1 has a greater pathto travel to the screencompared to ray 2. Thepath difference is ∆x.

Because of this path difference, the rays, which may be initially in phase,may arrive at P out of phase. If the phase difference ∆φ at P is 180o (π),then destructive interference occurs. The distance ∆x must thus be equal toλ/2. If the phase difference at P is 0, then constructive interference and thedistance ∆x = λ.



λ=θ nsind

λ+

=θ2

1n2sindn = 0,1,2,...

The phase difference arises fromdifference in path length ∆x = dsinθ:

When dsinθ = nλ, ∆φ = n2π

– an integral number of 2π.

( )( )φ∆+−ω=

−ω=

kxtsinAykxtsinAy

2

1

d

1

2

BPAPx −=∆

P

θ

θ≈∆ sindx

θ

A

B

θλ

π=φ∆

ω∆=∆λ

π

ωπ

λ=

∆

∆

λπ

ω=

sind2

tx22t

x2

c

2.8.1 Interference

If the screen is a largedistance away fromsource, then rays 1and 2 areapproximately parallel,therefore:


n = 0,1,2,...

∆φ = ∆(ωt)

ωt


( )dD1ny

dDny

Dy

dn

Dytan;

dnsin

λ+=

λ=

=λ

=θλ

=θ

What is the distance ∆y between the bright fringes if thescreen is placed a distance D from the sources?

For small θ in radians,sin θ = tan θ

d

dDy

λ=θ∆

λ=∆

The distance between any two fringesis thus:

The angularseparation is tan θ:Note: these formulas apply forlarge value of D compared to d.

d 1

2

Brightfringe

Constructiveinterference

Destructiveinterference

λ=θ nsindn = 0,1,2,...

λ+

=θ2

1n2sindn = 0,1,2,...

leads to an increase inamplitude of the wave

leads to a decrease inamplitude of the wave

Production ofinterferencefringes

D

Brightfringe

Brightfringe

n = 0

n = 1

n = 1

n = 0

n = 0

Dark

Dark

y

θ

Eωt

Eωt

θ

θ

2.8.2 Fringe spacing

for the nextadjacent fringe

distance from centre tonth bright fringe

Intensity I is the amount ofpower per square metre.For an E field, the averageintensity is given by:

Note that the intensity isproportional to the squareof the amplitude Eo.

2ooav E

21cI ε=

137Waves and Optics


For two waves with the same amplitude Eo, then the interference maximawill have an amplitude 2Eo and the interference minima will have zeroamplitude. What is the amplitude (and intensity) of the interference fringepattern as a function of angle θ?

θ=∆ sindx

Brightfringe

D

Brightfringe

Brightfringe

n = 0

n = 1

n = 1

n = 0

n = 0

Dark

Dark

y

d θ

θ

θ

Note: these angles are also equalto θ since both rays are “parallel”for screen large distance from slits

Fringes ofequal intensity

For light, the amplitude A is taken to be theamplitude of the E vector. Intensity is what the eyeresponds to and is a measure of power (or energy) density.

API = Intensity (Wm−2) is proportional to

the square of the amplitude Eo.

If the two waves of amplitude Eo arrive at the screen out of phase by anangle ∆φ, then it can be shown that the resultant amplitude is:

2cosE2E oR

φ∆=

It can be seen for ∆φ = 0, the maximum amplitude of theresultant ERmax is twice that of each component wave Eo andfor minima, ∆φ = π and the amplitude is zero.

The intensity is proportional to the square of the amplitude. Thus, thevariation of intensity with θ is given by:

θ

λ

π=

θ

λπ

=

φ∆=

sindcosII

sindcosI4

2cosI4I

2max

2o

2o θ

λ

π=φ∆ sind2

∆φ is in radians

Imax is the maximum intensity of the resultant interference pattern. Io is themaximum intensity of each individual component wave.

Note: This formula actually says thatthe peaks all have the same intensitysince dsinθ = nλ. In practice, this isnot the case due to diffraction.

2.8.3 Fringe intensity



θ=∆

θ=

=

+=∆

cosd2x

, thusandcosABd

,butAB2

BCABx

θ=

λ=∆

cosd2

mx where m is an integer. For near normal incidence, cos θ = 1 (i.e.θ = 0). However, one must be very careful about changes ofphases at interfaces since this will influence the condition forconstructive or destructive interference.

Interference maxima occur when thepath difference is an integral numberof wavelengths:

The path difference taken by Ray 2compared to Ray 1 is simply:

For a wave travelling from material n1 to n2, a phase change of π occurson reflection if n1 < n2. No phase change when n1 > n2.

For a film with a thickness d, and light at normal incidence, interferencemaxima occur when:

d2mx

=

λ=∆ m = 0,1,2...

This applies if neither or both have a half-cyclephase shift on reflection. If one or the other raysundergoes a phase shift, then this becomes thecondition for destructive interference.

d221mx

=

λ

+=∆

This is the condition for destructive interferenceif neither or both waves half a half-cyclereflection phase shift. If one or the other rays dohave such a phase shift, then this becomescondition for constructive interference.

Interference minima

In thisexample,n1<n2<n3

d

A

B

CD

θ

1 2

Partiallytransmittedray

Phasechange.

Incidentray

n1

n2

n3

Nophasechange

Phasechange

Nophasechange

2.8.4 Thin film interference

139Waves and Optics


When light travels through a medium,it does so at a lower velocity (v)compared to that in a vacuum (c). n

cv =

Velocity oflight in vacuum

Refractive indexof medium

Now

( )( ) ctxn

nc

v

=

=

=

timedistance

The product (n)(x) is the distance through which a light wave would travelif it were travelling with a speed c during time interval t. That is, since thevelocity of the light in the medium (of thickness x) is v, which is less thanc, then if the light wave were to travel through a vacuum with velocity c,then the distance travelled would be greater for the same time period t.

(n)(x) = optical path length (o.p.l.)

If the distance is thethickness x of themedium, then:

Light wave takestime t to traversedistance t throughmedium at velocity v

Light wavetravels a greaterdistance, nt, in avacuum withspeed c in thesame time t

x

v

nx

c

Why bother with optical path length?Because when working out whether thepath difference is an integral number ofwavelengths (for constructiveinterference) we need to work out theoptical path difference. That is, the pathdifference that two light waves wouldhave if they were both travelling through avacuum with speed c.

x

nx

n = 1n > 1

n = 1

When light waves strike“denser” medium, velocity andwavelength both decrease,frequency remains constant.

2.8.5 Optical path length



1. What colour does an oil film 300 nm thick appear tothe eye when illuminated at normal incidence?

( )nm600

103002

d2mx

9

=

×=λ

=

λ=∆

−

Solution:

Constructive interference at thiswavelength, therefore the filmwill appear reddish

2. A double slit interferometer uses light of 620 nm. A thin film oftransparent material (n = 1.6) is placed in the path of one of the beams.A shift of 50 bright fringes is observed. Determine the thickness of thefilm.

( )( )

( )m5.51x

1062050x6.0

50x16.1x1n

xxnx

9

r

r

µ=

×=

λ=−

−=

−=∆

−

Now, with film in position,the change in optical pathlength is ∆x = (nx − x) andthis corresponds to 50fringes. Thus, ∆x = 50λ.

Solution:

2.8.6 Examples

141Waves and Optics


2.9 Diffraction

Summary

θλ

π=α

α

α

=

sina22

2sin

II 2

2

max Single slit diffraction pattern

Phase difference

( )

α

α

θ

λ

π= 2

22

max2

2sinsindcosII Double slitdiffraction pattern

a22.1sin λ

=θ≈θCircular aperturediffraction pattern



Constructive and/or destructive interference arises from the superpositionof waves. Previously, we examined the interference pattern generated bytwo coherent point sources of light.

Phase difference:

θλ

π=φ∆ sind2

θ

λ

π= sindcosII 2

max

d

θ=∆ sindx

Bright

D

bright

Bright

n = 0

n = 1

n = 1

n=0

n=0

Dark

Dark

θ

dDy λ

=∆

The term diffraction describes the interference pattern arising from thesuperposition of an infinite number point sources of light waves and ariseswhen the light waves interact with some obstruction in the light path.Diffraction effects become more pronounced when the dimensions of theobject are comparable with the wavelength of the light.A precise mathematical treatment of diffraction is very complicated. Asimpler, but qualitative explanation may be found using Huygen’s principle.

2.9.1 Interference and diffraction

Constructiveinterference

Destructiveinterference

λ=θ nsindn = 0,1,2,...

λ+

=θ2

1n2sindn = 0,1,2,...

A plane wave is equivalent to that produced by an infinitenumber of stationary point sources

Consider an infinite number of point sources arranged in a straight line:

Spherical waves leave each source withvelocity v. At a time t, the wave fronts havemoved through a distance r = vt. Thus,circles of radius r give the position of thespherical wave front at a time t after leavingthe source.

r = vt

The envelope of these circles is a lineparallel with the line of the sources (whetherthis line be straight or curved, it doesn’tmatter). In this example, the the enveloperepresents a plane wave.

v

143Waves and Optics


Consider what happens when an obstruction is placed inthe path of the wave:

Waves tend to “bend” around the obstruction. What is the amplitude (andhence intensity) of the distorted wave at a distant point P?

Note: the presence of theobstacle has removedpositions of some of thespherical wavefronts and theresultant, at a distance point Pis different to that without theobstacle.

Obstacle

In this region, (theunobstructed path)spherical waves addconstructively to form aplane wave front.

Near the edges, someportions of the sphericalwavefronts are missing.The obstacle has removedsome of the spherical wavefronts.

P

If P is very distantfrom the aperture,then the rays whicheventually meet at Pare approximatelyparallel.

Conditions forFraunhoferdiffraction. WhenP is very close tothe aperture, thenthe resultingdiffraction patternis different and iscalled Fresneldiffraction.

the diffraction pattern

r = vt

2.9.3 Diffraction - circular aperture



slita

At a large distance from slit, rayson the screen are virtually parallel(not shown parallel here) andarrive at P at some angle 90-θ.

Now consider the wave as it passes through the slit and beyond. Let the slitbe represented by a total of N point sources within the slit.

Let P be the point of thefirst minimum (destructiveinterference). The firstminimum occurs when thepath difference ∆x = λ/2.

If the spacing between thepoint sources is da, theproduct ∆nda is the distancebetween the nth and (n + 1)th

sources (starting from n = 0).The condition for destructiveinterference is:

( ) θ∆=λ sinnda2

Difference in path lengthto central point is zero soa central point is amaximum.

If, for a “certain” θ, the waves from sources n = 0 and n = 2 satisfy thiscondition (∆n = 2), then, the intensity at point P will be diminished due tothe destructive interference between these two rays. But, a ray leavingfrom, say n = 1 will not necessarily satisfy this condition hence there maybe some light at P emanating from the source n = 1.

BUT, the ray leaving from n = 1 and going to P will cause destructiveinterference with that leaving from n = 3 (∆n = 2) since the pathdifference for this pair is also λ/2 (remember that the light rays arevirtually parallel as they travel to P). Hence, the light from n = 1 iscancelled by that leaving from n = 3. This same argument applies to all ofthe point sources from A to C if there is always a pair of rays, separatedby ∆n da which results in complete destructive interference at P.What is the value of θ for the first minimum?

P

D

∆x

View

ing

scre

enAθ

da B

“parallel” rays

C

90-θ

2.9.4 Diffraction - single slit

145Waves and Optics


a = 0.02 mm

θ=λ sinam

The angular position of the first, and all subsequent, minima are found from:

m = 1,2,3….. but not m = 0

the centralmaximum

slit width

The interference maxima occur (only approximately) midway between theminima and the central maximum has twice the width of the others.

Note: as a decreases (slitgets narrower), the angle θincreases (pattern spreadsout further). As slit getswider (a increases) theangle θ decreases(conditions approach thatof geometrical optics - nospreading out of rays).

For the first minimum each side of the central maximum, the angle θ isgenerally small and thus, with m = 1, we have:

θ≈λ

aHalf-width ofcentral maximum.

As θ increases from 0, then it is evident that rays from sources at theextreme ends of the slit (∆n = N) will first meet the condition for destructiveinterference (i.e. their path length will first differ by λ/2, the path differencesfor all other sources being smaller). Thus, it may be at first thought that thecondition λ/2 = asinθ (since a = Nda) will apply to the first minimum.

However, although these two rays will interfere destructively at P, the otherrays will not since they are not paired with any corresponding ray with apath difference of λ/2. It is not until the source located at half way, “B”matches with the source at the edge “A” do sources in between have amatching ray at ∆x = λ/2. Thus, the condition for the first absolute minimumis:

θ=λ sin

2a

2

This formula gives the direction of the minima(and hence the maxima) but says nothing about theintensity distribution in the diffraction pattern.

a = 0.08 mm

a = 0.04 mm



How to get the intensity of the diffraction pattern as a function of θ?Each point within the slit acts as a source of waves. Proceeding down theslit, each wave will be progressively more out of phase than the source atthe top of the slit until we reach the condition for the first minimum and thecycle of phase shifts repeats. Between each source, there is an equal phasedifference ∆α. The total phase difference between the top and bottomwaves is:

Addition of individual E vectors, each with a phase difference∆α, shows that the resultant amplitude variation, as a functionof θ, is:

Thus, since I is proportional to theamplitude E squared:

2

2sin

EE oα

α

=

θλ

π=α

α

α

=

sina22

2sin

II 2

2

maxNote: This equation is indeterminateat α = 0 (i.e. parallel rays from the slitto the central maximum θ = 0).However, it can be shown (usingcalculus) that I converges to Imax atthis condition.

θλ

π=α sina2

2π/λ x path difference

At what angle θ is intensity = 0? When α/2 is a multiple of π.

+

2πθ=λ

π=

θλ

π=

α

sinamm

sina2

as before

At what θ is the intensity a maximum? When α/2 is an odd multiple of π/2.

Check on directions:

Only approximately correct. Presence of α/2 in the denominator affects the value ofI as well as the sine function in the numerator.

Note: this is different to the expression given previously forthe addition of two waves of amplitude Eo and phasedifference φ. Here we have an “infinite” number of pointsources separated by a phase difference ∆α. The resultantamplitude E is expressed here in terms of the total phasedifference a between the first source and the last.

147Waves and Optics


We have previously examined interference from two coherent point sourcesand saw that the condition for constructive interference was:

and (without proving) that the intensity of the fringe pattern (which we nowcall the “diffraction pattern”) is written:

θ

λ

π= sindcosII 2

max

Now, for a single slit, minima occur at:

λ=θ nsind n = 0,1,2,...

m aλ θ= sinAs the slit gets narrower (a decreases) the angle θ forthe first minimum increases (pattern spreads outfurther). When a = λ, (at m = 1) sinθ = 1 and θ = 90o.

Θ = 90ο

The intensity distribution for two such slits is that previously calculated fortwo point sources separated by a distance d. However, if the slits are nowmade a little wider than λ, then the intensity distribution for the two pointsource condition is modulated by the single-slit diffraction pattern.

Single-slitdiffractionpattern

Double-slitdiffractionpattern

( )

θλ

π=α

α

α=

sina22

2sinII 2

2

max ( )

α

α

θ

λ

π= 2

22

max2

2sinsindcosII

a = 0.04 mm

Diffraction envelopeor “modulation”determined by α

Interference finestructure termdetermined by d

a = 0.04 mm

d = 0.50 mm

d = 0.25 mm

2.9.5 Diffraction - double slit



For plane waves arriving at a circular aperture, the resulting far-fielddiffraction pattern is similar to that obtained from a single slit but consistsof circular, rather than straight, fringes. The angular separation between thethe central bright spot and the first minimum is given by:

a22.1sin λ

=θ≈θ

Wavelengthof light

The factor 1.22 comes aboutbecause it is the smallest root of afirst-order Bessel function. Forcircular geometries, themathematics are most convenientlyhandled with Bessel functions.

The entrance to the lens of the eye can be considered a circular aperture. Iflight from two point sources is incident on the eye, the diffraction patternfrom each impinges on the retina. Experiment shows that the eye can sensea minimum change in intensity of about 20%. Thus, two sources can beresolved when the intensity between the maxima falls to within about 80%of the peak intensity.This condition is met when the central maximum of one pattern coincideswith the first minimum of the second. This is called the Rayleighcriterion. If dθ is the angle subtended at the eye by the two sources, thenthe resolving power, defined as 1/θ, can be calculated from:

Dda

a22.1sina

≈

θ≈

λ=θ

since θ isusually small

a

θ

iris

Ret

ina

Pointsources

Io 0.81(Io)

θ

D

d

2.9.6 Diffraction - circular aperture

149Waves and Optics


2.9.7 Example

Solution:

1. A microscope has an objective lens of diameter 4 mm and a focallength of 3.2 mm. Calculate the minimum distance between two objectswhich may just be resolved using this lens if the objects are illuminatedwith light of wavelength (a) 500 nm and (b) 620 nm and if the object isplaced at the focal point of the lens.

( )

( )

( )nm605

1089.10032.0d

1089.1sin

1062022.1sin004.0

nm620nm488

10525.10032.0dDd

10525.1sin

22.110500sin004.0

22.1sinanm500

004.0amm0032.0f

4

4

9

4

4

9

=

×=

×=θ

×=θ

=λ

=

×=

=

×=θ

×=θ

λ=θ

=λ

=

=

−

−

−

−

−

−


(a)

(b)


2.10 Polarisation

Summary

θ=2

o cosII Malus’ law

1

2p n

ntan =θ Brewster’s law

151Waves and Optics


Since the most common effects oflight are due to the action of theelectric field E, we shall only show thedirection of E in these diagrams toavoid clutter. At all times, anaccompanying B field is assumed tobe also present.

A light ray is an imaginary linedrawn perpendicular to the Evectors and indicates the directionof travel of the waves.

For a “single” ray of light, if the direction of the E vectors remainsconstant, then the light is said to be polarised in this direction.

It is often convenient to resolvethe E vectors in a light ray intohorizontal and vertical components.For linearly polarised light (θ is aconstant) of constant intensity, themagnitude of the componentsremain constant Direction of

polarisation

θ

Ecosθ

EsinθE

When the direction of polarisationchanges in a regular periodicmanner, the light is said to beelliptically polarised. For ellipticallypolarised light, the magnitudes ofthe components change as themagnitude and direction of Echanges. Circularly polarised light isjust a special case of ellipticallypolarised light and occurs when themagnitude of the E vector remainsconstant but its direction changes.

θ Ecosθ

EsinθE

Path tracedout by tip ofE vector

2.10.1 Polarised light

Light waves are transverse waves. The varying E and B fields have adirection which is at right angles (ie. transverse) to the direction ofpropagation of the wave. The transverse nature of light waves lead to somevery interesting physical effects. The term polarisation is the transverseproperty of the light wave which leads to these effects.



A more complete understanding of the nature of polarised light can behad when one considers the nature of unpolarised light.Most light sources emit light as a result of de-excitation of atoms. Although the light emittedfrom one particular atom may be polarised at aparticular direction, it is unlikely that the plane,or direction, of polarisation of any one atomwill be the same as any other (except in alaser). Thus, most light consists of a collectionof light rays which have all possible directionsof polarisation. Further, the emission of lightfrom one particular atom occurs over a veryshort time period.

For a beam of unpolarised light, all the momentary polarised rays whichmake up the beam can all be resolved into a single vertical and a singlehorizontal component - which are equal in magnitude at any one time dueto the randomness of the atomic oscillators which produce the waves inthe first place - and the very short time in which these rays appear after oneanother in the beam.The magnitude of the components change rapidly and irregularly. Further,unpolarised light does not consist of continuously varying sinusoidal Efields. Thus, the phase of the two components varies randomly (but together)in time.

E E

=

The significance of this is that in a ray of unpolarised light, the direction ofE would be in one direction for an instant, and then another randomdirection in the next instant. In practice, the time intervals between thesechanges in direction of polarisation are so small that the light appears toconsist of E fields of all possible directions at once.

2.10.2 Unpolarised light

153Waves and Optics


Consider a beam of unpolarised light passing through a sheet of polarisingmaterial. The light can be resolved into vertical and horizontal componentsof equal magnitude. The component that is aligned with the molecules in thematerial are absorbed, the other component is transmitted.

A polarising filter consists oflong needle-like molecules allaligned in a particulardirection.

Polarisedlight

Unpolarised lightresolved into verticaland horizontalcomponents.

The direction in which light istransmitted is called thepolarisation direction and isoften marked with an arrow.

2.10.3 Sources of polarised light

154

1

2p

p2

p2p1

21p1

1

2

2

1

nntan

cosn

)90sin(nsinn

90;

nn

sinsin

=θ

θ=

θ−=θ

=θ+θθ=θ

=θ

θ

Brewster’slaw

≈15% reflectedpolarised light

Glassblock

Plane of incidence

≈75%refracted

Incidentbeam

unpolarisedlight

90o

θpn1

n2

Now consider reflection from a surface. The incoming E vector causescharged particles within the atoms of the surface to oscillate resulting inradiation emission. Now, when the incoming light ray strikes an opticallymore dense material (n2 > n1) at an angle θp the polarising angle, E vectorsparallel to the plane of incidence are absorbed and refracted but not reflected.Only the E vectors perpendicular to the plane of incidence are reflected. Thereflected ray is polarised in a direction perpendicular to the plane ofincidence. The polarising angle occurs when the angle between the reflectedand refracted ray is 90o.

The Physics Companion

The overall intensity of thelight is reduced by a halfsince one component isabsorbed and the other istransmitted.


Unpolarised light is incident on a polarising material and polarised lightof intensity half of that of the unpolarised light is transmitted. If thispolarised light is then incident on a second sheet of polarising material,then the intensity of the beam transmitted through the second sheetdepends on the angle of the second polariser with respect to the first.

θ=

θ=

θ=

2o

22o

2o

cosII

cosEE

cosEE

Malus’ law

The intensity of the transmitted beam is amaximum at θ = 0 and zero at θ = 90o.

Since I ∝E2

If Io is the intensity of thepolarised light from the first sheet,then the magnitude of thetransmitted beam through thesecond sheet is:

If a third sheet ispositioned after thesecond and orientedat 90o to the firstsheet, then:

Polarisedlight

E

Eo

transmitted

Eosinθabsorbed

Io

The first sheet isusually called the“polariser” and thesecond sheet the“analyser”.

(1)

(2)

(3)

θ=2

o2 cosII

E3 = Eocosθsinθtransmitted

θθ=22

o3 sincosII

I is amaximum at45, 135, 225and 315o

2.10.4 Malus’ law

155Waves and Optics

Eocosθ


2.10.5 Example

1. Polaroid sunglasses work by reducing glare associated with polarisedlight reflecting from surfaces. For incident sunlight on a pond,determine (a) the angle at which the reflected light is completelypolarised, and (b), the reduction in intensity of the reflected light for aperson wearing sunglasses who sees the reflected light where thepolarised material is at an angle of 30° to the direction of polarisationof the light.

Solution:

θp

Refracted light

Incidentlight

Reflectedlight Io

(a) From Brewster’s law:

n = 1.33

n = 1

°=

=θ

=θ

−

1.53133.1tan

nntan

1p

1

2p

%75

30cosII

cosII

2

o

2o

=

=

θ=

(a) From Malus’ law:

i.e. 25% reduction in intensity



Electricity

Part 3


3.1 Electricity

Summary

221

dqqkF =

EqF 1=

2rrqkE =

AQk4E π=

EA=φ

( ))v(n)q(vnqAI 222111 −−+=

dtdqi =

EdqW

=

RIV

=

RIVIP 2==

AR l

ρ=

dA

VQC oε==

l

2

oNAL µ=

2CV21U =

2LI21U =

21

21AB

21AB

RRRRR

R1

R1

R1

+

=

+=

21AB RRR +=

Force between twochargesForce on a charge ina fieldElectric field - pointcharge

Electric field - pointcharge

Electric flux

Electric current

Electric potential

Ohm's law

Power - resistor

Resisitivity

Capacitance

Inductance

Energy -capacitor

Energy -inductor

Resistors-series

Resistors -parallel



3.1.1 Electric charge

Electrical (and magnetic) effects are a consequence of a property ofmatter called electric charge. Experiments show that there are two typesof charge that we label positive and negative. Experiments also show thatunlike charges attract and like charges repel.

221

dqqkF =

Distancebetween charges

Magnitude ofthe charges

Constant9 × 109 m2C−2

the permittivity of free space= 8.85 × 10−12 Farads/metre

The charge on a body usuallyrefers to its excess or netcharge. The smallest unit ofcharge is that on one electron e= 1.60219 × 10−19 Coulombs.

• Repulsive forces betweenlike charges are negative

• Attractive forces betweenunlike charge are positive

If the two charges are in somesubstance, eg. Air, then the Coulombforce is reduced. Instead of using εo, wemust use ε for the substance. Often, therelative permittivity εr is specified. o

rε

ε

=ε

Material εεεεrVacuum 1Water 80Glass 8

+-

Fq

q

d

The force of attraction or repulsion canbe calculated using Coulomb’s law:

o41kπε

=

- q2q1

+E

F

1. Imagine that one of thecharges is hidden from view.

2. The other charge stillexperiences the Coulombforce and thus we say itis acted upon by anelectric field.

3. If a test charge experiencesa force when placed in acertain place, then anelectric field exists at thatplace. The direction of thefield is taken to be that inwhich a positive test chargewould move in the field.

EqFdqkE

dqqkF

1

22

221

=

=

=

Let

Thus

Note: The origin of thefield E may be due to thepresence of manycharges but themagnitude and directionof the resultant field Ecan be obtained bymeasuring the force F ona single test charge q.

159Electricity


An electric field may be represented by lines of force. The total number oflines is called the electric flux. The number of lines per unit cross-sectional area is the electric field intensity, or simply, the magnitude ofthe electric field.

Uniform electricfield betweentwo chargedparallel plates

Non-uniformfield surroundinga point charge

• Arrows point in direction ofpath taken by a positive testcharge placed in the field

• Number density of linescrossing an area A indicateselectric field intensity

• Lines of force start from apositive charge and alwaysterminate on a negativecharge (even for an isolatedcharge where thecorresponding negativecharge may be quite somedistance away).

R

+

2rrqkE =

AQk4E π=

+ + + + + + + +

- - - - - - - - - - -

Q

Q

Area of a sphere radius R

φ=

ε=

π=

π=

=

∝

∝

π=

o

2

2

2

qkq4

R4RkqEA

RkqE

NEAANE

R4A

2

But

Thus

independent of Rbut proportional toN

by definition

electric flux

Electric flux

k=1/4πεo

How to calculate electric flux (e.g. around a point charge)Note: For an isolatedcharge (or chargedobject) the terminationcharge is so far awaythat it contributes littleto the field. When thetwo charges are closetogether, such as inthe parallel plates,both positive andnegative chargescontribute to thestrength of the field.For the plates, Q isthe charge on eitherplate, a factor of 2 hasalready been includedin the formula.

3.1.2 Electric flux



1. Conductors

3. Semiconductors

2. Insulators

Valence electrons areweakly bound to the atomiclattice and are free to moveabout from atom to atom.

Valence electrons tightlybound to the atomic latticeand are fixed in position.

In semiconductors, valenceelectrons within the crystalstructure of the material not asstrongly bound to the atomiclattice and if given enoughenergy, may become mobileand free to move just like in aconductor.

Atoms consist of a positivelycharged nucleus surrounded bynegatively charged electrons.Solids consist of a fixedarrangement of atoms usuallyarranged in a lattice. The positionof individual atoms within a solidremains constant becausechemical bonds hold the atoms inplace. The behaviour of the outerelectrons of atoms are responsiblefor the formation of chemicalbonds. These outer shell electronsare called valence electrons.

+-

-

-

-

-

-

- Valanceelectrons

Valence electrons onlyshown in these figures

3.1.3 Conductors and insulators

Electrons, especially in conductors, are mobile charge carriers (theyhave a charge, and they are mobile within the atomic lattice).

161Electricity


Let there be n1 positive carriers per unitvolume and n2 negative carriers per unitvolume. Charge carriers move with drift velocities v1 and −v2. In time ∆t,each particle moves a distance l = v1∆t and l = v2∆t.The total positive charge exiting from the right (and entering from the left)during ∆t is thus:

Total chargeCoulombs

Volume

The total negative charge exiting from the left is Q- = n2(−q2)(−v2∆t)A. Thetotal net movement of charge during ∆t is thus:

Mobile charge carriers may be either positively charged (e.g. positiveions in solution), negatively charged (e.g. negative ions, loosely boundvalence electrons). Consider themovement during a time ∆t ofpositive and negative chargecarriers in a conductorof cross-sectional area A andlength l placed in an electric field E:

Charge on onemobile carrier

No. of charge carriersper unit volume

tA)v(n)q(tAvnqQQ 222111 ∆−−+∆=+−+

( )AtvnqQ 111 ∆=+

-v2

+v1

Cross-sectionalarea A

l

E

-

-

--

-+

++ +

+

3.1.4 Electric current

The total charge passing any given point in Coulombs per second is calledelectric current:

( ))v(n)q(vnqAI

A)v(n)q(AvnqtQQ

222111

222111

−−+=

−−+=∆

+−+

Current density J = I/AAmps m−2 orCoulombs m−2 s−1

1 Amp is the rate of flow ofelectric charge when one

Coulomb of electric chargepasses a given point in an

electric circuit in one second.

In metallic conductors, the mobilecharge carriers are negativelycharged electrons hence n1 = 0.

Note that the Amp is a measure of quantity of charge per second andprovides no information about the net drift velocity of the charge carriers(≈ 0.1 mm s−1).

dtdqi =

Lower case quantities refer toinstantaneous values. Upper caserefers to steady-state or DC values.

In general,



Electric current involves the net flow ofelectrical charge carriers, which, in ametallic conductor, are negativelycharged electrons. Often in circuitanalysis, the physical nature of theactual flow of charge is not important -whether it be the flow of free electronsor the movement of positive ions in asolution. But, in the 1830s, no one had heard of the

electron. At that time, Faraday noticed that whencurrent flowed through a wire connected to achemical cell, the anode (positive) lost weight andthe cathode (negative) gained weight, hence it wasconcluded that charge carriers flowed through thewire from positive to negative.

We now know that positive and negative ions in the solutionmove towards the anode and cathode in opposite directionsand electrons in the wire go from negative to positive.

There are two types of chargecarriers: positive and negative. Forhistorical reasons, all laws andrules for electric circuits are basedon the direction that would betaken by positive charge carriers.Thus, in all circuit analysis,imagine that current flows due tothe motion of positive chargecarriers. Current then travels frompositive to negative. This is calledconventional current.

If we need to refer to the actual physicalprocess of conduction, then we refer tothe specific charge carriers appropriateto the conductor being considered.

I+

Positive mobilecharges travelin this direction

This happens dueto the movement ofboth positive andnegative ionswithin the solutioninside the cell.

3.1.5 Conventional current

163Electricity


When a charge carrier is moved from one point to another in an electricfield, its potential energy is changed since this movement involved aforce F moving through a distance.

d

+

+q

q

If q+ is moved against the field, then work isdone on the charge and the potential energyof the charge is increased.

The work done is thus:

The work per unit charge iscalled the electrical potential:

Joules

Joules/Coulomb(Volt)

If a charged particle is released in the field, then work is done on theparticle and it acquires kinetic energy. The force acting on the particle isproportional to the field strength E. The stronger the field, the larger theforce – the greater the acceleration and the greater the rate at which thecharged particle acquires kinetic energy.

qEF =

qEdFdW

=

=

EdqW

=

3.1.6 Potential difference

orNewtons/Coulomb

In a uniform electric field, the potentialdecreases uniformly along the fieldlines and is a potential gradient.

A uniform electric field E exists between two parallel charged plates since apositive test charge placed anywhere within this region will experience adownwards force of uniform value. The electric field also represents apotential gradient.If the negative side of the circuit isgrounded, then the electricalpotential at the negative plate is zeroand increases uniformly through thespace between the plates to the topplate where it is +V.The potential gradient (in volts permetre) is numerically equal to theelectric field strength (Newtons perCoulomb) but is opposite in direction.

+ + + + + + + +V E

0

0V/2

V

- - - - - - - - - d

EdV =



A voltage source, utilising chemical or mechanical means, raises theelectrical potential of mobile charge carriers (usually electrons) within it.There is a net build up of charge at the terminals of a voltage source. Thisnet charge results in an electric field which is channelled through theconductor. Mobile electrons within the conductor thus experience anelectric force and are accelerated.However, as soon as these electronsmove through the conductor, theysuffer collisions with other electronsand fixed atoms and lose velocity andthus some of their kinetic energy. Someof the fixed atoms correspondinglyacquire internal energy (vibrationalmotion) and the temperature of theconductor rises. After collision,electrons are accelerated once moreand again suffer more collisions.

Note: Negatively charged electrons move inopposite direction to that of electric field.

E

Heat

v

v

v

v

v

v

vv

v

3.1.7 Resistance

Alternate accelerations and decelerations result in a net average velocity ofthe mobile electrons (called the drift velocity) which constitutes an electriccurrent. Electrical potential energy is converted into heat within theconductor. The decelerations arising from collisions is called resistance.

Ohm’s lawRIV

=

Experiments show that, for a particular specimen of material, when theapplied voltage is increased, the current increases. For most materials,doubling the voltage results in a doubling of the current. That is, the currentis directly proportional to the current: VI ∝The constant of proportionality is calledthe resistance. Resistance limits thecurrent flow through a material for aparticular applied voltage. Units: OhmsThe rate at which electrical potentialenergy is converted into heat is thepower dissipated by the resistor. Sinceelectrical potential is Joules/Coulomb,and current is measured inCoulombs/second, then the product ofvoltage and current gives Joules/secondwhich is power (in Watts).

but

thus RIP

IRVVIP

2=

=

=

165Electricity


Experiments show that the resistance of a particular specimen of material(at a constant temperature) depends on three things:• the length of the conductor, l• the cross-sectional area of the conductor, A• the type of material, ρThe material property which characterisesthe ability for a particular material toconduct electricity is called the resistivity ρ(the inverse of which is the conductivity σ).

AR l

ρ=

Material ρρρρ ΩΩΩΩm @ 20 oCSilver 1.64 × 10− 8

Copper 1.72 × 10− 8

Aluminium 2.83 × 10− 8

Tungsten 5.5 × 10−8

The resistance R (in ohms) of a particular length l of materialof cross-sectional area A is given by:

3.1.8 Resistivity

but

thus

The number density of mobile chargecarriers n depends on the material. If thenumber density is large, then, if E (and hencev) is held constant , the resistivity must besmall. Thus, the resistivity depends inverselyon n. Insulators have a high resistivity sincen is very small. Conductors have a lowresistivity because n is very large.

nqvE

A

AnqvAIE

VE

AIV

AI

IRV

=ρ

ρ=

ρ=

=

ρ=

ρ=

=

∑

l

l

l

Sum of the positive andnegative mobile charge carriersΣnqvA = n1q1v1 + n2(−q2)(−v2)A

The quantityI/A is calledthe currentdensity J.

For a metal, only one type ofmobile charge carrier

Now,

hence

For a particular specimen of material, n, q, A and l are a constant.Increasing the applied field E results in an increase in the drift velocity vand hence an increase in current I.

E

+

++

+

++

++

+

The resistivity of a pure substanceis lower than that of one containingimpurities because the mobileelectrons are more likely to travelfurther and acquire a larger velocitywhen there is a regular array ofstationary atoms in the conductor.

Presence of impurity atomsdecrease the average drift velocity.


The units of ρ areΩm, the units of σare S m−1.


Consider an applied voltagewhich generates an electric fieldE within a conductor of resistanceR and of length l and area A.

Evidently, if the area A is increased, there will be more mobile chargecarriers available to move past a given point during a time ∆t under theinfluence of the field and the current I increases. Thus, for a particularspecimen, the resistance decreases with increasing cross-sectional area.

Now, the field E acts over a length l.

If the applied voltage is keptconstant, then it is evident that if l isincreased, E must decrease. Thedrift velocity depends on E so that ifE decreases, then so does v, andhence so does the current.

