531 PHYS - LN3. EM radiation Source: oscillating charge. E r. B r

EM radiation

Source: oscillating charge

E r

E r

B r

General Principles

¾ All bodies emit radiation continuously whatever their temperature.

¾ The predominant frequency (“color”) depends on the temperature.

¾ Mostly infrared at room temperature.

¾ The ability of a body to radiate is closely related to its ability to absorb radiation.

¾ At thermal equilibrium the rate of radiation is equal to the rate of absorption of a body.

Blackbody

¾ Ideal body

¾ Absorbs all radiation regardless of the frequency Blackbody Radiator

¾ Cavity with tiny hole (for example, a rectangular box)

¾ Filled with radiation at equilibrium

¾ Filled with standing waves at equilibrium with wavelengths

WIENWIEN’’S DISPLACEMENT LAWS DISPLACEMENT LAW

As a body is heated the wavelength of maximum energy As a body is heated the wavelength of maximum energy density is reduced and is given by

density is reduced and is given by λ λ

T T = 2.898 x 10 = 2.898 x 10

m K m K

This gives rise to the concept of colour temperature where the This gives rise to the concept of colour temperature where the temperature and colour are linked

temperature and colour are linked WIEN

WIEN’’S RADIATION LAWS RADIATION LAW Wien

Wienput forward the empirical law ρput forward the empirical law ρ((T T ) = ) = AAλλ^--55e e ^--B/B/λλTTwhere where AAand and BB are experimentally determined constants

are experimentally determined constants

Although this function works well at short wavelengths it does n Although this function works well at short wavelengths it does not give ot give good results at long wavelengths

good results at long wavelengths

Example

ExampleHow Hot is the Sun?How Hot is the Sun?

Consider the Sun as a blackbody. The peak radiation occurs at 500 nm

Use

Use WienWien’’ssdisplacement lawdisplacement law

λλ_maxmaxTT==2.8982.898xx1010^-3-3mmKK Thus

Thus T

T ==2.8982.898xx1010^--33//λλ_maxmax==2.8982.898xx1010^--33mmK/(500K/(500xx1010^--99m)=5800m)=5800KK Apparent surface temperature of the Sun is about 5800K Apparent surface temperature of the Sun is about 5800K

¾¾ RayleighRayleighcalculated the number of modes of vibration in the cavitycalculated the number of modes of vibration in the cavity

¾¾ He found that the number of possible modes per unit volume in the He found that the number of possible modes per unit volume in the wavelength range

wavelength range λλto to λλ++ddλλwas given by dnwas given by dn_λλ=(8=(8ππ//λλ⁴⁴) d) dλλ

¾

¾ Thermodynamic arguments indicated that each mode had Thermodynamic arguments indicated that each mode had kTkT of of energy

energy

¾

¾ The energy density in wavelength range The energy density in wavelength range λλto λto λ++ddλλwas calculated was calculated to be

to be ρρ(T(T) = 8 π) = 8 πk T λk T λ^--44

¾

¾ This is called the This is called the RayleighRayleigh--Jeans LawJeans Law

¾

¾ Note that at short wavelength the function blows upNote that at short wavelength the function blows up

¾

¾ This is called the ULTRAVIOLET CATASTROPHEThis is called the ULTRAVIOLET CATASTROPHE

Rayleigh

Rayleigh- -Jeans Law Jeans Law

¾

¾ Max Planck (1900) used thermodynamic arguments to derive Max Planck (1900) used thermodynamic arguments to derive Wien

Wien’’ssLaw for short wavelengthsLaw for short wavelengths

¾¾ He then used similar arguments to derive He then used similar arguments to derive RayleighRayleigh’’ssLaw at Law at long wavelengths

long wavelengths

¾

¾ These were combined to giveThese were combined to give

¾

¾ ρρ_λλ=A=Aλλ^--55/(e/(e^B/B/λλTT --1)1)

¾¾ At short wavelength e At short wavelength e B/B/λλTT>> 1 and so expression reduces >> 1 and so expression reduces to to WienWien’’ssLawLaw

