11

11STST CHAPTER CHAPTERSpecial theory of Special theory of

relativityrelativity

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ZCT 104/3EZCT 104/3E

ENGLISH TEACHINGENGLISH TEACHING 5 CHAPTERS WILL BE COVERED5 CHAPTERS WILL BE COVERED

1.1. Special theory of relativitySpecial theory of relativity

2.2. Wave nature of particlesWave nature of particles

3.3. Particle nature of wavesParticle nature of waves

4.4. Introduction to Quantum MechanicsIntroduction to Quantum Mechanics

5.5. Atomic modelsAtomic models

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LECTURE 1LECTURE 1Failure of Newtonian mechanicsFailure of Newtonian mechanics

Newton

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Revision: Revision: Still remember Still remember Newton's Newton's 3 3 lawlaw of motion of motion??

1.1. An object at rest will always be in the An object at rest will always be in the wrong place wrong place

2.2. An object in motion will always be An object in motion will always be headed in the wrongheaded in the wrong direction direction

3.3. For every action, there is an equal For every action, there is an equal and opposite criticismand opposite criticism

=b, just joking=b, just joking

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Newtonian view of Newtonian view of space and timespace and time

Space and time are absoluteSpace and time are absoluteTime flow independently of the state of Time flow independently of the state of motion of any physical system in 3-D spacemotion of any physical system in 3-D spaceIn essence, time and space do not mix. The In essence, time and space do not mix. The state of motion of a physical system does not state of motion of a physical system does not affact the rate of time flow within the systemaffact the rate of time flow within the system

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Inertial framesInertial frames

Inertial frames of reference is one in which Inertial frames of reference is one in which an object subject to no forcess moves in an object subject to no forcess moves in straight line at constant speedstraight line at constant speed

E.g. of inertial frames: the lab frame and E.g. of inertial frames: the lab frame and the constant-speed car framethe constant-speed car frame

Newtonian law of invariance (or called Newtonian law of invariance (or called principle of Newtonian relativity): All principle of Newtonian relativity): All inertial frames are equivalent, and the law inertial frames are equivalent, and the law of of mechanicsmechanics must be also take the same must be also take the same methematical form for all observers methematical form for all observers irrespective of their frame of referencesirrespective of their frame of references

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Example of inertial frames of reference

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Example of form invarianceExample of form invariance

In the aeroplane with constant speed In the aeroplane with constant speed wrp to the ground, Newton second wrp to the ground, Newton second law takes the form of law takes the form of F’ = m a’F’ = m a’

In the lab frame, Newton 2In the lab frame, Newton 2ndnd law is law is F F = ma= ma

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The laws of mechanics must be The laws of mechanics must be the same in all inertial frames the same in all inertial frames

of referenceof reference Although the ball Although the ball

path is different in path is different in both inertial both inertial reference frames, reference frames, both observers both observers agree on the validity agree on the validity of Newton’s law, of Newton’s law, conservation of conservation of energy and others energy and others physical principlesphysical principles

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Galilean Galilean transformationtransformation It relates the kinematical It relates the kinematical

quantities, such as position, quantities, such as position, velocity, acceleration velocity, acceleration between two inertial framesbetween two inertial frames

S: stationary frame (uses x,y,z,t as their coordinates)

S’: moving wrp to S with constant speed u away from S (uses x’,y’,z’,t’ as their coordinates)

Galilean transformation for the coordinates (in 1-D):

x’ = x – vt, y’ = y, z’=z, t’ = t Galilean addition law for

velocity (in 1-D): v’x = vx - v Simply a daily experience

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Galilean transformation and Galilean transformation and Newtonian view goes hand-in-Newtonian view goes hand-in-

handhand Galilean transformation assumes the Galilean transformation assumes the

notion of absolute space and time as notion of absolute space and time as hold by Newton, i.e. the length is hold by Newton, i.e. the length is independent of the state of motion. independent of the state of motion. So is the flowing rate of time.So is the flowing rate of time.

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ExampleExample Apply GT on th previous Apply GT on th previous

exampleexample The trajactories of the ball The trajactories of the ball

seen in the two frames, S’ seen in the two frames, S’ (van) and S (ground (van) and S (ground observer) is related by GT as observer) is related by GT as perper

20

2 20 0

LT : ' ; ' ; '

1ˆ ˆ ˆ ˆ( ) ( )

2

(parabola trajactory of projectile seen in S)

ˆ ˆ ˆ ˆ' '( ') ' '( ') ' ( ) ' ( ) '

1 1ˆ ˆ ˆ[ ] ' '

2 2

x y

x

x x y y

t t y y x x vt

r x t x y t y u t x v t gt y

r x t x y t y x t u t x y t y

u t u t x v t gt y v t gt

'

(vertical straight line seen in S')

y

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ExampleExample Using Galilean transformation of corrdinates, one can show Using Galilean transformation of corrdinates, one can show

that observer in S and S’ measure different cordinates for the that observer in S and S’ measure different cordinates for the ends of a stick at rest in S, they agree on the length of the ends of a stick at rest in S, they agree on the length of the stick. Assume the stick has end coodintaes stick. Assume the stick has end coodintaes x =a x =a and and x = a + l x = a + l in S. in S.

Doraemiyan (S’) Doraemiyan (S’) measure the end measure the end points of the stick at points of the stick at the same time, t’.the same time, t’.

Using Using xx’ = ’ = xx – – vt’vt’:: xx’(head) = ’(head) = xx(head) – (head) –

vt’ = a - vt’vt’ = a - vt’; ; xx’(end) = ’(end) = xx(end) – (end) – vt’ = vt’ =

((a+la+l)) - vt’ - vt’xx’(end) - ’(end) - xx’(head) = ’(head) =

xx(end) - (end) - xx(head) = (head) = llx(h) =a x(t)= l +a

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Galilean transformation when Galilean transformation when applied on light means speed applied on light means speed

of light is not constantof light is not constant

Frame S’ travel with velocity v relative to S. If light travels with the same speed in all directions relative too S, then (according to the classical Galilean velocity-addition) it should have different speeds as seen from S’.

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Maxwell theory of light is Maxwell theory of light is inconsistent with Newton’s law inconsistent with Newton’s law

of invarianceof invariance Consider a gadanken case: in an inertial frame

moving at the speed of light, the electromagnetic (EM) wave is ``frozen’’ and not waving anymore

Maxwell theory of EM wave will fail in the light-speed frame of reference

Galilean transformation is inconsistent with Maxwell theory of light

Newtonian law of invariance fails for EM in the light-speed frame

Galilean transformation is going to fail when v is approaching the speed of light – it has to be supplanted by Lorentz transformation (to be learned later)

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Ether and Michelson-Morley Ether and Michelson-Morley ExperimentsExperiments

In early 19In early 19thth century, it’s thought century, it’s thought (incorrectly) that(incorrectly) that

there exist an omi-pervasive medium called there exist an omi-pervasive medium called Ether in which light propagates at a speed Ether in which light propagates at a speed of 3x10of 3x1088m/s (analogue to sound propagate m/s (analogue to sound propagate in the mechanical medium of still air at in the mechanical medium of still air at speed 330m/s)speed 330m/s)

Thought to be the `absolute frame of Thought to be the `absolute frame of reference’ that goes in accordance with reference’ that goes in accordance with Newtonian view of absolute space and timeNewtonian view of absolute space and time

The effect of the ehter on speed of light can The effect of the ehter on speed of light can be experimentally measured be experimentally measured

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Ether and Michelson-Morley Ether and Michelson-Morley ExperimentsExperiments

If exists, from the viewpoint of the light If exists, from the viewpoint of the light source, the Ether wind appears to `drift’ with source, the Ether wind appears to `drift’ with a relative speed of a relative speed of u u wrp to Earth (One wrp to Earth (One assumes that ether frame is fixed wrp to the assumes that ether frame is fixed wrp to the Sun, hence one expects Sun, hence one expects u u ≈≈ 1010-4-4cc))

Consider a moving souce giving out two Consider a moving souce giving out two beams of light in different direction (say, 90 beams of light in different direction (say, 90 degree to each other)degree to each other)