+V

0 V

El

1. Variation with area

2. Variation with length

A

0 l

E

+

++

+

++

++

+

v

3.1.9 Variation of resistance

lEV =

3. Variation with temperatureIncreasing the temperature increasesthe random thermal motion of the atoms in the conductor thus increasingthe chance of collision with a mobile electron thus reducing the averagedrift velocity and increasing the resistivity. Different materials respond totemperature according to the temperature coefficient of resistivity α.

RT = resistance at TRo = resistance at To (usually 0 oC)

( )[ ]( )[ ]ooT

ooTTT1RR

TT1−α+=

−α+ρ=ρ

Material αααα (C−−−−1 ΩΩΩΩ−−−−1)Tungsten 4.5 × 10−−−− 3

Platinum 3.0 × 10−−−− 3

Copper 3.9 × 10−−−− 3

167Electricity

This formula applies toa conductor, not asemi-conductor.


Consider a chemical cell:

Chemical attractions causepositive charges to build up at(a) and negative charges at (b)

rather than“electrostatic”

Electrostatic repulsion due tobuild-up of positive charge at (a)eventually becomes equal to thechemical attractions tending todeposit more positive chargesand system reaches equilibrium.

The term Emf (electromotive force) is defined as the amount of energyexpended by the cell in moving 1 Coulomb of charge from (b) to (a) withinthe cell. “force” is poor choice of words

since Emf is really “energy”(Joules per Coulomb)

At open circuit, Emf = Vab

Excess ofpositive charge

Excess ofnegative charge+ -E

++

+

+

+ --

-- -

(a) (b)

Electricfield

3.1.10 Emf

Now connect an external load RL across (a) and (b). The terminal voltageVab is now reduced.

The circuit has been drawn to emphasizewhere potential drops and rises occur.

+ -(a) (b)

RL

Internalresistanceof cell

Vab is drop inpotential acrossload resistance RL

RinEmf

Vab

I

heat

heat

Positive current carriers givenenergy by chemical action

But, the continuous conversion of chemical potential energy to electricalenergy is not 100% efficient. Charge moving within the cell encountersinternal resistance which, in the presence of a current I, means a voltagedrop so that:

Assume positive carriers –conventional current flow.

Loss of positive charge from(a) reduces the accumulatedcharge at (a) and hencechemical reactions proceed andmore positive charges areshifted from (b) to (a) withinthe cell to make up for thoseleaving through the externalcircuit. Thus, there is a steadyflow of positive charge throughthe cell and through the wire.

At “closed circuit”, Emf = Vab+ IRin



dA

QV

EdVA

QE

o

o

ε

=

=

ε

=

But, for parallel plates holding a totalcharge Q on each plate, calculations showthat the electric field E in the regionbetween the plates is proportional to themagnitude of the charge Q and inverselyproportional to the area A of the plates.For a given accumulated charge +Q and−Q on each plate, the field E isindependent of the distance between theplates.

Now,

Q in these formulas refersto the charge on ONEplate. Both positive andnegative charges contributeto the field E. A factor of 2has already been includedin these formulas.

For a point charge in space, E dependson the distance away from the charge:

2o d

q4

1Eπε

=

Consider two parallel plates across which is placed a voltage V.When a voltage V is connected across the plates,current begins to flow as charge builds up on eachplate. In the diagram, negative charge builds up onthe lower plate and positive charge on the upperplate. The accumulated charge on the two platesestablishes an electric field between them. Sincethere is an electric field between the plates, there isan electrical potential difference between them.

A charged particle releasedbetween the plates willexperience an accelerating force.

+Q

-Q

E

∆VCapacitor

d

A

thus

++ + + + +

VE

- - - - - - -Uniformelectric field

3.1.11 Capacitance

Capacitance is defined as the ratio of the magnitude of the charge on eachplate (+Q or −Q) to the potential difference between them.A largecapacitor willstore morecharge forevery voltacross it thana smallcapacitor.

Units: Farads

If the space between the plates is filledwith a dielectric, then capacitance isincreased by a factor εr. A dielectric is aninsulator whose atoms become polarisedin the electric field. This adds to thestorage capacity of the capacitor.

dAC

QdA

Q

VQC

o

o

ε=

ε

=

= dA

QVoε

=but

Permittivity of freespace = 8.85 × 10−12

Farads m−1

169Electricity


If a capacitor is charged and the voltage source V is then disconnectedfrom it, the accumulated charge remains on the plates of the capacitor.

Since the charges on each plate areopposite, there is an electrostaticforce of attraction between thembut the charges are kept apart by thegap between the plates. In thiscondition, the capacitor is said to becharged. A voltmeter placed acrossthe terminals would read the voltageV used to charge the capacitor.

+ + + + +

- - - - - - -

dVAQk4E π=

If the plates are separated bya dielectric, then the field E isreduced. Instead of using εo,we must use ε for thesubstance. Often, the relativepermittivity εr is specified.

or

ε

ε

=ε

Material εεεεrVacuum 1Water 80Glass 8

When a dielectric is inserted in a capacitor, the molecules of the dielectricalign themselves with the applied field. This alignment causes a field ofopposite sign to exist within the material thus reducing the overall net field.For a given applied voltage, the total net field within the material is smallfor a material with a high permittivity ε. The permittivity is thus ameasure of how easily the charges within a material line up in the presenceof an applied external field.

3.1.12 Capacitors

o41kπε

=

In a conductor, charge carriers not only align butactually move under the influence of an appliedfield. This movement of charge carrierscompletely cancels the external field all together.The net electric field within a conductor placed inan external electric field is zero!

When a capacitor is connectedacross a voltage source, thecurrent in the circuit is initiallyvery large and then decreases asthe capacitor charges. Thevoltage across the capacitor isinitially zero and then rises as thecapacitor charges.

v

t

iV

C

i

t

vC

iC



lNIB oµ=

Number ofturns

Length ofsolenoid

In a conductor carrying a steady electric current, there is a magnetic fieldaround the conductor. The magnetic field of a current-carrying conductormay be concentrated by winding the conductor around a tube to form asolenoid. Application of

Ampere’s law gives:

Magnetic fieldstrength(Units: Tesla)

dtdiNA

dtdV

o lµ−=

Φ−=

When the current in the coil changes, the resulting change in magnetic fieldinduces an emf in the coil (Faraday’s law).

Magneticfield

Magnetic flux

Cross-sectionalarea of coil

AB Φ=

Ix x x x x x

. . . . . .

Fingers in direction of current, thumb is direction of field.

A

l

The induced voltage tends to opposethe change in current (Lenz’s law).

Permeability offree space (or air)in this example.

But, this is the voltage induced in each loop of the coil. Each loop lieswithin a field B and experiences the changing current. The total voltageinduced between the two ends of the coil is thus N times this:

dtdiNAV

2

ototal lµ−=

Inductance: determines what voltage is induced within thecoil for a given rate of change of current.l

2

oNAL µ=

where

3.1.13 Inductance

Units Henrys

171Electricity


In a circuit with an inductor, when the switch closes, a changing currentresults in a changing magnetic field around the coil. This changingmagnetic field induces a voltage (Emf) in the loops (Faraday’s law) whichtends to oppose the applied voltage (Lenz’s law). Because of the selfinduced opposing Emf, the current in the circuit does not rise to its finalvalue at the instant the circuit is closed, but grows at a rate which dependson the inductance (in henrys, L) and resistance (R) of the circuit. As thecurrent increases, the rate of change of current decreases and themagnitude of the opposing voltage decreases. The current reaches amaximum value I when the opposing voltage drops to zero and all thevoltage appears across the resistance R.

Rate ofchange ofcurrentthrough theinductor.

The voltageinduced by thechangingcurrent.

Inductance (Henrys)

When the switch is closed, the rate of change of currentis controlled by the value of L and R. Calculationsshow that the voltage across the inductor is given by

i

Switch

vL

VR

dtdiLvL −=

LRt

L Vev−

=

t

vL

V

i

t

I

Switch is closed

VR

t

V

The minus sign indicates that if thecurrent is decreasing (di/dt is negative)then the voltage vL is positive (ie samedirection as V) and vice versa.

3.1.14 Inductors



2

2

Q

0

Q

0

CV21U

CQ

21

dqCqdqvU

UvdqPdtdtdqi

viP

=

=

==

==

=

=

∫∫

3.1.15 Energy and power

Energy storedin a capacitor

In a circuit with a capacitor, energyis expended by the voltage source asit forces charge onto the plates of thecapacitor. When fully charged, anddisconnected from the voltage source,the voltage across the capacitorremains. The stored electric potentialenergy within the charged capacitormay be released when desired bydischarging the capacitor. To find theenergy stored in a capacitor, we canstart by considering the power usedduring charging it.

Power Lower caseletters refer toinstantaneousquantities.

Energy

Establishing a current in an inductor requires energy which is stored in themagnetic field. When an inductor is discharged, this energy is released.The energy stored can be calculated from the power consumption of thecircuit when a voltage is applied to it.

LRt

L

LRt

R

LRt

Vev

Vev

Iei

−

−

−

−=

=

=

VR

t VL

tV

−V

Power

Energy storedin the inductor

The energy is released when thecurrent decreases from I to 0 (i.e.when the circuit is broken.)

2

I

0

t

0

LI21U

idiL

dtdtdiLiU

dtdiLiiv

dtdiLv

=

=

=

−=

−=

∫

∫

Energy

Final steadystate current

i

Switch

VR

LVoltage induced in theinductor when currentis switched on

173Electricity


3.1.16 Circuits

Resistors in seriesA BR1 R2

V1 V2

R1

Resistors in parallel

21

21AB

21AB

RRRRR

R1

R1

R1

+

=

+=

21AB

21AB

21AB

IIIVVVRRR

==

+=

+=

21AB

21ABIIIVVV

+=

==

A B

R2

V1

V2

Capacitors in series

charge onone plate

Capacitors in parallel

321total

321

33

22

11

total

321

321

C1

C1

C1

C1

C1

C1

C1QV

CQV;

CQV;

CQV

VQC

VVVVQQQQ

++=

++=

===

=

++=

===

( )

total

321

321

332211

3

33

2

22

1

11

total

321

321

C

CCCVQ

CCCVCVCVCVQ

CQ

V;CQV;

CQV

VQC

VVVVQQQQ

=

++=

++=

++=

===

=

===

++=+

+Q1 +Q2 +Q3

C1 C2 C3

−Q1 −Q2 −Q3

V

+Q1

+Q2

+Q3

V

−Q1

−Q2

−Q3



1st law: Current into a junction = current out of a junctionI = I1+I2

I2

I1

2nd law: In any loop in a circuit, the sum of the voltagedrops equals the sum of the Emf’s

In the circuit shown,calculate R1, and thecurrent through thesection A-B

1. Divide the circuit up into current loops and draw an arrow whichindicates the direction of current assigned to each loop (the directionyou choose need not be the correct one. If you guess wrongly, thenthe current will simply come out negative in the calculations).

150 Ω

R1

4 V

6 V30 Ω

A

B

I1

I2

20 mA

2. Consider each loop separately:

3. Solve simultaneous equations for unknown quantities

( ) ( )( ) ( )( ) ( ) ( )( )( )

( ) ( )( ) ( )( )Ω=

−−+=

−=

−=

−+=−

−+=+

125R3003.03002.0R02.04

A03.0I6.0180I

3002.030I150I630I3002.0R02.04

1

1

2

2

22

21

substitute back into (1)

Current goingthe wrong wayis negative.

I

I

I2 is going in theopposite directionto I1 through thesection ABtherefore −ve0.030.02

4. From 1st law:

In loop #2, I1 isgoing in theopposite directionto I2 through thesection ABtherefore −ve

A

3.1.17 Kirchhoff’s laws

Example:

Solution:

Current going theright way through avoltage source ispositive.

(1)

(2)

A05.003.002.0IAB

=

+=

175Electricity


3.1.18 Examples

Increasing the distance d results in a decreasein field strength E since in the exampleshown here,V is a constant and V = Ed. Thisdecrease in E must result in a decrease in theaccumulated charge Q on each plate since

+

+ + + + +V

E

- - - - - - -

d

AQEoε

=

A decrease in accumulated charge for a given applied voltagemeans a decrease in capacitance C in accordance with:

What happens to the excess charge? If d isincreased, and the external voltage V is maintained across the capacitor,then there is a momentary increase in the opposing voltage across thecapacitor. That is, the excess charge flows back into the voltage source.

2. Explain what happens when the distancebetween a parallel plate capacitor acrosswhich is maintained a steady voltage Vis increased.

dAC oε=

Solution:

Power transmitted is VI = 11 kW. Some of this electrical energy is dissipatedas heat within the line at a rate of P = ∆VI where ∆V is the voltage drop fromone end of the line to the other (not the 11 kV which is the voltage drop fromthe line to earth). The power dissipated can also be calculated from P = I2Rwhere I is the current in the line and R is its resistance Doubling the voltageand halving the current would reduce the power dissipated in the line by afactor of four and increase that available to the consumer by the sameamount. Thus, it is better to transmit at high voltage and a low current.

1. A power station supplies 11kV at 1 amp along a transmission lineof resistance R. Calculate the power transmitted and the power lostto heat in the transmission line. Explain why it is efficient totransmit power at high voltage and low current.

Solution:



3.2 Magnetism

BqvF ×=

AB Φ=

BqmvR =

BIF ×= lθ= sinIABTorque

∫= 2rrxId'kB l

2x

rr vq'kB =

r2I

B oπ

µ=

a2NIB oµ

=

r2'IIF o

π

µ=

l

lNIB oµ=

Id.B oµ=∫ l

HB oµ=

MBB oo µ+= Magnetisation

Magnetic field intensity

Solenoid

Ampere's law

Magnetic field - flat circular coil

Magnetic field - two parallel wires

Magnetic field - long straight wire

Biot-Savart law

Magnetic field - moving charge

Torque from electric motor

Force on a current carrying conductor

Radius of curvature of movingcharge in magnetic field

Magnetic induction

Force on a moving charge in magnetic fieldSummary

177Electricity

HM χ= Susceptibility


A magnetic field is said to exist at a point if a force is exerted on amoving charge at that point. The force (N) acting on a moving charge(C) is perpendicular to both the direction of the field and the velocity(m/s) of the charge.

Over and above anyelectrostatic force

Right-hand rule

Experiments show that the magnitude ofthe force acting on a charge moving in amagnetic field is proportional to:

B is called the magnetic inductionor the magnetic field.

Tesla

Thumb - velocityFingers - fieldPalm - force

3.2.1 Magnetic field

CmNs

AmN

BsinvqF φ= BqvF ×=

TheMagneticfield is avectorfield. BF

v

vsinφ

φ

F B

v

or

q

A magnetic field may be represented by lines of induction. The magneticflux is proportional to the total number of lines.

Total numberof lines

The number of flux linesper unit cross-sectional areais a description of the magneticfield and is called the magnetic induction Bor magnetic flux density (Tesla).

Uniform magnetic field -lines are equally spaced.

1 Tesla = 1 Weberper square metreA

B Φ=

Weber

Note: not lines of forcesince, unlike the electricfield, the magnetic force isperpendicular to thedirection of the field.

Spacing between fieldlines indicates fieldstrength.

N∝Φ



x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

1. A positively charged particle isgiven velocity v in a directionperpendicular to a uniform magneticfield B. 2. A force F = qvB is

exerted on theparticle downwards.

3. Application of forcechanges the direction ofmotion of the particle.

4. If the motion of the particle iscompletely within the field,then particle travels in a circleof radius R with constanttangential speed |v|.

5. The force F is a centripetal force:

BqmvR

Rvm

qvBF2

=

=

=

mqB

=ω

Frequency

Radius of path

If the direction of motion is notperpendicular to the field, thenthe velocity component parallelto the field remains constant andparticle moves in a helix

3.2.2 Charged particle in magnetic field

v

F

It is a peculiar property of magnetic field lines that they always form closed loops.Electric field lines may start on an isolated positive charge and terminate on anotherisolated negative charge. Magnetic field lines do not start and finish on isolatedmagnetic poles even though we may draw them as starting from the North pole of amagnet and finishing on the south pole. Magnetic field lines actually pass throughthe magnet to join up again.

NSMathematically, this isequivalent to saying that:

0Ad.B =∫Gauss’ Law for magnetism.

Magnetic poles

179Electricity


Consider the movement of both positive and negative charge carriers in aconductor perpendicular to a magnetic field B

Positive charge moving with driftvelocity v1 to the right is actedupon by upward force F1 = qvB

Negative chargemoving with driftvelocity v2 to the leftis acted upon byupward force F2 =(−q)(−v)B

Cross-sectionalarea A

+-

F2

F1

v2

v1

Let n1 and n2 be the number of positive and negative charge carriers perunit volume. The total number of charge carriers N in a length l and cross-sectional area A of the conductor is:

( ) ( )( )

BIF

BIBJA

BAvqnvqnBvqAnBvqAnF

AnAnN

222111

222111

21

×=

=

=

+=

+=

+=

l

ll

lll

ll

Total force on allcharge carriers bothpositive and negativeCurrent densityJ

Note: Here we have considered the movement of both positive andnegative charge carriers within a conductor. If current flows due to themovement of only one type of charge, (e.g. electrons in a metal), then,from the macroscopic point of view, this is exactly equivalent to the equalmovement of only positive charge carriers in the opposite direction.

3.2.3 Force on a current-carrying conductor

l

The resultant force on the loop is zeroThe resultant torque (or moment) is:

θ=

θ=

=

=

sinbIB

sin2b)IB(2

r)IB(2Fr2Mx

l

l

l

θ

BF

B

r

I

I

I

Ix

b

F

B

l

But, the product lb = the area A ofthe loop, hence:

The product IA is called the magnetic moment ofthe loop and is (unfortunately) given the symbol: µ.

θ= sinIABTorque


x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x


A magnetic field can be created in two ways: (a) by the movement of chargecarriers in a conductor (i.e. an electric current) and (b) by a changing electricfield in an insulator (or empty space).

Consider the motion of a positivecharge moving with velocity v.Experiment shows that at everypoint P, the B vector lies in aplane perpendicular to thatdefined by r and v

The electric fieldlines radiateoutwards from thecharge.

B

B

B

B

v

rPθ+

The direction of field isgiven by right-hand rule:Thumb - velocityFingers - field

E moves from 1 to 2.Strength of E field at 2 thus increases.In a space where the electric field E ischanging, a magnetic field B is created.

B+

E +E

v

1

2

vVertical components of E alongthe line of v are shown. At thatpoint, the B field is horizontal.

2rsinqv'kB

rvq'kB

θ=

= 2x

r Unit vector in

direction of r

= 1 × 10−7 Wb A−1 m−1π

µ=

4'k o

3.2.4 Source of magnetic fields

The lines ofinduction arecircles in planesperpendicularto the velocity

v

+

181Electricity

Permeability offree space

(a)

(b)


Consider an element of conductor length dl and area A carrying n chargecarriers with drift velocity c. The total charge moving through a volumeelement is:

( )( )

( )

2

2

2

rsinId'k

rsinvnqAd'k

rsinvdQ'kdB

AdnqdQ

θ=

θ=

θ=

=

l

l

l

VolumeNumber densityof charges

since nqvA = I

Now,

∫=

=

2

2

rrxId'kB

rrxId'kBd

l

l

In vector notation:

Biot-Savart Law

Add field from allelements to gettotal field B.

+-v2

v1

dl

A

This law demonstrates how a steadycurrent I in a conductor produces amagnetic field B. It is one of the twoways in which a magnetic field maybe created. The other way is by thepresence of a changing electric fieldin space.

3.2.5 Biot-Savart law



(1) A long straight wire

(3) Flat circular coil (N coils)

r2I

B

rI'k2B

oπ

µ=

=

( )

a2NI

B

ax2

NIaB

o

2322

2o

µ=

+

µ=

π

µ=

4'k o

BI

rPerpendiculardistance from wire.

Direction of B givenby right-hand rule.(2) Force between two parallel wires of length l

Each conductor experiences a forceConsider the force on conductor A

r2'II

F

r2I

B=

'BIF

o

o

π

µ=

π

µ

=

l

l

but

I

r

B

I'F

A

Right-hand ruleshows that wirescarrying currentsin same directionattract and inoppositedirections, repel.

Used as practicaldefinition of theAmpere.

B

@ x = 0

flat

x r

a

I

The contribution from eachmoving charge comprising thecurrent I results in a steadymagnetic field in space whichencircles the conductor anddiminishes as 1/r.

3.2.6 Biot-Savart law - Applications

Permeability offree space

183Electricity

µo = 4π × 10−7 Wb A−1 m−1


lNIB oµ=

Number ofturns

Length of solenoid

The magnetic field of a current-carrying conductor may be concentrated bywinding the conductor around a tube to form a solenoid.

Singleconductor

. . . . . Strandedconductor

Single loop of a stranded conductor: i.e. a solenoid

Application ofAmpere’s law yields:

For a long solenoid, the field B is uniform across the cross-section withinthe solenoid. The field decreases slightly near the ends of the solenoid (asshown by the wider spacing between the flux lines in the above diagram.

Fingers in directionof current,thumb is directionof field.

I×××× ×××× ×××× ×××× ×××× ××××

. . . . . .

NS

Path ofintegration

B = 0

3.2.7 Ampere's law

( )

Id.B

Ir2Br2I

B

o

o

o

µ=

µ=π

π

µ=

∫ l

for a long straightconductor.

Circumference ofcircle at distance r

Ampere's law is an alternative statement to Biot-Savart law:

Ampere's law states that the line integral of the magneticinduction field around any closed path is proportional to thenet current through any area enclosed by the path.

Current enclosedby the path

B

I

r

The line integral is a specialtype of integral in whichonly the component of thefunction being integratedacting in the same directionof the direction of the pathof integration is summed.

a

b c

d

B

The dot product B.dl is zerofor the path segments aband cd since those pathsegments are perpendicularto B. The only contributionto the line integral are thesegments ad and bc.



A compass needle itself consists of a North and South pole, it being a smallmagnet free to rotate on a spindle. A compass needle placed in the vicinityof the coil would tend to align itself with the field surrounding the coil.Now, outside the coil, the alignment results in the familiar observation thatlike poles repel and unlike poles attract. Inside the coil, the compass needleis still aligned with the field but we can no longer say that like poles repeland unlike poles attract. Rather, it is more scientifically appropriate to saythat when two magnetic fields interact, they tend to align themselves. Thistendency to alignment exerts a mutual torque (or moment) between thebodies producing the fields. This is called the magnetic moment.

I×××× ×××× ×××× ×××× ×××× ××××

. . . . . .NS

Experiments show that:

• a charge moving perpendicular tomagnetic field lines experiences amagnetic force.

• a magnetic field is produced by amoving charge. 2r

sinqv'kB θ=

These two phenomena are a consequence of the natural tendency ofmagnetic fields to align themselves. Consider the magnetic field createdby the moving charges in a solenoid coil:

The ends of such a magnetised coil are commonly labelled North andSouth to indicate the direction of the field. These labels have come aboutsince it is the direction in which the solenoid would tend to align itself withthe magnetic field of the earth if it were free to rotate.

N

N

N

N

3.2.8 Magnetic moment

φ= sinqvBF

185Electricity


Consider a current carrying wire in a uniformmagnetic field. We know that such a wire wouldexperience a force upwards as shown in thediagram. How does this force come about?

B

I

r

F

NThe total field (given by the vector sum ofthe flux lines) is stronger at the bottom andweaker at the top.

If we think of flux lines being under tension then it is easy to imagine theincreased tension in the bottom lines of flux tending to push upwards onthe wire.

×

Alternately, imagine the current-carrying wire is attached to a pivot pointby a rigid beam. The magnetic moment arising from the tendency of thetwo magnetic fields to align themselves acts around this pivot point. Thewire itself is thus subjected to an upwards force as shown. The force is amaximum where the moment arm r is the greatest.

×

Fields alreadyaligned. No resultantmoment.

Fields oppositelyaligned. No resultantmoment.

F

×B

×B

F

×

F

B

r

F

×

B

r

×

F

×

F×

×

maximum r =maximum forceapplied to wire.

r

pivotpoint

MomentB

B

B

B

B

B

B

B

B

3.2.9 Magnetic force

I into page



•The magnetic induction B•The magnetic field intensity H

Permeability of free space(i.e. a vacuum)mo = 4π × 10−7 Wb A−1 m−1

Now, a magnetic field can be described by either of two vectors:

B and H have different physical significance.The magnetic induction B is a field vectorwhich determines the magnetic force actingon a moving charged particle.

a magnetic “effect”

The constant µo shows that the resulting magnetic induction B produced byan external source of magnetic field (such as a permanent magnet or movingcharge) depends upon the “ease” with which the surrounding materialpermits the creation of magnetic field lines.

a magnetic “cause”

µo is called the permeability of the material. Those materials which tend tofacilitate the setting-up of magnetic field lines have a higher permeabilitythan others.

B and H for asolenoid

Amperesmetre−1

Teslas(Wb m−2)

lNIB oµ=

lNI

BHo

=

µ=

3.2.10 Permeability

HB oµ=where:

The magnetic field intensity H is also a fieldvector, but describes the magnetic fieldgenerated by a moving charged particle.

When matter is placed in the region surrounding a current-carryingconductor, the magnetic field is different to that which exists when theconductor is in a vacuum due to the magnetisation of the material.

1. Uniform field 2. Presence of materialmay concentratefield

This behaviour arises due to the interaction of the externally applied fieldand the internal field generated by the orbiting (and also spinning)electrons within the material

187Electricity


The behaviour of materials when placed in a magnetic field is a consequenceof three types of behaviour:• paramagnetism• diamagnetism• ferromagnetism

Internal magnetic fields line up andreinforce the external field.

Internal fields add up in such a way soas to oppose the external field.

Internal fields are strongly interacting with each other lining up intomagnetic domains, even when there is no external field present.

Atoms whose shells are not completely filled.

Most prominent in atoms whose shells are completely filled.

3.2.11 Magnetic materials

Internally generated magnetic fields of the material add to the external field,i.e. that which would exist in the absence of any material through whichmagnetic flux passes and is described by the quantity µoH. The additionalfield produced is proportional to the total magnetic moment per unitvolume in the material. In an atom, the magnetic moment arises when atoms(whose orbiting electrons generate a magnetic field) attempt to line up withthe externally applied field.The total magnetic moment per unitvolume is called the magnetisation M: V

M totalµ=

The additional field due to magnetisation is given by µoM. Thus, when theconductor is completely surrounded by material, the magnetic field B in thematerial is: MBB oo µ+=

Additional field due tomagnetisation ofmaterial.External field

generated from currentBo = µoH

Total field within thematerial when currentflows in conductor.

Now, the magnetisation M is induced by theexternal field, and thus:

HMBM oχ=

∝Magnetic susceptibility

( )( )

HH1HHMHB

o

o

oo

µ=

χ+µ=

χ+µ=

µ+µ= ( )

( )χ+=

µ

µ=µ

χ+µ=µ

1

1

or

o

Relativepermeability

thus:

Material χ

Oxygen 1.9 × 10−6

Aluminium 2.2 × 10−5

Platinum 2.6 × 10−4

Paramagnetic materials

The result is that themagnetic field withinthe material is greaterby a factor µr = 1+χthan if there were avacuum present.



Ferromagnetic materials have permeabilities much larger than that offree space.

Iron is the most significantmagnetic material hencethe term ferromagneticbut the term also refers toother elements such asNickel and Cobalt.

This large relative permeability arises due to anappreciable net magnetic moment due to partially-filled 3rd electron shells.

The net magnetic moments interact within the material even when thereis no external field present.These interactions cause neighbouring moments to align themselvesparallel to each other in regions called magnetic domains.With no external field, the orientations of the domains are random. When afield H is applied, the domains attempt to align themselves with the field.Those domains already in alignment with the field tend to grow at theexpense of others, not aligned, which shrink. As H is increased, a point isreached where all domains are aligned with the field.

B

µr = 1000-10000

3.2.12 Ferromagnetism

The permeability of aferromagnetic materialvaries as the magneticfield intensity H is varied.For a given value of H, Bdepends on µ because His great enough to forcedomains to alignthemselves with H ratherthan their preferredcrystalline orientation, or,m depends on B Becausethe permeability dependsupon what fraction of themagnetic domains havealigned their momentswith the magnetic field.

B

H

Domains whose axes are mostnearly aligned with applied fieldgrow in size.

All domains aligned withfield (saturation). Anyincrease in H results inan increase in B in thesame manner as anynon-ferromagnetic (eg.paramagnetic) material.

A BH curve shows how flux density of an unmagnetised sample offerromagnetic material rises as the value of the field H is increased.

189Electricity


3.2.13 Example

Most magnetic materials will show some residual magnetism when the field isreturned to zero:

When the magnetic field intensity H arises from an alternating current, the flux densityB in a ferromagnetic material tends to lag behind the magnetic field intensity H whichcreates it. This is called hysteresis.

Only ferromagnetic materials have residual magnetism and thus show hysteresiseffect. The area within hysteresis loop is an indication of energy loss thus differentmaterials may be compared experimentally.

B

H

Remanent magnetism

H returned to zero

Demagnetisingfield required toremove remanentmagnetism Hc

Remanent magnetism

Domain growth inlarge applied fieldbecomes irreversible

The hysteresis loop represents an energy loss as the field has to be reversed tonegate the residual magnetism arising from the irreversible alignment of domains athigh values of H.

Solution:

1. A solenoid has length 20 cm, diameter 1 cm and 1000 turns. Whatcurrent must flow in the coil to develop a total flux of 1 × 10−6 Wb atthe centre of the solenoid?

( )

( )( )( )A206.2

1000104005.02.0101

NAI

NI

AB

72

6o

o

=

×ππ

×=

µ

φ=

µ=

φ=

−

−

ll

Hysteresis



3.3 Induction

Summary

dtdEmf Φ

−= Faraday's law

dtdiMinduced −=E Mutual inductance

EmfdtdN

dtdiL −=

Φ= Self inductance

lANL

2

oµ=

lANL

2

oµ=

Solenoid

Toroid

191Electricity


When the slider is moved to theright with velocity v, the(positive) charges in themoving wire experience a forceF = qvB upwards thus causing a(conventional) current to flowanti-clockwise in the loop.

dtdEmf

EmfvB

dtdsB

dtd

dsBBdAd

BA

Φ−=

=

=

=Φ

=

=Φ

Φ=

l

l

l

When the conductor moves to the right a distance ds, the cross-sectional area swept out by the movement is: dA = lds.

Now, the magnetic flux Φ isgiven by:

Emf

vBqPE

qvBPEqvBFqvBF

=

=∆

=∆

=

=

l

lll

Velocity v

The induced Emf in the circuit is equal to the rate of change of magneticflux through it.

Faraday’s Law

Sign conventions:• If the current is clockwise,

the Emf is positive.• If the direction of B is away

from the observer (as shownabove) then the increase inΦ is positive.

Proportional to the number of lines of magneticinduction in the B fieldChange in flux when wire moves a distance ds

I

l

ds

v

Note: The slider must "cut"the magnetic field B.

3.3.1 Faraday’s law

Consider a loop of conducting wire in a magnetic field. One side of theconducting loop is made as a slider which can be moved.


x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

Emf


dtdEmf Φ

−=From Faraday’s law, we have:

An Emf is induced in a conductor whenever there is a change of magneticflux. Now, a change of flux can be brought about in two ways:

• the conductor may move in themagnetic field and “cut” lines of B.

• a varying magnetic field can cause aninduced Emf in a stationary conductor.

If the current I is changing, then the flux density also changes. If thecurrent increases, more lines of flux move outwards from the conductor toencircle the conductor.

Large I(into page)

Small I(into page)

× ×I I+

B B

.

changes to

A second conductor placednearby experiences thischanging flux and a voltageis induced within it. If theconductor forms part of acircuit, then current flows(out of page in thisexample). Same effecthappens if second conductoris moved to the left.

Magnetic field lines from the increasing current in thefirst conductor appear to move to the right, which isthe same as saying that the second conductormoves towards the left. That is, charges in thesecond conductor move through space towards theleft relative to the lines of B. By Right Hand Rule,charges experience a force out of page, i.e. currentin second conductor is out of the page.Alternately, we can say that in the space surroundingthe second conductor, there is a changing magneticfield as the flux lines move outwards and becomecloser together. A changing magnetic field creates anelectric field perpendicular to it and the direction ofmotion of the B flux lines. This electric field E (i.e. aninduced Emf) then acts upon the stationary chargesin the second conductor causing them to move (if theconductor is part of a circuit). Same effect if secondconductor is moved inwards towards the firstconductor. If the change in B is “linear”(dB/dt=constant) then the induced current I2 isconstant.

Consider the secondconductor:

E

∆B ∆I1

I2

.

3.3.2 Electromagnetic induction

I (out of page)

193Electricity


1. Consider the example of induction given earlier:

The direction of the induced current in the moving wire is such that thedirection of the resulting magnetic force is opposite in direction to itsmotion. The work done in maintaining the motion v is dissipated as heat inthe circuit (here shown as a resistor) by the passage of current through it.

The (positive) charges inthe moving wire experiencea force F = qvB upwardsthus causing a current toflow anti-clockwise.However, a current carryingconductor within a magneticfield experiences a magneticforce F = Il × B, which by theright hand rule, is towards theleft in this example.

2. Now consider the case where a changing magnetic flux induces acurrent in a stationary conductor (which forms part of a circuit):

The induced current I2 creates amagnetic field B2 of its own. If theoriginal field from the first conductor is increasing

uniformly (dB/dt isconstant), then the the

current I2 is steady and themagnetic field B2 is also steady.

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

I

l

ds

v

F

Note, the direction of the induced magnetic field is such that it opposes thechange in flux through a loop in the circuit. If the field B1 is increasing, theinduced field inside the circuit loop is in the opposite direction tending tooppose the increase. If the field B1 is decreasing, current in the secondcircuit flows in the opposite direction to that shown above and the field B2 isin the same direction, tending to oppose the decrease.

∆B1

∆I1

I2

B2

3.3.3 Lenz’s law

Lenz’s Law



If a varying current flows in coil #1, a varying magnetic field is produced.This varying magnetic field can then in turn induce an Emf (and thus acurrent) in another coil #2.

dtdiMinduced −=E

A constant which depends onthe nature of the coils. Themutual inductance.

A portion of the flux set upby current in coil #1 linkswith coil #2. Thus, there isa flux passing through eachturn in coil #2.The product N2Φ2 iscalled the number of fluxlinkages.The mutual inductance M is defined as the

ratio of the number of flux linkages in coil #2to the current in coil #1: i1

1

22i

NM Φ=

But, if the current varies with time, then so does the number of fluxlinkages, (both i and Φ are functions of t) thus:

2

22

1

221

Edt

dNdtdiM

NMi

−=

Φ=

Φ=

the inducedEmf in coil #2

The mutual inductance can be considered as the induced Emf in coil#2 per unit rate rate of change of current in coil #1.

For a given mutual inductanceM, the greater the rate ofchange of i, the greater theinduced Emf in coil #2.

M = 1 volt per ampere per second

1 Henry

F2

N2

N1

Lower case i signifies aninstantaneous value of avarying current.

3.3.4 Mutual inductance

195Electricity


As the current increases in a single conductor, lines of induction arecreated and move outwards as the current increases.

As the lines of induction moveacross the conductor material, theygenerate a voltage within theconductor itself.

The self-induced current represents a self-induced voltage opposite to thatapplied to the wire. It is quite small for a single conductor, but can be madesignificant when the conductor is formed into a coil, or solenoid since as thecurrent builds up in first loop, the changing magnetic field induces a voltagein second loop which tends to oppose the applied voltage and so on.

The self-induced voltage isopposite to that whichproduces the original current.

Consider left side of conductor: Consider right side of conductor:

Magnetic field linesmove outwards tothe left which is thesame as the chargesin the conductormoving to the right.By right hand rule, force on chargescauses induced current to flow out ofpage which opposes original currentflowing into page.

By right hand rule, force on chargescauses induced current to flow out ofpage which opposes original currentflowing into page.