¾

¾ At long wavelength e At long wavelength e B/B/λλTT~ 1 +~ 1 +B/B/λλTT and so this reduces to and so this reduces to ρρ_λλ== ((A/B A/B ))T T λλ^--44 i.e. Rayleighi.e. Rayleigh’’ssLawLaw

s Radiation Law s Radiation Law

’ ’

Planck

Planck

Consider a cavity with walls kept at a constant temperature T.

emission from walls ¨ e.m. radiation¨ emission & absorption→ equilibrium ν ν ν ν

ρ

ρ ρ ν

ρ ω ρ ρ

ν

ω ν ν

ω

d to from range freq.

the in density energy d

density energy spectral or

where

; d

d 0

0

+

≡

=

=

=

∫

∫

In order to comply with the 2

In order to comply with the 2^ndnd law of law of thermodynamic,

thermodynamic,

ρρ_νν(and (and ρρ_ωω) is a universal function of frequency ) is a universal function of frequency and temperature, Not a function of the shape of and temperature, Not a function of the shape of the cavity

the cavity

We may consider a simple rectangular cavity uniformly filled wit We may consider a simple rectangular cavity uniformly filled with a h a dielectric and having perfectly conducting walls .

dielectric and having perfectly conducting walls .

First, one must count the number of waves inside a black cavity of dimension 2a×2a×L

x

z y

L

2a 2a V=(2a)(2a)(L) To find

To find ρρ_νν or ρor ρ_ωω we need to know what we need to know what standing wave, with what frequencies exist in standing wave, with what frequencies exist in the cavity. And what is the average energy.

the cavity. And what is the average energy.

The allowed standing waves are called The allowed standing waves are called NORMAL MODES

NORMAL MODES

According to Maxwell's equations, the electric field

According to Maxwell's equations, the electric field E(x,y,z,t)E(x,y,z,t)must must satisfy the wave equation

satisfy the wave equation

(1) t 0

E c

E 1 ₂

2 2

2 =

∂

− ∂

∇ r r

The field must be satisfy the following boundary condition at ea

The field must be satisfy the following boundary condition at each wall;ch wall;

(*) 0 n E r × r =

Where n is the normal to the particular wall under consideration Where n is the normal to the particular wall under consideration

(4) A ) ck dt (

A d

(3) u k u

have we (1), in substitute and

(2) z)A(t) y, (x, u E Let

2 2

2 2 2

r r r r

r r

−

=

−

=

∇

=

Where

Where kkis a constant. Equation (4) has the general solution is a constant. Equation (4) has the general solution

(

)

A

A= ₀ ω +φ

Where

Where AA₀₀and and φφare arbitrary constants and ωare arbitrary constants and ω==ckck. .

The solution given by (5) corresponds to a standing wave The solution given by (5) corresponds to a standing wave configuration of

configuration of e.me.m. field within the cavity.. field within the cavity.

In fact the amplitude of oscillation at a given point of the cav In fact the amplitude of oscillation at a given point of the cavity is ity is constant in time. A solution of this type is called an

constant in time. A solution of this type is called an e.me.m. . modemodeof the of the cavity.

cavity.

Equation (3) is Helmholtz eq. , under B.C. gives the following ; Equation (3) is Helmholtz eq. , under B.C. gives the following ;

( ) ( ) ( ) ( ) ( ) ( ) ( ) k z sin ( ) k z cos ( ) k z sin

e u

z k sin z k cos x k sin e u

z k sin z k sin x k cos e u

z y

x z z

z y

x y y

z y

x x

x

=

=

=

Satisfy eq.(3) for any value of

Satisfy eq.(3) for any value of ee_xx,e,e_yy,e,e_zz, provided that ;, provided that ;

2 z 2 y 2 x

2

k k k

k = + +

(6) (6)

(7) (7)

The solution (6) already satisfy the B.C. (*) on the three plane The solution (6) already satisfy the B.C. (*) on the three planes s x=0, y=0, z=0

x=0, y=0, z=0

If the condition that eq. (*) should also be satisfied on the ot If the condition that eq. (*) should also be satisfied on the other her walls of the cavity, we have;

walls of the cavity, we have;

L k n a , 2 k m a , 2

k_x l

π

π

π

=

=

=

l, ml, mand and nnare positive integers ¨are positive integers ¨represent the number of nodes represent the number of nodes that the standing wave modes has along

that the standing wave modes has along x, yx, yand and zzrespectivelyrespectively

L

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡ ⎟

⎠

⎜ ⎞

⎝ +⎛

⎟⎠

⎜ ⎞

⎝ +⎛

⎟⎠

⎜ ⎞

⎝

= ⎛

2 2

2 2 2

n , m ,

l L

n a

2 m a

2

c lπ π π

ω The freq.