Since the light source is moving through the Since the light source is moving through the omi-perasive ehter medium, the different omi-perasive ehter medium, the different directions of the two beams of light would directions of the two beams of light would mean that these two beams will move with mean that these two beams will move with different velocities when viewed in the different velocities when viewed in the frame of moving sourceframe of moving source

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Experimental setupExperimental setup Both arms has same Both arms has same

length length LL According to the ether According to the ether

wind concept:wind concept: For arm 1, the speed of For arm 1, the speed of

light to is light to is c-vc-v as it as it approaches M2, approaches M2,

c+vc+v as it is reflected from as it is reflected from M2M2

c’ = c - v

v

c’ = c + v

c v

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Experimental setupExperimental setup For arm 2, the speed of For arm 2, the speed of

light to-and-fro M1 is light to-and-fro M1 is

v

cc’= √ (c2 - v2)

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The two light beams start out in phase. When The two light beams start out in phase. When they return and “recombined” by semi-they return and “recombined” by semi-transparent mirror Mtransparent mirror Mo o interference pattern will interference pattern will be formed due to their difference in phase, be formed due to their difference in phase, cct /t /, where , where tt = = tt11 - - tt22≈≈LvLv22//cc3 3 is the is the time difference between the light beams when time difference between the light beams when return to Mreturn to Moo (figure a) (figure a)

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Now, when the whole set-up is rotated Now, when the whole set-up is rotated through 90through 90, arms 1 and 2 exchanges role, arms 1 and 2 exchanges role

As a result, the interefence pattern will As a result, the interefence pattern will be shifted as the time difference between be shifted as the time difference between the beams after rotation now becomes the beams after rotation now becomes 22tt

The number of interefence fringes shifted The number of interefence fringes shifted can be estimated via: can be estimated via:

no. of fringe shift = 2no. of fringe shift = 2cctt 22 LvLv22//cc2 2

≈≈0.40 0.40 (taking (taking v v ≈≈ 1010-4-4cc, , L = L = 11 m, 11 m, = = 500nm)500nm) Very precise experimentVery precise experiment

2222

Fig. (b) shows Fig. (b) shows expected fringe shift expected fringe shift after a rotation of the after a rotation of the interometer by 90 interometer by 90 degreedegree

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But MM sees only NULL result – no change But MM sees only NULL result – no change in the interference patternin the interference pattern

How to interprete the null result? How to interprete the null result? If Maxwell theory of light is right (as EM If Maxwell theory of light is right (as EM

wave) the notion of ether as an medium in wave) the notion of ether as an medium in which light is propagating has to be which light is propagating has to be discardeddiscarded

Put simply: ether is not shown to existPut simply: ether is not shown to exist Einstein put it more strongly: the absolute Einstein put it more strongly: the absolute

frame of reference (i.e. the ether frame) frame of reference (i.e. the ether frame) has to be discardedhas to be discarded

NULL resultNULL result

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PYQ (past year question), KSCP PYQ (past year question), KSCP 2003/042003/04

What were the consequences of the negative result of What were the consequences of the negative result of the Michelson-Morley experiment?the Michelson-Morley experiment?

I. It render untenable the hypothesis of the ether I. It render untenable the hypothesis of the ether II. It suggests the speed of light in the free space is the II. It suggests the speed of light in the free space is the

same everywhere, regardless of any motion of same everywhere, regardless of any motion of source or observersource or observer

III. It implies the existence of a unique frame of reference III. It implies the existence of a unique frame of reference in which the speed of light in this frame is equal to in which the speed of light in this frame is equal to cc

A.A. IIIIII only only B.B. I,III,II C. I, IIIC. I, III D.D. II, , II, IIIII, IIIE. E. Non of the above Non of the above

Ans: BAns: B

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Principle of special relativityPrinciple of special relativity

Einstein believes that pure Einstein believes that pure thought is sufficient to thought is sufficient to understand the worldunderstand the world

The most incomprehensive The most incomprehensive thing in the universe is that thing in the universe is that the universe is the universe is comprehensiblecomprehensible

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Classical EM theory is Classical EM theory is inconsistent with Galileao inconsistent with Galileao

transformation transformation

Their is inconsistency between EM Their is inconsistency between EM and Newtonian view of absolute space and Newtonian view of absolute space and timeand time

Einstein proposed SR to restore the Einstein proposed SR to restore the inconsistency between the two based inconsistency between the two based on two postulates:on two postulates:

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Postulates of SRPostulates of SR

1.1. The laws of physics are the same in The laws of physics are the same in all inertial reference frames – a all inertial reference frames – a generalisation of Newton’s relativitygeneralisation of Newton’s relativity

2.2. The speed of light in vacuum is the The speed of light in vacuum is the same for all observers independent of same for all observers independent of the motion of the source – constancy the motion of the source – constancy of the speed of lightof the speed of light

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Postulate 2 simply means that Postulate 2 simply means that Galilean transformation cannot be Galilean transformation cannot be applied on light speed. It also explains applied on light speed. It also explains the Null result of the MM experimentthe Null result of the MM experiment

Speed of light is always the same Speed of light is always the same whether one is moving or stationary whether one is moving or stationary wrp to the source – its speed doesn’t wrp to the source – its speed doesn’t increase or reduced when the light increase or reduced when the light source is movingsource is moving

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The notion of absolute frame of reference is discardedThe Newton notion that time is absolute and flows independently of the state of motion (or the frame of reference chosen) is radically modified – the rate of time flow does depends on the frame of reference (or equivalently, the state of motion). This being so due to the logical consequence of the constancy of the speed of light in all inertial frame

Einstein’s notion of space-time drastically Einstein’s notion of space-time drastically revolutionarizes that of Newton’srevolutionarizes that of Newton’s

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PYQ (past year question), Final PYQ (past year question), Final 2003/042003/04

Which of the following statement(s) is (are) Which of the following statement(s) is (are) true?true?II.. The assumption of the Ether The assumption of the Ether frame is inconsistent with the experimental frame is inconsistent with the experimental observationobservationII.II. The speed of light is constantThe speed of light is constantIIIIII.. Maxwell theory of Maxwell theory of electromagnetic radiation is inconsistent electromagnetic radiation is inconsistent with the notion of the Ether framewith the notion of the Ether frameIVIV Special relativity is inconsistent Special relativity is inconsistent with the notion of the Ether framewith the notion of the Ether frame

A.A. III,IV III,IV B.B. I, II, III I, II, III C. C. I, II, III,IVI, II, III,IVD.D. I, II I, II E. E. I, II,IVI, II,IV ANS: E, my own questionANS: E, my own question

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Simultaneity is not an absolute Simultaneity is not an absolute concept but frame dependentconcept but frame dependent

Simultaneity in one frame is not guaranteed in another frame of reference (due to postulate 2)

Two lightning bolts strike the ends of a moving Two lightning bolts strike the ends of a moving boxcar. (a) The events appear to be boxcar. (a) The events appear to be simultaneous to the stationary observer at O but simultaneous to the stationary observer at O but (b) for the observer at O’, the front of the train is (b) for the observer at O’, the front of the train is struck before the rearstruck before the rear

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Try to calculate it yourselfTry to calculate it yourself

The breakdown of simultaneity means that the two lights from A’ and B’ are not arriving at O’ at the same time. Can you calculate what is the time lag, i.e. tA-tB, between the two lights arriving at O’? t is the time measured in the O frame.

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Time dilation as a Time dilation as a consequence of Einstein’s consequence of Einstein’s

postulatepostulate In frames that are moving wrp to In frames that are moving wrp to

the stationary frame, time runs the stationary frame, time runs slowerslower

Gedanken experiment (thought Gedanken experiment (thought experiment)experiment)

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Gedanken ExperimentGedanken Experiment Since light speed Since light speed cc is invariant is invariant

(i.e. the same in all frames), it is (i.e. the same in all frames), it is used to measure time and spaceused to measure time and space

A mirror is fixed to a moving A mirror is fixed to a moving vehicle, and a light pulse leaves vehicle, and a light pulse leaves O’ at rest in the vehicle. O’ at rest in the vehicle.

(b) Relative to a stationery (b) Relative to a stationery observe on Earth, the mirror and observe on Earth, the mirror and O’ move with a speed O’ move with a speed vv. .