Magnetic field linesmove outwards tothe right which is thesame as the chargesin the conductormoving to the left.

3.3.5 Self-inductance

Large I (into page)

Small I (into page)

× ×I I+

B B

iNL Φ

=

If the current I changes withtime, then so does Φ and thus:

Emf−=

Φ=

Φ

=

dtdN

dtdiL

dtdidtdN

L

The self inductance is the self-induced Emf(or back Emf) per unit rate of change ofcurrent. The direction of the self-inducedEmf is found by Lenz’s Law.

InstantaneouscurrentInductance

A fluxΦ passes through each turn. Thenumber of flux linkages for all coils isNΦ. The number of linkages per unit ofcurrent is called the self inductance.



Inductance of a solenoid:

( )

l

l

l

l

ANL

iiNAN

iNL

AiNBA

iNB

2

o

o

o

o

µ=

µ=

Φ=

µ=

=Φ

µ=Length of coil

Instantaneouscurrent Number of turns

Field withinlong solenoid

Cross-sectional areaof solenoid loop

Inductance of a toroid:

The case of a toroid is somewhat simpler thanthat of the solenoid because the magnetic field isconfined wholly within the toroid.

and is uniform if the radius of thetoroid is large with respect to theradius of the turns.

The length l is the circumference of the toroid and thearea A is the cross-sectional area of the loops.

lANL

2

oµ=

The use of µo here signifies the inductance of a toroid with an air (or strictlyspeaking, vacuum) core. The inductance of a toroid may be significantlyincreased when the coil is wound on a material with a high permeability.

Note: Inductancedepends on N2

3.3.6 Solenoids and toroids

197Electricity


Consider a moving conductor in a magnetic field:Slider moves tothe right by theapplication ofan externallyapplied force F.

Charges inthe wire moveto the right,magneticfield createdencircling(clockwise)the charges

+

B

Looking in from the left,charge experiences an upwardsforce F = qvB due to thetendency of the magnetic fieldsto align themselves.

1.

2.

3.

4.

Charges move upwardsthrough wire (electriccurrent) under the influenceof F = qvB. Anothermagnetic field encircles themoving charges.

Looking down from top, chargesexperience a sideways force F = qvBdue to the tendency of the magneticfields to align themselves. This force Facts against the original force F whichis moving the slider (Lenz’s law).

5.

+

B

F

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x v

F

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

B

v

(Magnetic field approach)

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

B

v

3.3.7 Electromagnetic induction



Final steady-statecurrent

1. A coil has 1000 turns, length 300 mm and diameter 40 mm wound on acardboard former. Determine the inductance of the coil and the energystored within it when carrying a current of 1.5 A.

( )

J006.0

5.1103.521U

mH3.53.0

02.01000104

ANL

23

227

2o

=

×=

=

π×π=

µ=

−

−

l

300 mm

40 mm

If the current in the coilchange at the rate of 5 A s−1,calculate the emf generatedwhich opposes the emfproducing the change.

( )mV105

51021dtdiLV

3

=

×=

−=

−

Solution:

3.3.8 Examples

1. The shaded area moves through thefield and a current is induced in thatportion of the disk directeddownwards from (a) to (b)

2. Neighbouring portions of the disk arenot in the field but provideconducting paths for the current toflow back to (a)

3. The circulation of charge in thismanner are called eddy currents.

xx

xx

x

xx

xx

x

xx

xx

x

xx

xx

x

xx

xx

x

xx

xx

x

(a)

(b)

ω

I

2. Consider a rotating, conducting (non magnetic) disk through whichpasses a magnetic field. Will there be an induced current in the disk?

The downwards current from (a) to (b) experiences a sideways force F = IlBwhich opposes the rotation of the disk (Lenz’s law). Hence eddy currentsrepresent an energy loss (due to I2R heating) which may or may not bedesirable.

Solution:

199Electricity


3.4 Magnetic circuits

Summary

NIFm = Magnetomotive force

Φ=

mm

FR

AR m

l=µ

lmFH =

Reluctance

Permeability

Magnetic field intensity



lNIB oµ=

Number of turns

Length of solenoid

The magnetic field of a current-carrying conductor may be concentrated bywinding the conductor around a tube to form a solenoid.

Application of Ampere’slaw yields:

For a long solenoid, the field B is uniform across the cross-section withinthe solenoid.

The solenoid concentrates the magnetic field within the space enclosed byits windings. Although shown only for one line in the above figure, all linesof magnetic induction form complete loops, but the spacing of the linesoutside the solenoid is very large (small value of B compared to inside coil).

Ix x x x x x

. . . . . .

NS

The complete path taken by a line of magneticflux is called a

magnetic circuit.

Lines of magnetic flux only appear when there is current in the coil.Magnetic flux lines, can be imagined to be established through the actionof a magnetomotive force. The magnitude of the magnetomotive force isevidently dependent on the current in the solenoid and the number of turns.

Magnetomotive force (mmf)units: ampere-turns

This holds even for a single coil of wire. Thus, the length term l is amodifying effect of the solenoid and is not included in the definition of themagnetomotive force

3.4.1 Magnetomotive force

NIFm =

201Electricity


Lines of magnetic induction, or just magnetic flux, are established by amagnetomotive force (mmf).

Lines of induction form complete loops which we may regard as a magneticcircuit. For a given magnetomotive force, the strength of the resultingmagnetic induction field B depends upon the permeability µ of the region(either space or matter) where the field is created.The term reluctance Rm is given to the resistance encountered by amagnetomotive force in creating a particular magnetic circuit.

Φ=

mm

FR

AR m

l=µ

Since Rm is a property of a particular magnetic circuit, and µ, thepermeability, is a property of the region where the magnetic circuit exists,it is not surprising to find a link between these two quantities:

Length of the magnetic circuitCross-sectional area of themagnetic circuit

Magnetomotive force(Ampere-turns)

Flux (Webers)

Now, since B = µH and B = Φ/A, then:

l

l

l

m

m

m

FH

HA

AR

HARA

=

=Φ

=Φ

units: Ampere-turns/Weber

3.4.2 Reluctance

NIFm =

For ferromagnetic materials, the permeability µ of the material dependsupon the percentage of domains which are aligned with the applied field H.That is, the reluctance is a function of the mmf since µ can no longer beconsidered independent of the region but depends on Fm.This is an example of a nonlinear magnetic circuit.

AR m

µ=

l



Now, the reluctance of a magneticcircuit is inversely proportionalto the cross-sectional area, andalso, for an air-filled coil, thepermeability of the air insidethe coil is the same as the airoutside. Thus, for an air-filled coil,the total reluctance for the complete magnetic circuit is approximatelyequal to that given by just the interior portion of the coil itself since theinterior area A is much smaller than the exterior A outside the coil.

The operation of electrical machinery depends upon the associated magneticfields.

• the path of magnetic field lines can be controlled by the use ofhighly permeable materials (e.g. ferromagnetic materials) toconstruct the desired magnetic circuits appropriate to theapplication of the machinery involved.

• the magnetomotive force required to develop a desired magneticfield strength can be determined from:

Φ=

mm

FR sometimes called themagnetic Ohm’s law.Consider an air-filled solenoid:

Lines of induction are concentrated inside the coil (small cross-section)but are widely spaced outside the coil (large cross section).

AlR m

µ=

The magnetomotive force required to establish a required magneticflux in an air-filled coil can thus be calculated from: Φ= mm RF

large A: small Rm

small A:large Rm

3.4.3 Magnetic circuits

Consider an iron-filled solenoid:For an iron-core coil, the veryhigh permeability of the coreallows a far greater flux within thecore for the same mmf even though the interior area is relatively smallcompared to the exterior air part of the circuit. The permeability of the coreis so high that the return path through the surrounding air now contributesthe most to the reluctance ofthe magnetic circuit. The situationwhere the return path is not specificallydirected is difficult to analyse. Inpractice, the return path for the circuit is directed in some way.

203Electricity


If pole pieces have a cross-sectional areagiven by (m × n), then the effectivecross sectional area = (m + l)(n + l)

The magnetic circuit shown to the rightconsists of two sections in series since themagnetic lines of induction pass throughone section and then the other.

The total reluctance of the circuit is the sum of the individual reluctances.

AR m

µ=

l

If the material of a particular section isferromagnetic, then a value for Bmust be calculated and a BH curvemust be consulted to determine avalue for m.

Different parts of themagnetic circuit mayhave different cross-sections and lengths.

FLUX IS COMMON TO ALL PARTSOF SERIES CIRCUIT

Since in a series circuit, the same flux Φ goesthrough all sections, and since Fm = ΦRm, then, the total mmf = sum ofeach mmf for each section of circuit.Also, the total mmf = Hl, thus:

mFH Σ=l

Total length of circuit

3.4.4 Magnetic circuits

.....RRRR 3m2m1mmT ++=

.....FFFF 3m2m1mmT ++=

In some electric machines, it isnecessary to have an air gap in themagnetic circuit to permit themechanical rotation or movement of various parts etc. An air gap willincrease the reluctance of a magnetic circuit. The air gap may be in fact agap within an otherwise low reluctance circuit made from brass, paper orsome other non-magnetic material.

Air gap

Polepiece

Polepiece

l

Fringingresults in theflux density inthe air gapbeing less thanthat within theadjacent polepieces due tospreading outof the flux.

Empirical correction to allowfor the effects of fringing.

In a parallel magnetic circuit, themagnetic flux branches so that somelines go through one section and somethrough the other.MMF IS COMMON TO ALLPARTS OF PARALLEL CIRCUITEasy solutions if structure is symmetricsince flux divides equally.



I = 2A

1. Calculate the flux in the magnetic circuit shown below if the coil has150 turns, carries a current of 2 A and is wrapped around a magneticcore of cross-sectional area 0.003 m2. Also calculate the reluctance ofthe magnetic circuit.

B (T) H (A t m-1)0.12 1000.54 3000.94 70012.15 1600

Magnetic material

( )

( )

13

mm

1

m

WbAt102.18500162.0300

FR

Wb00162.0003.054.0

A

T54.0BmAt300

300H

HAt3002150

NIF

−

−

×=

=

Φ=

=

=Φ

Φ=

=

=

=

=

=

=

=

1

l

• Magnetic pathlength = 1m

• Cross-sectionalarea = 0.003 m2

Magnetomotive force

Magnetic field

Reluctance

3.4.5 Example

Solution:

205Electricity


3.5 R-C & R-L Circuits

Summary

RCt

R

RCt

c

Vev

e1Vv

−

−

=

−=

RCt

R

RCt

c

Vev

Vev

−

−

−=

=

R-C circuit: Charging

R-C circuit: Discharging

( )

LRt

L

LRtR

Vev

e1Vv

−

−

=

−=

LRt

L

LRt

R

Vev

Vev

−

−

−=

=

R-L circuit: Charging

R-L circuit: Discharging

∫= dtvRC1v inout

dtdvRCv in

out =

Integrator

Differentiator



( )

( )

RCt

o

RCt

o

RCt

RCt

f

RCt

RCt

eIi

ieI

eRV

dtdq

e1Qq

e1VCq

eVCq1

RCt

VCqVCln

RCtVClnqVCln

VClnRC

tqVCln

−

−

−

−

−

−

=

=

=

=

−=

−=

=−

−=−

−=−−

−=−−

fQVC =

Both current incircuit and chargeon capacitor arean exponentialfunctions of time

Io is the initialcurrent in thecircuit

At fully chargedcapacitor hascharge Qf andpotentialdifference = V

Small letterssignifyinstantaneousvalues

( ) constantRC

tqVCln

RCdt

qVCdq

RCq

RV

dtdq

RCq

RVi

CqiR

vvVCqv ;iRv

CR

CR

+=−−

=

−

−=

−=

+=

+=

==

dtdqi =but

1st order differentialequation - integrateboth sides

3.5.1 R-C circuit analysis

CR

(a)

(b)

V

Consider a series circuitcontaining a resistor R andcapacitor C. If connection (a) ismade, the initial potentialdifference across the capacitor iszero, the entire battery voltageappears across the resistor. Atthis instant, the current in thecircuit is:

RVIo =

As the capacitor charges, thevoltage across it increases andthe voltage across the resistorcorrespondingly decreases (andso does the current in the circuit).After a “long” time, the capacitoris fully charged and the entirebattery voltage appears acrossthe capacitor. The current thusdrops to zero and there is nopotential drop across the resistor(since there is no current).When connection (b) is made,the capacitor dischargesthrough the resistor. Thecurrent is initially high andthen over time drops to zero.

207Electricity

When t = 0,q =0.


RCt

R

RCt

c

RCt

f

Vev

e1Vv

e1Qq

−

−

−

=

−=

−=

VC

QCv

q

f

c

=

=

Charging

cR

cRvVv

vvV−=

+=

The product RC is calledthe time constant of thecircuit. It is the time inwhich the current (andthus vR) would decrease tozero if it continued todecrease at its initial rate.

RC693.02lnRCt

RCt2ln1ln

RCt

21ln

h

h

h

=

=

−=−

−=

vR

t

V

RC

vC

t

V

RC

but

and

thus

since:

RCV

dtdv

0t@

eRCV

dtdv

c

RCt

c

=

=

=

−

initial slope

The half-lifeof the circuitis the timefor thecurrent todecrease tohalf of itsinitial value.

Discharging

RCt

R

RCt

c

RCt

cR

Vev

Vev

Qeq

Cq

dtdqR

CqiR

vv0

−

−

−

−=

=

=

+=

+=

+=

integrating

vC

t

V

vR

t

−V

RCt

RCt

oo

RCt

o

h

h

e21

eI2I

eIi

−

−

−

=

=

=

3.5.2 Time constant and half-life



C

Rvin vout

As capacitor charges, voltage acrossit increases. When capacitor is fullycharged, all of Vin appears across itand no voltage drop across resistor.As time constant becomes smallerthan the period T of the input pulse,the capacitor has time to charge anddischarge fully.

For low pass filter, we want RC << T to pass through lowfrequencies. i.e. small time constant

C R RC

0.05 µF

0.01 µ F

0.002 µF

47 kΩ

47 kΩ

47 kΩ

2.35 × 10−3

4.7 × 10−4

9.4 × 10−5

decr

easi

ng ti

me

cons

tant

vout

vin

Input signal variesperiodically from 0 to +v

t

3.5.3 R-C low pass filter

209Electricity

Consider a square wave input signal.For a small time constant or a lowfrequency input signal, the capacitorcharges up quickly and so the outputsignal looks like the input signal.This circuit acts like a low pass filter,high frequencies are shorted toground through the capacitor.

Input:

Output:


As capacitor charges, voltage acrossit increases and voltage acrossresistor decreases. As time constantbecomes smaller than the period Tof the input pulse, the capacitor hastime to charge and discharge fullyand voltage across the resistordecreases to zero.

For high pass filter, want RC >> T to pass through highfrequencies. i.e. large time constant

3.5.4 R-C high pass filter

decr

easi

ng ti

me

cons

tant

vout

vin

t

-vout

C R RC

0.05 µF

0.01 µ F

0.002 µ F

47 kΩ

47 kΩ

47 kΩ

2.35 × 10−3

4.7 × 10−4

9.4 × 10−5

C

R

vin vout


For a large time constant thecapacitor takes a long time tocharge up and so the output signallooks like the input signal. Thiscircuit blocks low frequency inputsignals.

Input:

Output:


Let connection (a) be made. Because of the selfinduced Emf, the current (and hence VR) in thecircuit does not rise to its final value at theinstant the circuit is closed, but grows at a ratewhich depends on the inductance (Henrys) andresistance (Ohms) of the circuit.

( )( )

LRt

L

LRtR

LRt

LR

Vev

e1Vv

e1Ii

dtdiLiR

vvV

−

−

−

=

−=

−=

+=

+=

RL

RvVv

iRv−=

=

Charging

VR

t

V

VL

t

V

L/RL/R

The quantity L/R is thetime constant of thecircuit and I is the finalcurrent.

LRt

L

LRt

R

LRt

LR

Vev

Vev

Iei

dtdiLiR

vv0

−

−

−

−=

=

=

+=

+=

Discharging When connection (b) is made the current (and VR)does not fall to zero immediately but falls at a ratewhich depends on L and R. The energy required tomaintain the current during the decay is providedby the energy stored in the magnetic field of theconductor

VR

tVL

tV

-V

3.5.5 R-L circuits

R

(a)

(b)

V

L

211Electricity


L

Rload

Unsmootheddc source

e.g. output fromrectifier circuit

If source voltage rises aboveits average value, then anyincrease in current is opposedby inductor and energy isstored in magnetic field ofinductor.

If source voltage falls belowits average value, then storedenergy is released frominductor to oppose anydecrease in current.

Result is that currentthrough load resistor ismaintained as near constantlevel.

3.5.6 R-L filter circuits

Low pass (Choke) circuit

∆Vin ∆VoutL

RHigh pass filter

High frequencies areblocked by the inductor, lowfrequencies are passedthrough.



∫

∫

∫

∫

=

<<

≈

=∴

=

=

=∴=

=

dtvRC1v

vvvv

dtvRC1v

iRv

idtC1v

idtqidtdq

Cqv

inout

rc

inR

Rout

R

out

out

Integrating circuit

orwhen RC is large

thus

Output voltage signal is theintegral of the input voltage signal.

vin

vout

thus

now,

and thus

Now

3.5.7 Integrator/Differentiator

dtdvRCv

vvvvdt

dVRCv

dtdv

C

dtdqI

IRv

inout

inCRC

Cout

C

out

=

≈∴>>

=

=

=

=

Differentiating circuit

For small RCcapacitorcharges upquickly.

Output voltage is the derivativeof the input voltage.

vin

vout

vin

vout

C

Rvin vout

C

R

vin vout

213Electricity


3.5.8 Example

1. A conventional ignition system in a motor vehicle consists of aninduction coil, the current to which is periodically switched on and offthrough mechanical “contact breaker points”. A high voltage is inducedin the secondary side of the coil when the contact breaker opens andcloses. If the resistance of the primary side of the circuit is 8 Ω and theinductance of the coil is 2.4 H, calculate the following quantities:

L12 V

Contact breaker

Sparkplug

Primary

SecondaryR

(a) The initial current when the contact breaker is just closed.(b) The initial rate of change of current when the contact breaker is

just closed.(c) The final steady-state current(d) The time taken for the current to reach 95% of its maximum value

( )

1

tLR

tLR

sA54.25.18

ILR

dtdi

0t@

eLRI

dtdi

A5.1I8I12

IRV0i,0t@

e1Ii

−

−

−

=

=

=

=

=

=

=

=

==

−=

Solution:

( )

s9.0tte05.0

e15.15.195.0

33.3

t4.2

8

=

=

−=

−

−



3.6 AC Circuits

Summary

( )tsinVv o ω=

po

rms V707.02

VV ==

po

rms I707.02

II ==

C1XCω

=

LXL ω=

rmsrmsR IVP =

Capacitive reactance

Inductive reactance

Rms current

Rms voltage

Rms voltage

Reactive power

φ= cosIVP rmsrmsav Average (active) power

rmsrmsIVS = Apparent power

( )

−=φ

−+=

RXX

tan

XXRZ

CL

2CL

2

Impedance

222in

out

CR1

1VV

ω+

=

1CR =ω

Low pass filter

1CR

CRVV

222in

out

+ω

ω= High pass filter

3db point

215Electricity


The induced voltage is directly proportional to therate at which the conductor cuts across themagnetic field lines. Thus, the induced voltage isproportional to the velocity of the conductor in thex direction (Vx = Vsinθ). The velocity componentVy is parallel to the field lines and thus does notcontribute to the rate of “cutting”.

θ= sinVv oinduced

maximum (peak) voltage Vo induced at θ = π/2Vp depends on:

• total number of flux lines throughwhich the conductor passes

• angular velocity of loop• no. turns of conductor in loop

• fingers: direction of field• thumb: direction of Vx• palm: force on positive charge carriersThus, current is coming out from the page.

( )tsinVvto ω=

θ=ωsince

then

Instantaneousvoltage

Peak voltage

radians

V0

ωt

1 cycle

2π

RH rule:

θ

θ

ω

V Vy

Vx

B

Velocity

3.6.1 AC Voltage

Consider a constant angular speed (ω)• at (0) motion of conductor is parallel to B, hence

induced voltage = 0• at (1), conductor has begun to cut magnetic field

lines B, hence some voltage is induced.• at (2), conductor cuts magnetic field lines at a greater

rate than (1) and thus a greater voltage is induced.• at (3), conductor cuts magnetic field lines at

maximum rate, thus maximum voltage is induced.• From (3) to (6), the rate of cutting becomes less.• At (6), conductor moves parallel to B and v = 0.• From (6) to (9), conductor begins to cut field lines

again but in the opposite direction, hence, inducedvoltage is reversed in polarity.

Time for one cycle iscalled the period.

01

23

6

9B

ω

ω magnet



The instantaneous voltage across theresistor is:

The instantaneouscurrent in theresistor is:

tsinRVRvi

oω=

=

The maximumcurrent in theresistor is whensinωt = 1 thus: tsinIi

RVI

o

oo

ω=∴

=

iv

+

-

ωt2π

( )

tsinPp

RIP

tsinRI

RtsinI

Ri

vip

2o

2oo

22o

2o

2

ω=

=

ω=

ω=

=

=

• power is a function of sin2 ωt• power is sinusoidal in nature

with a frequency of twice theinstantaneous current andvoltage and is always positiveindicating powercontinuously supplied to theresistor.

Instantaneous power:

Both instantaneous voltage andcurrent are functions of (ωt). Thus,they are in “phase”.

R VR

3.6.2 Resistance

tsinVv or ω=

v

217Electricity

Maximum value of V

Max or peakpower whensinωt = 1

θ or ωt

PoPower intoresistor

Current

Io+

-

2π


The area under the power vs time function is energy. Thus, it is possible tocalculate an average power level which, over one cycle, is associated withthe amount of energy carried in one cycle of alternating power.

Po

2πθ or ωt

Instantaneouspower

Pav

This energy would be that given by an equivalent dc, or steady-state,voltage and current over a certain time period compared to that from analternating current and voltage for the same time period.

2RI

dsin2

RI

dsinI2R

di2R

Rdi21

2

dp

P

2o

2

0

22

o

2

0

22o

2

0

2

2

0

2

2

0i

av

=

θθπ

=

θθπ

=

θπ

=

θπ

=

π

θ

=

∫

∫

∫

∫

∫

π

π

π

π

π

Average power:

thisintegralevaluatesto just π

What equivalent steady-state currentwould give the same average power as analternating current?

Let:

Thus:

rmsrmsav

2rmsav

orms

2o

av

VIP

RIP2

II

R2

IP

=

=

=

=Now:

or

Equivalent steady-state values which give thesame power dissipation as the application of analternating current with peak values Vp and Ip

0

0rms

V707.02

VV

=

=

0

0rms

I707.02

II

=

=

In AC circuits, V and I without subscriptsindicate rms values unless stated otherwise.

For Resistor circuit only. Seelater for LCR series circuit.

This result is only forsinusoidal signals.

∫π

θπ

=

π=

2

0iav

av

dp21P

2PArea

3.6.3 rms voltage and current



dtdvCi

dtdvC

dtdq

CvqVQC

=

=

=

=

The AC source supplies an alternating voltage v. Thisvoltage appears across the capacitor.

In general,

C is a “constant”

thus

Instantaneous current is proportional to the rate of change of voltage.

differentiatingw.r.t. time

(q,v are instantaneousvalues and thusfunctions of t)

The instantaneous current is amaximum IP when the rate ofchange of voltage is amaximum. Also, themaximum voltage Vponly appears across thecapacitor after it hasbecome chargedwhereupon the currentI drops to zero. Thus,maximums and minimums inthe instantaneous current leadthe maximums and minimumsin the instantaneous voltage byπ/2.

i

v

ωt

+

-

2π

( )

( )

( )

[ ]

C

o

o

o

oo

o

o

o

o

XIV

C1

2tsinIi

CVI2

tsinCV

tcosCVi

tsinVdtdCi

tsinVv

=

=ω

π+ω=

ω=

π+ωω=

ωω=

ω=

ω=

Capacitivereactance (Ω)

Now,

since

dtdvCi =

@ i = Ip

Maximums incurrent incapacitorprecedemaximums involtage across it.

Can be peak or rmsbut not instantaneous

Capacitive reactance is the opposition toalternating current by capacitance. Theopposition tendered depends upon the rate ofchange of voltage through the circuit.

thus

3.6.4 Capacitive reactance

C

v

VC

219Electricity


What is capacitive reactance? How can a capacitor offer a resistance toalternating current.

+

+ + + + +V E

- - - - - - -i

t

When the switch is first closed,it takes time for the charge Q toaccumulate on each plate.Charge accumulation proceedsuntil the voltage across thecapacitor is equal to the voltageof the source. During this time,current flows in the circuit.

Consider a capacitor connectedto a DC supply so that thepolarity of the applied voltagecan be reversed by a switch.

-i

t

When the polarity is reversed, the capacitor initially discharges and thencharges to the opposite polarity. Current flows in the opposite directionwhile reverse charging takes place until the voltage across the capacitorbecomes equal to the supply voltage whereupon current flow ceases.

++ + + + +

VE

- - - - - - -

Now, if the switch were to be operated very quickly, then, upon charging,the current would not have time to drop to zero before the polarity of thesupply voltage was reversed. Similarly, on reverse charging, the reversecurrent would not have time to reach zero before the polarity of the sourcewas reversed. Thus, the current would only proceed a short distance alongthe curves as shown and a continuous alternating current would result. Thefaster the switch over of polarity, the greater the average or rms ac current.Thus, the “resistance” to ac current is greater at lower frequencies andlower at high frequencies.



Let the inductor have no resistance. Thus, any voltage that appears acrossthe terminals of the inductor must be due to the self-induced voltage inthe coil by a changing current through it (self-inductance).

At any instant, vL+ v = 0 by Kirchhoff

( )

( )

( )

( )

( )

π−ω

ω=

ωω

−=

ω=

ω=

ω=

ω−=

−=

∫

2tsin

LV

tcosLV

i

dttsinL

Vi

tsinL

Vdtdi

tsinVdtdiL

tsinVdtdiLv

o

o

o

o

o

p

L

( )

π−θ−=θ

2sincos

instantaneous vL isproportional to rate ofchange of current

iv

ωt

+

-

2π

L

o

o

oo

XIV

L

LV

I

=

=ω

ω

=

Inductive reactance (Ω)

Now, i will be a maximum Ip when 12

tsin =

π−ω

Can be peakor rms but notinstantaneous.

changes in current ininductor lags changes involtage across it

Inductive reactance is the opposition toalternating current by inductance. Theopposition tendered depends upon the rateof change of current through the circuit.

VL

For high frequencies, the magnitude of the induced back emf is large and thisrestricts the maximum current that can flow before the polarity of the voltagechanges over. Thus, the reactance increases with increasing frequency.

Note: The inductor acts like a voltage sourcewhose polarity depends on whether i (or V) isincreasing or decreasing. As in all Kirchhoffcircuits, the emf’s are written on the left side of theequation, and voltage drops on the right.

3.6.5 Inductive reactance

Note:

L

v

221Electricity


[ ]0

sin4

IV

dcossin2

IV

dvi21P

t2sinIV21

tcostsinIVvip

20

2oo

2

0

oo

2

0av

oo

oo

=

θπ

=

θθθπ

=

θπ

=

ω=

ωω=

=

π

π

π

∫

∫

Instantaneous power:

Averagepower:

For a capacitor, the product VrmsIrms is called the reactive power. Thereactive power is an indication of the power alternately supplied anddischarged from the capacitor.

Reactive power is given the units var to distinguish this type of powerfrom Watts which is reserved for dissipative or active power.

Volts Amps Reactive

In the case of a resistor, the powerdissipated Pav =VrmsIrms and is theaverage of the instantaneous powerfluctuations over one complete cycle.However, for a capacitor, the averageof the instantaneous powerfluctuations is now zero.

( )( )tcosIi

tsinVv

o

o

ω=

ω=For a capacitor:

Capacitorcharging

Capacitordischarging

p

-p

v

i

2π

The reactive power is directlyproportional to the power supplied toand obtained from the capacitor.

3.6.6 Reactive power (capacitor)

rmsrmsR IVP =



[ ]0

sin4

IV

dcossin2

IV

dvi21P

t2sinIV21

tcostsinIVvip

20

2oo

2

0

oo

2

0av

oo

oo

=

θπ

−=

θθθπ

−=

θπ

=

ω−=

ωω−=

=

π

π

π

∫

∫

Instantaneous power

Averagepower

The reactive power is :

units: var

and is an indication of the poweralternately supplied to, andobtained from, an inductivereactance. If there is someresistance in the circuit, such asfrom the windings of the inductor,then energy loss occurs and thereis some active power involved.

Θ = ωt

In the case of a resistor, thepower dissipated Pav =VrmsIrmsand is the average of theinstantaneous powerfluctuations. However, for aninductor, the average of theinstantaneous powerfluctuations is now zero.

( )( )tcosIi

tsinVv

o

o

ω−=

ω=For an inductor:

3.6.7 Reactive power (inductor)

rmsrmsR IVP =

223Electricity

Power intoinductor

Power out ofinductor

vp

-p

v

i

2π


Lv CR

A varying voltage v from thesource will cause a varyinginstantaneous current i to flow inthe circuit. Because it is a seriescircuit, the current must be thesame in each part of the circuit atany particular time t.

• For the resistance, changes in vR are in phase with those of i• For the inductor, changes in vL precedes those of i by π/2• For the capacitor, changes in vC follow those of i by π/2

+

-

ωt

vR

vLvC

vtotal

2π

From Kirchoff, vTotal = vR + vL + vC at any instant. Note that each of thesevoltages do not reach their peak values when VTotal reaches a maximum,thus |VoTotal| <> |VoR| + |VoL| + |VoC|. Also, since the rms value of anyvoltage = 0.707 Vo, then |Vrmstotal| <> |VrmsR| + |VrmsL| + |VrmsC|

Algebraic addition generally only applies to instantaneous quantities (canbe applied to other quantities, e.g. peak or rms, if in current and voltageare in phase - such as resistor circuit only).Note also that the resultant of the addition of sine waves of same frequency resultsin a sine wave of same frequency.

CAN ONLY ADDINSTANTANEOUS VALUES

CANNOT ADD PEAK (OR RMS)VALUES BECAUSE THEY AREOUT OF PHASE

- Except for a resistive circuit.

3.6.8 LCR series circuit



In LCR series circuits, how then may the total or resultant peak (or rms)values of voltage and current be determined from the individual peak (orrms) voltages? A VECTOR approach is needed (can use complex numbers).

Consider the axes below which indicates either peak (or rms) voltages VR,VC and VL:

VL

VC

VR

VTotal

• VL always precedes VR by π/2thus VL is upwards on thevertical axis

• VC always follows VR by π/2thus VC is downwards on thevertical axis

φ

Current is common point ofreference in series circuit.

peak or rms

The resultant peak or rms voltage VT is the vector sum of VR + VL + VC

The angle φ is the phase angle of the resultant peak (or rms) voltage w.r.t.the peak (or rms) common current and is found from:

R

CLV

VVtan

−=φ

For an AC series circuit:• same current flows in all components.• vector sum of rms or peak voltages must equal the applied rms

or peak voltage• algebraic sum of instantaneous voltages equals the applied

instantaneous voltage

In complex number form:

( )CLRT VVjVV −+=

Complex numbers are a convenientmathematical way to keep track ofdirections or “phases” of quantities.

3.6.9 LCR circuit – peak and rms voltage

In a series circuit, the instantaneous voltageacross the resistor is always in phase with theinstantaneous current, thus, for the inductor andthe capacitor:

225Electricity


This represents the power dissipated in a resistorin an LCR circuit or consumed in some otherfashion (eg to drive an electric motor).

φ=

φ=

φ=

ωωπ

φ=

ωπ

=

∫

∫π

π

cosIVP

cos2

I2

V2cosIV

tdtsin2

cosIV

tdp21P

rmsrmsav

oo

oo

2

0

2oo

2

0av

For a resistor:

However, in general, there is a phasedifference between i and v :

The instantaneous power in circuit containing resistance andreactance:

( )( )φ+ω=

ω=

tsinVvtsinIi

o

o

( )( )

( ) ( )tcostsinIVsintsinIVcosp

sintcoscostsintsinIVtsintsinIV

vip

oo2

oo

oo

oo

ωωφ+ωφ=

φω+φωω=

φ+ωω=

=

( )tcostsinIV oo ωω−

Inductor, pL

Capacitor; pC

Instantaneouspower for resistor

Instantaneouspower for:

where φ may be either positive (capacitor) or negative (inductor).

( )

( )tsinIV

tsinPp2

oo

2o

ω=

ω=

units: Watts

Average power for the circuit:

Cos φ is called thepower factor.

Note: no phasedifference for resistor onits own.

or

The average power for thecapacitor and the inductor = 0,thus, this second integral = 0.

Because Pav is thepower that is actuallyused by the circuit, it iscalled the real or activepower.

[ ]∫π

ωφ

2

0LC tdp;psin

3.6.10 Power in LCR circuit

p

-p

into circuit

back from circuit

Pav

+



units: VA

Consider an LCR circuit in which the rms voltage across thesource and the (common) current are measured with meters.

Even if we know Vrms and Irms,we cannot determine theaverage (ie. the active) powerconsumed by this circuit sincewe need to also measure cos φ

VT

R

CLV

VVtan

−=φ

But, knowing the rms voltage across the resistor, inductorand capacitor, then the angle φ can be obtained from:

Peak or rmsvalues

By Kirchhoff’s law, we know that the peak or rms value of the appliedvoltage is the vector sum of the peak or rms voltage across each of thecomponents. Thus, the magnitude of the peak or rms applied voltage VT is(by Pythagoras):

2CL

2RT VVVV −+=

Multiply both sides by |I|22

CL22

RT IVVIVIV −+=

Thus, if |V| and |I| refer to rms values, then the power as calculated by theproduct of measured values of Vrms Irms is called the apparent power Sand is a combination of the active and reactive power within the circuit.

The apparent power is the total average power that needs to be suppliedby the power source (ie the energy company). Some of this power (cosφ)is used by the circuit (the active power). Some of this power (sinφ - thereactive power) is alternately supplied to and received from the circuit asthe reactive components alternately charge and discharge.

Although the reactive power is notconsumed by the circuit, associatedtransmission losses in alternate transfers ofenergy lead to undesirable inefficiencies.

A similar analysisapplies to parallelcircuits wherevoltage is thecommon point ofreference.

This formula applies toseries and parallel circuits.

active reactive

Note: The apparent power S is not simplythe sum of Pav and PR due to the phasedifference between V and I.

3.6.11 Apparent power

φ= cosIVP rmsrmsav

rmsrmsIVS =

L CR

A

227Electricity


φ= cosIVP rmsrmsavThe active power is the average power actuallyconsumed or dissipated by an LCR circuit.The apparent power is that which must besupplied to the circuit. rmsrmsIVS =

Even though an LCR circuit only uses a portion of the power supplied to it,it is desirable to reduce the transmission losses associated with energytransfers to and from the reactive parts of the circuit (i.e. a power factor of100% is desirable).

SP

cos av=φ

In an inductively reactive circuit the phase angle φ is negative and the powerfactor is termed lagging. In a capacitively reactive circuit, the phase angle ispositive and the power factor is termed leading. Most industrial circuits areeither capacitively or inductively reactive. To reduce transmission losses,capacitors and inductors may be purposely added to the overall circuit toensure that the power factor is as close to 100% as possible.Power factor correction involves connecting either a capacitor or inductorin parallel with the load so as to bring the power factor to 100%. When thisoccurs, the additional energy alternately transferred to and from the sourceis now exchanged between the load and the power factor correctioncomponent.Example: Consider a 1 kW (active) load with a 70%lagging (i.e. inductively reactive) power factor. Fromthe point of view of the source, the apparent power is:

For a lagging power factor, we require a capacitor to be connected inparallel with it to bring the power factor up to 100%. What value ofcapacitor is required? Since, in this example, the circuit is inductivelyreactive, it is simply necessary to use a capacitor whose reactive power isequal to the reactive power of the inductively reactive load

VA14287.0

1000S

=

=

6.45

cos7.0 1

−=φ

φ=−

The reactive power is the power alternatelyexchanged to and from reactive components:

Thus, the power factor is:

φ= sinIVP rmsrmsreactive

CV

XV

IVP

2rms

C

2rms

rmsrmsR

ω=

=

=

where Vrms is the rms voltage across the capacitor(same as voltage across load since capacitor is inparallel with it) and Irms is the rms current (as yetunknown) through the parallel capacitor.

