• More than 99% of what we know about the universecomes from observing electromagnetic waves• Other sources• The Earth, meteorites• Samples returned from solar wind, Moon, etc.• Close up observations of nearby planets• Neutrinos

Wave Equations summarized:• Waves look like:• Related by:• Two independent solutions to these equations:

0

0

, , , sin

, , , sin

x y z t kx t

x y z t kx t

E E

B Bck

0 0

0 0

or

y z

z y

E cB

E cE

E0

B0

E0

B0

0 0E cB

• Note that E, B, and direction of travel are all mutually perpendicular

• The two solutions are called polarizations

• We describe polariza-tion by telling which way E-field points

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83.00 10 m/sc

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Wavelength and wave number• The quantity k is called

the wave number• The wave repeats in time• It also repeats in space

0

0

sin

sin

kx t

kx t

E E

B B

ck 1

2

f T

f 2k

• EM waves most commonly described in terms of frequency or wavelengthc

k

2

2f

c f

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The Electromagnetic Spectrum• Different types of waves are classified

by their frequency (or wavelength)c f

Radio WavesMicrowavesInfraredVisibleUltravioletX-raysGamma Rays I

ncre

asin

gf I

ncre

asin

g

RedOrangeYellowGreenBlue

Violet

VermillionSaffron

ChartreuseTurquoise

Indigo

Know these, in order

These tooNot these

• Boundaries are arbitrary and overlap

• Visible is 380-740 nm

Dealing with WavesBefore we tackle electromagnetic waves in 3D,let’s think about some sort of ordinary waves in 1D• Like, waves on a string

• Fixed boundary conditions: (0) =0, (L)=0• We want to work in infinite length limit, and find the contribution from all waves• One way to do this is to make the string finite, then take L later on• Though this can be done, it is a bit frustrating because

of the boundary conditions• You only get standing waves• Can’t really talk about direction of the wave

To get around this problem, a common trick isto use periodic boundary conditions• We demand that:• the two ends match: (0) = (L),• and their derivatives match: ’(0) = ’(L),

• This allows waves with a direction to them:

L

Breaking up complicated waves• Typical wave equation:• We need periodic boundary conditions:

, cos ,x t kx t

0, ,t L t cos coskL t t

• This places a restriction on k• Cosine and sine have period of 2• kL must be a multiple of 2

L

2

0, 1, 2,

kL n

n

• Any solution of wave equation can be written as sum of waves of this type

• Each component of the wave has an energy E(k)• The total energy is the sum of the contributions

• The energy per unit length is then just:

totk

E E k n

E n k

2k

L

1tot

n

EE n k

L L

2 n

kE n k

Taking the Infinite Length Limit:• We want the energy per unit length for an infinite string:

• In the limit, this is an integral:

lim2

tot

Ln

E kE n k

L

0

1lim

2 kn

kE n k

k

E(k)

2

totE dkE k

L

• We now want to go to 3D:• Steps are similar

L

L

L

, ,x y Zk k kk

2 ,

2 ,

2 .

x x

y y

z z

k L n

k L n

k L n

3

332

totE dE

L

kk

3

32

du E

kk

Statistical MechanicsThe application of statistics to the properties of systems containing a large number of objects

The techniques of statistical mechanics:• When there are many possibilities, energy will be

distributed among all of them• The probability of a single “item” being in a given

“state” depends on temperature and energy

BE k TP E e

23 51.3806 10 J/K 8.6173 10 eV/KBk

gravity

Gas molecules in a tall box:

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Statistical Mechanics: One Mode of EM Field • We need E(k) for each value of k• Quantum Mechanics: Electromagnetic waves have energy:• The probability of each of these possibilities is:• The probabilities must sum to one:

,

0,1,2,

E n

n

Bn k TnP e Bn k T

nP Ce

0

1 nn

P

0

Bn k T

n

C e

0

1Bn k T

n

Ce

• The average amount of energy in the mode k is then given by:• These sums can be done:*