The freq. ωωof the mode will be;of the mode will be;

If NIf N_ννrepresent the number of modes in the cavity with freq. between represent the number of modes in the cavity with freq. between 0 0 and νand ν

Next thing would be to calculate the number of modes between the Next thing would be to calculate the number of modes between the frequency range of (

frequency range of (0,0,νν). ).

In wave number units this is the range

In wave number units this is the range (0, 2(0, 2πνπν/c)./c). The number of The number of modes and the density can be calculated by treating each modes and the density can be calculated by treating each combination of the numbers (

combination of the numbers (kk_xx ,k,k_yy ,,kk_zz) as a point in the positive ) as a point in the positive octant of the Cartesian coordinate system. The unit cell of the octant of the Cartesian coordinate system. The unit cell of thecube cube is defined by the lengths (

is defined by the lengths (ππ/2a, /2a, ππ/2a, /2a, ππ/L) The number of nodes in /L) The number of nodes in this cube is

this cube is

k_z

k_x

k_y

density) (mode

c

8 V

d / ) dN (

p ₃

πν

ν

ν

The factor 2 is for the two perpendicular directions of polariza The factor 2 is for the two perpendicular directions of polarizationtion

where

where VV isis thethe totaltotal volumevolume ofof thethe cavitycavity The corresponding

The corresponding number ofnumber ofmodes per unit volume and per unit modes per unit volume and per unit frequency range is

frequency range is

c V 3 8 L) a)(

)(2 a (2

c 2 3 .4 8 1 mode 2

one of volume

space freq.

in octant positive in

volume

N ₃

3 3

πν π

π π

π πν

ν ⎟ =

⎠

⎜ ⎞

⎝

⎛

=

=

Since energy density is given by;

Since energy density is given by;

E ) (

v

p ν

ρ =

Where <E> is the average energy in each mode Where <E> is the average energy in each mode

By assuming a continuous spectrum (high number of modes), the By assuming a continuous spectrum (high number of modes), the average energy in a given temperature T can be obtained directl average energy in a given temperature T can be obtained directly y from the

from the BoltzmannBoltzmannstatistics;statistics;

The probability

The probability dpdp that the energy of a given cavity mode lie that the energy of a given cavity mode lie between E and

between E and E+dEE+dEis express by is express by dpdp=C exp[=C exp[--(E/(E/kTkT)],)],where where CCis a is a constant.

constant.

The average energy of the mode <E> is therefore given by;

The average energy of the mode <E> is therefore given by;

kT dE e

dE e

E E

0 kT / E 0

kT / E

=

=

∫

∫

∞

−

∞ −

Each

Each ““vibrationvibration”” or normal or normal mode can take a continuous mode can take a continuous range of energy

range of energy

¨

Thus we would get the energy density Thus we would get the energy density

c kT E 8

) (

p

2 v

ν πν

ρ = =

WRONG WRONG

Right away we know that the previous so called Rayleigh Right away we know that the previous so called Rayleigh-- Jeans equation must be wrong since

Jeans equation must be wrong since therethere is isnono limit limitfor forthethe energy

energy density ifdensity if TT →→ ∞∞ ..

Thus we would get the energy density Thus we would get the energy density

c kT E 8

) (

p

2 v

ν πν

ρ = =

Quantum hypothesis:

2 3

2 3

1

1

1 2 3

1

1

/ / /

/ / /

/

/ /

h kT h kT h kT

h kT h kT h kT

nh kT

h kT nh kT

h e h e h e

e e e

nh e h

e e

ν ν ν

ν ν ν

ν

ν ν

ν ν ν

ε

ν ν

− − −

− − −

∞ −

∞ −

+ + + ⋅⋅⋅

= + + + + ⋅⋅⋅

= =

−

∑

∑

Let the allowed energies be E

Let the allowed energies be E₁₁, E, E₂₂, E, E₃₃, , …………

Then

Then εε_nn= n h = n h νν and the relative probability of obtaining an energy and the relative probability of obtaining an energy ΕΕ_jj is exp(

is exp( --EE_jj/ / kTkT) (i.e., a Boltzmann) (i.e., a Boltzmanndistribution)distribution)