φ= sinIVP rmsrmsR

Irms is the total current

3.6.12 Power factor



The total opposition to current in an AC circuit is called impedance. Fora series circuit, the impedance is the vector sum of the resistances andthe reactances within the circuit

IVZ =

Peak or rms valueswhich are vectors.

IVX

IV

X

IVR

LL

CC

R

=

=

=

Now, from a consideration of the voltages:

XL

XC

R

Z

φ

( )

−=φ

−+=

RXX

tan

XXRZ

CL

2CL

2

phase difference between the maximumsin the current and the voltage

For an RC series circuit:

CR1RX

tan C

ω=

−=φ

For an RL series circuit:

RLR

Xtan L

ω=

=φ

If XC > XL, circuit is capacitively reactive.If XC < XL, circuit is inductively reactive.If XC = XL - then resonance.

IV

Z

Z

XXR

IV

IV

I

VI

V

VVVV

2CL

2

2CL

2

2RT

2CL

2RT

=

=

−+=

−+=

−+=

Dividethrough by I

Can multiply and dividemagnitudes but mustadd as vectors

since

3.6.13 Impedance

229Electricity


3.6.14 Example

1RL

1R

LR

L1

121

RL1

12

1

RL1

1LRI

IRVV

IRVXRI

IZV

2

22

2

22

2

22

2

22

222in

out

out

2L

2

in

=ω

=ω

ω+

=

ω+

=

ω+

=

ω+

=

=

+=

=

L

R

Solution:

1. Determine the expression for Vout/Vin for the following high pass filtercircuit. Also determine the condition for

∆Vin ∆Vout


21

VV

in

out=

This condition is known as the 3 db point and is often usedfor defining the frequency range or bandwidth of the filter.


3.7 Electromagnetic waves

Summary

dtdid.B oooφ

εµ+µ=∫ L Displacement current

dtdid.B oooφ

εµ+µ=∫ l

Emfdtdd.E =Φ

−=∫ l

o

qAd.Eε

==φ ∫0Ad.B =∫

Maxwell's equations

Bv

1Eooεµ

= Electromagnetic field

oo

2 1cεµ

= Speed of light in vacuum

nvc= Refractive index

2

o

2o

BE

B1E

uuu

µε=

+=

or

Energy density

EB1Ioµ

= Intensity

Average intensity

231Electricity

εo = 8.85 × 10−12 F m − 1

µo = 4π × 10−7 Wb A−1 m−1

00o

av BE2

1Iµ

=

EAid ωε= Displacement current


When the switch is first closed, ittakes time for the charge Q toaccumulate on each plate. Chargeaccumulation proceeds until thevoltage across the capacitor is equal tothe voltage of the source. During thistime, current flows in the wire. Note,no current passes through the air spacebetween the plates (since there are nomobile charge carriers there).

Consider an air-filled capacitor connected to a DC supply through a switch.

++ + + + +

VE

- - - - - - -

+Q

-Q

Now, let us apply Ampere’s law to various parts of the circuit. While thereis a current flowing in the wire, there is of course a magnetic field whichencircles the wire the strength of which can be calculated according toAmpere’s law.

B

IV

Id.B oµ=∫ l

The line integral of B around the closedpath is proportional to the current I which“pierces” the surface bounded by thepath. Note that I flows through the flatsurface (shaded) enclosed by the circle B.

Now, what happens if we stretch the surface so that it lies within the gap ofthe capacitor? The current which pierces the surface is now zero, whichimplies that the magnetic field encircling the wire is now zero!! Somethingmust be wrong with Ampere’s law!

I

no current I in air gap -no mobile chargecarriers present.

It shouldn’t matter which surfacewe choose! The line integral(around the line which defines thesurface) comes out the same.Therefore there must be a currentI!! But how can there be a currentthrough the air in the gap ???

3.7.1 Charging a capacitor



Maxwell argued that although there is no actual conventional current inthe gap (since the material inside the gap is an insulator - eg. air, with nomobile charge carriers to “carry” the current), one could imagine that thereis indeed a current arising from the small displacement of charges withinthe atoms and molecules of the gap material.

+ -

+ -

+ -

+ -+ -

+ -+ -

+ -

+ -

+ -

E

1. As current flows in the wires, build up of charge on either plate ofthe capacitor establishes a field E within the gap. This field startsfrom zero and reaches a maximum when the capacitor is fullycharged.

2. The field E acts upon the air molecules within the gap causing themto become polarised as electrons within the atoms are attractedtowards the positive plate, and positively charged nuclei becomeattracted to the negative plate.

3. The charges within the molecules are displaced but cannot movevery far. The charges actually move while the field E is increasing(or decreasing). When the field E is steady, the charges remainstationary in their displaced positions.

4. Any movement of charge is an electric current, thus, when the fieldE is changing, there is a current in the gap as the charges within themolecules are displaced from further from their equilibriumpositions. This current is called the displacement current. Thedisplacement current only flows when the field E is eitherincreasing or decreasing.

Here’s how it works:

The amount of lining up of molecules inthe insulating material (or “dielectric”) ischaracterised by the permittivity of thematerial. Materials with a high permittivityhave molecules that line up rather easilyin the electric field. When this alignmentoccurs, a series of internal fields arecreated which has the effect of reducingthe total field between the plates.

3.7.2 Displacement current

+ -

233Electricity


Maxwell argued further that the effects of a displacement current existeven when there is a vacuum between the plates! He reasoned that thechanging electric field E between the plates results in there being anencircling magnetic field within the gap - one didn’t actually require thereto be a movement of charges to produce a magnetic field, all that wasneeded was a changing electric field!

As capacitor charges up, electric field E in thegap increases. The electric field is provided bythe build up of charge on the plates of thecapacitor rather than the presence of actualcharge carriers in motion.

+ -

E

The changing electric field within the gap iselectrically and magnetically equivalent to theexistence of a displacement current within the gap.

( )do iid.B +µ=∫ l

Thus, for the example of a charging capacitor where there is a changingcurrent in a circuit, Ampere’s law still holds as long as the displacementcurrent id is included: Note the use of lower case i to signify

changing or “dynamic” quantities.

If we choose a surface which is pierced by the wire (i.e. conductor) thenthe current is just i and id =0. If we choose a surface which is pierced by thefield E, then the current is id and i = 0. In any case, i = id.

Now, consider the electric flux φ between the plates.

dtddtdqii

EAq

EA

Ad.E

o

d

o

o

φε=

==

ε=

φε=

=

=φ ∫for a constant E over area A

Thus, Ampere’s law becomes:

φε+µ=∫ dt

did.B ool

This equation states that amagnetic field B may becreated by either a current i(steady or changing) and/or achanging electric flux.by Gauss’ law

3.7.3 Ampere’s law


Displacementcurrent

Conductioncurrent

For parallel plates of area A and gap d, thedisplacement current is: EAid ωε=


Let us summarise the various equations of electricity and magnetism:

dtdid.B oooφ

εµ+µ=∫ l

Faraday’s law:

Emfdtdd.E

=

Φ−=∫ l

Ampere’s law:

Gauss’ law:(Electric charge)

o

q

Ad.E

ε=

=φ ∫

0Ad.B =∫

Gauss’ law:(Magnetism)

These four equations are known asMaxwell’s equations and can beused to quantify all aspects ofelectricity and magnetism, includingthe existence and properties ofelectromagnetic waves.Note the similarity between Ampere’s lawand Faraday’s law. These laws are notquite symmetric. Ampere’s law says that amagnetic field can be created by either asteady (or changing) current in aconductor, or a changing electric field inspace. Faraday’s law says that an electricfield can only be created by a changingmagnetic field. This asymmetry arisesbecause of the fact that magnetic fieldlines always form closed loops. An electricfield line may originate on one isolatedcharge and terminate on another.Magnetic field lines always close uponthemselves. There is no such thing as anisolated magnetic “charge”. The fact thatisolated electric charges can exist meansthat an electric current is a physicalpossibility as the charges move from oneplace to another. This leads to the µoi termin Ampere’s law. The impossibility of any“magnetic” current means that this term ismissing in Faraday’s law. In fact, this isexactly what the first two equations aresaying. In the electric field, field lines mayoriginate from a charged particle within thesurface and then pass out through asurface and terminate on isolated chargessome distance away. Gauss’ law formagnetism says that for any closedsurface in space, the total magnetic flux iszero meaning that all magnetic flux linesjoin up with themselves.

3.7.4 Maxwell’s equations

235Electricity


Consider an electric field which leaves the charge which created it andmoves through space with a velocity v. The field is on the point of enteringan area A.

Now, consider Ampere’s law:

Eyvdtdd.B

oo

oo

εµ=

φεµ=∫ l

In a time interval dt, the fieldhas travelled a distance vdt. Theelectric flux through the area Ahas changed. Eyv

dtd

EyvdtdEA

=φ

=φ

=φ

The line integral is the product of B and thelength y (since y lies along the same direction asB and after time dt, B now exists only on the righthand edge of the area), thus:

Bv

1E

EvBEyvBydtdd.B

oo

oo

oo

oo

εµ=

εµ=

εµ=

φεµ=∫ l

Consider now the passage of amagnetic field B through an area A.

vBEByvEy

dtdd.E

=

=

Φ−=∫ l

The change inmagnetic flux is:

Byvdtd

ByvdtdBA

=Φ

=Φ

=Φ

An application of Faraday’s law yields:

Faraday’s law and Ampere’s lawsay that a moving electric field isaccompanied by a magnetic fieldand a moving magnetic field isaccompanied by an electric field.This can only happen when v = c:

18oo

2

oo

ms103c

1c

vBBv

1E

−

×=

εµ=

=εµ

=

Note: we areconsidering themovement of thefield only, thereare no chargespresent hence noI term inAmpere’s law.

E

A = y v(dt)vdt

y

E

y

B

E

Bvdt

y

The fields move with the velocityof light: c = 3 × 108 ms−1.

3.7.5 Electric and magnetic fields

v

236

εo = 8.85 × 10−12 F m − 1µo = 4π × 10−7 Wb A−1 m−1



How can an electric field move through space?

Previously we considered an electric field moving through space. Nocharges were present, hence, in Ampere’s law, we only needed to considerthe displacement current. The conduction current i was set to zero.

Consider an electric charge initially at rest which then accelerates andreaches a constant velocity. An electric field (and associated increasingmagnetic field) moves along in space with the charge. The fields are attachedto the charge.At some instant, point P, some distance from thecharge, initially experiences the electric field E.When the charge q accelerates, the field E movesalong with (i.e. it is attached to) the charge.

The “change in E” signal is a E vector field which is not attached to thecharge. It is a field which is “launched” into space when the charge isaccelerated. The field travels through space with an accompanying magneticfield of its own at velocity c. The “change in E” signal is a movingelectromagnetic field. A new “change in E” field is transmitted each time thecharge accelerates or decelerates. Electromagnetic waves are travelling Eand B fields in space and are created by accelerating charges.

Some time later, the charge reaches a constantvelocity and the charge and P are no longer inline with each other. The direction of E asexperienced by the point P will have changed.A magnetic field also exists at point P inaccordance with Ampere’s law. Let usconcentrate on the change in the direction of E:

B’E

v

P

BE

The point P does not experience the change in direction of E instantaneously.When the charge accelerates to a new velocity, it “transmits” a “change in E”vector outwards. When the point P receives this “change in E” signal, it isadded to the original E at P to give the new direction of E at P.

Original directionNewdirection

“change in E”

q

P

E

PE

q

3.7.6 Electromagnetic waves

Rest

Moving

237Electricity


The most common type of travelling electromagnetic fields are those whichvary continuously in a smooth periodic manner. Previously we examined anelectric field “pulse” and its accompanying magnetic field pulse. Nowconsider a smoothly varying electric field travelling through space:

Snapshot of a periodic electric fieldmoving with velocity c

This field periodically changes in magnitude and direction.

E

E

B

An electric field travelling through space is accompanied by a magneticfield with a direction normal to the field and the velocity. Since in thisexample the electric field varies in magnitude periodically with time, thenthe magnetic field also varies periodically with time:

Because of the periodicity, the moving electric and magnetic fields aretogether called an electromagnetic wave. These “field” waves have all theproperties of mechanical waves, displaying diffraction, interference,reflection and refraction. The range of frequencies most commonlyencountered is termed the electromagnetic spectrum.

3.7.7 Periodic electromagnetic waves

238

( )( )

18

ooms1031c

ctBtE

−

×≈εµ

=

=

t

t

( ) tsinEtE 0 ω=

( ) tsinEtE 0 ω=

( ) tsinBtB o ω=



For an electromagnetic wave travelling in an insulator, or dielectric, thepermittivity and permeability of the material has to be taken intoconsideration. The velocity of the wave in a dielectric becomes:

εµ

=

εµεµ

=

1

1voorr

nvc

rr

=

εµ=

the refractive indexof the medium.

µr ≈ 1 for mostmaterials

In a perfect conductor, the electricfield is completely cancelled by theredistribution of charges whichoccurs and the wave is reflected.

3.7.8 Electromagnetic waves in a dielectric

239Electricity

For very high frequencies (such as that of visiblelight) values of ε can be substantially less thanthe steady state value.

or

or

µµ=µ

εε=εwhere

For electromagnetic waves incident on a real conducting surface, most ofthe wave is reflected. That which is absorbed, diminishes in magnitudeexponentially with depth into the material. The depth at which themagnitude of the fields has dropped by ≈63% is called the skin depth δ.

µσπ=δ

f1

At relatively lowfrequencies:

At relatively highfrequencies:

µ

ε

σ=δ

2

+

+

E

Atombecomespolarised inthe presenceof E

X rays (and also gamma rays) have very short wavelengths and very highfrequencies. Electrons within a material cannot respond fast enough to thesefrequencies to allow the usual energy dissipative mechanisms to occur. Instead, Xrays lose energy by ejecting electrons from the material through which they travel(ionisation). But, the electrons most affected by this ionisation process are those inthe inner most shells of an atom (K and L). Since there are relatively few electrons inthese innermost shells, X rays do not have a high probability of interacting with themand hence their high penetrating ability.

When an electromagnetic wave isincident on a surface, part of the energyin the waves is absorbed, part of it may bereflected, and part transmitted through it.

and thus:

In a dielectric, the amount of radiationabsorbed depends upon the permittivity whichin general, is frequency dependent. The field Ecauses atoms in the material to becomepolarised. Losses arise as the polarisationdirection changes in the presence of anoscillating field E from the wave.

FrequencyHz Conductivity

Depends on frequency. The higher thefrequency, the lower the value of ε.

----

---

--

-

--

-

---

-

-

-

-


Consider the energy required to charge a capacitor. The energy is storedwithin the field between the plates of the capacitor.

2E CV

21U =

( )AdE21

dEdA

21U

EdVdAC

2o

22oE

o

ε=

ε=

=

ε=Now,

and

Thus:

The energy density is the energy contained within the electric field perunit volume (J m−3). Here, the volume occupied by the field E is theproduct Ad. Thus:

If the space between theplates of the capacitor isfilled with a dielectric, thenwe use ε = εr εo

2EE E

21

AdUu ε==

This formula says that the energy density ofan electric field is proportional to the squareof the amplitude of the field.

Energy density of amagnetic field.

BE

oo

B2

oo

2oBo

22oE

uu

1c

uc

c2u21

cB21u

BcE

=∴

µε

=

µε=

µε=

ε=

=Now,

but

The energy carried by anelectromagnetic wave is dividedequally between the E and B fields.

The total energy density of anelectromagnetic wave is thus:

2

o

2o

BE

B1E

uuu

µε=

+=

or

2B LI

21U =

The energy required to “charge” an inductor of length l, cross sectionalarea A and N number of turns is:

2

oB

2

o

2

o

2

oB

B2

1u

AB2

1

NlBNA

21U

µ=

µ=

µ

µ=

l

l

lNIB oµ=

l

2

oNAL µ=

The energy is stored with themagnetic field:

where

Al isvolumeof field

3.7.9 Energy in electromagnetic waves



The total energy density of an electromagnetic wave is : 2oEu ε=

The energy density is proportional to the square of the magnitude of the Efield and is the energy (J) per unit volume of space containing the field.When an electromagnetic wave travels through space, energy is transportedalong with it (i.e. within the field). The intensity is power (i.e. rate ofenergy transfer) transmitted per unit area.If, during time ∆t, an electromagnetic wave passes a particular point inspace, then the length of the “volume” containing the field is l = c∆t. If weconsider a 1m2 area perpendicular to the direction of travel, then theintensity of the electromagnetic field is obtained from:

uc

tct

uA1A

tu

A1V

tuArea

PI

=

∆∆

=

∆=

∆=

=

l

EB1

1EB

cEI

o

ooo

2o

µ=

µεε=

ε=

Substituting the energydensity u, we have:

oo

1cεµ

=

In terms of magnetic fieldintensity H, and taking intoaccount the vector nature of Eand H, the intensity I isexpressed:

HES ×=

The use of theperpendicular areais a hint: a since Eand B are vectors,and it is theperpendicular areathrough which theintensity is beingcalculated, theintensity I will be avector also and willinvolve a crossproduct.

The vector S is thePoynting vector.

3.7.10 Intensity of electromagnetic wave

241Electricity

since

If E and B both vary sinusoidally with time,and E0 and B0 give a maximum intensity I0,then:

( )tsinBE1I 2oo

oω

µ=

Average intensity ofelectromagnetic wave

To find the average power, or average intensity,we integrate the sin2 function over 2π::

ooo

2

0

2oo

o

2

0i

av

BE2

1

dsin2BE1

2

dI

I

µ=

θθπµ

=

π

θ

=

∫

∫

π

π

this integralevaluates to π

Or, working in terms of the E field only:

2avo

2ooav

cE

E21cI

ε=

ε=

2o

2av E

21E =

rms value of E


3.7.11 Examples

242

Solution:

1. Calculate the magnitude of the magnetic field in an electromagneticwave if the magnitude of the electric field is measured to be 10 V m−1.

T1033.3103

10cEB

8

8

−

×=

×

=

=

Solution:

2. The radiation output from the sun is estimated to be 4 × 1023 kW. (a)Calculate the radiation intensity at the edge of the atmosphere if thedistance from the sun is taken to be 1.49 × 1011 m. (b) Determine theaverage magnitude of the E and B fields of this radiation. Compare Bwith the magnitude of the Earth’s magnetic field BE

( )2

211

23

av

mWk4.1

1049.14

104API

−

=

×π

×

=

=

(a) (b)

( )( )

T10B

T1042.2103

726B

mV726E

E1031085.8104.1

cEI

4E

6

8av

1av

2av

8123

2avoav

−

−

−

−

≈

×=

×

=

=

××=×

ε=


Earth’s magnetic field is twoorders of magnitude larger.


Mechanics

Part 4


4.1 Scalars and vectors

Summary

2V

2H RRR +=

H

VRR

tan =θ

kjiR C,B,A=

kjiA 111 CBA ++=

kjiB 222 CBA ++=

222 CBAR ++=

θ=• cosBABA

212121 CCBBAA ++=•BA

θ=

×=

sinC BABAC


Magnitude of vector

Angle between vectors

Vector components

Magnitude of resultant

Unit vector

Vector dot product

Vector cross product


We draw an arrow at θ = 60o and length 5 units on the x and y coordinateaxes to represent the force acting on the body.

5 N

60oA good example of a vector quantity is aforce. Let’s look at a force of 5 N pullingupwards on a body at an angle of 60°. Thevector that represents this force consists of amagnitude and a direction.

Can we replace this force with two forces FXand FY which act in the vertical andhorizontal directions, and which when actingtogether at the same time, produce the exactsame motion of the bodyas the single 5Nforce actingat 60o?

FY

FX

The answer is of course, yes! All we dois lay out on the x and y axes arrowswhich correspond to the force at θ = 60o.

y

x

These horizontal and vertical forces, whichwhen acting together have the same effect asthe original force, are called the forcecomponents of F. We can call thesecomponents FH and FV, or FX and FY, itdoesn’t matter, as long as we know whichcomponent is which. The magnitude of thesecomponents are simply:

F

FH

FV

y

x

F

θ

θ

θ=

θ=

sinFF

cosFF

V

H

4.1.1 Vectors

Physical quantities that have adirection associated with themare called vectors.Examples are: displacement,velocity, acceleration, angularacceleration, torque, momentum,force, weight.

Examples are: temperature,energy, mass, electric charge,distance.

Physical quantities that do nothave a direction associated withthem are called scalars.

245Mechanics


Dividing up a vector into components is a useful procedure because itallows us to easily calculate the effect of a combination of vectors.

F

P The combined effectof the two forces canbe expressed as asingle force which wecall the resultant R.

To find the resultant of F and P, we simplyadd the vertical components of each togetherto obtain the vertical component of theresultant, and do the same for the horizontalcomponents.

The magnitude of resultant R of thesetwo components is given byPythagoras’ theorem:

y

x

F

FH

FV

PPV

PH

2V

2H RRR +=

The direction, or angle, of theresultant is found from:

H

VRR

tan =θ

RV and RH arethe componentsof the resultantforce R.

R

RV

RH

RRV

RH

R

θ θ

4.1.2 Addition of vectors

VVV PFR +=

HHH PFR +=

There is a graphical method ofdetermining the magnitude anddirection of vectors. We draw vectors ina head to tail manner. The resultant Ris found by joining up any gap betweenthe first tail and the last head.



The division of a vector into horizontal and verticalcomponents is a very useful concept. The idea alsoapplies in three dimensions, where a vector hascomponents along the x, y and z axes.

x y

z

R

That is, the direction of the components ofthe vector are that of the correspondingcoordinate axis. Let the magnitude ofthese components be A, B and C.

Rather than saying a vector R hasa magnitude of 5 units and acts at 30o to the x axis, 20o to the y axis, and80o to the z axis, we need a good concise method of expressing thisinformation.

A B

C

Let the vectors i, j and k have amagnitude of 1 unit, and havedirections along the x, y and z axesrespectively. Now, the componentsof our vector can be written:

Why do this? It is a good way tokeep the components of a vectororganised. Indeed a great way ofwriting a vector in terms of itscomponents is: kjiR CBA ++=

This is a vector equation. We express a vector as the resultant of itscomponent vectors. The magnitude of the component vectors are the scalarsA,B and C. When the unit vector i (which points in the x direction) ismultiplied by the scalar magnitude A, we obtain Ai which is the vectorcomponent in the x direction and so on for the product Bj and Ck. Themagnitude of R is given by:

kjiR C,B,A=

When two vectors are to be added, we simply add together thecorresponding magnitudes of the i unit vector, the j and the k unit vectors.

kjiA 111 CBA ++=

kjiB 222 CBA ++=

( ) ( ) ( )kjikjikjiBA

212121

222111CCBBAA

CBACBA+++++=

+++++=+

MagnitudeDirection

222 CBAR ++=

4.1.3 The unit vectors

247Mechanics


When force is multiplied by distance, we obtain the work done by theforce. But the force and the distance must be acting in the same direction.

s

FFsW =

But, what if the forceacts at an angle to thedisplacement?

s

F

It is only the component of forcewhich is in the same direction as thedisplacement which contributes to thework done by the force. Hence, tofind the work done, we need tomultiply the distance times thehorizontal component of the force.

θ= cosFsW

Even though force anddisplacement are both vectors, theproduct of the horizontal componentand the distance is a scalar.

The product of two vectors involving the component of onewith respect to the other is called the vector dot product:

s

F

Fcosθθ

sF •=WThe dot product gives a scalarquantity. More formally, the dotproduct of two vectors A and B

is given by:

θ=• cosBABAA

θ

BA •

B

The product of themagnitudes of these twovectors is the dot product

If kjiA 111 CBA ++=

kjiB 222 CBA ++=

212121 CCBBAA ++=•BAthen

4.1.4 Vector dot product



When a force is applied to a body so as to causea rotation, we say that the application of theforce results in there being a moment appliedabout the axis of rotation.The magnitude of the moment isgiven by the product of the forcetimes the distance from the axis ofrotation.

r

F

rFT =

The moment, or torque only depends on the perpendicular distancebetween the axis of rotation and the line of action of the force. Thus, if theforce is applied at an angle, the moment is given by the vertical componentof the force times the distance.

The product of the vertical component and the perpendicular distanceyields a new vector perpendicular to both.

θ=

×=

sinC BABACB

Bsinθ

θ

A

C

F

Fsinθ

θ

θ= sinFrT

r

The resulting momentis a vector quantity.

The direction is given by theright hand rule. If the fingerscurl in the direction from A toB, then the direction of theresulting vector AxB is givenby the thumb. Thus, AxB isopposite in direction to BxA.

4.1.5 Vector cross product

249Mechanics


Quantities like displacement,velocity, acceleration, angularacceleration, torque, momentum,force, all have both a magnitude anda direction associated with them,they are vector quantities.

Quantities such as temperature,energy, mass, electric charge donot have a direction associatedwith them. For example, a massmoving with a velocity has thesame amount of kinetic energy nomatter what the direction of thevelocity might be. They are scalarquantities.

A vector dot product gives ascalar quantity.

A

θ

B

θ=• cosBABA

Acosθ

θ=

=

sinBACxBAC

B

Bsinθ

θ

A

C

A vector cross product gives avector quantity.

Dot and cross productsWhy does the dot product give ascalar quantity and the cross producta vector quantity? This is a hardquestion to answer. The dot productA.B gives the same result no matterwhich coordinate axes are used torepresent the two vectors A and B.For the cross product, the twomagnitudes of the two vectorcomponents being multiplied are atright angles to each other. Theresulting product is a number, whichin two dimensional space, would bea scalar. But, in three dimensions,the cross product behavesmathematically like a vector.Mathematically, it appears that thisnew vector points in a directionnormal to the plane formed by theother two vectors. That is, the vectornature of the cross product is aconsequence of the threedimensional character of our spaceand laws of mathematics!

4.1.6 Scalars and vectors (summary)



4.1.7 Example

Solution:

15sin80=14.77

12sin45=8.48

12cos45=8.48

15cos80=2.6

15

12

62.2908.113.6tan

74.1208.113.6R

08.11)48.8(6.2H

3.648.877.14V

22

=θ

=θ

=

+=

=−−=

=−=

∑

∑

1. Find the difference A-B between the two vectors shownbelow:

A

B

80o45o

12 15

251Mechanics


4.2 Statics

Summary

0F

0F

V

H

=

=

∑

∑

0M =∑

Static equilibrium



Equilibrium is a state of balance.When an object (or body) is at rest,then we say that the body is in staticequilibrium and the vector sum of allforces acting on a body is zero..

When a body is moving with aconstant velocity, then the bodyis in dynamic equilibrium.

To make it easy to determine the vector sum of theforces, it is convenient to divide them up intohorizontal and vertical components.

0F

0F

V

H

=

=

∑

∑This is the ForceLaw of Equilibrium

The force law of equilibrium is useful fordetermining the magnitude and directions ofunknown forces on a body at rest.Static equilibrium also means that the body is notrotating. Hence, for static equilibrium, the sum of themoments acting on the body must equal zero.

0M =∑

When determining moments acting on a body, we must choose clockwise or anti-clockwise as being positive and stick to it. It doesn’t matter which direction is chosen,some text books say clockwise is positive, others anti-clockwise.

0F =∑If the bodyis at rest

4.2.1 Equilibrium

When two forces are inequilibrium, they must be equaland opposite in direction.When three forces are inequilibrium, their lines of actionintersect at a common point (theyare concurrent) and act in thesame plane (they are coplanar).When four forces are in equilibrium,the resultant of any two of themmust be equal and opposite to theresultant of the other two.

253Mechanics


Newton’s 3rd law states that every action is accompanied by anequal and opposite reaction.

W =mg

N =mg

When an object rests on asurface, the gravitationalforce W acts downwards.The reaction to thisdownwards force is whatwe call the normal forceacting upwards. Thenormal force N is equaland opposite to W.Surface

When the object rests on thesurface, the atoms within thesurface actually deflect from theirequilibrium positions in the crystalstructure of the material. Thisdeflection is much like that of adeflected spring. The deflectionresults in their being a resistingforce much like that when we pushour hands down and compress aspring. Every surface deflects avery small amount when placedunder load. This deflection results inthe reaction force.

N = W V

= mgcosθ

Surface

WV

W H = mgsinθ

W = mg

On an inclined surface, thenormal force has amagnitude equal andopposite to that of thevertical component of theweight W.

θ

4.2.2 The normal force



The normal force is one example of a reaction to an imposed load. Considera truss of a bridge which has a pin joint at one end and a sliding joint onthe other.

The pin allows rotation of thetruss but prevents linearmovement in the x-y directions.

The rollers ensure that no forces aretransmitted in a direction parallel tothe supporting surface and allows thetruss to expand or contract accordingto a change in load or temperature

Now, the truss can be considered to be a rigid body (comprised ofindividual members). In two dimensions, the laws of equilibrium can onlybe used if the body is supported by no more than two reaction forces. Ifthere are more than two (or three in three dimensions), then the reactionforces are indeterminate since it is generally not possible to determinewhat proportion of the load is carried by each support. Thus, in the case ofthe truss, we need to show that the pin and roller supports correspond totwo and only two effective support or reaction forces.

y

x

Since the pin allows nomovement, there are reactionforces FH and FV acting toprovide a total reaction FL

The reacting force at a rollersupport is always perpendicularto the supporting surface (in thisexample, vertically upwards)

FH

FV

FRFL

4.2.3 Reactions

255Mechanics


Earlier, we showed how the supports offered to a body subjected togravitational and other loads can be shown as reaction forces. This is theprocedure for drawing a free-body diagram.

A diagram of a body which is isolated from all itssupports and shows only those forces actingdirectly on it is called a free body diagram.

This isimportant

It so happens that part of an object may be represented as afree body diagram as long as the internal forces acting onthe isolated member from the part removed are shown.

Surface

W

N

Loading system Free body diagram

Compression intruss memberpushes on the pinjoint.

Tension in trussmember pulls on thepin joint.

Vertical andhorizontalreaction forces

Part of atruss

Free bodydiagram of pinjoint

Surface

N

WV

WH

4.2.4 Free body diagram



The equations of equilibrium can be used to findthe resultant of a force system. Consider theforces on the beam shown below: Can all these forces be

replaced by a single forcepositioned in such a way sothat the reaction forces at eachend of the beam areunchanged?

Yes!

10 N8 N

12 N

To do this, we imagine that the resultant is the sum of the individual forces.In this case, all the forces are vertical, and thus, taking upwards to bepositive, we can see that the magnitude of the resultant is:

N1481210R

=

−+=

Upwards.

But where should this 14 N upwards force be located? We takemoments about a point X. Taking anticlockwise as positive, we have:

4 m

2 m1 m

R

x?

14

( ) ( ) ( )m3x

41228110x14=

+−+=Note, an anticlockwise moment is takenas positive since here, we choseupwards as a positive direction forforces, and to the right, as a positivedirection for distance.

X

4.2.5 Resultants

We don’t show the reaction forces at the end of the bar since we want theresultant of the applied forces. If the reaction forces were included, theresultant would be zero because the bar is in static equilibrium.

257Mechanics


1. A ladder leans against a smooth wall.George, who has a mass of 81.5 kg,has climbed up the ladder and isresting at a point C as shown. Hisfriend Fred pushes against the ladderat D with a force 240 N to steady it.Calculate the force exerted on thewall and on the floor by the ladder.

The first step in solving this problem,and all mechanics problems, is to drawa free-body diagram.

C

D

A

B

1.5

4 m

1 m

Now, start with the equations of equilibrium:

V

V

F800

0F

=

=∑

240RF0F240R

0F

H

H

H

−=

=++−

=∑

Can’t do any more than this, so goon to verticals.

( ) ( )

N50240290F

290RR418005.1240

0M

H

A

=

−=

=

=+

=∑

Well, this is fine for the vertical reaction atA, but we still need some more informationto determine R. Let’s try summing themoments, say about “A”.

C

DA

B

1.5

4 m

1 m

R

F

W

P

800

240

4.2.6 Example

Solution:



4.3 Moment of inertia

Summary

A

axx ∑=

A

ayy ∑=

Y

22

I

axAX

=

=∑

X

22

I

ayAY

=

=∑

2CXX AdII +=

AIk =

Centroid of an area

2nd moment of anarea or moment ofinertia

Transfer formula

Radius of gyration

259Mechanics


Gravity acts on all the atoms in a body, but it is convenient (for the purposesof mechanical analysis) to imagine that the gravitational force acts at asingle point. The centre of mass is that point in the body through which theforce of gravity can be said to act. The position of the centre of mass for anobject can be calculated by considering its dimensions. Even complicatedshaped bodies can be readily analysed.To do this, let us imagine that a body is reducedin thickness to that of a sheet of paper. Theproblem is thus reduced to two dimensions. Thepoint which used to be the centre of mass is nowcalled the centroid of the area. Centroid

The two dimensional surface area can be considered to be composed of aseries of small elements, each of an area “a” and thickness “t” and density ρ.

The weight of one smallelement is thus:

taw ρ=

The total weight is:

tAg

gta

wW

ρ=

ρ=

=

∑

∑

Total area.

Now, the centroid of an area is the centre of mass fora thin object. If the object is supported at its centre ofmass, it is in static equilibrium.

Let us take moments about the pointO. For equilibrium:

O

a

W

w

y

x

x

A

axx

axtxtA

wx

...xwxwxwxW 332211

∑

∑

∑

=

ρ=ρ

=

++=

A

ayy ∑=

If the object were tipped up on itsside, then:

These equationsgive the coordinatesof the centroid of thearea A. We haveneglected the factorg for conveniencehere.

4.3.1 Centroid and centre of mass



π

=

π

=

3r4y

3r4x

x

y

r

x

y

b

a

b

x

y

a

2by

2ax

=

=

3by

3ax

=

=

4.3.2 Centroids of areas

The centroid of a composite area can be calculated by dividing thecomposite area into simple areas and taking the moment about a convenientfixed point. We then treat each simple area as an element where:

∑= axAx

∑= ayAy

The moment of an area is equal tothe sum of the moments of itscomponent areas.

Ayayay

Axaxax

2211

2211

+

=

+

=

Ayayay

Axaxax

2211

2211

−

=

−

=

The moment of anarea with arearemoved is themoment of thewhole area minusthat of the arearemoved.

a1

a2

A

x2

x1

O O

x1

a1 a2

π3r4

x2

261Mechanics


A very important quantity for determining the stress carrying capacity ofbeam is the 2nd moment of an area - often called the moment of inertiaof an area and given the symbol I.The 2nd moment of an area isdetermined in a similar way to thefirst moment of an area except thedistances are squared.

Y

22

I

axAX

=

=∑

X

22

I

ayAY

=

=∑

This is themoment ofinertia w.r.t.the y axis.

This is themoment ofinertia w.r.t.the x axis.

For beams, a large moment ofinertia about a particular axis is ameasure of its load carryingcapacity. For example, an I sectionbeam has a larger moment ofinertia than a simple square sectionof the same cross sectional areabecause its mass is concentrated atlarger distance away from the axis.

The distances X and Y arenot the centroid butrepresent locations called theradius of gyration.

X X

Y

Y

4.3.3 2nd moment of an area

Note, since we are taking aboutmoments of inertia, (i.e the secondmoment of an area), the radius ofgyration does not correspond to thecoordinates of the centroid (the firstmoment of an area) about an axis sincethe second moment of an area containsa a sum of distance squared terms.

AIk

AkI 2X

=

=

The distance X (or Y) is the radius ofgyration about the y (or x) axis and isusually given the symbol k. Theradius of gyration is the perpendiculardistance from the axis at which thetotal mass or area of the body may beconcentrated without changing itsmoment of inertia.