0

nn

E nP

k 1Bk T

Ee

k

I don’t care about these details

0

0

B

B

n k T

n

n k T

n

ne

e

How to do these sums:

• Recall: geometric series:• Take derivative of this expression:• Multiply by y

0

0

B

B

n k T

n

n k T

n

neE

e

k

Bn k Ty e

2 3

0

11

1n

n

y y y yy

1 2 3

20

1 11 2 3 4

1 1n

n

dny y y y

dy y y

20 1

n

n

yny

y

• Let:• Then:

20 0

1,

1 1

b

b b

bb

k Tn k T n k T

k T k Tn n

ee ne

e e

2

1

1 11

b bb

b bb

k T k T k T

k T k Tk T

e e eE

e ee

k

Black Body Radiation

• Now put it all together: Energy density for a thermal distribution of EM fields:

3

32

du E

kk

1Bk TE

e

k

32

3 30

4

12 2 Bk T

du E k dk

e

kk

• Wait: There are two polarizations for every wave number k:• Two modes, each with the same energy

2 22

0

1

1Bk Tk dk

e

• For light, recall:• Let

ck 22

0

1

1Bck k T

cku k dk

e

Bx kc k T4 3

20 1

Bx

k Tc x dxu

c e

4

15

42

315Bk T

uc

Units: J/m3

Black Body Radiation: Spectrum

• This is the correct expression for the total energy density• Sometimes, we will only measure the density over a fixed range of k

• We want energy density per unit wave number:• Experimentalists rarely do it this way

• More common ways of measuring it would beenergy per unit frequency or energy per unit wavelength

• Let’s work it out per unit wavelength to see how this works:

22

0

1

1Bck k T

cku k dk

e

42

315Bk T

uc

3

2

1

1Bck k T

du ck

dk e

2k 2k

2

2dk

d

du du dk

d dk d

3

2 2

1 2

1Bck k T

ck

e

2

25

16

1Bc k T

du c

d e

Fudging the minus sign

Wien’s Law:

• The spectrum depends on the temperature:• We can find the peak wavelength:

2

25

16

1Bc k T

du c

d e

322.8978 10 m K

4.96511 B

cT

k

• Color gives you a good idea of the temperature

• Colors go from dull red (cool) to electric blue (hot)

2900 K

4500 K

10,000 K

20,000 K

5500 K

Sample Problem:32

2.8978 10 m K4.96511 B

cT

k

The graph at the right shows the light received from five stars(a) Which star is the hottest?(b) Which two stars have the same

surface temperature(c) What is the temperature of the

green star?

• The overall size of the curve depends on the size and distance of the star• The peak/color of the star depends on the temperature• Red star peaks at smallest wavelength / highest temperature• Blue and black peak at the same wavelength• Green curve peak is around 705 nm

3

9

2.8978 10 m K

705 10 mT

4110 K

Stefan-Boltzman Law• We have formulas for the energy density• For stars, we want to know rate at which energy escapes

• Watts per square meter

How much power comes out of a hole of area A in a black body at temperature T?• Energy density is u• It is light – moving at velocity c• Half of it is moving the wrong way (½)• The half that is moving the right way is only moving

partly in the right direction (½)• The resulting total power per unit area (flux): • is called the Stefan-Boltzmann constant

14 ucF

42

360Bk T

cc

4TF

8 2 45.670 10 W/m /K

• We can similarly find the flux per unit wavelength:

Units: W/m2

2 2

25

4

1Bc k T

d c

d e

F

Total Luminosity• A star can be treated very crudely as a sphere with surface temperature T and radius R • The total power (called luminosity) coming from a star is:

L A F 2 44L R T• Relative luminosity of

two stars:

2 4L R T

L R T

• Real stars don’t follow Planck’s Law exactly• Nonetheless, we can solve luminosity equation

for T, called the effective temperature