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Ligth triangleLigth triangle

Consider the Consider the geometry of the geometry of the triangle of the lighttriangle of the light

We can calculate the We can calculate the relationship between relationship between tt, , tt’ and ’ and vv

l l 22 = = ((cct/2 t/2 ))22

= d = d 22 ++ ((uu t/2 t/2 ))22

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Due to constancy of light postulate, both Due to constancy of light postulate, both observer must agree on observer must agree on cc::

Speed of light = total distance travelled divide Speed of light = total distance travelled divide by time takenby time taken

For observer in O’, For observer in O’, cc = = 2 d 2 d //t’t’ For observer in O, For observer in O, cc = = 2 l / 2 l / t,t, where where

l l 22 = d = d 22 ++ ((uu t/2 t/2 ))22

Eliminating Eliminating ll and and d, d, t= t= t’t’,, where where

uu22/c/c22

Lorentz factor, always > or equal 1, so that Lorentz factor, always > or equal 1, so that t t > = > = t’t’

Lorentz factor, Lorentz factor,

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Proper timeProper time

Try to discriminate between two kinds of Try to discriminate between two kinds of time interval: time interval:

t’ ,t’ , proper time that measures the time proper time that measures the time interval of the two events at the same point interval of the two events at the same point in space (e.g. light emitted and received at in space (e.g. light emitted and received at the same point in the vehicle)the same point in the vehicle)

Proper time is the time measured by a clock Proper time is the time measured by a clock that is stationary wrp to the events that it that is stationary wrp to the events that it measuresmeasures

Note that proper time is always ``shorter’’ Note that proper time is always ``shorter’’ compared to improper timecompared to improper time

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The elapsed time The elapsed time t t between the same events between the same events in any other frame is dilated by a factor of in any other frame is dilated by a factor of compared to the proper time interval compared to the proper time interval t’t’

In other words, according to a stationary In other words, according to a stationary observer, a moving clock runs slower than an observer, a moving clock runs slower than an identical stationary clockidentical stationary clock

Chinese proverb:Chinese proverb: 1 day in the heaven = 10 years in the 1 day in the heaven = 10 years in the

human plane human plane 天上方一日，人间已十年天上方一日，人间已十年

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ExampleExample

The watch of a student in the class is running at a The watch of a student in the class is running at a rate different than that of a student ponteng class rate different than that of a student ponteng class to lumba motosikal haram. to lumba motosikal haram. The time of the student on the bike’s is running at The time of the student on the bike’s is running at a slower rate compared to that of the student in a slower rate compared to that of the student in the classthe classOnc can imagine that when the watch on the Onc can imagine that when the watch on the arms of the motocyclist ticks once in a second (as arms of the motocyclist ticks once in a second (as is concluded by the local, or rest, observer, i,e, is concluded by the local, or rest, observer, i,e, the motocyclist), the student in the class (non-the motocyclist), the student in the class (non-local observer) find the watch of the motocyclist local observer) find the watch of the motocyclist ticks at 1.000001 second per second.ticks at 1.000001 second per second.

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To recapTo recap

t= t= t’ t’ ;; proper time interval, proper time interval, tt’ ’ ≤≤ tt The rate of time flowing in one frame is different The rate of time flowing in one frame is different

from the others (frames that are moving with a from the others (frames that are moving with a constant speed relative to a give frame)constant speed relative to a give frame)

The relationship between the time intervals of the The relationship between the time intervals of the two frames moving at an non-zero relatively two frames moving at an non-zero relatively velocity are given by the time dilation formulavelocity are given by the time dilation formula

One must be aware of the subtle different between One must be aware of the subtle different between which is the proper time and which is the improper which is the proper time and which is the improper oneone

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ExampleExample

When you are measuring the time interval When you are measuring the time interval between your heartbeats (on your bed in between your heartbeats (on your bed in you bedroom) using your watch, you are you bedroom) using your watch, you are measuring the measuring the proper time intervalproper time interval

Say a doctor who is in a car traveling at Say a doctor who is in a car traveling at some constant speed with recpect to you is some constant speed with recpect to you is monitoring your heartbeat by some remote monitoring your heartbeat by some remote device. The time interval between the device. The time interval between the heartbeat measured by him, is improper heartbeat measured by him, is improper time because he is moving wrp to youtime because he is moving wrp to you

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PYQ, Semester Test I, PYQ, Semester Test I, 2003/042003/04

Suppose that you are travelling on board Suppose that you are travelling on board a spacecraft that is moving with respect a spacecraft that is moving with respect to the Earth at a speed of 0.975to the Earth at a speed of 0.975cc. You . You are breathing at a rate of 8.0 breaths per are breathing at a rate of 8.0 breaths per minute. As monitored on Earth, what is minute. As monitored on Earth, what is your breathing rate?your breathing rate?

A. A. 13.313.3 B.B. 2.88 2.88 C.C.22.222.2 D.D. 1.77 1.77 E.E. Non of the above Non of the above

ANS: D, Cutnell, Q4, pg. 877ANS: D, Cutnell, Q4, pg. 877

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SolutionSolution Suppose that you are travelling on board a spacecraft that is Suppose that you are travelling on board a spacecraft that is

moving with respect to the Earth at a speed of 0.975moving with respect to the Earth at a speed of 0.975cc. You . You are breathing at a rate of 8.0 breaths per minute. As are breathing at a rate of 8.0 breaths per minute. As monitored on Earth, what is your breathing rate?monitored on Earth, what is your breathing rate?

= 1/(1 – = 1/(1 – uu22//cc22))1/21/2 = 1/(1 – 0.975 = 1/(1 – 0.97522))1/21/2 = 4.5 = 4.5 Use Use t = t = t’ t’ Given local interval between breaths Given local interval between breaths t’ = t’ = 1/8 = 1/8 =

0.125 min per breath (proper time interval)0.125 min per breath (proper time interval) t = t = t t ==4.5 x 4.5 x .125 = 0.563 min per .125 = 0.563 min per

breathbreath t’ = t’ = 1.77 breath per min (as seen by the 1.77 breath per min (as seen by the

spaccraft observer)spaccraft observer) To an oberver on the spacecraft, you seem to To an oberver on the spacecraft, you seem to

breath at a slower ratebreath at a slower rate

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Example (read it yourself)Example (read it yourself)

A spacecraft is moving past the Earth A spacecraft is moving past the Earth at a constantat a constant speedspeed 0.920.92cc. The . The astronaut measures the time interval astronaut measures the time interval betweenbetween successive ``ticks'' of the successive ``ticks'' of the spacecraft clock to be 1spacecraft clock to be 1.0.0 s. What is s. What is the time interval that an Earth the time interval that an Earth observerobserver measures between ``ticks'' measures between ``ticks'' of the astronaut's clock? of the astronaut's clock?

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SolutionSolution

t’ = t’ = 1.0 s is the proper time interval 1.0 s is the proper time interval measured by the astronautmeasured by the astronaut

Earth observer measures a greater Earth observer measures a greater time interval, time interval, t,t, than does the than does the astronaut, who is at rest relative to astronaut, who is at rest relative to the clockthe clock

The Lorentz factor The Lorentz factor uu22//cc22-1/2-1/2 = = 0.920.9222-1/2-1/2 = 2.6 = 2.6

Hence, Hence, tt = = t’ = t’ = 2.6 x 1.0s = 2.6 s2.6 x 1.0s = 2.6 s

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Example: Muon decay Example: Muon decay lifetimelifetimeA muon is an unstable elementaryA muon is an unstable elementary particle whichparticle which has has

a lifetime a lifetime 00 = 2.2= 2.2 microsecondmicrosecond ( (proper time, proper time, measuredmeasured in the muon rest frame) and decays into in the muon rest frame) and decays into lighter particles.lighter particles.