Moments of inertia andmoments of areasIn this chapter, we have talked aboutmoments of areas, and moments of inertiaof areas. We must not get these confusedwith the moment of inertia of a rotatingbody. The two moments of inertia aresimilar, but one applies to an area, theother to a body. Thus when talking aboutmoments of inertia, we must specify whatwe are talking about. Hence, unless thecontext is clear, we say “moment of inertiaof an area” or “moment of inertia of abody” about a particular axis.



3bh

12bh4

12bh3

12bh

4hbh

12bh

AdII12bhI

33

33

23

21CXX

3

CX

==

+=

+=

+=

=

3bhI

12bhI

3

X

3

CX

=

=

12bhI

36bhI

3

X

3

CX

=

=

4rI

4

CXπ

=

CY

4

X

4CX

I8rI

r98

8I

=π

=

π−

π=

XC

X

b

hr

b

XC

X

r XC

XThe units of 2nd moment of area ormoment of inertia of an area are m4

4.3.4 Moments of inertia of areas

The moment of inertia of a composite areacan be determined using the moments ofinertia of the component simple areas.• Determine the moment of inertia of the area

component with respect to a parallel centroidal axis.• Add the product of the area component and the distance from the

centroidal axis to the axis XX squared.

2CXX AdII +=

A1

A2d2= h/3

d1

X X

h

b b

1

12bh

36bh3

36bh2

36bh

9h

2bh

36bh

AdII36

bhI

33

33

23

22CXX

3

CX

==

+=

+=

+=

=

2

Total 333

X bh125

12bh

3bhI =+=

Transfer formula

263Mechanics

12


4.3.5 Example

Solution:

( )( )( )( )( )

( )( )( )( )( )

46xx

46

23

2xx

46

23

1xx

m1073.5I

m1074.3

040.0072.003.008.012

08.003.0I

m1099.1

072.0095.003.011.012

03.0110.0I

−

−

−

×=

×=

−+=

×=

−+=

( )( )( ) ( )( )( )( ) ( )

m072.003.008.003.0110.0

04.003.008.0095.003.0110.0y

=

+

+

=

1. Determine the moment of inertia ofthe cross-sectional area shown:

110 mm

30 m

m

80 m

m x x

y

30 mm


Find location of centroidof the area.


4.4 Linear motion

Summary

atuv +=

sa2uv 22∆+=

2ta21uts ∆+=∆

gsinv2

t of

θ−=

dtdsv =

dtdva =

bcabac vvv +=

Velocity

Relative motion

Acceleration

Kinematics

Time of flight

265Mechanics


Velocity is a vector quantity that is a measure of a body’s displacement persecond. The term speed is used as a common name for the magnitude of abody’s velocity. Bodies do not always move with a constant velocity. Theaverage velocity is found from the total distance travelled over a timeperiod.

tsvav

∆

∆=

The smaller the time intervalselected, the more representative isthe calculation of the instantaneousvelocity at a particular time t. dt

dstslimv

0t

=

∆

∆=

→∆

4.4.1 Velocity and acceleration

dtdv

tvlima

0t

=

∆

∆=

→∆

Acceleration is a measure of the magnitude of the change in velocity over atime period - that is, the rate of change of velocity with time.

When the velocityincreases in time, it iscalled an acceleration.When the velocitydecreases with time, itis called adeceleration.

The rate of change of velocity maynot necessarily be constant. That is,the acceleration of a body may varywith time. If the time period is madesmall enough, then the instantaneousacceleration is

It is often convenient to describe the motion of one body in terms ofthe motion of another.This is easily done because velocity is a vector, all we need do isadd the components. The general equation is:

bcabac vvv +=

The velocity of“a” w.r.t.ground “c”

The velocity of “b”w.r.t. ground “c”

The velocity of “a”with respect to “b”.That is, the velocity of“a” as seen from “b”.

This is a vectorequation - cannotjust add themagnitudes of thevelocities.

Relative motion



vdsdt

dtdsv

=

=

Consider the mathematical definitions of velocity and acceleration:

velocityacceleration

This is called a differential equation becauseit contains differentials (ds and dv).

Let us assume acceleration, a, is a constant and at time t = 0, thedisplacement s = so and the velocity v = u, the initial velocity.

Cat

dtav

dtadvdtdva

+=

=

=

=

∫

atuvuC

uv,0t@

+=

=∴

==

( )

[ ]

( )

sa2uv

uvssa2

uv21

v21ssa

dvvdsa

dvvdsa

22

2o

22

v

u

2o

s

s

v

u

2

o

∆+=

−=−

−=

=−

=

=

∫ ∫( )

( )

2

o2

s

s

t

0

ta21uts

ssat21ut

dsdtatu

dsdtatudtds

atuv

o

∆+=∆

−=+

=+

=+

=

+=

∫∫

1. 2. 3.

4.4.2 Kinematics

dvvdsa

vdsdvdtdva

=

=

=

Substituting dt = ds/v

s

t

Constantvelocity,accelerationa = 0u

dtdsv ==

221

o atutss +=−

so

Velocity is nota constant. Ifaccelerationis a constant,then curve isa parabola.

dtdsv =

221

o atutss +=−s

t

so

The slope of a plot of velocityvs time is the acceleration.

dtdva =

Acceleration is aconstant. Velocitychanges linearlywith time.

v

t

u

atuv +=

The integral of thevelocity vs time isthe distance.

∫==

=

dtvs

dtvdsdtdsv

267Mechanics


Consider the path of a soccer ball as itleaves the ground at an angle θ andvelocity vo.

Range

h

x

y

θ

vo

The key to analysing projectile motion is to divide the motion intohorizontal and vertical components.

Horizontal Vertical

θ= cosvv oxo θ= sinvv oyo

If we ignore air resistanceand other losses, then vxremains constant until theball strikes the ground at theend of the travel. That is, theacceleration of the projectilein the x direction is zero.

The vertical component vychanges at the rate of g = −9.81m s−2 due to gravity. vy does notremain constant during theflight of the ball. That is, theacceleration of the projectile inthe y direction is −9.81 m s−2.

( )tcosvx o θ=

( ) 221

o gttsinvy +θ=

−9.81221 atuts +=From:

Note, when t = time of flight, thedistance y (the distance above orbelow the ground) - is zero ifstarting and ending at groundlevel.

Note, when t = time offlight, x is the range.

with u = vox

4.4.3 Projectile motion



( )tcosvx o θ=

( ) 221

o gttsinvy +θ=

Now, the time of flight tf depends only on the vertical motion.

When starting from ground level, y = 0, the time for the projectile to hit theground occurs when y = 0 again. Hence:

( )

gsinv2

t

gtsinv

gttsinv

of

f21

o

2f2

1fo

θ−=

−=θ

−=θ

Note that the time of flight depends on the angle θ. This is importantbecause the range depends on both θ and t.

Substituting t for time of flight thus gives the range:

( )

g2sinv

cossingv2

gsinv2

cosvx

2o

2o

oo

θ−=θθ

−=

θ−θ=

The maximum range occurs whendx/dθ = 0

[ ]

°=θ

θ=θ

=

θ+θ−−

=θ

45cossin0

cossingv2

ddx 22

2o

The maximum height hoccurs when dy/dt = 0:

( )

gsinv

t

0

gtsinvdtdy

gttsinvy

o

o

221

o

θ−=

=

+θ=

+θ=

When g = −9.81is substituted, tcomes outpositive

i.e. Velocity in thevertical direction = 0

When g = −9.81 is substituted,tf comes out positive

269Mechanics


1. A bus leaves a bus stop after picking up passengers and accelerates at arate of 2 m s−2 for 3 seconds, travels at constant velocity for 2 minutes,decelerates at a rate of 5 m s−2 and then comes to a stop at the next busstop. Calculate the distance between the bus stops.

(a) a = 2 m s−2, u = 0, t = 3 s (b) a = 0 ms−2, u = 6.0 m s−1,v = 6.0 m s−1, t = 2 min

(c) a = −5 m s−2, u = 6.0 m s−1, v = 0

( )

( )1

22

221

221

sm6v

9220as2uv

m9

32

atuts

−

=

+=

+=

=

=

+=

( )m720

1206

atuts 221

=

=

+=

( )m6.3s

s5260

as2uv2

22

=

−+=

+=

Total distance = 9 + 720 + 3.6 = 832.6 m

4.4.4 Examples

Solution:

2. A boat travelling east at 20 km h−1 while another is going North at 10km h−1. Calculate the velocity of the first boat as seen by the second?

1

2

20 km/h 10 km/h

2121 vvv +=

-vM

vG

vGM

h/km4.221020V 22

12

=

+=

°=φ

=φ

6.262010tan

φ

Note, if we had used1212 vvv +=

We would have found VMG which is thevelocity of the second boat w.r.t. the first.

2112 vvv −=

Solution:



4.5 Forces

Summary

maF =

NF µ=

0F

0F

y

x

=

=

∑

∑

0Mx =∑

mgW =

Newton’s first law

Weight force

D’Alembert’s principle

d’Alembert’s principle

Sliding or tipping

271Mechanics


So far we have talked about the motion of objects or bodies, that is, velocityand acceleration, without any consideration as to what is the cause of thismotion. Accelerations of bodies arise due to the application of forces,specifically, unbalanced forces. The magnitude of the acceleration isproportional to the magnitude of the unbalanced force. The direction of theacceleration is in the same direction as the unbalanced force.

W

N

P F

Consider a body at rest on a frictionless surface:

If all the forces on the bodyare balanced, then the bodywill remain at rest.

Let one of the applied forces P beremoved.

The forces on the body are not allbalanced. In the vertical direction,W balances N. But in the horizontaldirection, F is unbalanced.The magnitude of the resultingacceleration is proportional to themagnitude of the force F andinversely proportional to the massof the object.

The direction of the accelerationis in the same direction as theunbalanced force.

maFmFa

=

=

or

P balances FN balances W

4.5.1 Newton’s laws

N

a

WF

Consider a freely falling body of mass m. Neglectingair resistance, the only force acting on the body is theforce of gravity. That is, the unbalanced force actingon the body is its weight W.

WExperiment shows that the resulting acceleration, nomatter what the mass of the body, is approximately9.81m s−2, the actual value depending on the locationof the body on the earth. This value of thegravitational acceleration is given the symbol g. mgW

mWg

=

=



Consider a car towing a trailer. The car accelerates. Since the trailer isattached to the car by a draw bar, the trailer accelerates at the same rate asthe car. From the point of view of someone standing on the footpathwatching the trailer, the trailer is observed to experience an unbalancedforce via the draw bar.This unbalanced force causesthe trailer to accelerate. Themagnitude of the accelerationdepends on the force applied tothe draw bar by the car and themass of the trailer.Now, imagine you are driving the car which istowing the trailer and while accelerating, youlook in the rear vision mirror. What do you see?A trailer behind you of course.However, from your point of view in the accelerating car, you do not seethe trailer accelerating or decelerating, the trailer is, from your point ofview, motionless. It stays exactly in the same position at the rear of thecar.From your point of view, the trailer is in equilibrium, no unbalanced forcesare acting on it, since if they were, it would be accelerating away from you.From the point of view of someone on the footpath, there is an unbalancedforce, that in the draw bar, causing the trailer to accelerate.From the point of view of you in the car, we saythat the force in the draw bar acting on the traileris balanced by an inertia force.The inertia of a body manifests itself as aforce when the body accelerates. Theinertia force is always equal and opposite tothe unbalanced force producing theacceleration.The inertia force allows the motion of bodies being acted upon by forces tobe analysed using equations of static equilibrium. This method of analysisis called d’Alembert’s principle.

N

W

PI

4.5.2 Inertia

273Mechanics


Consider the masses attached to a stringwhich passes over the pulleys as shown:

In these systems of bodies, it isconvenient to assume that the friction inthe pulleys and the weight of the pulleysis negligible compared to the otherforces involved and thus can beneglected. The tension T in the string isthus a constant throughout its length.

In this system, B movesdownwards and A moves upwards.

Careful consideration shows that thedisplacement of A is twice that of B. To seethis, imagine that A moves downwards adistance ∆s. A point on the string on the leftside of pulley B moves upwards ∆s.

Pulley B rotates around a pivot point X and dueto the reduced leverage of the centre connectionto B, the mass B moves upwards ∆s/2Since the displacement of B is half that of A,then the velocity and acceleration of B (if any)will be half that of A.

To determine which way masses moves, we use the following systematicapproach. Let mass B be held stationary. The tension in the string isproduced by the weight of A, which in this case, is 20(9.81)=196.2 N. Nowlet A be held stationary. The tension in the string is now provided byweight B. But B is supported by 2T, hence, T is (80(9.81))/2=392.4 N.Since the greatest tension in the string is produced when weight B pulls onthe system (with A stationary) then the direction of motion is that B movesdownwards.The direction of motion is in the same direction as the direction of “pull”by the body which is free to move with all the others held immobile whichproduces the maximum tension.

B

X

∆s

TT

A

B20kg

80kg

T

4.5.3 Ropes and pulleys



If two bodies are pressed together by a forcenormal to the contacting surfaces, then motion, orattempted motion, of one body with respect to theother in a direction parallel to the contact surfaceis resisted. This resisting force is called friction.In order to move one of the bodies, the maximumvalue of the frictional force must be overcome.

4.5.4 Friction

The maximum value of friction force is a measure of resistance to slidingof the two surfaces. For most contacting surfaces, the resistance to slidingdepends on:• The magnitude of the normal component of the

force of contact N between the two surfaces• The nature of the contacting materials• The finish and state of the contacting surfaces• Whether or not the surfaces are moving relative

to one another

All these thingscan be groupedtogether and givena value called thecoefficient offriction.

NF µ=

Frictionforce

Coefficientof friction

Normal force

This mathematical expression is calledAmontons’ law after Amontons who in1699 performed many experiments onthe frictional properties of materials.

Note that the friction forcedoes not depend on thearea of contact between thebodies.

Even the best-prepared surfaces are not entirely smooth. When two bodiesare in contact, the asperities of each of them interlock to some extent. Tomove one of the bodies sideways, these asperities or protrusions must besheared off. The coefficient of static friction is usually higher than that ofsliding friction due to this initial shearing action on the microscopic scale.

Interlockedasperities

In the case of static friction, Amonton’slaw gives the maximum value offriction force that can be achieved justprior to sliding. Below this force, thefriction force is equal to the sidewaysforce since the body remains at rest.

W

F

P

N

275Mechanics


The inertia force and d’Alembert’s principle can be used to solvedynamics problems using equations of static equilibrium.

0F

0F

y

x

=

=

∑

∑

These equations are appropriate when allforces act through the centre of gravity -that is, the forces are concurrent.For many force systems, the actual pointsof application of forces are important andwe must consider the equilibrium ofmoments as well as the forces:

( )( ) ( )( )bPaW

0Mx

=

=∑

W

F P

N

W

F

P

N

a

bx

4.5.5 Non-concurrent forces

In the case of a block resting on a flatsurface, we have the force P acting on thecentre of gravity, but the resulting frictionforce F acts along the surface. Both P andW produce moments around the point x.The forces F and N have no momentabout x since the line of action of theseforces passes through the point. When themoments about x are balanced, then theblock is on the verge of tipping.In many practical cases, it is desirable to determine whether a body will tipor slide in response to a sideways force P. For tipping, the maximum valueof P that balances the moment given by W can be readily determined. Forsliding, the maximum value of P that balances the maximum value offriction force (which is given by Amonton’s law) can also be readilydetermined. The lesser of these two values of P indicates which event willhappen first upon an increasing value of P.



In 1687, Newton proposed the law of universal gravitation. Application ofthis law was in accordance with the observed motions of the planets andKepler’s laws.

221

dmmGF =

The constant of proportionality, G, is called theuniversal gravitational constant and has the valueG = 6.673 × 10–11 N m2 kg–2

The main idea of the law of universal gravitation is that every object in theuniverse attracts every other object. Precisely why this is so is not fullyunderstood. The gravitational force is one of the four fundamental forcesof nature, the others being the electrostatic force, and the weak and strongnuclear forces.The gravitational force acts at a distanceby some unknown mechanism. It issometimes convenient to regard theeffects of the gravitational force on abody in terms of a gravitational field.The motion of a particular object ofmass m within a gravitational field of theEarth mE at a distance d from the Earthcan be readily calculated.

( )[ ]81.9m

1038.6

1061067.6m

dmGm

dmmGF

26

2411

2E

221

=

×

××=

=

=

−

Radius of the Earth= 6.38 × 106 m

Newton proved that when using thelaw of universal gravitation, the mass of abody can be thought of to exist at a singlepoint at the centre of the body. This allowsus to say that to calculate the gravitational force on an object on the Earth’ssurface, we assume that the whole mass of the Earth is concentrated at itscentre and the distance d becomes the radius of the Earth.The Moon is an average 381340 km from the Earth. The Moon is withinthe gravitational field of the Earth and so the force exerted on the Moon bythe Earth is:

( )( ) 3

28

2411

1075.2m

1081.3

1061067.6mF

−

−

×=

×

××=

The Moon is falling towards theEarth with an acceleration of2.75 × 10−3 m s−2. Its orbitalvelocity keeps the Moon fromactually falling into the Earth.

Acceleration due togravity at the Earth’ssurface

4.5.6 Gravitation

Mass of the Earth= 6 × 1024 kg

277Mechanics


FI

FI

( )N8.864

181.980NmamgN

FWN I

=

+=

+=

+=

The only thing to rememberabout this method is that theinertia force FI is:

And acts in a directionopposite to that of theacceleration of the body.

maFI =

1. A person with mass 80 kg enters alift on the 12th floor of a buildingand presses the ground floor button.Determine the force exerted by theperson on the floor of the lift when:(a) The lift is stationary(b) The lift descends with an

acceleration of 1 m s−2

(c) The lift comes to a stop with adeceleration of 1 m s−2

Solution:(a) Draw a free-body diagram of the person:

W

N

Since the person is at rest,there is no inertia force.

( )N8.784

81.980mgWN

=

=

=

=

(b) Draw free-body diagram with inertia forcedirected opposite to acceleration.

W

N

a

( )N8.704

181.980NmgmaNWFN I

=

−=

=+

=+

(c) Draw free-body diagram with inertia forcedirected opposite to acceleration.

W

N

a

The equations ofequilibrium are simply:

0F

0F

y

x

=

=

∑

∑

A free-body diagram is arepresentation of an object orbody which shows all forcesacting upon it. The body issupported only by forces and isin contact with nothing else at all.

4.5.7 Examples



2. A trailer has a mass of600 kg and has a rollingresistance of 700 N whichrepresents rolling friction andair resistance. Calculate:

(c) Force P in the draw bar when car decelerates at 1.5 m s−2

3. If the mass of the car is 800 kg, and has a rolling resistance of 900 N,calculate the force T required to be applied to the road by the wheels ofthe car for the accelerations and decelerations given previously.

PR0a =

N700RP ==

( )( )N1300

1600700FRP I

=

+=

+=

( )( )N200P

7005.1600PRFP I

−=

=+

=+

PR

aFI

aP

FIR

P actually isdirectedopposite tothat shown

T

P

Ra = 1 m s−2

FI

( )( )N3000

18009001300TTFRP I

=

++=

=++

( )( )N500

5.1800900200TRPFT I

−=

−+−=

=++T

P

R

FI

a = 1.5 m s−2

That is, opposite indirection to that shownabove.

(a) Force P in the draw bar when the car moves with constant velocity

(b) Force P in the draw bar when car accelerates at 1ms-2

279Mechanics


4. Calculate the mass of B required to give body A an accelerationof 2.4 m s−2 upwards.

By inspection of the diagram, the magnitudes ofdisplacement, velocity and acceleration of B arehalf that of A. Hence, we need body B to have adownwards acceleration of 2.4/2 = 1.2 m s−2.Now, considering the motion of A, we have

A

W = 588.6 N

T

a =

2.4m

s-2 ( )

N1444.260

maFI

=

=

=

N6.732T6.588144T0

0Fy

=

−−=

=∑

Turning now to body B:

( ) ( ) ( )

kg170m2.1465m6.8

81.9m2.1m6.7322WFT20

0F

I

Y

=

=

−+=

−+=

=∑

Final answer

B

W

T T

B

W

2T

a =1

.2m

s-2

FI FI

A

B60 kg

? kg

T

a

Solution:



5. A box containing a new refrigerator isstanding on the flat bed of a deliverytruck. The block has a mass of 300 kgand is 1.2 m high and 0. 6 m wide.The coefficient of friction is 0.31. Ifthe truck stops suddenly, will the boxslide or tip?

(b) Sliding

a

W

FI

N

( )N3.912

294331.0NF=

=µ=

( ) ( )N5.1471F3.29436.0F

0M

I

I

a

=

=

=∑

(a) Tipping

FI

N

W = 2943 N

a

0.6

0.3

Inertia force has to be atleast this value for box tobe on the point of tipping.

Inertia force has to be at least this valuebefore box will slide.

Since the inertia force will reach the value of friction force first as theacceleration (and hence FI) increases, then the box will slide and will not tip.

Solution:

a v

281Mechanics

6. Calculate the gravitational force between two 1 kg masses placed1 cm apart.

Solution:

( )

N10673.6

01.0110673.6

dmm

GF

7

211

221

−

−

×=

×=

=


4.6 Rotational motion

Summary

RmR

mvF 2

2t

C ω==

dtdθ

=ω

dtd

dtd

2

2ω

=θ

=α

αθ+ω=ω

α+ω=ω

α+ω=θ

2

t

t21t

2o

2o

2o

θ= Rs

ω= Rvt

α= Ra t

Rv

a2

tn =

Angular velocity

Angular acceleration

Equations of motion

Linear displacement

Tangential velocity

Tangential acceleration

Normal acceleration

Centripetal force



The radian is the angle swept out by an arc oflength equal to the radius of the circle. Thecomplete circumference of a circle can becalculated from 2πR. The circumference isnothing more than an arc which sweepsthrough 360o. Thus, dividing the circumferenceby the radius gives us the number of radians ina complete circle.

4.6.1 Rotational motion

(about 57.3o)

r

1 radian

2π radians = 360o

Angulardisplacement

Lineardisplacement

R

When a body moves in a straight line, this is called linear motion. Anothercommon form of motion is when a body moves along the path of a circle.This is called rotational motion.

Tangentialvelocityθ

Consider a body moving in a circular path.Since the angular displacement ismeasured in terms of the angle (inradians) swept out by the body, then therate of change of angular displacementwith respect to time must be called theangular velocity and has the units rad s−1.

dtdθ

=ωAll points on a rotating (rigid) bodyhave the same angular velocity.

The rate of change of angular velocity w.r.t. time is called angularacceleration and has the units: rad s−2 and is given the symbol α

dtd

dtd

2

2ω

=θ

=α

r

Translational motion Angular motion

as2uv

atuv

at21uts

22

2

+=

+=

+=

αθ+ω=ω

α+ω=ω

α+ω=θ

2

t

t21t

2o

2o

2o

Always bestto convert toradians,radians persecond, etcwith theseformulas.

Equations of motion for rotational motion are very similarto those of linear motion:

283Mechanics


4.6.2 Equations of motion

The radian is the angle which is swept out by an arc of length equal to theradius of the circle. Let the arc length be given the symbol s. Hence, thetranslational distance along the arc, for one radian, is equal to the radius R.

θ= Rs

The linear or tangential velocity and tangential acceleration of a point onthe rotating body is found by differentiating:

ω=

θ=

=

θ=

RvdtdR

dtdsv

Rs

t

t

α=

ω

=

=

ω=

RadtdR

dtdva

Rv

t

t

tvt

The velocity isalways tangent tothe curvature.

a

This is the tangentialacceleration.

vt1

vt2

∆vnvt2

vt1

∆θ

Consider the motion of an object travelling in a circle with a constantangular velocity. For an angular displacement ∆θ, the tangential velocitychanges direction from vt1 to vt2.

∆vn represents the velocitydifference and it is directedinwards towards the centre ofrotation. This constant changein direction of the tangentialvelocity represents anacceleration, normal to thetangential velocity, anddirected inwards towards thecentre of rotation.

For an increment ∆θ, theacceleration is :

tva n

n∆

∆=

tsv

∆

∆=

∆vn

vt

vt

Rv

a

tRsv

tv

a

vv

Rs

2t

n

t

nn

t

n

=

∆

∆=

∆

∆=

∆=

∆

vt

vt

R ∆s

∆θ

∆θ

and

Similar triangles

If the angular velocity of the object is changing, then theobject accelerates in a tangential direction as well as in thenormal direction.



We have seen that motion of a body ina curved or circular arc results in anacceleration which is directed inwardstowards the centre of rotation, evenwhen the body has a constanttangential speed.

vtRva

2t

n =

Now, since this is a real acceleration, itmust arise due to the application of a force.

Consider a stonebeing whirled aroundusing a string. The normalacceleration is vt

2/r. This accelerationmust arise due to the application of an unbalanced force.The force must also be in the same direction as the acceleration. Hence,there is an unbalanced force directed inwards towards the centre ofrotation! We call this force a centripetal force.The force is proportional to the massand the acceleration, thus:

RmF

RvR

mvF

2C

t

2t

C

ω=

ω=

=

It is very convenient to include an inertia force in rotation problems sothat equilibrium conditions may may calculated. The inertia force FI isequal and opposite to that of the centripetal force. The inertial force isoften called the centrifugal force.

FC

FI

but

thus

4.6.3 Centripetal force

285Mechanics


The rotation of planets around the sun generates centrifugal forces whichbalance the inwards pulling gravitational force. The closer the planet is tothe sun, the stronger the gravitational force (by the inverse square law) andthe faster the orbital velocity must be to maintain its orbit.

Now, in circular motion, thecentripetal accelerationacts inwards towards thecentre of rotation and isequal to:

Rva

2E

n =

For the motion of a satellite of mass m1 around the Earth, the sameprinciples apply and we have:

2S

2E

3E

2E

E2

E2E

ES

4m

GTR

TR4m

Rmm

G

π=

π=

Kepler’s third law states that thesquare of the period of the orbit of a planet is proportional tothe cube of planet’s distance from the sun. Thus, since:

Alternately, since all the terms on the right side of the equation above areconstant, then the ratio of the mass of a planet with the mass of the sun canbe calculated:

an

nEam

2ES

Rmm

G

R

mS

mE

vE

Consider theEarth and theSun:

2p

2m

3m

4

mG

TR

π=

Now, by forming the ratio:S

p

2p

3p

2m

3m

mm

T

R

TR

=

Motion of moon or satellitearound the planet

Motion of planetaround the sun

Mass ofplanet

Mass of sun

4.6.4 Planetary motion

we can measure the massof a planet.

RGmv

Rvm

RmmGF

E

21

2E1

=

==

Orbitalvelocity

then:

and:TR2RvE

π

=ω=

Period

Since G is a constant which can be determined byexperiment, the mass of the sun can be determinedfrom observations of R and T of any planet.



1. What is the maximum speed that a car may travel at without skiddingon a level curve of radius 100 m for a coefficient of friction of 0.2(raining) and 0.6 (dry)?

Rvg

Rmvmg

mgFR

mvF

m100r

2

2

2

I

=µ

=µ

µ=

=

=

( )

1

1

2

hkm87

sm26.24v100v81.96.0

−

−

=

=

=

( )

1

1

2

hkm50

sm14v100v81.92.0

−

−

=

=

=

Raining:

Dry:

Centrifugal(inertia) force

Frictionforce

Car will slide when the centrifugal forcebecomes equal to the friction force (which isthe centripetal force in this case).

N

RmvF

2=

mg

µN

4.6.5 Example

Solution:

287Mechanics


4.7 Rotation

Summary

FrT =

α= IMI

∑=2mrI

2CX MdII +=

MIK =

Torque

Moment of inertia about axis

Inertia moment

Transfer formula

Radius of gyration



Consider a wheel mounted on a shaft. The weight ofthe wheel is balanced by N. A horizontal forceapplied to the centre bearing of the assembly willsimply cause the wheel to translate. However, if theforce is applied to a point off-centre, such as at therim, then the wheel will rotate about its axis.

W

N

F

r

α

The application of a force acting through a perpendicular distance in thismanner is called a moment and sometimes referred to as a torque. Themagnitude of the accelerating moment, or torque is:

FrT =

Perpendicular distance between theline of action of the force and theaxis about which rotation occurs.

4.7.1 Moment of inertia

When the disk rotates under the action of anaccelerating moment, it acquires angularacceleration. As in all our problems on dynamics,we introduce an inertia force, or in this case, aninertia moment, to act in the opposite sense to theaccelerating moment. This allows rotational dynamicsproblems to be treated using equations of equilibrium.

What is the value of MI? Since it acts in the opposite sense to M, then themagnitude is just Fr. However, we can express this inertia moment interms of the angular acceleration just as we expressed the inertia force intranslational problems using the product FI =ma.We divide the rotating mass into a large number ofelements, each with a mass m. Each mass element hasthe same angular acceleration α, but is located at aparticular radius from the axis of rotation r.Now, the tangential acceleration of each mass element is: α= raThus, the inertia force acting on each mass element is: α== mrmaFIThe first mass element is located a distance r1 fromthe centre of rotation, then the inertia moment forthis mass element is: α=

=

21

1I1I

mr

rFM

The total inertia moment MI is the sum of all these:

[ ]α=

=

∑

∑2

II

mr

rFMThe quantity is called the moment of

inertia of the rotating bodyand has the units kg m2.

∑2mr

α= IMI

m

r αtα

W

N

αMI

M

289Mechanics


The moment of inertia is closely related to the 2nd moment of an area.The only difference is that here we are dealing with the masses rather thanareas. The moment of inertia of a body about some axis can be determinedby dividing the body into elemental masses and summing the resultingmoments.

dmr

mrI

2

2

∫

∑=

=

Formulas for moments of inertia of bodiesof simple shapes are usually to be found inengineering data books.

2X

2C

MR23I

MR21I

=

=

( )22C rRM

21I +=

2X

2C

MR57I

MR52I

=

=

22X

2C

MdM121I

M121I

+=

=

l

l

4.7.2 Moment of inertia of common shapes

R C C

X X

C C

R

C

C

X

X

X X

C C

d

l

R

r

M = total mass

Sphere

Solidcylinder

Hollowcylinder

Rod

Units: kg m2

The moment of inertia of an area has unit m4. The moment of inertia of abody is an entirely different thing and has units kg m2.



The total moment of inertia of an irregularly shaped body can be oftendetermined by dividing the body into simple shapes and using a transferformula. For each simple shape, the moment of inertia about its ownparallel centroidal axis can be found in reference tables. This moment ofinertia is then transferred to the desired axis by:

2CX MdII +=

Moment ofinertia withrespect to thedesired axis ofrotation

Moment of inertiawith respect to aparallel centroidalaxis of the body

Totalmass

Perpendiculardistance betweenparallel axes

C

C

X

X

22X

2C

MdM121I

M121I

+=

=

l

l

d

l

For example, in the preceding table, it can be seen that the moment ofinertia about some distant axis XX is equal to the moment of inertia of therod about its central axis plus the total mass times the distance to the axissquared.

Units: kg m2

4.7.3 Composite bodies

Parallel axis theorem

291Mechanics


The moment of inertia of a body is a measure of its resistance to angularacceleration just as the mass of a body is a measure of its resistance to atranslational acceleration.

∑=2mrI

Now, in our derivation of the formula for calculating the moment of inertia,we agreed that a body may be subdivided into a series of elemental masseseach with its own radius from the centre of rotation and that the totalmoment of inertia is found from:

For example, for a solid disk, themoment of inertia around its central axisevaluates to

Now let us imagine another disk which hasthe same mass, but all of this mass isconcentrated at a band of material at the rimof the disk. The radius of this disk is selectedsuch that the moments of inertia of the twodisks are the same and is given the symbol KThe moment of inertia is (by inspection):

2C MKI =

This radius K has a special meaning and is called the radius of gyration. Itis the radius at which the total mass of a body may be evenly concentratedso as to obtain the same moment of inertia.

MIK =

If a large moment of inertia is required (such as in a flywheel), then theradius of gyration should be made as large as possible. This can be donewithout increasing the radial dimensions of the disk by concentrating themass at the outer edge – i.e. by making the edge thicker.

All the mass M is ata distance K fromthe axis of rotation

4.7.4 Radius of gyration

K

2MR21



2

MIK

4MI

2

X

l

l

=

=

=

Wrong!

Doing it this way ignores the effect ofthe distribution of the mass within thebody about the axis in question. The r2

factor in the definition of moment ofinertia means that mass at a greaterdistance away from the axis makes ahigher contribution to the overall valueof I compared to mass closer to the axis.

3

MIK

31M

4lM

12MI

2

22

X

l

l

l

=

=

=

+=

C

X

X

d = l/2

l

c.m.

Note, the radius of gyration is furtheraway from the axis of rotation than thecentre of mass showing that mass at agreater distance away from the axiscontributes more to the rotationalinertia of the body.

4.7.5 Calculation of moment of inertia

Right

Now, the centre of mass of an object is that point at which the total mass ofthe object can be said to be concentrated. We might think therefore that themoment of inertia of a body with respect to some axis can be obtained bysimply multiplying the total mass by the distance from the centre of massto that axis. Consider the following calculation for the moment of inertia ofa beam about its end:

293Mechanics


In analysing rotational motion, it is convenient to use equilibriumequations. Thus, we consider an inertia moment which acts in theopposite sense to the accelerating moment - or torque.

Now, the inertia moment is equal to:

α= IMI

Notice the resemblance to the inertiaforce in translational problems. Themoment of inertia I is the rotationalequivalent of mass. The angularacceleration is the rotational equivalentof translational acceleration.

The rotational inertia of abody depends not only on itsmass but on how this mass isdistributed around the axis ofrotation. The radius ofgyration is a measure of thisdistribution. A large radius ofgyration means that the mass ofa rotating body has isdistributed more towards theouter edges of the structure. Asmall radius of gyration meansthat the mass is concentratedmore towards the centre ofrotation.

In problems involving rotation, we employ a rotational dynamicequilibrium by taking into consideration an inertia moment :

α+=

=

∑

∑IM0

0M

Sum of all accelerating anddecelerating moments (egfriction moments)

Inertiamoment

In these formulas, we must becareful to use signs correctly. Wecould settle on any particularsystem as long as we areconsistent. For example, wecould agree that clockwisemoments are positive. Or, wecould say that the direction ofany accelerating moments orapplied torques are positive andfriction and inertia moments arenegative.

W

N

α

MI

M

4.7.6 Dynamic equilibrium



Plane motion, such as rolling without slipping, canbe treated as a combination of rotational andtranslational motion. Consider a wheel beingaccelerated by a pull P through its centreof mass.

F is a friction force which is a reaction to the pull P acting on the centre ofmass. FI is the inertia force given by:

In this example, the action of force P causes thedisk to move to the right with an acceleration a.An inertia force FI = ma acts towards the left.

α== MRMaFI

MaFI =

An inertia moment Iα balances the appliedtorque arising from the friction force Fthrough perpendicular distance R.

NmaPFFP0

0F

I

H

µ+=

−−=

=∑

α=

−=

=∑

IFRMFR0

0M

I

C Note, the horizontalforces P and FI passthrough the centre ofmass and thus do notcontribute a momentabout the centre of mass.

Now, the disk has an angular acceleration α.The tangential acceleration of a point on therim of the disk, such as point A, is:

α

N

W

P

F

Iα

FI

a

A

R

C

But, the acceleration of A with respect to the centre of mass C is exactlythe same in magnitude as the acceleration of C with respect to the point A.That is, the horizontal translational acceleration of the centre of mass of thedisk. Hence, the translational inertia force FI can be expressed in terms ofthe angular acceleration:

α= Ra

Translation

Rotation

4.7.7 Rolling motion

295Mechanics


2. A mass of 5 kg on theend of a rope woundaround a drum resultsin a moment that isjust sufficient toovercome friction inthe bearings. The 300kg drum then rotatesat a uniform angularvelocity. Determinethe coefficient offriction if the shaftdiameter is 0.15 m.