The correction was, however, not at all an easy task to find but The correction was, however, not at all an easy task to find but required totally new concepts to be taken into account. Instead required totally new concepts to be taken into account. Instead of of assuming totally continuous spectrum the introduction of the lig assuming totally continuous spectrum the introduction of the light ht quantum lead to the cutoff that was also intuitively needed.

quantum lead to the cutoff that was also intuitively needed.

How?

) n exp(

) n exp(

n kT )]

kT / h ( n exp[

)]

kT / h ( n kT exp[

n h kT

) kT / nh kTexp(

1

) kT / nh kT exp(

nh E

0 n

0 n 0

n 0 n

0 n

0 n

α α α ν

ν ν ν

ν ν

−

= −

−

⎟ −

⎠

⎜ ⎞

⎝

⎛

=

−

= −

∑∑

∑

∑

∑

∑

∞

=

∞

=

∞

=

∞

=

∞

=

∞

=

[

]

d ln kT d ) exp(

1 ln 1 d kT d ) n exp(

d ln

kT d _n0 α

α α α α α

α α

α = − −

−

− −

=

−

−

=

∑

kT where α =hν

1 ) kT / h exp(

h 1

exp h ) exp(

1

) kT exp(

= −

= −

−

−

= −

ν ν α

ν α

α α

1 ) kT / h exp(

E h

= − ν

ν

Note that for

Note that for hhνν//kTkT→→0, 0, exp(exp(hhνν//kTkT) ) →→1+1+hhνν//kTkT, and therefore <E> , and therefore <E> →→kTkT Similarly,

Similarly,<E> <E> →→0 as0 ashhνν//kTkT→→∞∞

Now the energy density also becomes

Now the energy density also becomes diffdifferenterent

1 e

h c E 8 ) ( p

kT h 3

2

−

=

=

ν

ν ν πν

ρ

1 e

1 c

2

3

−

=

=

ω

π

ω π

ρ ρ ^h

ν

= ρ

This

This PlanckPlanck formulaformula hashas alsoalso beenbeen experimentallyexperimentally verifiedverified In term of

In term of ρρ_ωω;;

NoteNote

1 /kT h for e

c h 4

1 /kT h for c kT

4

/kT h - 3

2 3

2

>>

<<

ν ν ν

ν ν

ν

¾

¾ Planck result fits experiment perfectlyPlanck result fits experiment perfectly

¾¾ RR--J is accurate in infrared but diverges in ultravioletJ is accurate in infrared but diverges in ultraviolet

¾¾ WienWienworks well except in the infraredworks well except in the infrared

One important and interesting fact is to notice that the average One important and interesting fact is to notice that the average number of photons

number of photonsin each mode isin each mode is WhatWhat’’s s ρρ_TT((λλ) ?) ?

We know

We know νν= = cc//λλso so ddνν= = --((cc//λλ²²))ddλ λ and define and define ρρ_TT((λλ))ddλλ= -= -ρρ_TT(ν(ν))ddνν

so so ρρ_TT(λ(λ) = ) = --ρρ_TT((νν))ddνν//ddλλ= ρ= ρ_TT((νν))cc//λλ²²

1 )]

kT /(

hc exp[

d hc

8 1 )]

kT /(

hc exp[

d c

h d 8

)

( _{3 2 5}

T = −

= −

λ λ λ

π λ

λ λ

λ λ π λ ρ

Note: Both Stefan

Note: Both Stefan’’s Law & s Law & WienWien’’ssdisplacement law can be derived displacement law can be derived from the Plank formula

from the Plank formula

1 ) kT / h exp(

1 h

q E

= −

= ν ν

Exercise Exercise

Determine the wavelength of maximum emission for the human Determine the wavelength of maximum emission for the human body (37

body (37°°C), assuming a BlackC), assuming a Black--Body distribution of the emitted EM Body distribution of the emitted EM radiation.

radiation.