Fast muons (say, travelling at Fast muons (say, travelling at v = v = 99%99%cc) are created ) are created in the interactions of very high-energy particlesin the interactions of very high-energy particles as as they enter the Earth's upper atmosphere.they enter the Earth's upper atmosphere. Assume Assume v = v =

0.990.99cc In the muon rest In the muon rest

frame, the frame, the distance travelled distance travelled by muon before by muon before decay is decay is

D’ = D’ = (0.99(0.99cc))00

= = 650 m650 m

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A muon travelling at 99% the speed of lightA muon travelling at 99% the speed of light. . has ahas a Lorentz factor Lorentz factor = 7.09 = 7.09 Hence, to an observer in the rest frame (e.g Hence, to an observer in the rest frame (e.g

Earth)Earth) the lifetime of the muon is no longer the lifetime of the muon is no longer 00 = 2.2= 2.2 s but s but

x x = = 7.09 x 2 microseconds = 15.6 7.09 x 2 microseconds = 15.6 ss

Thus the muon would appear to travel for Thus the muon would appear to travel for 15.6 15.6 microsecondmicrosecondss before it decaysbefore it decays

The distance it traversed as seen from EarthThe distance it traversed as seen from Earthis is DD = (0.99= (0.99cc) ) x x 15.6 15.6 s = 4,630 km (c.f. s = 4,630 km (c.f. D’ D’ = = 650 m650 m ))

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Muon are detected at a much Muon are detected at a much lower altitude lower altitude

Observation has Observation has verified the relativistic verified the relativistic effect of time dilation – effect of time dilation – muons are detected at muons are detected at a distance of 4700 m a distance of 4700 m below the atmospheric below the atmospheric level in which they are level in which they are producedproduced

Hence the dilated Hence the dilated muon lifetime is muon lifetime is confirmed confirmed experimentallyexperimentally

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5050

Length contractionLength contraction Length measured differs from frame to frame – Length measured differs from frame to frame –

another consequence of relativistic effectanother consequence of relativistic effect Gedanken experiment again!Gedanken experiment again!

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Two observers: O on Earth, O’ traveling to and Two observers: O on Earth, O’ traveling to and fro from Earth and alpha centauri with speed fro from Earth and alpha centauri with speed uu

Total distance between Earth - alpha centauri Total distance between Earth - alpha centauri – Earth, according to O (Earth observer), = – Earth, according to O (Earth observer), = LL00

O sees O’ return to Earth after O sees O’ return to Earth after tt00

Observer O’ in a spaceship is heading Observer O’ in a spaceship is heading C with C with speed speed uu and returns to Earth after and returns to Earth after tt’ ’ according to his clockaccording to his clock

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Use some simple logics…Use some simple logics…

In O: 2In O: 2LL00 = u = utt00

In O’: 2In O’: 2LL0’0’ = = uutt0’0’

Due to time dilation effect, Due to time dilation effect, tt0’ 0’ is is shorter than shorter than tt0 0 , i.e. , i.e. tt0 0 > > tt0’ 0’

tt0 0 is related to is related to tt0’0’ via a time dilation via a time dilation effect, effect, tt0’0’ = = tt0 0 // , hence , hence

LL0’ 0’ / / LL0 0 = = tt0’0’ //tt00 = 1 / = 1 / , or , or

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LL0’0’ = = LL0 0 / / LL00 is defined as the proper length = length of is defined as the proper length = length of

object measured in the frame in which the object measured in the frame in which the object (in this case, the distance btw Earth object (in this case, the distance btw Earth and and C) is at restC) is at rest

LL0’0’ is the length measured in the O’ frame, is the length measured in the O’ frame, which is moving wrp to the “object” – here which is moving wrp to the “object” – here refer to the distance between E- refer to the distance between E- CC

The length of a moving objecte is measured to The length of a moving objecte is measured to be shorter than the proper length – length be shorter than the proper length – length contractioncontraction

5454

If an observer at rest wrp to an If an observer at rest wrp to an object measures its length to be object measures its length to be LL0 0 , , an observer moving with a relative an observer moving with a relative speed speed uu wrp to the object will find wrp to the object will find the object to be shorter than its rest the object to be shorter than its rest length by a foctor 1 / length by a foctor 1 /

5555

A stick moves to the A stick moves to the right with a speed right with a speed vv (as (as seen in a rest frame, O)seen in a rest frame, O)

(a) The stick as viewed (a) The stick as viewed by a frame attached to by a frame attached to it (O’ frame, it (O’ frame, LLpp = proper = proper length)length)

(b) The stick as seen by (b) The stick as seen by an observer in a frame an observer in a frame O. The length measured O. The length measured in the O frame (in the O frame (LL) is ) is shorter than the proper shorter than the proper length by a factor 1/ length by a factor 1/

Example of moving rulerExample of moving ruler

5656

Length contraction only Length contraction only happens along the direction of happens along the direction of

motionmotionExample: A spaceship in the form of a triangle Example: A spaceship in the form of a triangle flies by an oberver at rest wrp to the ship (see fig flies by an oberver at rest wrp to the ship (see fig (a)), the distance (a)), the distance xx and and yy are found to be 50.0 m are found to be 50.0 m and 25.0 m respectively. What is the shape of the and 25.0 m respectively. What is the shape of the ship as seen by an observer who sees the ship in ship as seen by an observer who sees the ship in motion along the direction shown in fig (b)?motion along the direction shown in fig (b)?

5757

SolutionSolution The observer sees the horizontal length of the The observer sees the horizontal length of the

ship to be contracted to a length of ship to be contracted to a length of L = LL = Lpp//= = 50 m50 m√√(1 – 0.950(1 – 0.95022) = 15.6 m) = 15.6 m The 25 m vertical height is unchanged The 25 m vertical height is unchanged

because it is perpendicular to the direction of because it is perpendicular to the direction of relative motion between the observer and the relative motion between the observer and the spaceship. spaceship.

5858

An observer on Earth sees a spaceship An observer on Earth sees a spaceship at an altitude of 435 moving downward at an altitude of 435 moving downward toward the Earth with a speed of 0.97toward the Earth with a speed of 0.97cc. . What is the altitude of the spaceship as What is the altitude of the spaceship as measured by an observer in the measured by an observer in the spaceship?spaceship?

ExampleExample

5959

Draw the diagram yourselfDraw the diagram yourself

As a useful strategy to solve physics As a useful strategy to solve physics problem you should always try to problem you should always try to translate the problems from text into translate the problems from text into diagramatical form with all the diagramatical form with all the correct labellingcorrect labelling

6060

SolutionSolution

One can consider the altitude see by theOne can consider the altitude see by thestationary (Earth) observer as the stationary (Earth) observer as the proper length (say, L'). The observer in proper length (say, L'). The observer in the spaceship should sees a contracted the spaceship should sees a contracted length, L, as compared to the proper length, L, as compared to the proper length. Hence the moving observer in length. Hence the moving observer in the ship finds the altitude to be the ship finds the altitude to be

L = L' / L = L' / = 435 m x [1- (0.97) = 435 m x [1- (0.97)22]]-1/2-1/2 = 106 = 106 mm

6161

PYQ, KSCP 03/04PYQ, KSCP 03/04 How fast does a rocket have to How fast does a rocket have to

go for its length to be contracted go for its length to be contracted to 99% of its rest length?to 99% of its rest length?

Ans: Ans: Rest length = proper length Rest length = proper length = = LLPP = length of the rocket as = length of the rocket as seen by observer on the rocket seen by observer on the rocket itself itself

L L improper length = length of the improper length = length of the rocket as seen from Earth rocket as seen from Earth oberveroberver

Always remember that proper Always remember that proper length is longer than improper length is longer than improper lengthlength 2

0.99 1

0.141c

p

L v

L c

v

6262

Lorentz TransformationLorentz Transformation All inertial frames are equivalentAll inertial frames are equivalent Hence all physical processes Hence all physical processes

analysed in one frame can also be analysed in one frame can also be analysed in other inertial frame and analysed in other inertial frame and yield consistent resultsyield consistent results

A transformation law is required to A transformation law is required to related the space and time related the space and time coordinates from one frame to coordinates from one frame to anotheranother

6363

An event observed in two An event observed in two frames of reference must yield frames of reference must yield consistant results related by consistant results related by

transformatin lawstransformatin laws

6464

Different frame uses different Different frame uses different notation for coordinates notation for coordinates

(because their clocks and ruler (because their clocks and ruler are differentare different O' frame uses O' frame uses {{x',y',z‘x',y',z‘;;t‘t‘}} to denote the to denote the

coordinatecoordinates of an event, s of an event, whereas O frame whereas O frame uses uses {{x,y,zx,y,z;;tt}}