Free body diagram

( )

( )

( )1.0

81.930027.294NN27.294F

215.0F07.22

rFMNm07.22

45.005.49TRM

0M

N05.49mgT

F

F

=µ

µ=

µ==

=

=

=

=

=

=

==

∑MMF

T

R

N

4.7.8 Examples

Solution:

1. Determine the moment of inertia about thecentral axis of a 8 spoked wagon wheel:

1. Rim

2. Spokes

3. Hub

( )

( )( )2

22

22

m kg 15.7

75.08.09.115.0I

kg9.1105.075.08.0980M

=

+=

=

−π=

( )

( )( )2

22

22

m kg0365.0

075.015.06.25.0I

kg6.205.0075.015.0980M

=

+=

=

−π=

( )

( ) ( )2

22

2

m kg 439.0

45.015.12

6.015.1I

kg15.16.0025.0980M

=

+=

=

π=

(one spoke)

Total I :( )

2m kg 88.7

0365.08439.015.7I

=

++=

( )

m 482.086.215.19.11

88.7K

=

++

=

ρ = 980 kg m−3Solution:

0.05

0.750.80.15

0.075 0.45

0.6Spokes0.025 mradius



4. A motor mechanic presses the end of a toolagainst a 200 mm diameter, 20 mm thickgrindstone which is rotating at 3000 rpm. Theforce applied to the grindstone at its outer edgeis 100 N. The power to the grindstone issuddenly shut off. Calculate how long it takesfor the grindstone to come to a rest (ignoringfriction at the grindstone centre bearing).

( )( )( )

2

22

2

mkg0063.0

1.0200002.01.021

MR21I

=

π=

=

( ) ( )

s62t

t158760230000

srad1587

0063.01.0100IT

2

2

=

−+

π=

−=α

α=−

α=

−

Solution:

α

Ια

T

ρ = 2000 kg m−3

ω = 3000 rpm

3. To determine the moment of inertia of thedrum shown in (2), a mass of 25 kg is attachedto the rope around the rim and allowed to fall.The friction moment MF is 22 N m. If the massis observed to drop a distance of 4 m in 4seconds, calculate the moment of inertia aboutthe central axis.

2

2

2

2

srad11.145.05.0

rams5.0a

4a5.04

at21s

−

−

=

=α

α=

=

=

=

( ) ( )N75.232

5.02581.9300FmgT I

=

−=

−=

25 kg

T

FIa

W

( ) ( )2

F

mkg5.74I

11.1I2245.075.232IMTR0

0M

=

+=

α−−=

=∑

RTR

MF

MI

α

Solution:

297Mechanics


4.8 Work and energy

Summary

FsW =

ksF −=

2ks21W −=

mghPE =

2ks21PE =

2mv21KE =

221 IKE ω=

θ= TWFvP =

ω= TP

Work

Spring restoring force

Work done on/by a spring

Gravitational potential energy

Strain potential energy

Kinetic energy

Rotational kinetic energy

Rotational work

Power

Rotational power



• Energy exists in many forms• There is a flow, or transfer, of energy when a change of form takes place• Heat and work are words which refer to the amount of energy in transit

from one place to another.• Heat and Work cannot be stored.

4.8.1 Work and energy

We generally speak of two types of mechanical energy:

Potential energy Kinetic energyThe energy associated with the positionof a body in the gravitational field orthat stored within a body placed understress.

The energy associated withthe velocity of a body.

Kinetic and potential energy may be converted from one type to another.The transfer occurs through the mechanism of work.The application of anexternal force actingthrough a distancesignifies a transfer ofenergy which we callwork. Work is doneby forces on a body.

The magnitude of thework done W on abody by force F is theproduct of the force Fand the distance sthrough which it acts.

If the direction of force issame as that of directionof the displacement “s”,then the work done by Fis positive.

FsW =

The units of work areNewtons metres. Sincework is energy in transitfrom one form toanother, the units ofwork are the same asenergy: Nm = Joules

Work may be positive or negative. When the direction of the force is in thesame direction as the displacement of the body, the work is positive.

F F

s

W = mg

N

The forces W and N do not do any work as the block is moved horizontallyalong the surface by the force P and there is no displacement in the verticaldirection. Work is done BY forces ON bodies. When the force acts in thesame direction as the displacement of the body, the work done is positive.When the force acts in the opposite direction as the displacement of thebody, the work done is negative.

Note, we use thesymbol W for both Workand Weight. Thecontext should make itclear which is meant.

299Mechanics


Is the work positive or negative? Wespeak about a spring doing work on thebody which is deforming it. The spring’srestoring force does negative work whenbeing compressed or stretched since thedirection of the restoring force is always opposite to that of thedisplacement. When the spring is released, it does positive work on thebody to which it is attached since now, the force provided by the spring isin the same direction as the displacement of the body.

A good example of a variable force is that offered by a compressed orstretched spring.

F 2F

Freelength

-s−2s

When a spring is compressed, it provides a resisting or restoring forcewhose magnitude is proportional to the amount of compression. Theconstant of proportionality is called the spring constant or springstiffness.

ksF −=

As the spring is compressed, it offers an ever increasing restoring force. If,in compressing the spring from a displacement = 0 to −s, we break thecompression up into a series of small steps over which we consider the therestoring force to be more or less constant, then adding the work done foreach of these small steps together gives us the work done by the spring.

If the steps are made very very small,then mathematically, we can state:

∑ ∆= sFW

2

s

0

s

0

ks21W

ksds

FdsW

−=

−=

=

∫

∫−

−

The constant ofproportionality if calledthe spring “stiffness”and has the units N/m

Work is thearea under aplot of force vsdisplacement.

The minus sign means that therestoring force F is opposite indirection to the displacement s.

F

−s

W

−k

4.8.2 Variable force



The spring isreleased andoffers a restoringforce to the rightwhich acceleratesthe mass.

ss (+)

F (+)

2

0

s

2

0

s

0

s

ks21W

ks21

dsksdsFW

=

−=

−==

−

−−

∫∫

The spring isreleased andoffers arestoring forceto the left.

ss (−)

F (−)

2

0

s

2

0

s

0

s

ks21W

ks21

dsksdsFW

=

−=

−== ∫∫

vmass

vmass

The spring iscompressed andoffers a resistingor restoringforce to theright.

The spring is stretchedand offers a restoringforce to the left.

ss (+)

F (−)

ss (−)

F (+)

2

s

0

2

s

0

s

0

ks21W

ks21

dsksdsFW

−=

−=

−== ∫∫

2

s

0

2

s

0

s

0

ks21W

ks21

dsksdsFW

−=

−=

−==

−

−−

∫∫vmass

vmass

P

P

Up: (+) Down: (−) Left: (−) Right: (+)

4.8.3 Work done by a spring

301Mechanics


Gravitational potential energy

Kinetic energy

The energy associated with theposition of a body.

The energy associated with thevelocity of a body.

Gravitational potential energy ofa body may be increased by raisingit a distance h. When a bodymoves to a lower level, itsgravitational potential energy isreduced.

Consider an unbalanced force P applied to amass initially at rest on a frictionless surface.The mass is given an acceleration a and movesa distance s and acquires a velocity v.

2

2

mv21KE

masFsUas2v

=

=

=

=

mghPE =

Elastic potential energy is theenergy stored in a stretched orcompressed spring.

2ks21PE =

Elastic potential energy

The raising or lowering of a body isa force mg being applied through adistance h. Thus, the work orchange in potential energy is:

2ks21W =

Note: The work done by aspring when released ispositive and given by:

Work done by spring whenbeing compressed or stretched isnegative:

2ks21W −=

4.8.4 Energy

The correspondence between work and energy is a powerful tool in solvingmechanics problems. The procedure for applying the energy approach is:

Initial kineticenergy of abody

Positive work Negative workFinal kineticenergy+ – =

This equation is another way of saying that energy is neither created ordestroyed but only changed into another form by the mechanisms of eitherheat or work. This is the law of conservation of energy.



The moment of inertia of a body is a measure of its resistance to angularacceleration just as the mass of a body is a measure of its resistance to atranslational acceleration.

Translational RotationalDistance s θ

Velocity v ω

Acceleration a α

Mass m IWork W = Fd W = TθKinetic energy 2

21 mv 2

21 Iω

We divide the rotatingmass into a large numberof elements, each with amass m. Each masselement has the sameangular velocity ω, but islocated at a particular radius from the axis of rotation r.Now, the kinetic energy of each mass element is: 2

21 mvKE =

But, v = rω, thus the kinetic energy for a single mass elementbecomes:

The total kinetic energy is the sum of all these:

[ ] 22

22

mr21

mr21KE

ω=

ω=

∑

∑But, the quantity is the moment of

inertia of therotating body.

∑2mr

221 IKE ω=

m

r

v

ω

2221 mrKE ω=

Thus, the rotational kinetic energy is:

The work done on a rotating body can be considered in terms of themoment or torque “T” applied moving through an angular displacement θ

θ=

θ=

=

TWFRFsW

θ= Rssinceθ

R

F

s

4.8.5 Rotational kinetic energy

303Mechanics


Consider a mass m that has to be raised to a height h.

The amount of work to be done is W = mgh. Now, to raise this mass, wecan apply a force equal to mg through a height h, or we can use an inclinedslope and apply a smaller force F = mg sinθ over a longer distance s. Theproduct Fs comes out to be equal to mgh since the same amount of workhas to be done no matter how the mass is raised (ignoring any friction onthe slope). The difference between the two methods is that the secondmethod takes a longer time and thus requires less power.Power = time rate of doing work and is measured in Joules persecond, or Watts.

FvPdtdsF

dtdFsdt

dWP

=

=

=

=

if F is a constant

Power = work over time

Power = force times velocity

In rotational motion, power is stillthe time rate of change of doingwork, thus:

ω=

θ=

θ=

=

TPdtdT

dtdTdt

dWP

h

where T is torque

4.8.6 Power

s

mg sinθ

mgmg

A very powerful machine iscapable of doing a lot ofwork in a short time. Theexact same amount of workmay be able to be done bya less powerful machinebut the time taken to do thiswork will be longer.

A small engine can deliver large amountsof power if it spins quickly enough. Forexample, a small engine which develops 50Nm of torque at 10,000 rpm (1047 rad s−1)is producing 52 kW of power. The samepower may be had with a larger enginewhich develops more torque at a lowerrpm. In the above example, 50 kW ofpower occurs at at 2400 rpm if the torquedeveloped is 200 Nm. The engine in acargo ship may develop a huge amount oftorque but only be turning a few hundredrpm which, when multiplied togetherproduces an enormous power output.



4.8.7 Relativity

Einstein found that classical concepts of length, time and mass requiredmodification to satisfy the requirement for the general applicability ofphysical laws and the observed constant speed of light. In the theory ofspecial relativity, he proposed that:Length:

Time:

Mass:

2

2

cv1' −= ll

2

2

cv1

'tt

−

∆=∆

2

2o

cv1

mm

−

=

Rest mass (mass ofbody at rest relative tothe observer)

Relativistic mass(mass of objectmoving with velocity vrelative to an observer)

Time intervalmeasured by aclock in the movingframe of reference.Time interval

measured byclock stationarywith respect to v.

Length measured inthe moving frame ofreference.

Length asmeasured by astationary observerwith respect to v.

Note: as v approachesc, the relativistic massgoes to infinity. At v = 0,m = mo.

Note: as v approachesc, the length as itappears to thestationary observershrinks to zero.

Note: as v approachesc, the time as measuredby a stationary observergoes to infinity.

The total energy of a body is: 2oK

2T cmEmcE +==

Energy ofrest mass

Kineticenergyof motion

Totalenergy

A photon has zero restmass but it does havea relativistic mass andthus momentum. pmch

mchfE 2

==λ

==

When a body of rest mass mo is supplied with anincrement of energy ∆E (e.g. applied by an electricfield) the mass and the velocity both change. The change inkinetic energy of the body is equal to the energy incrementsupplied. When the velocity is much less than the speed oflight, the change in mass is very small and the change invelocity is very large and can be calculated from ½mov2. Ifthe velocity is of the same order a the speed of light, themass change becomes larger and the velocity changebecomes smaller. As the velocity approaches c, the massapproaches infinity, lengths shrink to zero and time slowsdown to a stop.

2o

2o

2

2

2o

2o

2K

vm21

cmc

cv1

mcmmcE

≈

−

−

=

−=

when v << c

305Mechanics


1. The driver of a car brakes hard to avoid collision with a pedestrian. Skidmarks on the road indicate that the car was brought to rest in 15 m.There is disagreement between the driver and other witnesses as to howfast the car was travelling initially. Tests with a similar car show thatthe coefficient of friction between the tyres and the road surface is 0.6.The mass of the car is 1600 kg. Determine the initial velocity of the car.

( ) ( )

1

1

2

221

FI

hkm8.47

sm28.13v

015160081.96.0v160021

0Nsmv

KEFsKE

−

−

=

=

=−

=µ−

=−

Note: Friction forcedoes negative worksince it is directedopposite to that of thedisplacement

2. A 10 kg mass falls onto a spring which compresses a distance 0.15 m.The spring stiffness is 7200 N m−1. What height above the free lengthof the spring was the mass positioned initially?

h

s

m = 10 kg

( )

( )( ) ( )m68.0h

15.072002115.0h81.910

0ks21shmg0

2

2

=

=+

=−++

Note: The mass starts and finishes fromrest hence its initial and final kinetic energyis zero. The force offered by the spring onthe mass does negative work since thespring force is in the opposite direction tothe displacement of the mass.

15 m

F

Initial KE

Final KEpositive

work

negativework

W

F

v

4.8.8 Examples

Solution:

Solution:



Let us first consider therotational motion. Thedisk is acted upon by amoment P×R which tendsto increase the angular velocity of the disc. The product of themoment and the angular displacement thus represents positive workdone on the disk.

3. A yo-yo is allowed to fall under its own weight. The string around therim causes the yo-yo to rotate while falling. The other end of the stringis held fixed. What is the translational acceleration of the yo-yo?Employ the energy method for the solution.

Note: In energyproblems, we doNOT show internalinertia forces infree bodydiagrams. Thesewere tools we usedfor the forcemethod of analysis.

g32a

ah2gh34v

mv43v

23m

21

RvmR

21mv

21mgh

I21mv

21PhPhmgh

2

22

2

222

22

=

==

=

=

+=

ω+=+−

ω W

v C

P

a

Let the yo-yo fall a distance s = h. The force W = mg is acting in thesame direction as h, thus, the positive work done is mgh. The force Tacts in the opposite direction through a distance h and negative work isdone. The initial kinetic energy of the disk is zero, and the final kineticenergy is the sum of the translational and rotational KE’s.

[ ]

PsWRsPRW

=

θ=

θ=

The yo-yo has theshape of a disk.

Answer

Solution:

307Mechanics


4.9 Impulse and momentum

Summary

matp=

∆

∆

FtI =

mvp =

Angular momentum =Angular impulse =

ωITt

Momentum

Rate of change of momentum

Impulse

Angular momentumAngular impulse

Conservation of momentum'22

'112211 mmmm vvvv +=+



Let us imagine a mass m moving withvelocity vo in a horizontal direction tothe right. A retarding force is appliedwhich eventually brings the mass to rest.

The concepts of impulse and momentum are yet more tools for solvingproblems in mechanics.Momentum is a property of a body that determines the length of time ittakes to bring it to rest. It is equal to the product of the mass and thevelocity of the body and is given the symbol p.

Initially, the momentum of the body is mvo. After a time interval ∆t, thevelocity has decreased to say v1. There has been a change in momentumduring the time ∆t because the velocity has changed ∆v = vo-v1. Thechange in momentum is:

matvm

tp

=

∆

∆=

∆

∆

But, the product ma is the force acting on the body. Thus we can say thatForce is actually the rate of change of momentum.

If a body is in motion, then of course the length of time taken to bring it torest depends on the magnitude of the force we use to slow it down.

m

v

F

Momentum is a vector quantity (since velocity is a vector quantity). Theunits of momentum are kg m s-1

vmp ∆=∆

The rate of change of momentum with time is:

4.9.1 Impulse and momentum

1. Momentum issomething that thebody possesses byvirtue of itsmotion.

2. We apply forcesfor a certain timeto the body tochange itsmomentum.

3. A larger force changesthe momentum morecompared to a smallerforce for the same timeof application.

Momentum is a measure of the time taken to bring a body to rest with agiven force. Thus, to bring a body to rest, we apply a force F for a time t.The product of force times time is called impulse.Momentum is somethingthat belongs to a body - it isa measure of its motion.

Impulse is something that weapply to a body to change itsmomentum.

FtI =

mvp =

309Mechanics


Initialmomentum

Positiveimpulses

Negativeimpulses

Finalmomentum+ – =

In solving mechanics problems by the impulse-momentum method, weconsider the initial momentum of a body, the positive and negativeimpulses upon it, and thus obtain the final momentum.

This equation must be applied with careful attention to the directions of theforces and velocities involved. We must decide on a positive direction forvelocities. Only components of force acting in the same direction as thevelocities contribute to the momentum and impulse given a body.

Impulse-momentum equation

4.9.2 Impulse-momentum equation

Momentum is not energy. Momentum is force × time, not force × distance.Momentum is a measure of the time required to bring a body to rest, not ameasure of the distance through which a force acts to bring a body to rest.

The momentum associated with the rotational motion of a body is calledangular momentum. Angular momentum is a property of a rotating bodywhich determines the time it takes to bring the body to rest.

In rotational motion, mass isreplaced by moment of inertiaand force by torque. Thus:

Initial angularmomentum

Positiveangularimpulses

Negativeangularimpulses

Final angularmomentum+ – =

Angular momentum =Angular impulse =

ωITt

Angular Impulse - momentum equation

Imagine that a force F is applied to a large shipmoving at a low velocity. The product of the masstimes the velocity of the ship is a measure of thetime through which the force F must be applied tobring it to rest. This results in the ship travelling acertain distance s during this time.If the initial velocity of the ship was doubled, theship has now got four times the initial kineticenergy and if the same force is used, then four times the distance would berequired since W = Fs. But, the ship has only twice the momentum, andthus the time through which the force is to be applied is only doubled.

F



Collisions which result in no dissipation of energy are called elastic.Collisions involving dissipation of energy are called inelastic.(a) two bodies move towards each other.

(b) they collide elastically imparting a force F to each other during atime t

(c) and then move away from each other

m2m1

v2’v1

’

( ) ( )'22

'112211

'22

'112211

mmmm

mmFtFtmm

vvvv

vvvv

+=+

+−=−+−+

Applying the impulse-momentum equation, we have:

Where we have written the velocity as vectorquantities to remind us to consider thedirections of the velocities involved.Note that the two impulses havecancelled out since the force actingbetween the bodies is equal and oppositein direction (by Newton’s third law).

FF

This law of conservationof momentum applies toinelastic collisions aswell as collisions whichare not head-on butoblique or glancing - aslong as the velocityvectors are treatedcorrectly.

A collision is a process in which momentum may be transferred from onebody to another but the total momentum of the system is unchanged.

m2m1

v1 v2

+

4.9.3 Conservation of momentum

311Mechanics


1. A 1600 kg car moving with velocity 24 km h−1 strikes the rear end of atruck which was moving in the same direction (to the right) but at 10km h−1. The two vehicles become locked together during the collision.Determine the velocity of the vehicles after impact and change in KE.

( )( ) ( )

11

212211

hkm3.12sm43.3v

v960078.2800067.61600vmmvmvm

−−

==

=+

+=+

1

2

1

1

sm78.26.3

10v

sm67.66.3

24v

−

−

=

=

=

=

( ) ( )( )

kJ105.565.66KE

kJ5.5643.380001600KE

kJ5.6678.2800067.61600KE2

21

F

2212

21

I

=−=∆

=+=

=+=

Energy lost to heat, deformationand sound during inelasticcollision

2. If the collision was completely elastic, calculate the velocity of the carand the truck after impact. (Hint: No energy lost in elastic collisions)

( ) ( )

( ) ( )

( ) ( )

12

11

21

21

22112211

22

21

222

1212

1F

2212

21

I

sm08.4'v

sm187.0'v

32912'v8000'v1600'v8000'v160078.2800067.61600

'vm'vmvmvm'v4000'v800

'v8000'v160066500

KEkJ5.66

78.2800067.61600KE

−

−

=

=

=+

+=+

+=+

+=

+=

=

=

+=

….. (1)

….. (2) Now we have two equations(1) and (2), two unknowns v'1and v'2, and these can besolved by substitution.

4.9.4 Examples

Solution:

Solution:



3. A yo-yo is allowed to fall under its own weight. The string around therim causes the yo-yo to rotate while falling. The other end of the stringis held fixed. What is the translational acceleration of the yo-yo?Employ the impulse-momentum method for the solution.

( )

g32a

atv

g32

tv

v23gt

MvMv21Mgt

Mv21Pt

RvMR

21PRt

Rv

MR21I

IPRtIPRtIMvtPMgMvPtWtMv

2

2

2

22

2

22

2

2

2

21

2

21

=

=

=

=

=−

=

=

=ω

=

ω=

ω=+ω

=−

=−+Linear

Angular

but

and

thus

Substituting:

but

thus

In this problem, equations for both linear andangular momentum may be used to determinethe linear acceleration of the disk. First, letthe disk fall for a time t. Then, write out theequations for both linear and angularmomentum separately and include theappropriate positive and negative impulses.

For linear momentum, thelinear impulses are W and Pacting for a time t. Forangular momentum, theangular impulse is themoment P acting through Raround C for time t.

ω W

v C

P

a

Solution:

313Mechanics


Propertiesof

Matter

Part 5


5.1 Fluids

atmg ppp +=

Summary

hgp ∆ρ=∆

11

22 F

AAF =

AWF∆

==γ2l

dFR

=γ

4pR∆

=γ

2pR∆

=γ

θcos2ρghRγ =

2222

2111 v

21ghpv

21ghp ρ+ρ+=ρ+ρ+

2211 vAvA =

dydvAF

=η

vR6Fv πη=

( )η

ρ−ρ=

9gR2

v s2

tl

lη∆π

=8pRQ

4

Absolute pressure

Hydrostatic pressure

Pascal’s principle

Coefficient of surface tension

Coefficient of surface tension

Surface tension - Bubble

Surface tension - Drop

Contact angle

Equation of continuity

Bernoulli’s equation

Viscosity

Stokes’ law

Terminal velocity

Flow rate through tube



5.1.1 Fluids

Intermolecular forces are negligibleand molecules move with rapid,random motion filling the spaceavailable to them.

Intermolecular attractive forces arestrong enough to bind molecules (butonly loosely).

GAS

LIQUID

Liquids and gases are fluids: • offers little or no resistance toshear forces

• takes the shape of its container

Density

Relative Density (or specific gravity - s.g.)

Vm

=ρ

waterofdensity substance ofdensity .d.r =

Pressure

Mass in kg

Volume m3

Material Densitykg m−−−−3

Air 1.2Water 1000Lead 113000

Gauge pressure (pg)

Atmospheric pressure (patm)

Absolute pressure is usuallyused in most physics formulae

A pressure gauge usually measuresthe pressure above or belowatmospheric pressure.

1 atmosphere = 760 Torr =1013 millibars = 101.3 kPa= 760 mmHg

AreaForcePressure =

atmg ppp +=

When dealing with fluids, some of the more important mechanical propertiesto be considered are the density, relative density, and the pressure.

(No units)

(Units: kg m−3)

(Units: N m−2 = Pa)



Pascal’s principle can be used to advantage in hydraulic systems usingpistons. A small force applied to a small diameter piston of area A1 can beconnected, by fluid pressure, into a larger force at a largerdiameter piston with area A2. The force applied can becalculated from the ratio of the areas in the two pistons:

There are two parts to hydrostatic pressure:

(1) The pressure due to the weight of the fluid itself.

5.1.2 Hydrostatic pressure

ghpAgAhAVgA

mgAFp

ρ=

ρ=

ρ=

=

=

Weight ofFluid /Area

Depth beneaththe free surface

Pressure only depends on densityand depth beneath the surface of thefluid and not on the volume of fluid.

What is thepressure actingon the bottomof thecontainer?h

h = 0

The pressure difference between two pointsat different depths is thus:

But, this is a gauge pressure (i.e.pressure above (or below) atmosphericpressure), thus the total absolutepressure is:

( )hgp

hhgpp 1212

∆ρ=∆

−ρ=−

ghpp o ρ+=

p1

p2

h2

h1

the pressure at the surface ofthe fluid (i.e. where h = 0)

• Pressure applied to an enclosed fluid is transmitted equally in alldirections throughout the fluid.

• The direction of pressure is always perpendicular to the areabeing considered but at any particular point in the fluid, is equalin all directions. (Pascal's principle)

(2) The pressure which is applied by some external means

11

22 F

AAF =

317Properties of Matter


1. Consider a glass of milk.

2. Consider a volume V ofliquid within the milk.

3. Liquid V is subjected to an upwards force whichequals the weight W of the liquid in V.

4. If the liquid V were now replaced with a solid of exactly thesame shape, then this solid would be subjected to exactly thesame force as was the volume V of liquid. This force is equalto the weight of the liquid displaced.

The force must come from the surrounding fluid sincethere is nothing else in contact with the volume V. Theforce must equal the weight of the volume V sinceotherwise the volume V would be moving up or down.

When a body is completely or partially immersed in a fluid, the fluid exertsan upwards buoyancy force equal to the weight of the fluid displaced.

Archimedes' principle

W

V

5.1.3 Archimedes’ principle

Q. Will a body immersed in a fluid sink or float?

A. Body will sink if mg > Fb Body will float if mg < Fb

Buoyancy force or"upthrust" =weight of volumeof liquid displaced

Fb

mg

Weightof body

Forces actthroughcentre ofmass ofbody A. The body sinks until its weight is

equal to the weight of thevolume of water displaced. Thegreater the density of the fluid(e.g. salt water), the higher thebody sits on the surface sinceless volume has to be displacedto match weight of body.

Q. What happens if a body is onlypartially submerged?



Long-range attractive forces andshort-range repulsive forces balance at theequilibrium position so net force is zero.

Attraction

Repulsion

DistanceEquilibriumposition

But, attractive forces are long-range forcesrepulsive forces are short-range, thus, a molecule B at thesurface feels an attraction from the molecule at layers 1 and2 since attractive forces are long range. RA = FA1 + FA2However, the molecule at the surface only feels therepulsion from the molecule directly beneath it (RR= FR1)since repulsive forces are short range. Thus, tocounterbalance all the “extra” attractive forces from thedeeper molecules, the surface molecule has to movedownwards and closer to layer 1 since the repulsive forceincreases with decreasing distance.

Forc

e

Surface

Layer 1

Layer 2

RA

RR

A

B

A BFRFA

5.1.4 Surface tension

Forces between atoms or molecules take the form of a repulsion that isvery strong at short distances and an attraction which diminishes instrength with larger distances. Atoms and molecules take up an equilibriumposition where the repulsive and attractive forces are balanced.

Consider two molecules in a liquid, one on the surface, and another in theinterior. Long-range attractive forces FA have a resultant RA zero on A,downwards on B. Short-range repulsive forces FR have a resultant RR zeroon A, upwards on B.



For any surface molecule, theshort-range repulsive forcescome from neighbours to theside and below but not fromthe top.Resultant long-rangeattractive force must act inopposite direction so as tobalance short-range forcesand so are also perpendicularto the surface (on average)

Gravity acts verticallydownwards.

Resultant short-rangeforces are perpendicularto the surface bysymmetry (on average)FR

FR

W

5.1.5 Contact angle

or: the surface must beperpendicular to the resultantlong-range attraction on thesurface molecules.

All other forces involved willpush the surface around untilthese conditions are met.These issues determine the wetting angle or contact angle of the liquid..

In a large pond,gravitational effects aredominant and pulls thefluid surface into ahorizontal plane. Gravityacts on all moleculessurface and interior.Gravity is thus a volumeeffect that depends ondimensions cubed. In asmall droplet, surfacetension effects aredominant and surface ispulled into a shape thatminimises the surfacearea of the drop (i.e. asphere). Surfacetension only acts onsurface molecules andthus is a surface effectand depends on lineardimension only.

Case 1: Wet contact

θ > 90o

F2

F1

FR

Case 2: Waterproof contact θ < 90o

F2Resultant forceon cornermolecule:a “local vertical”

F1

FR

Liquid pullswith resultantattraction F1

Molecules insolid pull with F1

Liquid molecules attractedmore by the solid than liquid.

Liquid molecules attracted more bythe liquid than solid molecules.

In both cases, the liquid is pulled into shape bysurface forces until the surface of the liquid isperpendicular to the resultant force Fr.



The coefficient of surface tension γ is defined as:

Work per change in areaF - Force (N)d =2l - length of contact between

body and liquid (m)W - work done in changing the surface area by ∆A∆A - change in surface area

F

lA

∆A

AW

F

∆=

=γ2l Note: a film has two

surfaces (the backand the front)

Examples: dFR

=γ

( )t2mg+

=γl

Additionalforce

Length incontact withfluid

(b)

(a)

units of γare N m−1

5.1.6 Coefficient of surface tension

Additional forcerequired to justbalance the force dueto surface tension

l

t

mg

∆p - difference of internalpressure of bubble tooutside pressure

( )

4pR

R22Rp 2

∆=

π

π∆=γ

Factor of 4 to account for twosurfaces (internal and externalsurfaces of bubble).

(c) Surface tension tends to compress the gas inside a bubble. Compressionproceeds until the increase in internal pressure balances surface tension

R

A liquid drop has only one(outside) surface thusfactor of 2 instead of 4

(d) Surface tension tends to compressthe liquid inside a drop.Compression proceeds untilincrease in internal pressurebalances surface tension.

( )

2pR

R2Rp 2

∆=

π

π∆=γ



ρ - density of liquidR - Radius of tubeθ - angle of contact (wetting angle)

Contact angle θ < 90o.Surface is pulled up by surfacetension forces FR.

θ

hR

FR FR

Contact angle θ > 90o.Surface is pulled down by surfacetension forces FR.

Effects of capillarity are seen more clearly in small-scale apparatus sincesurface tension is a surface effect while weight is a volume effect. Thecoefficient of surface tension is given in this case by:

θh

R

FRFR

5.1.7 Capillary action

Consider a narrow tube placed in a container of liquid. It is observed thatthe level of liquid in the tube is different to that of the level in thecontainer. The level in the tube may be higher or lower than thesurrounding liquid. It depends on the contact angle. A narrow tube inwhich this occurs is called a capillary.

θcos2ρghR

πR2πR

θcosρghγ

πR2

πRθcos

ρghF

pAθcosF

2

2R

R

=

=

=

=

=

l

Cos θ term is included since Fracts at an angle to the weightof the liquid.

Length of contact is thecircumference of the tube.

Thus:



Consider a mass element δm1 =ρA1v1∆t passing a point P1during a time interval ∆t. Workdone on the mass element δm1

The same amount ofmass δm2 = δm1passes point P2 duringtime interval ∆t.

( )

ρ

δ=

=

∆=

mp

VpdApW

1

11

11

Work done by the mass element δm:ρ

δ=

mpW 2

Absolutepressure

Change in kinetic energy of a mass element ∆m:

Change in potential energy of a mass element ∆m:

Conservation of energybetween P1 and P2:

( )12 hhmgU −δ=∆

( )21

22 vvm

21KE −δ=∆

( ) ( ) ( )

2222

2111

21

221221

v21ghpv

21ghp

vvm21hhmgppm

KEPEW

ρ+ρ+=ρ+ρ+

−δ+−δ=−ρ

δ

∆+∆=∆

Bernoulli’sequation

Net mechanical work done on system: ( )21 ppmW −ρ

δ=∆

p1h1

h2

A1

A2

v2

v1

p2

P1

P2

∆d

5.1.8 Bernoulli’s equation

Consider the flow of fluidthrough a tube of varyingcross-section:

The mass of fluidpassing point P2during time interval∆t is: tvAm 222 ∆ρ=δ

The mass of fluidpassing point P1 duringtime interval ∆t is:

( )tvAVm

11

11∆ρ=

∆ρ=δ∆ volume = cross-sectional area × ∆ distance = cross-sectional area × velocity × ∆ time

∆V1 is thevolume offluid thatpasses P1during ∆t

The fluid is incompressible (ρ1 = ρ2), and no fluid leaks out or is addedthrough the walls of the pipe (δm1 = δm2). Thus:

2211 vAvA =

The product Av is the volume flow rate Q in m3 s−1. Thevolume flow rate is a constant for incompressible fluids.

Equation of continuity

tvAtvA 2211 ∆ρ=∆ρ

In laminar flow, particles within a fluid move along smooth paths calledstreamlines. In a schematics diagram, the spacing of streamlines is anindication of the velocity of the fluid particles.



Consider a fluid held between two plates.The bottom plate is fixed, and aforce is applied to the topplate causing it to move.

5.1.9 Viscosity

“boundary layer”molecules stick to surface

Laminar flow:interior moleculesfollow streamlines

Top plate moving v

Bottom plate fixed

Velocityprofile

y

Force

As moleculeapproaches, loss inbond potential energy,gain in kinetic energy.

The bond ismomentarily formedaccompanied byenergy loss (heat).

A force is requiredto break bond andresults in increasein potential energy.

A force needs to be continually acting to maintain velocityof top plate. This force, when spread out over a cross-sectional area, is an applied shear stress τ.

y

v Velocitygradientdv/dy

AF

=τ

Consider the events that happen when onemolecule of the fluid passes another one:

The rate of change of velocity with increasingdistance from the bottom plate is a constant.i.e., the velocity gradient is a constant.

Experiment shows thatthere is often a linearrelationship between theapplied shear stress andthe velocity gradient.The constant ofproportionality (orslope) is called thecoefficient of viscosityη. When η is a constant,the fluid is called aNewtonian fluid.

dydvAF

=ηη depends ontemperature.

FA

dvdy

η

common unit: = poise

sPaormNs

2

2cmsdyn

units:

Substance Viscosity(Centipoise)@ 20 °°°°C

Water 1.002Methanol 0.597Lubricating oils 1 - 2

1 poise = 0.1 N s m−2

The velocitygradient issometimes calledthe strain rate.



A spherical body moving through a viscous medium experiences aresistive force which is proportional to its:

• velocity• radius• coefficient of

viscosity of themedium

Drivingforce Equal opposing

viscous dragforce Fv

Uniformvelocity

• Stokes’ law applies only for streamline conditions and when boundarylayer molecules remain stationary on the surface of the sphere.

vR6Fv πη=

5.1.10 Fluid flow

Experiments show that the resistive force is given by:

Stokes’ law

If a body falls through a viscous medium, it willaccelerate until the viscous resistive force and theupthrust due to buoyancy equals the weight of the body.It then reaches its terminal velocity.

Fv

Fb

mg

This termincreases withincreasingvelocity

Weight ofsphere

( )η

ρ−ρ=

9gR2

v s2

tl

Density of bodyDensity of fluid

Coefficient of viscosity

Radius ofbody

Terminalvelocity

Note: vt is proportional to R2

Buoyancyforce

Viscousforce

t3

s3 Rv6gR

34gR

34

πη+ρπ=ρπ l

Terminalvelocity

For the flow of a fluid in a narrow tube, the volume passing through thetube per second depends on:

• pressure difference ∆p• radius of tube (R)• length of tube (l)• viscosity of fluid η

lη∆π

=

=

8pRQ

tVolQ

4

Velocityprofile

Q

Assuming streamline flow:Note the strong dependenceon the tube radius (to thefourth power).



285

1. A lead sphere of mass 5 kg is placed into a beaker of mercury at roomtemperature. Calculate the %volume of the sphere that floats above thesurface of the mercury.