( )

K 310 T , K J 10 38065 . 1 k , s m 10 99879 . 2 c , s J 10 62608 . 6 h

1 e

hc 8 d

dE

1 23 1

8 34

kT / hc 5

=

⋅

=

⋅

=

⋅

=

= −

=

−

−

− +

−

λ λ

π ρ λ

( )

kT hc 5 1 kT) exp( hc 1 1 ...

1 0 e

hc 8 d 0 d d d

max max

310 T kT / hc 5

⋅

=

⎟⎟

⎟⎟

⎠

⎞

⎜⎜

⎜⎜

⎝

⎛

−

⋅

⎟⎟ =

⎠

⎜⎜ ⎞

⎝

⎛

⇔ −

=

=

λ λ

λ π λ λ

ρ

λ

The maximum in the energy distribution is obtaining solving:

The maximum in the energy distribution is obtaining solving:

μm 282 . kT 9 hc 5

~ 1 kT) exp( hc 1 1

kT 1 hc kT If hc

max max

max

max max

=

⋅

=

⎟⎟

⎟⎟

⎠

⎞

⎜⎜

⎜⎜

⎝

⎛

−

⋅

>>

⇔

>>

λ λ

λ λ λ

Exercise Exercise

A heater filament has a radius of 2 mm and a length of 200 mm. I A heater filament has a radius of 2 mm and a length of 200 mm. If its f its surface temperature is 2000 K what is the net radiated power?

surface temperature is 2000 K what is the net radiated power?

•• Radiated heat from object of temperature TRadiated heat from object of temperature T into surroundings into surroundings with temperature

with temperature TT₀₀is given byis given by RR = = eeσσAA((TT ⁴⁴--TT₀₀⁴⁴))

•• Since T Since T = 2000K and = 2000K and TT_{0 0} = 300 K the T = 300 K the T ⁴⁴term will be much larger term will be much larger than the

than the TT₀₀⁴⁴(check!) and so the rate of heat loss is(check!) and so the rate of heat loss is R = R = eeσσAAT T ⁴⁴

•

• Surface area of cylinder is given bySurface area of cylinder is given by

A =2A =2ππrrl = 2 x 3.14 l = 2 x 3.14 ××(2 x 10(2 x 10^-3-3m) ×m) ×0.2m = 2.51×0.2m = 2.51×1010^--33mm²²

•

• We will assume that e We will assume that e = 1, thus= 1, thus

R = 1 R = 1 ××(5.67×(5.67×1010^--88W mW m^--22KK^-4-4)( 2.5x10)( 2.5x10^-3-3mm²²)(2000 K))(2000 K)⁴⁴= 2.27 kW= 2.27 kW

Example: Sunlight falls at the rate of 1.4 kW/m

Example: Sunlight falls at the rate of 1.4 kW/m²² on the earthon the earth’’s s surface when the sun is directly overhead. The earth

surface when the sun is directly overhead. The earth’’s orbital s orbital radius is 1.5

radius is 1.5××1010¹¹¹¹ m while the sunm while the sun’’s radius is 7.0s radius is 7.0××1010⁸⁸ m. Find the m. Find the temperature of the sun

temperature of the sun’’s surface.s surface.

Solution:

Solution:

Intensity = Power/Area = P/A Intensity = Power/Area = P/A Power = IA = RA = (1.4x10

Power = IA = RA = (1.4x10³³W/mW/m²²)(4)(4π)(1.5x10π)(1.5x10¹¹¹¹m)m)²²

= 3.96x10

= 3.96x10²⁶²⁶WW

total power radiated by the sun total power radiated by the sun radiation rate from sun

radiation rate from sun’’s surface:s surface:

2 7 2

8 26 2

s

W/m 10 43 . ) 6 10 0 . 7 )(

4 (

W 10 96 . 3 r 4

P A

R P = ×

×

= ×

=

= π π

Let e=1 Let e=1

K 10 8 . 10 5

67 . 5 )(

1 (

10 43 . 6 e

T R ³

4 / 1 8 4 7

/ 1

×

⎟⎟ =

⎠

⎜⎜ ⎞

⎝

⎛

×

= ×

⎟⎠

⎜ ⎞

⎝

=⎛ ₋

σ