How to related How to related {{x',y',z',t‘x',y',z',t‘} to {} to {x,y,zx,y,z;;tt}?}? In Newtonian mechanics, we use In Newtonian mechanics, we use GalileanGalilean

transformation transformation However, as discussed, GT fails when However, as discussed, GT fails when uu cc

because the GT is not consistent with the because the GT is not consistent with the constancy of the light speed postulateconstancy of the light speed postulate

TheThe relativistic version of the transformation relativistic version of the transformation law is given by Lorentzlaw is given by Lorentz transformationtransformation

6565

O’O

I see O’ moving with a velocity +u

+u

I measures the coordinates of M as {x,t} Object

M

Two observers in two inertial Two observers in two inertial frames with relative motionframes with relative motion

I measures the coordinates of M as {x’,t’}

I see O moving with a velocity -u

-u

6666

Derivation of Lorentz Derivation of Lorentz transformationtransformation

Our purpose is to find the transformation that Our purpose is to find the transformation that relates {relates {x,tx,t} with {} with {x’,t’x’,t’}}

6767

Consider a rocket moving with a speed Consider a rocket moving with a speed uu (O' (O' frame)frame) along the along the xx'xx' direction wrp to the direction wrp to the stationary O framestationary O frame

A light pulse is emitted at the instant A light pulse is emitted at the instant t' t' == t t =0=0

when the two origins of the two reference when the two origins of the two reference frames coincideframes coincide

The lightThe light signal travels as a spherical wave signal travels as a spherical wave at a constant speed at a constant speed cc in both in both framesframes

After some times After some times tt, the origin of the wave, the origin of the wave centered at O has a radius centered at O has a radius rr = = ctct, where, where

rr 22 = = xx22 ++ y y22 ++ z z22

6868

FFrom the view point of O',rom the view point of O', a after some times fter some times t‘ t‘ the origin of the origin of the wave, centered at O' has a radius: the wave, centered at O' has a radius:

r'r' = = ct'ct' , (, (rr’’ ) )22 = = ((xx’)’)22 + + ((y’ y’ ))22 + + ((z’ z’ ))22

y'y'==yy, , z'z' = = zz ( (because the motion of O' is along the because the motion of O' is along the xxxx’’) ) axisaxis – – no change for no change for yy,,zz coordinates coordinates ((condition Acondition A))

The transformation from The transformation from xx to to xx’ (and vice versa) must be ’ (and vice versa) must be linear, i.e. linear, i.e. x’x’ x x ((condition Bcondition B) )

Boundary condition (1)Boundary condition (1): In the limit of : In the limit of vv c, c, from the from the viewpoint of O, the origin of O’ is located on the wavefront (to viewpoint of O, the origin of O’ is located on the wavefront (to the right of O)the right of O) x x ’ = 0 must correspond to ’ = 0 must correspond to xx = = ctct

Boundary condition (2):Boundary condition (2): In the same limit, from the viewpoint In the same limit, from the viewpoint of O’, the origin of O is located on the wavefront (to the left of O’, the origin of O is located on the wavefront (to the left of O’) of O’) xx = 0 corresponds to = 0 corresponds to x’x’ = - = -ct’ct’

Putting everything together we assume the form Putting everything together we assume the form x’x’ = = kk((x x - - ctct) ) to relate to relate xx’ to {’ to {x,tx,t} as this is the form that fulfill all the } as this is the form that fulfill all the conditions (A,B)conditions (A,B) and and boundary consdition (1) boundary consdition (1) ; (; (kk some some proportional constant to be determined) proportional constant to be determined)

Likewise, we assume the form Likewise, we assume the form xx = = kk((x’x’ + + ct ct ’) to relate ’) to relate xx to to {x ’,{x ’,t t ‘} as this is the form that fulfill all the ‘} as this is the form that fulfill all the conditions (A,B)conditions (A,B) and and boundary consdition (2) boundary consdition (2) ; ;

Methematical detailsMethematical details

6969

Hence, with Hence, with rr = = ctct , , rr’’ = = ctct ’, ’, xx = = kk((x’ + x’ + ct’ct’ ), ), x x ’ = ’ = kk((x - ctx - ct) we s) we solvolve e for for {{x'x',,t't'} in } in terms of {terms of {xx,,tt }} to obtain the desired to obtain the desired transformation law (do it as an exercise)transformation law (do it as an exercise)

2' ( )

1

x utx x ut

uc

2

2

2

/' ( / )

1

t u c xt t u c x

uc

Finally, the transformation Finally, the transformation obtainedobtained

7070

the constant the constant kk is identified as the Lorentz factor, is identified as the Lorentz factor, Note that, now, the Note that, now, the length and time interval length and time interval

measured become dependent of the state of motion measured become dependent of the state of motion (in terms of (in terms of ) – in contrast to Newton’s viewpoint) – in contrast to Newton’s viewpoint

LorentzLorentz transformation reduces to Galilean transformation reduces to Galilean transformation when transformation when uu <<<< cc (show this (show this yourself) yourself)

i.e. LT i.e. LT GT in the limit GT in the limit vv<<<<cc

Space and time now becomes Space and time now becomes state-of-motion dependent (via state-of-motion dependent (via

))

7171

How to express {x, t} in terms of {x’, t’}

We have related {We have related {x'x',,t't'} in terms of } in terms of {{xx,,tt } as per } as per

Now, how do we express {x, t } in terms of {x’, t’}

2' ( )

1

x utx x ut

uc

2

2

2

/' ( / )

1

t u c xt t u c x

uc

7272

O’ moving to the right with velocity +u is equivalent to O moving to the left with velocity -u ',x x u u

The two transformations above are equivalent; use which is appropriate in a given question

' ( ) ( ' ')x x ut x x ut

2 2' ( / ) ' ( / ) 't t u c x t t u c x

7373

Length contraction can be Length contraction can be recovered from the LTrecovered from the LT

Consider the rest length of a ruler as measured Consider the rest length of a ruler as measured in frame O’ is in frame O’ is L’L’ = = x’ = x’x’ = x’2 2 - x’ - x’1 1 (proper length) (proper length) measured at the same instant in that frame, measured at the same instant in that frame, hence hence t’t’2 2 = = t’t’11

What is the length of the rule as measured by O?What is the length of the rule as measured by O? The length in O, according the LT isThe length in O, according the LT is

L’L’x’ = x’x’ = x’2 2 - x’ - x’1 1 = = ((xx2 2 - x- x11) – ) – uu((tt2 2 -t-t11)] )] The length of the ruler in O is simply the distance The length of the ruler in O is simply the distance

btw btw xx2 2 and and xx11 measured at the same instant in measured at the same instant in that frame, hence that frame, hence tt2 2 = = tt11, hence , hence L’L’ = = LL

7474

Similarly, how would you Similarly, how would you

recover time dilation from the recover time dilation from the LT? LT?

Do it as homeworkDo it as homework

7575

Lorentz velocity Lorentz velocity transformationtransformation

O’

M is moving with a velocity +ux from my point of view Object

M

I see M moving with a velocity +ux’

I see O’ moving with a velocity +u

+u

O

Object M

O

How to relate the velocity of the object M as seen in the O’ (u’x) frame to that seen in the O frame (ux)?