Vm HgHg ∆ρ=

ρHg = 13600 kg m−3

ρPb = 11300 kg m−3

Hg

Pb

Let the volume of Hg displaced =∆V. Then the mass of fluid displaced is:Solution:

and this is equal to the mass of the sphere (5 kg). Thus, if V is the totalvolume of the sphere, then:

V

Vm

Pb

HgPb

ρ=

∆ρ=

5.1.11 Examples

2. Transmission oil is pumped through a 10 mm diameter pipe 1.2 m longunder a pressure difference of 200 kPa. What is the flow rate of oil(Litres/sec) if the oil is cold (η = 2 Pa.s) and then hot (η = 0.1 Pa.s)

( )( )

( )( )

1

43

hot

1

43

cold

4

smL409

2.11.08005.010200Q

smL4.20

2.128005.010200Q

8pRQ

tVolQ

−

−

=

×π=

=

×π=

η

∆π=

=

l

200kPa

1.2 mQ (L/sec)

Solution:

%831360011300

VV

Hg

Pb ==ρ

ρ=

∆



5.2 Solids

kxF =

ε=σ E

Summary

AF

=σ

AF

=τ

ll∆

=ε

γ

τ=G

VVB h

∆

σ∆=

VV∆

=ε

ll∆

∆

=νww

2kx21W =

( )xikdtdxkxF

ωλ+=

λ+=

Hooke’s law

Stress

Shear stress

Strain

Volume strain

Bulk modulus

Shear modulus

Poisson’s ratio

Strain energy

Elastic modulus

Viscoelasticity



5.2.1 Hooke’s law

k depends on the type of material andthe dimensions of the specimen.

F

x

2F

2x

Robert Hooke (in 1676) found (by doing experiments)that if a certain force was needed to stretch a bar by x,then double the force was needed to stretch the samebar by 2x. Mathematically, this was expressed byHooke as:

Hooke’s lawkxF =

Attraction

Repulsion

DistanceEquilibriumposition

Strength of thebond Fmax

Consider the forces acting between two atoms:Note that near the equilibrium position,the force required to move one atomaway from another is very nearlydirectly proportional to the displacementx in accordance with Hooke's law.

Thomas Young (in 1807) described Hooke’s relationship in such a waywhich did not rely on the geometry of a particular specimen.

l

ll

l

xEAF

AkE

xAkAkx

AF

kxF

=

=

=

=

=

Let

Start with Hooke’s law

Divide both sides by A, the cross-sectionalarea of the specimen

Multiply and divide by l, the length of the specimen

Stress Strain

E is a material property which describes theelasticity, or stiffness of a material and iscalled Young’s modulus. Esteel 210 GPa

EAluminium 70 GPaEglass 70 GPa

Force and distance are linearlyrelated (approximately) near theequilibrium position.



The units of stress arePa (same as pressure).

When forces tend to pull on a bodyand thus stretch or elongate it, tensilestresses are produced within thematerial.

Force per unit area(stress) acts on anelement of materialwithin the body

Tensilestress

σ

A

F

F

F

AF

=σ

5.2.2 Stress

Stress

Force

Area overwhich forceacts

Tensile stress

Compressive stressWhen forces tend to push on abody and thus shorten orcompress it, compressivestresses are produced within thematerial.

Compressivestress

σ

F

A

Compression and tension arecalled normal stresses.

Because the force producing the stress actsnormal to the planes under consideration.Symbol σ used for normal stresses.

Force acting parallel to areaproduces shear stress τ.

AF

=τ

F

F

F

A

Shear stress



F

F

l

∆l

Strain is the fractional change in lengthof a body subjected to a deformingforce.

ll∆

=ε

can be positive ornegative

Original length

Application of a deforming force causes atoms withinthe body to be shifted away or displaced from theirequilibrium positions. The net effect of this is ameasurable change in dimensions of the body.

Linear strain

Fx

h γ

Shearing angle γ tan γ = x/hor γ = x/h for small deflections

Referred to as theshear strain

Shear strain

γ

τ=G

Shearmodulus

When a solid issubjected to uniformpressure over its wholesurface, then thedeformation is describedby the volume strain.

Volume strainThe bulk modulus isthe ratio of the changein hydrostatic pressureover the volume strain

VVB h

∆

σ∆=

VV∆

=ε

σh

σh

σh

σh

5.2.3 Strain



It is observed that for many materials, when stretched or compressed alongthe length within the elastic limit, there is a contraction or expansion of thesides as well as an extension orcompression of the length.Poisson’s ratio is the ratio ofthe fractional change in onedimension to the fractionalchange of the other dimension.

w

∆w

∆l

l

F

ll∆

∆

=νww

5.2.4 Poisson’s ratio

Poisson’s ratio is a measure of how much amaterial tries to maintain a constant volumeunder compression or tension.Consider a bar of square cross section w × wplaced in tension under an applied force F.The initial total volume of the bar is:

l11 AV =

Where A1 = w2. After the application of load, the length of the barincreases by ∆l. The width of the bar decreases by ∆w. The volume of thebar is now calculated from:

( )( )

( )

( ) ( )( )νε−ε+≈

νε−ε+=

∆−ε+=

∆−∆+=

21A1A1

ww1w1

wwV

1

21

22

22

ll

l

ll

The change in volume is thus:( )

( )ν−ε=

νε−ε+−≈−

21A21AAVV

1

1112l

ll

For there to be no volume change, thenν has to be less than 0.5. ν > 0.5implies that the volume decreases withtension, an unlikely event. When ν =0.5, there is no volume change and thecontraction in width is quitepronounced (e.g. rubber). When ν = 0,the volume change is the largest andthere is no perceptible contraction inwidth. Most materials have a value of νwithin the range 0.2 to 0.4.

since ε

2 << 1

When the material contracts inwards (a so-called plane stress condition)under an applied tensile stress σT, there is no sideways stress induced in thematerial. If the sides of the material are held in position by external forcesor restraints (plane strain), then there is a stress σ induced, the value ofwhich is given by: Tνσ=σ

In terms of stresses and strains, in plane strain conditions (sides held inposition), there is an effective increase in the stiffness of the specimendue to the induced sideways stresses. Hooke’s law becomes:

ε

ν−

=σ 21E



Hooke’s law applies for the linear elastic region. When load isremoved, body returns to its original shape. If body is stretchedbeyond the elastic limit, it will only partially return to its originalshape and thus acquire a permanent set.

Stre

ss

Strain

Ultimatestrength

Fracture

Elastic limit oryield point

Proportionallimit

Elastic behaviourslope depends on E

Plasticbehaviour

Ductilematerial

5.2.5 Mechanical properties of materials

Stre

ss

Strain

Fracture

Elastic behaviourslope given by E

Brittle material

2kx21

kxdxW

kxFFdxdW

=

=

=

=

∫

When a solid is stretched or compressed by the application of an appliedforce, the application of the force F through a distance dx requires workto be done on the system.

Strain potential energy

When a solid fractures, the stored strain potential energy is convertedinto heat, kinetic energy, plastic deformation and surfaces - that is, thesurface energy of the cracked parts. Brittle materials generally shatterinto many surfaces which soak up the stored strain energy releasedduring fracture. Ductile materials tend to absorb the stored strainenergy in the fracture surfaces and also plastic deformation inside thematerial.



L

π=

L2xsinFF max

xL2

FL2xFF

max

max

π=

π=

but, for small values of θ, sinθ = θ

all constant for aparticularmaterial

kxF =

for smalldisplacements only

F+

F-

x

Sine wave

Fmax

Lo

kxsinAF =

AmplitudeFrequencyk = 2π/λ

L = λ/4

5.2.6 Linear elasticity

Thus:

Consider the shape of the force law between two atoms or moleculesin more detail. Its shape resembles that of a sine wave in the vicinityof the force maximum.

Let

F may be expressed in terms of force per unit area (or stress) which isgiven the symbol σ

xL2

maxπσ

=σ

Let the fractional change in displacement from the equilibrium position(the strain) be given by:

oLx

=ε

EL2

L maxo

=

πσ=

ε

σ

Substituting for x and transferring ε gives:

All materialproperties

Young’s modulusor “stiffness”

Hooke’s law

Shape of this partof the curve isnearly that of asine wave.



5.2.7 Viscoelastic materials

Most materials are neither completely elastic or completely fluid but fallwithin these two extremes. For materials which display an appreciable“fluid like” behaviour, even though they might appear to be solid, arecalled viscoelastic. Their mechanical properties are described in terms ofan elastic modulus and a viscosity. The viscous component usually affectsthe response of a material when subjected to a changing force. The elasticproperties usually are more important when materials are subjected to astatic force.The response of materials and systems can often be modelled by springs anddashpots. This allows both static and dynamic processes to be modelledmathematically with some convenience. Springs represent the solid-likecharacteristics of a system and is equivalent to the elastic modulus E.Dashpots represent the fluid-like aspects of a system and the dampingcoefficient λ is related to viscosity η.

k

kxF =

dtdxF λ=

dtdxkxF λ+=

λ

k

VoigtNewtonHooke

λ

Maxwell

dtdF

k1F1

dtdx

+

λ

=

k

λ

Consider the response of a parallelspring and dashpot to an sinusoidal force. The resulting displacement x isalso sinusoidal but is out of phase with the force.:

( )

( )

( )

( )ω=

ω=

=

=

ω

ω

φ+ω

ix

iexdtdx

exx

eFF

tio

tio

tio

Thus:

( )xikdtdxkxF

ωλ+=

λ+=

The product ωλis a loss termand is related toviscosity.

The spring stiffness isa storage term and isrelated to elasticmodulus.

The damping coefficientλ has the samerelationship to viscosity ηas the spring stiffness kdoes to the elasticmodulus E. The dampingcoefficient and stiffnessapply to a particularspecimen geometrywhere the modulus andviscosity apply to thematerial (i.e. E and η arematerial properties).

Units: N s m−1



Above absolute zero of temperature, atoms are vibrating around theirequilibrium positions with a very high frequency ≈ 1014Hz.

These high frequencyvibrations travel throughoutthe solid as elastic waveswith a velocity equal to thespeed of sound in thematerial.

Internal energy is the energycarried by the elastic lattice waveswithin the material. An increase ininternal energy resulting from achange in temperature correspondsto an increase in the amplitude ofthe lattice waves.

In classical thermodynamics, we might treat each vibrating atom as aharmonic oscillator (i.e. the atoms are connected to other atoms by linearsprings undergoing simple harmonic motion). For a single atom, theremay be oscillations in three directions and thus, the energy of vibration is:

kT3U =

For a mole of atoms, the total energy is:

RT3kTN3U A

=

=

Since R = NAk

But, CdTdU

= the molar specific heat.

Thus: R3C =

That is, the molar specific heat for all solids is a constant equal to 3R.This is true for most solids at reasonably high temperatures but is notobserved to hold at low temperatures close to absolute zero.

Latticewaves ina solid

Dulong-Petit law

R = 8.3145 J mol−1 K−1

5.2.8 Lattice waves

When a solid absorbs heat energy, its temperature rises.The energy being absorbed is converted to internalenergy. In solids, molecules do not translate or rotate(otherwise the material wouldn’t be solid!). Internal energyis the vibrational energy of the constituent atoms.

The energy of vibration of an atom (PEto KE to PE etc) is the same no matterwhat the molecular or atomic mass.The total energy depends on the totalNUMBER of atoms. The (mass) specificheat of solids with a low molecularweight is larger than that for solids witha high molecular weight because in theformer, there are a greater number ofmolecules in 1kg of material than thereare in the latter.



Matter is not continuous. The regular spacing of an atomic latticeintroduces a discreteness in the allowable frequencies, and hence energies,of lattice waves.Just as the energy of electromagnetic waves is carried by photons, theallowable energies associated with lattice waves are called phonons.

Quantum theory agrees with classical theoryabove the Debye temperature qD but agreeswith experimental observations at lowtemperatures.

When the energies of phonons are calculated, it can be shown that themolar specific heat changes with decreasing temperature and is inagreement with experimental observations.

Phonons are important because internalenergy is the energy of the phononswithin a solid. In metals, the conductionelectrons increase their kinetic energywith increasing temperature and thuscontribute to the specific heat. But themajor proportion of internal energyoccurs as phonon energy.

5.2.9 Phonons

U

T

Classical theoryC

dTdU

= (constant slope)

Quantumtheory

Experiment

qo ≈ 300 K

Thermal energy is transported from hot to cold regions within a solid byboth phonons and free electrons. The total thermal conductivity k is thesummation of these conductivities : ep kkk +=

Thermal conductivity in conductors (e.g. metals) arises almost entirely fromthe transport of energy by free electrons. Good thermal conductors are alsogood electrical conductors becauseof the large number density of freeelectrons. The two conductivitiesare physically related:

Heat conduction in insulators is entirely via phonons. Anything whichinhibits the travel of, or scatters, phonons leads to a decrease in thermalconductivity. Lattice imperfections scatter phonons and thus amorphoussubstances (glass) have a lower thermal conductivity than crystallinematerials. Increasing the temperature also results in increasing scatteringthus k usually decreases with increasing temperature.

TkLσ

=

L is called the Lorentznumber and is a constant.

Material k (W m−−−−1 K−−−−1)Aluminium 220Steel 54Glass 0.79Water 0.65Glass wool 0.037Air 0.034



5.2.10 Examples

Solution:

1. A cube of aluminium is subjected to a hydrostatic pressure of 4GPa.Determine the % change in length of the side of the cube.

2. A 60mm cube of copper is subjected to a shearing force of150 kN. The top face of the cube is displaced 0.25 mm withrespect to the bottom. Calculate:(a) the shearing stress(b) the shearing strain(c) the shear modulus

Solution:

0.25 mm

150 kN

GPa6.760042.0

4.319G

0042.06025.0

MPa4.31906.0

101502

3

==

==γ

=

×=τ

%3.5VV

VV1041046.7

VVB

910

h

=∆

∆

×=×

∆

σ∆=

%78.1

33.5

3VVV 3

=∆

∆=

∆=

∆

=

ll

ll

ll

l

BAl = 7.46 × 1010 GPa



5.3 Matter

Summary

Whfmv21 2

−= Photoelectric effect

tiV

xm2 2

2

δ

δψ=ψ+

δ

ψδ−

mvh

=λ de Broglie matter waves

Schroedinger equation

hfE = Photon energy

−=

λ 22 n1

21R1

Balmer formula

Einstein’s equation

π≥∆∆

2hxp Heisenberg uncertainty principle

k97.4hc

max =λ Wien’s displacement law

2oK

2T cmEmcE +==



Wien argued that since the moleculeswithin a heated body are vibratingwith thermal energy, then the Maxwelldistribution of velocities would resultin acceleration of charges within themolecules thus leading to the emissionof radiation with a characteristicintensity spectrum. Wien’s predictionsagreed well with experimental resultsat high frequencies but did not fit wellat low frequencies. Rayleigh and Jeansperformed more rigorous calculationsand found that the predicted radiation emission spectrum agreed well atlow frequencies but had an ever increasing intensity at higher frequencies -a feature they termed the ultraviolet catastrophe.

5.3.1 Radiation emission spectrum

Wavelength

Inte

nsity

4000 K

3000 K

5000 K

Radiation emissionspectrum

Experiments show that the distribution of intensity of radiation withwavelength emitted from a black body has a characteristic shape whichdepends upon the body’s temperature.Maxwell calculated the velocity distribution of molecules in an ideal gasand found that the distribution had a characteristic shape that alsodepended on the temperature of the gas.

hfE =

Hz

Planck’s constant: 6.626 × 10-34 J s−1

In 1901, Planck found that the emission spectrum could only be explained interms of Maxwell’s statistics and Boltzmann’s statistical interpretation ofentropy for the radiation emitted by electric oscillators (little was known aboutthe structure of the atom at this point) if the radiation emission was allowed toonly occur in discrete amounts which he called energy quanta. Planckdetermined that the energy distribution could be expressed as:


as long as the energy term E had a minimum value given by:

( )1e

E8kTE4 −λ

π=λψ k is Boltzmann’s constant: 1.38 × 10−23 J K−1

The maximum in the emission spectrum is found bydifferentiating Planck’s energy distribution with respectto λ. This yields:

k97.4hc

max =λ Wien’s displacement law


5.3.2 Photoelectric effect

In thermoionic emission, electrons canbe ejected from a hot filament as a resultof the kinetic energy imparted to them.These electrons can form an electriccurrent if an external field is applied.In 1887, Hertz observed that a currentcould be created if a metal was illuminatedby light of a sufficiently high frequency.Experiments showed that the current in this casecould only be produced if the frequency of the lightwas above a critical value (the threshold frequency). If thefrequency was below this value, no current was produced even if the lightwas of very high intensity. This effect was called the photoelectric effectand for many years, remained unexplained.When electrons are ejected, it is found thateven with no applied potential at thecathode, there is still a very small current.A small reverse voltage (the stoppingpotential) is needed to stop all thephotoelectric current.The explanation of the photoelectric effect was given by Einstein in 1905.Einstein postulated that light consisted of energy quanta in accordancewith Planck’s equation. If the energy of the incoming light was greater thanthe work function of the metal surface, then the excess energy would beimparted to the electron as kinetic energy. The maximum kinetic energy isgiven by:

oe

2

Vq

Whfmv21

=

−=

Photoelectriccurrent

Vo V– +

Charge on electron1.602 × 10-19 C

340

-

-

--

+

Photoelectriccurrent

Incidentlight


I2

I1

The stopping potential is a measure of themaximum kinetic energy of the ejectedelectrons. If the stopping potential is plottedagainst frequency, the slope of the resultinglinear function is Planck’s constant and theintercept is the work function.

f

Vo

eqW Threshold

frequency

The work function is a measureof the surface energy potential.It is on the order of a few eV formost metals.


5.3.3 Line spectra


Ever since the 18th century, it was known that the emission spectrumfrom a heated gas consisted of a series of lines instead of a continuousrainbow of colours. The position (or wavelength) of spectral lines wasknown to be unique to each type of element.When white light is shonethrough a cool gas, it is foundthat dark lines appear in theresulting spectrum, the positionof which correspond exactlywith the bright line spectraobtained when the gas is heated.Physicists in the 19th century used emission and absorption spectra toidentify gases in the atmosphere of the sun. In 1885, Balmer formulated anempirical equation that accounted for the position of the lines in the visiblepart of the hydrogen spectrum.

−=

λ 22 n1

21R1

In this formula, n is the line number and takes on the values 3, 4, 5, …Theformula predicts an infinite number of spectral lines which become closertogether with increasing value of n. At n = ∞, the wavelength is 364.6 nm,the limit of the Balmer series. Other series were since discovered in thehydrogen spectrum by letting the first term in the brackets equal 1, 3, 4, etc.

−=

λ

−=

λ

−=

λ

−=

λ

22

22

22

22

n1

41R1

n1

31R1

n1

21R1

n1

11R1 Lyman series

(ultraviolet)n = 2, 3, 4...

Paschen series(infrared)n = 4, 5, 6...Brackett series(infrared)n = 5, 6, 7….

Rydberg constant: 1.0973731 × 107 m−1

The existence of spectral lines could notbe explained by classical physics.Balmer’s equation demonstrated anorder, and an integral order at that, tothe position of lines within thefrequency spectrum. Balmer did notpropose any physical explanation for hisformula, but simply noted that itdescribed the phenomena almostexactly. The importance of the equationis that is provides a test for what was tobecome a model for atomic structure.

The Balmer formula applies only to thehydrogen atom. Rydberg proposed a moregeneral formula for the heavier elements.

n =

3

n =

4

n =

5

n =

6

n = ∞

660

490

437

413

365

Balmer series

λ (nm)

Balmer series(visible)n = 3, 4, 5...


5.3.4 Bohr atom

342

In 1897, Thomson demonstrated that cathode rays (observed to be emittedfrom the cathodes of vacuum tubes) were in fact charged particles which hecalled electrons. Thomson proposed that the atom consisted of a positivelycharged sphere in which were embedded negatively charged electrons.Rutherford subsequently found in 1911 that the electrons orbited at somedistance from a central positively charged nucleus. Rutherford proposed thatelectrostatic attraction between the nucleus and the electron was balanced bythe centrifugal force arising from the orbital motion. However, if this werethe case, then the electrons (being accelerated inwards towards the centre ofrotation) would continuously radiate all their energy as electromagneticwaves and very quickly fall into the nucleus.In 1913, Bohr postulated two important additionsto Rutherford’s theory of atomic structure:• Electrons can orbit the nucleus in what

are called stationary states in whichno emission of radiation occurs and inwhich the angular momentum isconstrained to have values:

π

==

2nhvrmL e

The 2π appears becauseL is expressed in terms ofω rather than f.


Lyman

Balmer

Paschen

n = 1

n = 2

n = 3n = 4 n = 5 n = 6

Ehf ∆=

• Electrons can make transitions from one state to anotheraccompanied by the emission or absorption of a single photon ofenergy E = hf thus leading to absorption and emission spectra.

Mechanicalmodel ofhydrogen atom

rvm

rq

41 2

e2

2e

o=

πε

r

As in the Rutherford atom, the centrifugalforce is balanced by Coulomb attraction:

with the additionthat:

π

=

2nhvrme

By summing the kinetic energy (from the orbital velocity)and the potential energy from the electrostatic force, the total energy of anelectron at a given energy level n is given by:

me

222o

4e

2e

nnh8

qZmE

ε

−=

from which the Rydberg constant may be calculated since ∆E = hf

Note, Z = 1 for the hydrogen atom where the energy of theground state is 13.6 eV. The energy levels for each state nrises as Z2, thus, the energy level of the innermost shell formulti-electron atoms can be several thousand eV.



5.3.5 Energy levels

The stationery states or energy levels allowed by the Bohr model of theatom are observed to consist of sub-levels (evidenced by fine splitting ofspectral lines). These groups of sub-levels are conveniently called electronshells, and are numbered K, L, M, N etc, with K being the innermost shellcorresponding to n = 1. The number n is called the principle quantumnumber and describes how energy is quantised.

n = 1K

The energy required to move an electronfrom an electron shell to infinity is calledthe ionisation energy. It is convenient toassign the energy at infinity as being 0since as an electron moves closer to thenucleus (which is positively charged) itspotential to do work is less, thus theenergy levels for each shell shown arenegative. For hydrogen, the ionisationenergy is -13.6 eV. The energies for thehigher energy levels is given by:

Hydrogen The electron-volt is a unit of energy.1 eV = 1.609 × 10−23J

At each value of n (i.e. at each energy level) the angular momentum cantake on several distinct values. The number of values is described by asecond quantum number l. The allowed values of l are 0, 1, … (n−1).Each value of l is indicated by a letter: l = 0 s

l = 1 pl = 2 dl = 3 fl = 4 gl = 5 h

A third quantum number m describes the allowable changesin angle of the angular momentum vector in the presenceof an electric field. It takes the values −l to 0 to +l.A fourth quantum number describes the spin of an electronwhere the spin can be either −1/2 or +1/2.According to the Pauli exclusion principle,no electron in any one atom can have thesame combination of quantum numbers.This provides the basis for the filling ofenergy levels.

n = 2

n = 3

n = 4

n = ∞

L

M

NO

2s(2)

2p(6)

3s(2)3p(6)4s(2)3d(10)4p(6)

4d(10)4f(14)

1s(2)

n = 3l = 0, 1, 2 = s, p, or dm = -2, -1, 0, 1, 2

5 values of m times two for spinthus 10 possible electrons

When all the electrons in an atom are in thelowest possible energy levels, the atom is said tobe in its ground state.

−13.6 eV

−3.39 eV

−1.51 eV

−1.85 eV−0.54 eV

0 eV

2n6.13E −=

For example the 3d energy levelcan hold up to 10 electrons:

thus:and:

For Hydrogen


5.3.6 X rays

X rays are produced by electrons being decelerated from an initial high speedby collisions with a target material. Electrons may be produced bythermoionic emission and given a high velocity by the application of anelectric field.When an electron (qe) collides with atarget material, it is rapidlydecelerated and a photon is emitted.The wavelengths of the photonsinvolved are mainly in the X rayregion of the electromagneticspectrum. The most rapiddecelerations result in the shortestwavelength photons. For othercollisions, the electron may loseenergy as photons of largerwavelength and also to heat byincreasing the vibrational internalenergy of the target. The result is acontinuous spectrum of photonenergies with a minimum wavelengthdependent upon the kinetic energy ofthe electrons.Incoming electrons may also ionise theatoms of the target by ejecting boundelectrons from within material. Someof these ejected electrons may comefrom the innermost energy levelswhich, in solid, can have energies inexcess of 100,000 eV. An outerelectron can fall into this vacancy andemit a photon in the process.X rays resulting from filling of K shellvacancies by an electron from the L shellare called Kα X rays. X rays from M toK shell transitions are Kβ, and thosefrom N (and higher) to K transitions areKγ. Similarly, transitions from M to Lare Lα, N (and higher) are Lβ.

λ

X ra

y in

tens

ity

Kβ

Kα

λmin

Continuousspectrum

When an X ray is incident on amaterial, electrons within the targetin the K and L shells can be ejected.Since there are relatively fewelectrons in these innermost shells,X rays do not have a high probabilityof interacting with them and hencetheir high penetrating ability.

n = 1 K

n = 2

n = 3

n = 4

n = ∞

L

M

NO

Kα

Kβ

Kγ

Lα

Lβ

Shortwavelengthhard X rays

Longwavelengthsoft X rays



5.3.7 Matter waves

The Bohr model of the atom strictly applies only to a single electron orbitinga nucleus and ignores interactions between electrons and other neighbouringatoms. Further, the theory does not offer any explanation as to why theangular momentum is to be quantised. Such explanations and treatments canonly be explained in terms of wave mechanics.In 1924 de Broglie postulated that matter exhibited a dual nature (just asdid electromagnetic radiation) and proposed that the wavelength of aparticular object of mass m is found from:

mvh

=λ

where mv is the momentum p of the object. The resulting waves are calledmatter waves. In the case of atomic structure, matter waves for electronsare standing waves that correspond to particular electron orbits.


Because h is a very small number, the wavelength of largeobjects is very small. For small objects, e.g. electrons, thewavelength is comparable to atomic dimensions.

π=

=π

2hnmvr

mvhnr2

For a particular radius r, a standing wave is obtained when the circumferenceof the path is an integral number of wavelengths: r2n π=λ

Thus, from theexpression for matterwaves, we obtain:

Bohr conditionfor stable statesince L = mvr.

The wave-particle duality of matter means that inherently, an electron isneither a wave or a particle but its motion can be quantified using themathematical equations appropriate to waves and particles. The wave natureof matter is often interpreted as being one of probabilities. The amplitude ofa matter wave represents the probability of finding the associated particle at aparticular position.Since matter is described in terms of a probability, there becomes an inherentlimitation in what we can know about the motion and position of a particlesuch as an electron. The Heisenberg uncertainty principle quantifies theseuncertainties. For momentum and position, the requirement is:

π≥∆∆

2hxp

Where ∆p and ∆x are the uncertainties associated with these quantities.The more we reduce the uncertainty in one, the more the uncertainty in theother increases.


The solution to the wave equation is the wave function Ψ. If V is afunction of x only, then the wave equation can be separated intotime-independent and time-dependent equations that can be readily solved.

5.3.8 Quantum mechanics

Vm2

pE2+=

The total energy of a system is the sum of the potential andkinetic energies. Expressed in terms of momentum, p, this isstated:

Thus: ( )t,xVm2

phf2+=

Let

tiE

xip

δ

Ψδ=

δ

Ψδ−=

Ψ is a variable, the formand value of which providesinformation about themotion of a wave/particle.

( )t

it,xVxm2 2

22

δ

Ψδ=Ψ+

δ

Ψδ−

Thus:

Schroedinger equation

The resulting solutions of these equations, when multiplied together, givethe wave function:

( ) ψ=ψ+δ

ψδ− EV

xm2 2

22

( )tEi

et =φ

( ) ( ) ( )txt,x φψ=Ψ

The wave function gives all the information about the motion of a particle,such as an electron in an atom. Ψ is a complex quantity, the magnitude ofwhich |Ψ| is interpreted as a probability density function which in turncan be used to determine the probability of an electron being at someposition between x and ∆x.Quantum mechanics is concerned with determining the wave function(i.e. solving the Schroedinger equation) for particular potential energyfunctions such as those inside atoms. It is found that valid solutions to thetime-independent wave equation occur only when the energy is quantised.The solutions correspond to stationary states.

hfE =since

The value of the potential function may depend onboth position and time. The form of V(x,t) is differentfor different arrangements of atoms (e.g. a singleisolated atom, an atom in a regular array of a crystal).

Solutions to the Schroedinger equation can be found for potential functionswhich are a function of both x and t. This enables time-dependentphenomena (e.g. the probability of transitions between energy levels in anatom) to be calculated and hence the intensity of spectral lines.



Radioactivity was discovered by accident in 1896 by Becquerel who foundthat the rays emitted by a sample of uranium were similar to X rays(discovered a few months before by Roentgen) but appeared spontaneouslyfrom the uranium sample. These new rays had a very high penetratingpower and could also cause ionisation of the air just like X rays. Soon after,many other naturally occurring radioactive elements were discovered, chiefamong these being radium and polonium by Marie Curie in 1911.Rutherford found that a sample of uranium emitted two kinds of rays whichhe called alpha (α) and beta (β) rays. In 1900, Villard found a third typeof ray from radium which were called gamma (γ) rays. One of the moststriking properties of these rays was their penetrating power into matter.Experiments showed that alpha rays were the least penetrating butproduced the greatest ionisation and gamma rays were the most penetratingbut caused the least amount of ionisation.

Experiments in the early 1900’s showed that a magnetic field could deflectα and β rays but not γ rays. Since the deflection of β rays was in the samedirection as that observed by electrons in a magnetic field, it was concludedthat β rays were in fact negatively charged electrons. Since the deflection ofα rays was in the opposite direction, it was concluded that α rays werepositively charged particles. Further experiments by Rutherford showed thatα particles were helium nuclei consisting of two protons and two neutrons.

α

β

γ

Alpha particles are the leastpenetrating because all their energygoes into ionisation of the materialwith which they interact. Gammarays are the least ionizing andhence retain their energy and arethe most penetrating.

5.3.9 Radioactivity

In 1913, experiments with radioactive materials indicated that the nucleusconsists of positively charged protons and negatively charged electrons (i.e.electrons in addition to orbiting electrons). In 1932, Chadwick showed thatthe nucleus contains additional particles called neutrons which had a masssimilar to that of a proton but no electric charge. The emission of a betaparticle (an electron) from nucleus was thought to result from thetransformation (rather than a separation) of a neutron to a proton +β.Experiments showed that to satisfy the law of conservation of energy, anadditional particle was also emitted along with the beta particle from thenucleus. This particle was called the neutrino, it has a very small mass andcarries no electric charge.



5.3.10 Half life

Radioactive decay is a random process. The probability that a particularradioactive nucleus (radionuclide) will decay in any selected time period isindependent of the state of neighbouring nucleii, the chemical state, pressureand temperature. If there are N radioactive nucleii present, then the numberthat decay per unit time period (say a second) is given by:

NdtdN

λ−=

λ is a constant called the decay constant, which is different for differenttypes of atoms. Notice that the decay rate depends on the total number ofnucleii, or atoms, present. The decay constant indicates the probability that asingle atom will decay in a unit of time. The larger the value of λ, the morerapidly the decay. Integrating the above equation, we obtain an expressionfor the number of remaining atoms present after a time t:

tot eNN λ−

=

A convenient measure of radioactive decay is thehalf-life t1/2. This is the time for one half of theradioactive nucleii present to decay.

λ=

λ−=

==λ−

69.0t

t21ln

21e

NN

21

21

t

o

t 21

The decay constant can be determined byexperiment. The number of disintegrations persecond (dN/dt) can be measured with ascintillation detector or a geiger counter(depending on the nature of the material beingstudied). Thus, since dN/dt is proportional to Nt,a plot of ln(dN/dt) has a slope −λ.

ot NlntNln +λ−=

At t = 0, Nt = No

( )

t

ot

to

to

NlndtdNln

Ndt

dNNN

NdtdN

NNN

∝

λ−=

−λ−=

λ−=

−=

t

dtdNln

λ

137

14

238

Ba

C

U 4.51 × 109 years

5568 years

2.64 mins

Units: sec−1

constant



By the 1930’s it was known that nuclear reactions whether induced bynaturally occurring radioactivity or artificially by bombarding elementswith α, β particles, γ rays or neutrons result in a change in the chemicalnature of the element being studied. For example, when gamma rays areused to bombard atoms of mercury, the resulting nuclear reaction occurs:

γ + 80Hg198 → 79Au197 + 1H1

γ ray mercury gold protonNuclear reactions may absorb or release energy, but the energy involved isfar greater than that associated with chemical reactions involving thebreakage and formation of chemicalbonds. In nuclear reactions, changesin mass result in large changes ofenergy in accordance with Einstein’sequation E = mc2.It is found that the sum of the masses ofindividual protons, neutrons and electrons isalways more than the mass of a completeatom. Consider a Carbon 12 atom 6C12

The dream of alchemists ofthe middle ages.

Mass of 6 protons & 6 electrons = 6.04695mass of 6 neutrons = 6.05199

Total mass of parts = 12.09894

Mass of C12 atom = 12.00000

Difference in mass = 0.09894

This difference in massrepresents an energy loss of92.1 MeV when the atomwas formed. This energy iscalled the nuclear bindingenergy.

In nuclear reactions, if the total binding energy of the products is greaterthan that of the reactants, then energy is liberated. That is, the mass of theproducts is less than the mass of thereactants. The distribution of theaverage binding energy of all theelements show that when two lightnuclei combine to form a heaviernucleus (fusion), or when a heavynucleus splits to form two lighternuclei (fission), energy is released.In stars, temperature is so high that hydrogen nuclei (positively chargedprotons) have enough kinetic energy to overcome their mutual repulsionand undergo fusion into helium nuclei releasing energy.

Mass number ZAver

age

bind

ing

ener

gy

per n

ucle

ar p

artic

le Heavyelements

Lightelements

Fe

1eV = 1.6 × 10−19 J

5.3.11 Nuclear energy

Mass number - relativemass of an atom withrespect to 1/12 the massof a carbon 12 atom.

AXZ

Atomic number- number of protonson the nucleus.

Chemicalsymbol



Nuclear particles (protons and neutrons) consist of three quarks. A protonconsists of two up quarks and one down. A neutron consists of two downand one up.

In the early 20th century it was believed that all matter consisted of protons,neutrons, electrons and neutrinos. Today, it is accepted that quarks,electrons and messenger particles responsible for the fundamental forcesof nature are the basic building blocks of matter. Quarks and electrons areordinary particles which contain matter. There are a number of differenttypes of quark:

• Up• Down• Strange• Charmed• Bottom• Top

Each type of quarkcomes in threedifferent “colours”: red,green and blue.

Quantum theory shows how everything in the universe, includingelectromagnetic waves and gravity, consist of particles. These particleshave a property called spin.

The four fundamental forces of nature (gravitational force,electromagnetic force, the weak nuclear force and the strong nuclearforce) can be thought of acting between two particles by the exchange ofmessenger particles.In the nucleus, quarks are held together by the strong nuclear force by theexchange of particles called gluons. Electromagnetic attraction is causedby the exchange of particles called photons. Gravity is thought of asoccurring by the exchange of gravitons.

Particles with spin 1/2 are those thatmake up the matter of the universe(quarks and electrons) whereasparticles with spin 0, 1 and 2 are themessenger particles which result in thefundamental forces between them. Theproperty of spin lead Dirac to proposethe existence of antimatter.

-qe+qe

Electron Positron

γ radiation

5.3.12 Nuclear physics

When antimatter and matter meet, the matter of them both isconverted into energy in the form of electromagnetic radiation.

The names given to the types ofquark are arbitrary, and simplyserve as labels to distinguishone type of quark from another.



determine an expression for the radius of the hydrogen atom when theelectron is in the ground state (n = 1) and calculate this radius given thevalues of me and qe below.