7676

2' ( ), ' ( )

udx dx udt dt dt dx

c

DerivationDerivation

By definition, ux = dx/dt, u’x = dx’/dt’ The velocity in the O’ frame can be

obtained by taking the differentials of the Lorentz transformation,

7777

CombiningCombining

21

x

x

u uuu

c

where we have made used of the definition ux = dx/dt

'

2

' ( )

' ( )x

dx dx udtu

udt dt dxc

2

dx dtu

dt dt dtdt u dxdtdt c dt

7878

Comparing the LT of velocity with that of GT

2

''

' 1

xx

x

u udxu

u udtc

Galilean transformation of velocity:

' ( )'

'x x

dx d x ut dxu u u u

dt dt dt

GT reduces to LT in the limit u << c

Lorentz transformation of velocity:

7979

Please try to understand the Please try to understand the definition of udefinition of ux x , u’, u’x x , u so that you , u so that you wont get confusedwont get confused

8080

LT is consistent with the LT is consistent with the constancy of speed of lightconstancy of speed of light

in either O or O’ frame, the speed of in either O or O’ frame, the speed of light seen must be the same, clight seen must be the same, c

Say object M is moving with speed of Say object M is moving with speed of light as seen by O, i.e. light as seen by O, i.e. uuxx == c c

According to LT, the speed of M as According to LT, the speed of M as seen by O’ is seen by O’ is

22

'1

1 11

xx

x

u u c u c u c uu c

u u cu uc u

c c cc

8181

That is, in either frame, both observers That is, in either frame, both observers agree that the speed of light they agree that the speed of light they measure is the same, measure is the same, cc = 3 x 10 = 3 x 1088m/sm/s

In contrast, according to GT, the speed In contrast, according to GT, the speed of light seen by O’ would be of light seen by O’ would be

'x xu u u c u

Which is inconsistent with constancy of speed of light postulate

8282

2

'1

xx

x

u uu

u u

c

To recapTo recap

the LT given in the previous the LT given in the previous analysis relates analysis relates u’u’xx to to uuxx in which in which O’ is moving with +u wrp to O,O’ is moving with +u wrp to O,

8383

From the view point of O’From the view point of O’

To express To express uuxx in terms of in terms of u’u’xx simply simply perform the similar derivation from perform the similar derivation from the view point of O’ such that O is the view point of O’ such that O is moving in the moving in the –u –u direction: direction: ..

2 2

''

'1 1

x xx x

x x

u u u uu u

u u u u

c c

' , ' ,x x x xu u u u u u

8484

Recap: Lorentz transformation Recap: Lorentz transformation relates relates

{x’,t’} {x’,t’} {x,t}; u’ {x,t}; u’xx uuxx

( ' ')x x ut

' ( )x x ut 2' ( / )t t u c x

2' ( / ) 't t u c x

2

'1

xx

x

u uu

u u

c

2

''

1

xx

x

u uu

u u

c

8585

ExamplExamplee A boy is slapped twice on the face by his old girlfriend. This A boy is slapped twice on the face by his old girlfriend. This

is happening in a hotel room (a rest frame we call O). is happening in a hotel room (a rest frame we call O).

O

t1

The two slapping occurs at t1 , t2 such that t = t2- t1 = 1 s, and x =0.

t2

To his new girlfriend in a car moving with speed u on the road (we call this moving frame O’), what is the time interval between the two slapping?

O’

u

8686

' '2 1

22 1 2 1

2

2

'

( ) ( / )( )

( / )

11 7.09

0.991

t t t

t t u c x x

t u c x

t s sc

c

The time t’ as seen by O’ in terms of t is simply related by

2' ( / )t t u c x Hence the time interval as measured by his new girlfriend in O’, t’ in terms of t is simply

This is nothing but just the time dilation effect calculated using LT approach

8787

Example (relativistic Example (relativistic velocity addition)velocity addition)

Rocket 1 is approaching rocket 2 on Rocket 1 is approaching rocket 2 on aa

head-on collision course. Each is head-on collision course. Each is moving at velocity 4c/5moving at velocity 4c/5 relative to an relative to an independent observer midway independent observer midway between the two. Withbetween the two. With what velocity what velocity does rocket 2 approaches rocket 1? does rocket 2 approaches rocket 1?

8888

C.f. In GT, their relative speed would just be 4c/5 + 4c/5 = 1.6 c – which violates constancy of speed of light postulate. See how LT handle this situation:

Diagramatical translation of Diagramatical translation of the question in textthe question in text

8989

Choose the observer in the middle as in the Choose the observer in the middle as in the stationary frame, Ostationary frame, O

Choose Choose rocket 1 as the moving frame O‘rocket 1 as the moving frame O‘ Call tCall the velocity ofhe velocity of rocket 2 rocket 2 as seen from rocket as seen from rocket

1 u’1 u’xx. This . This is the quantity we are interested inis the quantity we are interested in

Frame O' is movingFrame O' is moving in the +ve directionin the +ve direction as seen as seen in Oin O, so , so uu = +4 = +4cc/5/5

The velocityThe velocity of rocket 2 as seen from O is in the of rocket 2 as seen from O is in the

-ve direction, so -ve direction, so uuxx = = -- 44cc/5/5

Now, what is the velocity of rocket 2 as seen Now, what is the velocity of rocket 2 as seen from framefrom frame O', O', u u ’’x x = ?= ? (intuitively, (intuitively, u u ’’xx must be in must be in the negative direction)the negative direction)

9090

Use the LT

2

2

4 4405 5

'4 4 4115 51

xx

x

c cu u

u cu u c cc

c

i.e. the velocity of rocket 2 as seen from rocket 1 (the moving frame, O’) is –40c/41, which means that O’ sees rocket 2 moving in the –ve direction (to the left in the picture), as expected.

9191

PYQ, KSCP 2003/04PYQ, KSCP 2003/04A man in a spaceship moving at a velocity of 0.9A man in a spaceship moving at a velocity of 0.9cc with with

respect to the Earth shines a light beam in the same respect to the Earth shines a light beam in the same direction in which the spaceship is travelling. Compute direction in which the spaceship is travelling. Compute the velocity of the light beam relative to Earth using the velocity of the light beam relative to Earth using (i) (i) Galilean approachGalilean approach (ii) s (ii) special relativity approach pecial relativity approach [6 [6 marks]. marks]. Please define clearly all the symbols used inPlease define clearly all the symbols used in

your working. your working. AnsAnsO’ is the moving frame travelling at O’ is the moving frame travelling at vv = 0.9 = 0.9cc with respect with respect

to the Earth. Speed of the light beam as seen in the to the Earth. Speed of the light beam as seen in the frame O’ is frame O’ is uu’ = ’ = cc. O is the Earth frame. We wish to find . O is the Earth frame. We wish to find the speed of the light beam as seen from frame O, the speed of the light beam as seen from frame O, uu. .

(i) According to Galilean transformation, (i) According to Galilean transformation, uu = = uu’ + ’ + vv = = c c + + 0.90.9cc = 1.9 = 1.9c.c.(ii) Use (ii) Use

2 2

' 0.90.9

1 ' 1

u v c cu c v c

v cu c

c c

9292

Relativistic DynamicsRelativistic Dynamics By Einstein’s postulate, the onservational By Einstein’s postulate, the onservational

law of linear momentum must also hold law of linear momentum must also hold true in all frames of referencetrue in all frames of reference

m1u1

m2u2

m1v1 m2v2

Conservation of linear momentum classically means

m1u1 +m2u2 = m1v1 +m2v2

9393

Modification of expression of Modification of expression of linear momentumlinear momentum

Classically, Classically, pp = = mumu. In the other frame, . In the other frame, p p ’ = ’ = m m ’’u u ’; the mass ’; the mass m m ’ (as seen in frame O’) is the same ’ (as seen in frame O’) is the same as as mm (as seen in O frame) – this is according to (as seen in O frame) – this is according to Newton’s mechanicsNewton’s mechanics

However, simple consideration will reveal that in However, simple consideration will reveal that in order to preserver the consistency between order to preserver the consistency between conservation of momentum and the LT, the conservation of momentum and the LT, the definition of momentum has to be modified such definition of momentum has to be modified such that m’ is not equal to m. that m’ is not equal to m.

That is, the mass of an moving object, m, is That is, the mass of an moving object, m, is different from its value when it’s at rest, mdifferent from its value when it’s at rest, m00

9494

In other words…In other words… In order to preserve the consistency between In order to preserve the consistency between

Lorentz transformation of velocity and Lorentz transformation of velocity and conservation of linear momentum, the conservation of linear momentum, the definition of 1-D linear momentum, classically definition of 1-D linear momentum, classically defined as defined as ppclassicalclassical = = mu, mu, has to be modified to has to be modified to

ppsrsr = = mu mu = = mm00u u (where the relativisitic mass (where the relativisitic mass m = m = mm00 is not the same the rest mass is not the same the rest mass mm00

Read up the text for a more rigorous Read up the text for a more rigorous illustration why the definition of classical illustration why the definition of classical momentum is inconsistent with LTmomentum is inconsistent with LT

9595

Grafically…Grafically…

O

O’

vM

I see M is at rest. Its mass is m0, momentum, p’ = 0

I see the momentum of M as p = mv=m0v

9696

Two kinds of massTwo kinds of mass

Differentiate two kinds of mass: rest mass and relativistic mass

m0 = rest mass = the mass measured in a frame where the object is at rest. The rest mass of an object must be the same in all frames (not only in its rest frame).

Relativisitic mass m = m0 of an object changes depends on its speed

9797

Behaviour of Behaviour of ppSR SR as compared as compared to to ppclassicclassic Classical momentum is Classical momentum is

constant in mass, constant in mass, ppclassicclassic = = mm00vv

Relativisitic momentum Relativisitic momentum is is ppSRSR = = mm00vv

ppSRSR / / ppclassicclassic = = as as v v cc

In the other limit, In the other limit, ppSRSR / / ppclassicclassic = = as as v v << << cc

9898

ExampleExample

The rest mass of an electron is mThe rest mass of an electron is m00 = = 9.11 x 109.11 x 10-31-31kg. kg.

p = m0u

Compare it with that calculated with classical definition.

If it moves with u = 0.75 c, what is its relativistic momentum?

m0

9999

SolutionSolution

The Lorentz factor is The Lorentz factor is = [1-(u/c)= [1-(u/c)22] ] -1/2-1/2 = [1-(0.75c/c) = [1-(0.75c/c)22] ] -1/2-1/2=1.51=1.51

Hence the relativistic momentum is simply Hence the relativistic momentum is simply

p = p = x m x m0 0 x 0.75x 0.75cc = 1.51 x 9.11 x 10= 1.51 x 9.11 x 10-31-31kg x 0.75 x 3 x 10kg x 0.75 x 3 x 1088 m/s = 3.1 x 10m/s = 3.1 x 10-22 -22 kg m/s = Ns kg m/s = Ns

In comparison, classical momentum gives pclassical = m0 x 0.75c = 2.5 x 10-22

Ns – about 34% lesser than the relativistic value

100100

Work-Kinetic energy Work-Kinetic energy theoremtheorem

Recall the law of conservation of Recall the law of conservation of mechanical energy:mechanical energy:

Work done by external force on a system, W = the change in kinetic energy of the system, K

101101

K1

F

K2

F

s

= K2 - K1

W = F s Conservation of mechanical energy: W = The total energy of the object, E = K + U. Ignoring potential energy, E of the object is solely in the form of kinetic energy. If K1 = 0, then E = K2. But in general, U also needs to be taken into account for E.

102102

In classical mechanics, mechanical energy (kinetic + potential) of an object is closely related to its momentum and mass

Since in SR we have redefined the classical mass and momentum to that of relativistic version

mclass(cosnt) mSR = m0

pclass = mclass u pSR = (m0)u

We will like to derive K in SR in the following slides

E.g, in classical mechanics, K = p2/2m = 2mu2/2. However, this relationship has to be supplanted by the relativistic version K = mu2/2 K = E – m0c2 = mc2 - m0c2

we must also modify the relation btw work and energy so that the law conservation of energy is consistent with SR

103103

Force, work and kinetic energyForce, work and kinetic energy When a force is acting on an object with When a force is acting on an object with

rest mass mrest mass m00, it will get accelerated (say , it will get accelerated (say from rest) to some speed (say u) and from rest) to some speed (say u) and increase in kinetic energy from 0 to Kincrease in kinetic energy from 0 to K

K as a function of u can be derived from first principle based on the definition of:

Force,F = dp/dt,

work done, W = F dx,

and conservation of mechanical energy, K = W

104104

Derivation of relativistic Derivation of relativistic kinetic energykinetic energy

2

1 0 0 0

x u u

x

dp dp du dpudx udx udu

dx du dx du

2 2 2

1 1 10 0 0

x x x

x x x

dp dp dxW F dx dx dx

dt dx dt

Force = rate change of momentum

Chain rule in calculus

dxu

dtwhere, by definition, is the velocity of the

object

105105

Explicitly, p = m0u,

Hence, dp/du = d/du(m0u)

= m0 [u (d/du) + ]

= m0 + (u2/c2) 3] = m0 (1-u2/c2)-3/2

in which we have inserted the relation 3

3/ 22 22 2

2 2

1 1

1 1

d d u u

du du c cu uc c

integrate

2 2 2 20 0 0K W m c m c mc m c

3/ 22

0 201

u uW m u du

c

106106

E = mcE = mc22 is the total relativistic energy of an is the total relativistic energy of an moving objectmoving object

2 2 2 20 0 0K m c m c mc m c

The relativisitic kinetic energy of an object of rest mass m0 travelling at speed u

E0 = m0c2 is called the rest energy of the object. Its value is a constant for a given object

Any object has non-zero rest mass contains energy as per E0 = m0c2

One can imagine that masses are ‘energies frozen in the form of masses’ as per E0 = m0c2

107107

Or in other words, the total relativistic energy Or in other words, the total relativistic energy of a moving object is the sum of its rest of a moving object is the sum of its rest energy and its relativistic kinetic energyenergy and its relativistic kinetic energy

2 20E mc m c K

The mass of an moving object m is larger than its rest mass m0 due to the contribution from its relativistic kinetic energy – this is a pure relativistic effect not possible in classical mechanicsE = mc2 relates the mass of an object to the total energy released when the object is converted into pure energy

108108

Example, 10 kg of mass, if converted into pure Example, 10 kg of mass, if converted into pure energy, it will be equivalent to energy, it will be equivalent to

E = mcE = mc22 = 10 x (3 x10 = 10 x (3 x1088) ) 22 J = 9 x10 J = 9 x101717J J – – equivalent to a few tons of TNT explosiveequivalent to a few tons of TNT explosive

109109

PYQ, KSCP 2003/04PYQ, KSCP 2003/04

(i)(i) What is the rest mass of a What is the rest mass of a proton in terms of MeV? proton in terms of MeV?

Ans: Ans: (i)(i) 1.67 x 101.67 x 10-27-27kg kg xx (3 (3x10x108 8 m/s)m/s)22 = =

1.5031.503x10x10-10-10J = (1.503 J = (1.503 x10x10-10-10/1/1.6x10.6x10--

1919) eV = 939.4 MeV) eV = 939.4 MeV

110110

PYQ, KSCP 2003/04PYQ, KSCP 2003/04 What is the relativistic mass of a What is the relativistic mass of a

proton whose kinetic energy is 1 proton whose kinetic energy is 1 GeV?GeV?

Ans: Ans: rest mass of proton, rest mass of proton, mmpp = 939.4 = 939.4 MeVMeV 21 pK m c

21 / 1GeV/ 939.4MeV 1.06pK m c 1.06 1 2.06

2 2 2.06 939.4MeV = 1939.4 MeVpmc m c

111111

Reduction of relativistic Reduction of relativistic kinetic energy to the kinetic energy to the

classical limitclassical limit The expression of the relativistic The expression of the relativistic

kinetic energykinetic energy2 2

0 0K m c m c

must reduce to that of classical one in the limit u 0 when compared with c, i.e.