5.3.13 Example

1. The inwards force acting on an orbiting electron in the Bohr model ofthe atom arises from electro static attraction and is given by theCoulomb force law:

Solution:

This force is balanced by the centripetal force given by:

If, as Bohr postulated, that the angular momentum L = mvr can onlytake on values such that:

2

2e

o rq

41Fπε

=

rvm

F2

e=

π

=

2hnvrme

me = 9.11 × 10-31 kgqe = 1.602 × 10-19 Ch = 6.626 × 10-34 J s−1

( )( )( )

nm053.0m10298.5

106.11011.9

10626.61085.8

11

21931

23412

=

×=

××π

××

=

−

−−

−−

Hydrogen atom

2ee

22o

2

e

e2

2e

o

e

2e

2

2e

o

Zqmnh

r

rm2nh

rm

rZq

41

rm2nhv

rvm

rZq

41

π

ε=

π=

πε

π=

=πε

r

Coulomb force = Centripetal force

From Bohr condition

Radius of electron orbit for hydrogen (Z = 1)



5.4 Universe

Summary

oLLlog5.275.4M −=

MeV7.24e2HEH4 01

42

11 ++→

+

221

dmmGF =

Intensity

Magnitude

Nuclear fusion

Gravitational force

RGmv E

= Orbital velocity

2d4LIπ

=



5.4.1 Copernicus, Tycho and Kepler

Copernicus (1473-1543) believed that the discrepancies between theobserved motions of the planets and those predicted by the geocentric modelof Ptolemy could be explained by a heliocentric model (proposed by someearly Greek philosophers) but still containing epicycles and equants (indeedeven more than Ptolemy) along with uniform circular motion.

The planet moves along a small circlecalled an epicycle. The centre of theepicycle rotates about the Earth.

The planet moves along a circular pathwhose centre is offset from the position ofthe Earth giving it an eccentric orbit.

The planet moves along a circular pathwhose centre is offset from the position ofthe Earth. However, its velocity was aconstant with respect to another point C’which is as far off centre as the Earth fromC but in the other direction.

Equant

Eccentric

Epicycle

Earth

Planet

Epicycle

Earth

Earth

Planet

Planet

Deferent

CC'

C

Kepler (1571-1630), using the most accurate astronomical observations ofTycho Brahe, discovered that despite the attractiveness of uniform circularmotion, it simply just did not fit the motion of the planets as observed.

For Aristotle, the universe was a series of interconnecting spheres, theoutermost containing the the stars which was moved by God, and by whichmotion was transmitted to the planets and then to the Moon, the innermostsphere. At the centre of the universe was the Earth. Mechanical details ofaccounting for the motion of planets was perfected by Ptolemy (150AD)through the use of epicyles, equants and eccentrics the combination ofwhich explained the motion of heavenly bodies with uniform circularmotion.



On the basis of Tycho’s observations, Kepler dispensed with theconcept of uniform circular motion and discovered that:• The orbit of a planets is an ellipse where the sun is at one focus• the velocity of a planet along the orbit varies so that a line joining the sun

to the planet sweeps out equal areas in equal intervals of time• the square of the period of the orbit of a planet is proportional to the cube

of planet’s distance from the sun.With these three laws of planetary motion, Kepler dispensed withepicycles, equants, eccentrics and a host of other detail - and indeeddispensed with the long held requirement for uniform circular motion.

With Galileo’s observations of the heavens with the then newly inventedtelescope, the overwhelming evidence was for the acceptance of theheliocentric model of the universe. However, such was the nature ofGalileo’s personality that he came into conflict with the powerful catholicchurch which at that time had not yet incorporated the heliocentric modelinto religious doctrine. It took another 100 years or so before the heliocentricmodel gained general acceptance by theologians.

Planetmovesfastest

Planetmovesslowest

A A

For any interval intime, no matter whereon the orbit, theplanet sweeps outequal area.

Elliptical orbitof planet

5.4.2 Kepler’s laws

Although Kepler could explain the geometry of planetary motion, he couldnot explain why the planets moved. Kepler proposed that the sun, being atthe centre of the universe, emitted “rays” which swept the planets,including the Earth, around the sun. In 1674, Robert Hooke proposed thatall celestial bodies, (including the sun, planets and the Earth) have anattraction towards their own centres - a gravitational force. However,Hooke could not say what the magnitude of this force might be. IsaacNewton, in 1666, proposed that the force acting between any two bodiesexhibited an inverse square law. Newton proposed that this same forceextended to planetary bodies and that the force of attraction between thesun and the planets was this same inverse square law. Newtondemonstrated that such a law would lead naturally to Kepler’s three laws ofplanetary motion.



The heliocentric model of the universerequired the distance from the Earth to thestars to be a very great distance to accountfor the apparent lack of motion of the starsas seen from the Earth.Using a parallax method with the Earth'sorbit as the base, Bessel, in 1838, measuredthe parallax of the star 61 Cygni to be 0.30seconds of an arc which represented adistance of 3 parsecs or 11 light years.

Measurements of astronomical distancesusing parallax are limited by the resolutionof the measuring instruments to only thenearest stars (less than about 300 lightyears), the remainder showing no observableparallax. Further, astronomers of the timewere still uncertain about whether nebulae were inside or outside thegalaxy. In the early part of the 20th century, it was shown using the relativebrightness of maximum and minimum in pulsating stars called Cepheidvariables that the nebulae which could be resolved into stars were locatedfar outside the Milky Way and were galaxies in their own right. In 1926,Edwin Hubble, using the luminosity of Cepheid variables within theAndromeda galaxy, estimated that the distance to this galaxy was 720,000light years, a figure later corrected to 2.36 million light years when theprecise nature of the Cepheid variables observed by Hubble wereinvestigated further. Present day techniques can observe objects (quasars)about 6000 million light years away.

Differencesin the angleof observationat opposite endof Earth’s orbitallowedcalculation ofdistance to star

θ

Diameter of Earth’sorbit 3 × 1011 m

5.4.3 Size of the universe

If r = 1.49 × 1011 m is the mean radius ofthe Earth’s orbit around the sun, then thedistance d at which the angle θ subtendedis one second of an arc is called a parallaxsecond, or parsec (pc). 1pc = 3.086 × 1016

m. It is usually convenient to work inMegaparsecs. 1Mpc = 3 × 1022 m. Thelight year is the distance traveled by lightin one year. 1 ly = 9.461 × 1015 m.Therefore 1 pc = 3.261 ly. The au is themean radius of the Earth’s orbit and isequal to 1.49 × 1011 m.



In the period 1912 to 1929, astronomers undertook spectral analysis of thelight from distant galaxies. It was discovered that as a result of the Dopplereffect, the positions of spectral lines were displaced towards the red end ofthe spectrum, implying that the galaxies were moving away from the Earthat incredible speeds. In 1929, Hubble found that the shift in the spectrallines, and hence the velocity of recession of the galaxy, was dependentdirectly on the distance the galaxy was away from the Earth. That is,galaxies are moving away from each other, the further away, the greatertheir speed.

Hubble’s Law

The recession velocity of a galaxy is proportional toits distance away from the point of observation.

The constant ofproportionality is calledthe Hubble constant andis estimated to be within50 and 100 km s−1

megaparsec−1

Although it appears that allgalaxies are receding from us,this does not necessarily meanthat the Milky Way is at thecentre of the universe. Thesame expansion will be seenanywhere in the universe - justlike spots on the surface of aballoon being blown up. Anyspot will see the others receding,the further away they are, thefaster the recession.

Before 1920, it was generally accepted that the universe was static andunchanging. Einstein’s theory of general relativity actually predicted anexpanding universe so Einstein modified the theory by introducing acosmological constant which had the effect of balancing this expansionexactly with the force of gravity - thus leading to a static universe. In 1922,Alexander Friedmann showed that if it were accepted that on a large scale,the universe was of uniform appearance in all directions, then Theory ofGeneral Relativity actually correctly predicted an expanding universe -precisely that observed several years later by Hubble.

Using the present value of the Hubble constant, and working backwardsuntil “zero” distance, the estimated age of the universe is somewherebetween 10,000 to 20,000 million years. Other measurements using the ageof heavy elements and the age of galaxies put the age of the universe atsimilar values.

5.4.4 Expansion of the universe



The synthesis of expansion of the universe and the equations of generalrelativity by Friedmann lead to the conclusion of the existence of asingularity in space and time from which all matter was created. This iscalled the Big Bang theory of the universe. The theory proposes that theuniverse was created in a gigantic explosion with the creation of Hydrogenand Helium. Space rapidly expanded and the hydrogen and helium cooledas the density decreased and coalesced into stars and galaxies within whichthe heavier elements were made. Calculations of the age of the universeand the rate of expansion of the universe would result in a residualbackground radiation temperature of about 3 K at our present point intime. This radiation was detected by radio astronomy in 1965, providinggood support for the big bang theory.

Big Bang Theory• Allowed an expanding universe

in agreement with Hubble andFriedmann.

• The Big Bang represents thecreation of matter, space andtime.

• Predicts existence ofbackground radiation pervadingthe universe which has beenexperimentally observed.

5.4.5 Modern theories of the universe

In 1948 Bondi, Gold, and Hoyle presented the steady-state theory of theuniverse as an alternative to the big bang theory. They sought toincorporate the equations of general relativity into a scheme whichremoved the requirement for a singularity at the beginning of the universe -a feature which could never be opened to investigation and therefore, intheir view, objectionable. The key to their scheme was that the universe, ona large scale, is uniform in appearance in both space and time inaccordance with Friedmann’s model for an expanding universe. Thedecrease in the density of the universe caused by its expansion is exactlybalanced by the continuous creation of matter which formed new galaxiesto replace those which had moved on past our field of view. The amount ofmatter created each year was in the order of 1 particle per km3 which iswithin our experimental measure of uncertainty.

• Allowed an expanding universe inagreement with Hubble andFriedmann.

• Allowed an infinite past and aninfinite future for the universe (timehas no beginning and no end).

• The large-scale structure of theuniverse remains unchanged inaccordance with our observations.

Steady-State Theory



0 0 Big Bang

1 sec 10 × 109 10 × 10−10 Inflation

60 sec 1 × 109 1 × 10−10 H & He nuclei

1 × 106 yrs 3000 3 × 10−4 Neutral atoms

2 × 109 5 0.2 Milky Way

5 × 109 1.5 0.66 Solar system

4.6 × 109 Earth

12 × 109 −270 1 Present day

In accordance with the Big Bang theory, it is generally now accepted thatall matter, space and time was created from a single point known as asingularity in which the universe had zero size and infinite energy.Quantum theory predicts that matter in the form of a matter andantimatter can be formed from energy in accordance with the Einsteinequation E = mc2. Collisions between matter and antimatter at the big bangproduced large amounts of gamma radiation. A slight asymmetry betweenthe amount of matter and antimatter produced an overall net amount ofmatter in existence.As the universe expanded, it cooled and elementary particles (electrons,neutrons and quarks) were able to “condense” to form protons andneutrons and eventually the heavier elements.

Time Temp oC Relative EventSize

5.4.6 Creation of matter

The distribution of matter after the singularity depends on the assumptionsmade about the mean density of the universe. If the mean density is chosento be above a critical value, then theory predicts that gravitational forceswill eventually overcome those from expansion of the singularity and theuniverse will collapse (the Big Crunch) - a closed universe. If the meandensity is below a critical value, then the expansion of the universe willcontinue indefinitely - an open universe in which eventually all stars will dieout - the Big Chill.When the mean density of the universe is calculated, by first estimating themass of a galaxy by observing the motion of its stars and multiplying themass of each galaxy by the number of galaxies, the mean density of theuniverse is found to be very much below the critical value. This discrepancymeans that there appears to be a of substantial amount of invisible or darkmatter within the universe that is at the present time, unaccounted for.



Three minutes after the big bang, all the matter in the universe consisted ofabout 75% hydrogen and 25% helium nuclei. As the universe cooled,these nuclei began to capture electrons creating neutral atoms.Gravitational forces allowed these neutral atoms to coalesce into gasclouds. Gravitational forces from other parts of the universe caused somegas clouds to rotate and with continued collapse, the loss of gravitationalpotential energy resulted in a faster rate of rotation resulting in disk orspiral type galaxies such as the Milky Way.

At the formation of the Milky Way, the only elements were hydrogen andhelium. As the oldest stars experienced supernova explosions, the heavierelements were created and dispersed throughout the galaxy.

With increasing density of material within a forming galaxy due togravitational attraction, parts of the gas cloud begin to rotate on their own.Increasing collisions between molecules as they coalesce result in anincrease in temperature until the conditions for nuclear fusion are reached.The resulting heat generated by the fusion reaction gives rise to an increasein pressure which is balanced by gravitational collapse resulting in a stablesystem called stars.

Galaxies

Stars

The Milky Way contains about 100 × 109 stars and has the form of a flatdisk rotating about its centre. The size of the Milky Way is about 30,000parsecs (100,000light-years) indiameter. Thesun takesapproximately250 million yearsto travel oncearound the centerof the Milky Way.

Milky Way

Astronomers believethat clusters of galaxiesare the fundamentalunits of the universe.

5.4.7 Formation of stars and galaxies

Universe

Clusters of galaxies

Stars

Solar system

PlanetsMoons

Galaxies



The apparent brightness (Wm−2) of a star (i.e. thebrightness we see on Earth) depends on the star’sintrinsic luminosity (or radiant power output in Watts)and the distance at which the star is from us. It can beconveniently measured using a photometer.

The intensity or brightness, I, of the light emitted by a star is what the eyeresponds to and is a measure of radiant flux density in W m−2. Theintrinsic luminosity, L, of a star is a measure of the star’s radiant poweroutput (in Watts) and this is of significant interest to astronomers.

oLLlog5.275.4M −=

5.4.8 Luminosity

The intrinsic luminosity or radiant power (W) depends on the size of the star and itsintrinsic brightness or intensity (W m−2) at its surface. A large star with low intrinsicbrightness (i.e. a low temperature) may have the same intrinsic luminosity (poweroutput) as a small star of large intrinsic brightness.

If the distance d to a star is known, (by parallax measurements) and theapparent brightness I measured (with a photometer) then the intrinsicluminosity L of the star be calculated using the inverse square law.

The temperature of the sun can beobtained from spectroscopicmeasurements. The intrinsicluminosity is then given by theStefan-Boltzmann radiationemission law using the knownradius of the sun. At R = 0.7 × 109 mT = 5800K, Lo = 3.90 × 1026 W.

2d4LIπ

=

The absolute magnitude isconveniently expressed in termsof the ratio of the intrinsicluminosity of the star and thatof the sun. Mathematically, theabsolute magnitude of a star iscalculated from:

The brightness that a star would have if it were at a distance of exactly 10parsecs away is called (by definition) the absolute brightness or absolutemagnitude M and allows us to compare the brightness of stars independentof their distance from us.

Wavelength

Inte

nsity

4000 K

3000 K

5000 K

Radiation emissionspectrum

If the emission spectra of two stars arecompared and show the same shape (i.e.same colour) then we can say that theyare at the same temperature. This meansthat any difference in their total poweroutput is due to a difference in their size.



In main sequence stars, of which the sun is one, the source of energy is thenuclear binding energy associated with the fusion of hydrogen into helium.In the Hertzpsrung-Russell diagram, for main sequence stars, the largerstars are at the top left while the lower mass stars are at the lower right. Forstars with a size similar to the sun, the main sequence lasts forabout 1 × 1010 years. As the amount of hydrogen becomes depleted,gravitational forces collapse the star inwards until the temperature is highenough to cause the fusion of helium into heavier elements. The outerlayers of the star expand outwards and the star enters the red giant branch.

In its early stages a protostar star is formed by coalescence of the gascloud which has the appearance of a disk. Protoplanets may also begin toform. Formation of a hot core takes about 1 × 106 years. Approximately 70× 106 years later, contraction causes a rise in temperature which is enoughto initiate nuclear fusion and the star is then on the main sequence.During the red giant stage, thesurface temperature is about3000 K. There are periods ofinstability at this point where theouter atmosphere of the star isblown off as a planetarynebulae by the pressure frombelow. About 75000 years later,the core of the star collapses toform a very faint white dwarf.Gravitational collapse in a whitedwarf is countered by quantumeffects acting on their electronswithin it. If the star is moremassive than the sun, it movesalong through the red giant stagemore quickly and instead offorming a white dwarf the innerportions collapse to form a

Lum

inos

ity

Surface temperature × 1000 C30 20 10 6 4 3

Mainsequence

Whitedwarfs

Red giantbranch

Horizontalbranch

SunEvolutionof thesun

Hertzsprung-Russell diagram

5.4.9 Lifecycle of stars

neutron star or alternately a black hole. The shock wave from thiscauses the outer layers to explode as a supernova.



Nuclear fusion in main sequencestars results in the formation of ahelium nucleus from four protons.This does not happen in one step, butthrough a series of reactions.The energy (24.7 MeV) is released inthe form of gamma rays and kineticenergy of the helium nuclei. Positronscreated during the process areannihilated by combining with freeelectrons to produce 2 MeV of energyin the form of gamma rays. The totalenergy released is 26.7 MeV.

H

H

H

H

He

γ radiation

MeV7.24e2HEH4 01

42

11 ++→

+

Positrons are anti-electrons

Very large stars (approximately 5 times the mass of the sun) go throughtheir evolutionary stages very quickly compared to the Sun. As the supplyof hydrogen in such a star becomes depleted, the star contracts until thetemperature becomes high enough to allow the fusion of helium, which isnow in abundance, into heavier elements such as carbon and nitrogen. Theouter atmosphere of the star increases and it becomes a red giant. Nuclearfusion proceeds providing radiation pressure outwards to counteract theinward pull of gravity and elements up to the atomic mass of iron arecreated. Then, the star's core starts to contract rapidly since once iron isformed, fusion cannot proceed any further.The sudden collapse of the core releases a vast amount of gravitationalpotential energy which blows away all the outer parts of the star in aviolent explosion called a supernova. All the elements with atomicweights greater than iron are formed during this explosion.In the central core, protons and electrons are formed into neutrons and thecore becomes a neutron star. The gravitational collapse also causes thecore to spin rapidly and the charged particles in the vicinity emit radiationas a result which is detected by us as regular pulses and these objects arethus known as pulsars. Alternatively, the gravitational collapse is so largethat a black hole is formed.The Crab nebula is the remains of a supernova explosion in our galaxywhich was observed first hand by astronomers in 1054.

5.4.10 Energy and matter



During the formation of a star, heavier molecules, consisting metals andwater, may form what is called an accretion disk. The molecules in theaccretion disk cool and sometimes collide with each other and form largedust particles. These dust particles have their own gravitational field whichmay attract more dust particles. The process rapidly accelerates andprotoplanets are formed.

Larger planets have a gravity large enough to retain some of the originalgas molecules and form gas giants. Others form smaller rocky planetsinitially without an atmosphere. Inside the planets, heavy (more dense)elements gravitate towards the centre leaving the lighter elements nearthe surface.

As the planets begin to form,the evolving sun begins toradiate both energy in theform of electromagneticradiation and also chargedparticles called the solar wind.The solar wind pushes any remaining gas molecules and dust out of thesolar system leaving only the planets behind.

5.4.11 Solar system

Planet Mass Diameter Orbit radius Orbital Periodx1024kg km 109m eccentricity

Mercury 0.328 4800 57.7 0.206 88 daysVenus 4.83 12310 107.7 0.007 224.3 daysEarth 5.98 12757 149 0.017 365.3 daysMars 0.637 6790 226.9 0.093 687.1 daysJupiter 1900 142850 774.8 0.048 11.9 yearsSaturn 567 115000 1415.5 0.056 29.6 yearsUranus 88 47150 2859.3 0.047 84.0 yearsNeptune 103 44700 4480.3 0.009 164.8 yearsPluto 1.1 6000 5888.5 0.249 248.4 years

Mercury

Venus

Earth Mars

Jupiter Saturn Uranus Neptune Pluto

Sun

Rocky planetsGas giants Ice



Technically speaking, the sun is a gaseous body composed of hydrogen(75%) and helium (25%) with very small amounts of oxygen and otherelements. The surface temperature of the sun is about 5800 oC. The interiortemperature is very much higher than this (≈ 15,000,000 oC).Within this very hot interior,hydrogen nuclei (which arejust single protons) undergonuclear fusion to formhelium nuclei. The actualfusion reaction is complex,but overall, four hydrogennuclei combine to form onehelium nucleus. The mass ofa helium nucleus is less thanthe original four hydrogennuclei, and this difference inmass is converted to energy

The sun is a medium mass star presently within its main sequence. It isestimated to be about 4500 million years old, and it will be at least another2000 million years before the hydrogen fuel supply begins to run out. Themass of the sun is estimated to be approximately 2 × 1030 kg.

H

H

H

H

He

The conversion of 1g of hydrogen intohelium produces about 4 × 1026 W ofenergy. In the sun, approximately 7 ×108 tons of hydrogen are converted toabout 6.95 × 106 tons of helium persecond - the difference is converted toenergy.

γ radiation.

1,392,000 km

Earth

Sun

149 × 106 km

ME = 5.98 × 1024 kg

M= 333 MET= 5800 K

12,757 km

L = 3.90 × 1026 W

5.4.12 Sun



Photosphere is thedisk of the sun

Limb darkening isevidence of thegaseous nature of thesun

Chromosphere iscooler than thephotosphere.

Sunspots are seen as dark spots on the Sun'ssurface (first seen by Galileo in 1610). Theyappear dark by contrast with the brighter andhotter surrounding surface. They consist ofmatter welling up from the interior of the sunbeing affected by the sun’s magnetic field.They can last a few hours or several monthsand can occur in groups on on their own.They usually occur in waves going through amaximum number every 11 years or so.Galileo observed the rotation of sunspotsacross the face of the sun and concluded thatthe sun is rotating once every 26 days.

The Corona is atenuous atmosphereof molecules, atoms,ions. Can usually onlybe seen during a totaleclipse and extendsto about 25,000,000km from the sun.

Solar prominenceconsist of atoms andions ejected fromsunspots and passthrough thechromosphere

Faculae arevery brightareas usuallynear sunspotsand can beseen near thelimb.

Solar flares are usually associatedwith sunspots and faculae. They giverise to bursts of intense ultra violetradiation which cause disturbances inthe ionosphere of the Earth resultingin disruption of telecommunications.Electrically charged particles are alsoejected which interact with the Earth'smagnetic field and causing the auroraborealis. This constant stream ofelectrically charged particles from thesun is called the solar wind.

In the period 1826 and 1843, Heinrich Schwabe discovered the cycle of sunspotactivity in the sun. At the beginning of each cycle, sunspots occur in high latitudeson the sun and reach a maximum number within about 4 years. They then occur atlower and lower latitudes and sometimes die off all together within about 7 years ofreaching the maximum number.

5.4.13 Solar atmosphere



Sunspots look dark because they are at a lower temperature than thesurrounding regions. In fact, the temperature of sunspots is about 4,000 Kand can be up to 50,000 km in diameter. In addition to electromagneticradiation, the Sun also emits a stream of charged particles which we callthe solar wind. These particles travel with a velocity of about 450 km s−1.The solar wind consists of protons, electrons and helium nuclei. The solarwind is a different thing to cosmic rays. Cosmic rays arrive at the Earthfrom all directions and are thought to consist mainly of protons.

The solar wind, consisting of charged particles in motion, createdisturbances in the Earth’s magnetic field. These disturbances areparticularly pronounced during sunspot activity. Due to the shape of theEarth’s magnetic field, charged particles tend to twist in spirals about theEarth’s magnetic lines of force and are drawnto the magnetic poles where they can traveldownwards and cause ionisation of airmolecules which result in theaurora displays.

Nuclearreactionsoccur inthe core ofthe sun

γ radiationtravels outwardsto convectivecells andbeyond

Convective cellsgive rise tosunspots

Solarneutrinos

Solarwind

N

S

5.4.14 Solar wind



As gamma rays travel outwards from the centre of the sun, their energy iscontinuously absorbed and re-emitted at lower wavelengths. By the timethis radiation reaches the surface of the sun, it is mostly visible light.

Not all of the radiation from the sun strikes the Earth’s surface. That whichis not reflected back into space or absorbed in the atmosphere is mostly inthe visible and radio parts of the electromagnetic spectrum. Ultraviolet, Xray and gamma ray radiation is filtered out by the atmosphere.If a solar collector is placed out in the direct sun on the Earth’s surface,the plate receives both direct and diffuse radiant energy. If say 40% is lostdue to atmospheric absorption, and an extra 20% is obtained by diffuseradiation, then the radiation intensity at the Earth’s surface isapproximately 1 kW m−2.Not all frequencies of radiation from the nuclear reactions within the coreof the sun are emitted from the sun due to absorption within the sun itself.Wavelengths of light in the absorption spectrum of hydrogen and heliumand other elements appear as “dark” lines in the overall emission spectrumof the sun. One set of absorption lines could not be identified in 1815 whenthese spectra were first studied prompting the announcement of thediscovery of a new element helium - the “sun element”.

The sun can be regarded as a black body.That is, it emits the maximum amount ofradiation possible over all wavelengths.By the Stefan-Boltzmann law, the rate ofradiant energy emission is given by:

4ATQ σε=D

σ = 56.7 × 10−12 kW m−2 K−4

Stefan-Boltzmann constantε = 1 for a black body.

Assuming the surface area of the sun is 4πr2 where R = 0.7 × 109 m, then,setting T = 5800K we have:

( ) ( )W109.3

5800107.041107.56Q26

42912

×=

×π×=−

If the radius of the Earth’s orbit is 1.49 × 1011 m, then the intensity ofradiation received just outside the atmosphere is:

( )2

211

26

kWm4.1

1049.14

109.3I

−

=

×π

×

=

5.4.15 Solar radiation



Mass 5.98 × 1024 kgRadius 6.38 × 106 mDensity 5520 kg m−3

Escape velocity 11200 m s-1

Distance from Sun 1.49 × 1011 mRotation period 23.93 hoursLength of year 365.26 days

Tilt of axis 23.4o

Orbit eccentricity 0.017Age 4.5 billion years

Layer Depth (km)

Crust 0 – 40Mantle 40 – 2700Core 2700 – 5200Inner Core 5200 – 6371

Crust

MantleOutercore

Core

The core of the Earth is thought to besolid iron or iron and nickel - materialssimilar to that found in meteorites whichfall on the Earth from space. The outercore is thought to be liquid sincetransverse S waves from earthquakes donot appear to travel through it. The innercore is thought to be solid.The mantle is solid rock mainlyconsisting of the minerals olivine andpyroxene. Local melting occurs withinthe mantle which sometimes erupts asvolcanoes at the Earth’s surface

The underlying cause of activity within the crust (the formation of mountains,volcanoes, earthquakes etc) are thought to be a result of movement of plates whichare in constant motion. The plates are in contact with each other at margins and theirmovement creates upwelling of basalt from the mantle along constructive margins(usually on the sea floor) and downwards motion of basalt and piling up of the uppercrust in destructive margins. This movement gives rise to continental drift.

5.4.16 Earth

The crust, although very thin, can bedivided into two zones, the lower ofwhich is about 3–5 km thick andconsists of basalt. The upper zone is thecontinental crust having varyingcomposition mainly comprising granite.



In 1600, William Gilbert, court physician to Queen Elizabeth, proposedthat the Earth is a magnet and that a freely suspended magnet on theEarth’s surface would line up approximately in a north south direction.We now know that the Earth’s magnetic field arises due to electric currentsin the core and its shape is similar to that which would result if a short barmagnet were placed at the centre of the Earth. The direction of themagnetic poles (where the magnetic lines of force are vertical) are about15o from the Earth’s rotational axis. The magnitude of the field at theEarth’s surface is approximately 10−4 T (1 Gauss).The magnetic lines of force are not parallel with the surface of the Earth.The angle which the lines of force make with the surface is called theinclination. At the magnetic poles the inclination is 90o. At the magneticequator, the inclination is zero.

During periods of high sunspot activity, charged particles from the sun (thesolar wind) interact with the Earth’s magnetic field and travel in a spiralingpath according to the right hand rule. At regions of high magnetic fieldstrength, such as at the poles, they are mostly reflected back along the fieldlines and are thus trapped in what are called the Van Allen radiation beltsthus shielding us from potentially ionizing radiation.. Since at the polescharged particles are drawn in towards the Earth’s surface, they ionize airmolecules causing the aurora displays.

N

S

15o

The magneticfield of the Earth

extends outwardsfor about 5 timesthe radius of theEarth. At largerdistances, the

field is distortedby the charged

particlesof the solar

wind.Solarwind

θInclination

S

N

v

B

F

The component ofvelocity which is perpendicularto the field lines causes the particles totravel in a circle. The component of velocityperpendicular to the field lines causes positiveparticles to travel towards the north magneticpole. The overall path is a spiral.

5.4.17 Earth’s magnetic field



%VolNitrogen 78.09Oxygen 20.95Argon 0.93Carbon Dioxide 0.023-0.050Neon 0.0018Helium 0.0005Krypton 0.0001Hydrogen 0.00005Xenon 0.000008Methane 0.00017Nitrous Oxide 0.000003Carbon Monoxide 0.000005& trace amounts of Sulfur Dioxide,Nitrogen Dioxide, Ammonia andOzone

Composition of the atmosphere

The atmosphere is a relatively thin layer of gas which covers the Earth.80% by mass of the atmosphere exists in the lowest 12 km of theatmosphere which extends some 100 km into space.Ozone (O3) is produced in the stratosphere by ultraviolet light acting onoxygen molecules causing them to split and recombine into O3. Ozone isvery reactive and if it were not continually regenerated, would disappearwithin a few months. The continued absorption of ultraviolet radiation inthe stratosphere to produce ozone shields us from its harmful effects.Degeneration of the ozone layer by chloroflourocarbon (CFC) gases is amajor concern. Approximately 34% of the incident radiation on the Earth isreflected back into space by clouds (called the albedo effect). 19% isabsorbed by the atmosphere, and 47% is absorbed at the Earth’s surface.Water vapour, carbon dioxide,methane, nitrous oxides, and CFC’s aretransparent to visible electromagneticradiation but absorb infrared radiationbeing emitted from the Earth’s surface.The presence of these gases in theatmosphere causes the surface of theEarth to be warmer than it otherwisewould be. To some extent, thisgreenhouse effect is a naturalphenomenon. However, human activityduring the past 100–200 years hasincreased the level of greenhouse gasesto the extent that global warming is ofsignificant concern.

The ionosphere is created byionisation of air molecules byradiation from the sun. Bothionised atoms and free electronsare produced, both of whichserve to reflect radio waves ofcertain frequencies allowing longdistance radio communications.

Mesosphere

Stratosphere

Troposphere10203040506070

Mt Everest

Ozone layer

Ionosphere

300 kmMeteors

5.4.18 Atmosphere



5.4.19 Examples

2. Calculate the mass of the Earth given the radius of the Earth to be6.38 × 106 m and acceleration due to gravity as 9.81 m s−2.

( )

kg1098.510673.61038.68.9

GgRm

gmR

mmGF

24

11

26

2E

E

12E

E1

×=

×

×

=

=

==

−

Solution:

1. The planet Pluto orbits the sun at a mean distance of 5.9 × 1012m.Determine the period of Pluto’s orbit TP if the Earth takes one yearto orbit the sun at a radius of 1.49 × 10 11 m.

Solution:Kepler’s third law states that the square of theperiod of the orbit of a planet is proportional tothe cube of planet’s distance from the sun, sowe can use information about the Earth’s orbitto calculate the period of any other planet ifthe radius of the orbit is known.

( ) ( )yr249T

109.5

T

1049.1

1

p

312

2P

311

2

=

×

=

×

This calculation was first done byCavendish using a value of G that hemeasured between two known masses inhis laboratory.



LVwater = 22.56 × 105 J kg−−−−1

LFwater = 3.33 × 105 J kg−−−−1

αAl = 7.2 × 10−−−−5 oC−1

αsteel = 1.2 × 10−−−−5 oC−1

αconcrete = 10 × 10−−−−5 oC−1

αbrass = 20 × 10−6 oC−1

αCu = 16.5 × 10−6 oC−1

αglass = 7.75 × 10-6 oC−1

cwater = 4186 J kg−−−−1 K−−−−1

cFe = 540 J kg−−−−1 K−−−−1

cAl = 920 J kg−−−−1 K−−−−1

csteel = 450 J kg−−−−1 K−−−−1

cice = 2300 J kg−−−−1 K−−−−1

coil = 2100 J kg−−−−1 K−−−−1

ksteel = 50.2 W m-1 K-1

kAl = 205 W m-1 K-1

kCu = 385 W m-1 K-1

kplaster = 0.12 W m-1 K-1

kfibreglass = 0.039 W m-1 K-1

kair = 0.034 W m-1 K-1

kbrick = 1.33 W m-1 K-1

kglass = 0.79 W m−1 K−1

kwater = 0.65 W m−1 K−1

Mair = 0.02892 kg mol−1

MO2 = 0.032 kg mol−1

MH = 0.001008 kg mol−1

cpAir = 1.005 kJ kg−1 K−1

cvAir = 0.718 kJ kg−1 K−1

γair = 1.4patm = 101.3 kPa

Materials data

Io = 1× 10−12 W m−2

ρAg = 1.64 × 10−8 Ω m ρCu = 1.72 × 10−8 Ω mρAl = 2.83 × 10−8 Ω mρW = 5.5 × 10−8 Ω m

Esteel = 210 GPaEAl = 70 GPaEglass = 70 GPaEwood = 14 GPaEFe = 1.9 × 1011 PaEbrass = 100 GPaECu = 1.1 × 1011 PaBAl = 7.46 × 1010 PaBwater = 2.04 × 109 Pa

ρoil = 915 kg m−3

ρFe = 7800 kg m−3

ρwater = 1000 kg m−3

ρair = 1.2 kg m−3

ρHg = 13600 kg m−3

ρPb = 11300 kg m−3

ρglycerine = 1300 kg m−3

ρsteel = 7700 kg m−3

χO = 1.9 × 10−6 χAl = 2.2 × 10−5

χPt = 2.6 × 10−4

mE = 5.98 × 1024 kgRE = 6.38 × 106 mRMOrbit = 3.81 × 108 mREOrbit = 1.49 × 1011 mµr water = 0.999991εr water = 81µr air = 1.00000036εr air = 1.0006εr glass = 6

σAg = 6.2 × 107 S m−1 σAl = 3.7 × 107 S m−1 σwater = 1 × 10−3 S m−1

σsalt water = 4 S m−1

σglass = 1 × 10 −12 S m−1

2




Useful information

Speed of light c 2.99793 × 108 m s −1

Charge on an electron qe 1.6022 × 10-19 CBoltzmann’s constant k 1.38062 × 10-23 J K−1

Stefan-Boltzmann constant s 5.6697 × 10-8 W m−2 K−4

Planck’s constant h 6.626 × 10-34 J sRydberg constant R 1.0973731 × 107 m−1

Avogadro’s number NA 6.02217 × 1023

Universal gas constant R 8.3143 J mol−1 K−1

Mass of an electron me 9.1096 × 10-31 kgMass of a proton mp 1.673× 10-27 kgMass of a neutron mn 1.675× 10-27 kgAtomic mass unit amu 1.6602× 10-27 kgCoulomb force constant k 9 × 109 m2 C−2

Permittivity of free space εo 8.85 × 10−12 F m−1

Permeability of free space µo 4π × 10-7 Wb A−1 m−1

Acceleration due to gravity g 9.81 m s−2

Standard atmospheric pressure patm 101.325 kPaAbsolute zero −273.15oCGravitational constant G 6.673 × 10–11 N m2 kg–2

Electron volt eV 1.6022 × 10-19 JMass of the earth mE 5.983 × 1024 kgRadius of the earth rE 6.371 × 106 mMass of the sun mS 1.999 × 1030 kg

One atomic mass unit (amu) is 1/12th the mass of a carbon 12 atom byinternational agreement. The atomic weight of C12 is 12 amu and 1 amu is1.6602 × 10-27 kg.

The total sum of the atomic weights for a molecule of substance is called themolecular weight m.w. The molecular weight of water, H2O, is 16 + 2 = 18 amu.

The molecular weight expressed in grams contains one mole of molecules (ormore precisely, one gram-mole). 12 grams of C12 contains 6.02 × 1023 atoms. 18grams of H2O also contains 6.02 × 1023 molecules. This is Avogadro’s number.

The molar mass Mm is the mass of one mole of molecules (or atoms) expressed inkilograms. The molar mass of H2O is 0.018 kg.


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