2 20

0

lim ( )2 2classical

relativisticu c

p m uK

m

112112

Expand Expand with binomial with binomial expansion expansion

For u << c, we can always expand For u << c, we can always expand in terms of (u/c)in terms of (u/c)2 2 asas

1/ 22 2 4

2 2 41 1 terms of order and highe

2r

u u u

c c c

2 2 20 ( 1)K mc m c c

22

0 2

11 ... 1

2

um c

c

20

2

m u

i.e., the relativistic kinetic energy reduces to classical expression in the u << c limit

113113

ExampleExample An electron moves with speed u = 0.85c. Find An electron moves with speed u = 0.85c. Find

its total energy and kinetic energy in eV. its total energy and kinetic energy in eV. CERN’s picture: the circular accelerator CERN’s picture: the circular accelerator

accelerates electron almost the speed of lightaccelerates electron almost the speed of light

114114

SolutionSolutionDue to mass-energy equivalence, sometimes we express the mass of an object in unit of energy

Electron has rest mass m0 = 9.1 x 10-31kg The rest mass of the electron can be

expressed as energy equivalent, via m0c2 = 9.1 x 10-31kg x (3 x 108m/s)2 = 8.19 x 10-14 J = 8.19 x 10-14 x (1.6x10-19)-1 eV = 511.88 x 103 eV = 0.511 MeV

115115

SolutionSolution First, find the Lorentz factor, First, find the Lorentz factor, = 1.89 = 1.89 The rest mass of electron, mThe rest mass of electron, m00cc22, is , is

0.5 MeV0.5 MeV Hence the total energy is Hence the total energy is

E = mcE = mc22 = = mm00cc22)= 1.89 x 0.5 MeV = )= 1.89 x 0.5 MeV = 0.97 MeV0.97 MeV

Kinetic energy is the difference Kinetic energy is the difference between the total relativistic energy between the total relativistic energy and the rest mass, K = E -and the rest mass, K = E -mm00cc22= = (0.97 – 0.51)MeV = 0.46 MeV(0.97 – 0.51)MeV = 0.46 MeV

116116

Conservation of Kinetic energy Conservation of Kinetic energy in relativistic collisionin relativistic collision

Calculate (i) the kinetic energy of the Calculate (i) the kinetic energy of the system and (ii) mass increase for a system and (ii) mass increase for a completely inelastic head-on of two completely inelastic head-on of two balls (with rest mass balls (with rest mass mm00 each) each) moving toward the other at speed moving toward the other at speed uu//c = c = 1.5x101.5x10-6 -6 (the speed of a jet (the speed of a jet plane). plane). MM is the resultant mass after is the resultant mass after collision, assumed at rest.collision, assumed at rest.

mm00 mm00

M

u u

117117

SolutionSolution (i) (i) KK = 2 = 2mcmc2 2 - 2- 2mm00cc22 = 2( = 2(mm00cc22

(ii) (ii) EEbefore before = = EEafter after 2 2mm00cc2 2 = = McMc2 2 M =M = 2 2mm00 Mass increase Mass increase M M = = M M - 2- 2mm0 0 == 2( 2(mm00 Approximation: Approximation: uu//c c = …=1.5x10= …=1.5x10-6-6 ≈≈1 + ½ 1 + ½

uu22//cc2 2 (binomail expansion) (binomail expansion) M M ≈≈ 221 + ½ 1 + ½ uu22//cc22mm00

Mass increase Mass increase M M = = M M - 2- 2mm00

≈≈ ((uu22//cc22mm00 = 1.5x10= 1.5x10-6-6mm00

Comparing Comparing K K with with McMc22: the kinetic energy is : the kinetic energy is not lost in relativistic inelastic collision but is not lost in relativistic inelastic collision but is converted into the mass of the final composite converted into the mass of the final composite object, i.e. kinetic energy is conservedobject, i.e. kinetic energy is conserved

In contrast, in classical mechanics, momentum In contrast, in classical mechanics, momentum is conserved but kinetic energy is not in an is conserved but kinetic energy is not in an inelastic collisioninelastic collision

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2 4 2 4 22 2 2 4 2 20 0

0 02 2 2 2

21

m c m c EE m c m c

u E c pc

2 2 22 2 2 4 2 2 2 2

0 0 2 2;

u c pE m c p m u

c E

2 2 2 2 40E p c m c Conservation of

energy-momentum

In terms of relativistic In terms of relativistic momentum, the relativistic total momentum, the relativistic total energy can be expressed as energy can be expressed as followedfollowed

Relativistic momentum and relativisitc Energy

119119

In SR, both relativistic mass-In SR, both relativistic mass-energy and momentum are energy and momentum are

always conserved in a collision always conserved in a collision (in contrast to classical (in contrast to classical

mechanics in which KE is not mechanics in which KE is not conserved in inelastic collision)conserved in inelastic collision)

120120

Example: measuring pion mass Example: measuring pion mass using conservation of using conservation of momentum-energymomentum-energy

pi meson decays into a muon + massless neutrino pi meson decays into a muon + massless neutrino If the mass of the muon is known to be 106 MeV/If the mass of the muon is known to be 106 MeV/cc22, ,

and the kinetik energy of the muon is measured to be and the kinetik energy of the muon is measured to be 4.6 MeV, find the mass of the pion 4.6 MeV, find the mass of the pion

121121

SolutionSolution

2 2 2 2 4

2 2 4 2 2 2 4 2 2

2 2 2

Relationship between Kinetic energy and momentum:

Conservation of relativistic energy:

E p c m c

E E E

m c m c c p m c c p

m c m c p p

Momentum conservation: p p

2

22 2 2 2 4

22 2 4

Also, total energy = K.E. + rest energy

;

E K m c

p c K m c m c

p c K m c m c

122122

22 2 2 2 4

2 2 4 2 2

2 22 4 2 2 4 2 2 4

2 2 2

22 22 2

Plug into

2

106MeV 106MeV4.6MeV+ 4.6MeV 2 4.6MeV

111MeV 996MeV=143MeV

p c K m c m c

m c m c c p cp

m c K m c m c K m c m c

K m c K K m c

c cc c

123123

Binding energyBinding energy The nucleus of a deuterium The nucleus of a deuterium comprises of one neutron and one comprises of one neutron and one proton. Both nucleons are bounded proton. Both nucleons are bounded within the deuterium nucleuswithin the deuterium nucleus

Neutron, mnproton, mp

Deuterium, md

Nuclear fusion

Initially, the total Energy = (mn+ mn)c2

After fusion, the total energy = mdc2 + U

U

Analogous to exothermic process in chemistry

124124

U is the energy that will be released U is the energy that will be released when a proton and a neutron is fused when a proton and a neutron is fused in a nuclear reaction. The same in a nuclear reaction. The same amount of energy is required if we amount of energy is required if we want to separate the proton from the want to separate the proton from the neutron in a deuterium nucleusneutron in a deuterium nucleus

U is called the binding energyU is called the binding energy

125125

U can be explained in terms energy-mass U can be explained in terms energy-mass equivalence relation, as followedequivalence relation, as followed

For the following argument, we will ignore KE For the following argument, we will ignore KE for simplicity sakefor simplicity sake

Experimentally, we finds that mExperimentally, we finds that mnn + m + mp p > m> mdd By conservation of energy-momentum,By conservation of energy-momentum, E(before) = E(after)E(before) = E(after) mmnncc22 + m + mppcc2 2 + 0 = m+ 0 = mddcc22 + U + U Hence, U = (mHence, U = (mpp + m + mnn)c)c2 2 - m- mddcc22 = = mcmc22

The difference in mass between deuterium and The difference in mass between deuterium and the sum of (mthe sum of (mnn + m + mnn)c)c22 is converted into the is converted into the binding energy that binds the proton to the binding energy that binds the proton to the neutron togetherneutron together

126126

ExampleExample mmnn= 1.008665u; m= 1.008665u; mpp= 1.007276u;= 1.007276u; mmdd= 2.013553u; = 2.013553u; u = standard atomic unit = mass of 1/12 of the u = standard atomic unit = mass of 1/12 of the

mass of a mass of a 1212C nucleus C nucleus = 1.66 x 10= 1.66 x 10-27-27kg kg = 1.66 x 10= 1.66 x 10-27 -27 x cx c22 J = 1.494 x 10 J = 1.494 x 10-10-10 J J = 1.494 x 10= 1.494 x 10-10-10/(1.6x10/(1.6x10-19-19) eV ) eV = 933.75 x 10= 933.75 x 1066 eV = 933.75 x MeV eV = 933.75 x MeV

•Hence the binding energy

U mc = (mp + mn)c2 - mdc2

=0.002388u = 2.23 MeV

127127

FissionFission

Such as Such as The reverse of nuclear fusion is nuclear The reverse of nuclear fusion is nuclear

fissionfission An parent nuclide M disintegrates into An parent nuclide M disintegrates into

daughter nuclides such that their total mass daughter nuclides such that their total mass mmi i < M. < M.

The energy of the mass deficit equivalent, The energy of the mass deficit equivalent,

Q = (M - Q = (M - mmii)c)c22 = = mcmc2 2 called disintegration called disintegration energy will be releasedenergy will be released

236 90 143 192 37 55 0U Rb Cs+3 n

128128

SR finishes here…SR finishes here…