chapter 1 thermal radiation and planck’s postulate 1.2 thermal radiation thermal radiation: the...

Chapter 1 Thermal radiation and Planck’s postulate

1.2 thermal radiation

Thermal radiation: The radiation emitted by a body as a result of temperature. Blackbody : A body that surface absorbs all the thermal radiation incident on

them. Spectral radiancy : The spectral distribution of blackbody radiation.)(TR

:)( dRT represents the emitted energy from a unit area per unit time between and at absolute temperature T. d

1899 by Lummer and Pringsheim


The spectral radiancy of blackbody radiation shows that:

(1) little power radiation at very low frequency

(2) the power radiation increases rapidly as ν increases from very

small value.

(3) the power radiation is most intense at certain for particular

temperature.

(4) drops slowly, but continuously as ν increases

, and

(5) increases linearly with increasing temperature.

(6) the total radiation for all ν ( radiancy )

increases less rapidly than linearly with increasing temperature.

max

)(,max TR

.0)( TR

max

dRR TT )(0


Stefan’s law (1879): 4284 /1067.5, KmWTR o

T Stefan-Boltzmann constant

Wien’s displacement (1894):

Tmax

1.3 Classical theory of cavity radiation

Rayleigh and Jeans (1900):

(1) standing wave with nodes at the metallic surface

(2) geometrical arguments count the number of standing waves

(3) average total energy depends only on the temperature

one-dimensional cavity:

one-dimensional electromagnetic standing wave

)2sin()2

sin(),( 0 tx

EtxE


for all time t, nodes at .......3,2,1,0,/2 nnx

ancnanaax

x

2//22

0

standing wave

:)( dN the number of allowed standing wave between ν and ν+dν

dcadndN

dcadncan

)/4(2)(

)/2()/2(

two polarization states

n0

))(/2( dcad

)/2( cad


for three-dimensional cavity

dcadrcar )/2()/2(

the volume of concentric shell drrr

dc

Vd

c

adrrdN

dc

ad

c

av

c

adrr

23

23

32

23222

884

8

12)(

)2

(4)2

()2

(44

The number of allowed electromagnetic standing wave in 3D

Proof:

nodal planes

)2sin()/2sin(),(

)2sin()/2sin(),(

)2sin()/2sin(),(

2/cos)2/(

2/cos)2/(

2/cos)2/(

0

0

0

tzEtzE

tyEtyE

txEtxE

zz

yy

xx

z

y

x

propagation direction

λ/2

λ/2


for nodes:

.....3,2,1,/2,,0

.....3,2,1,/2,,0

.....3,2,1,/2,,0

zzz

yyy

xxx

nnzaz

nnyay

nnxax

222

2222222

/2

)coscos(cos)/2(

cos)/2(,cos)/2(,cos)/2(

zyx

zyx

zyx

nnna

nnna

nanana

dcadrcannnr

racnnnacc

zyx

zyx

)/2()/2(

)2/()2/(/

222

222

dcadcadN

dNdrrdrrdrrN2323

22

)/(4)/2)(2/()(

)(2/4)8/1()(

considering two polarization state

dcVdN 23)/1(42/)(

:/8)( 32 cN Density of states per unit volume per unit frequency


the law of equipartition energy:

For a system of gas molecules in thermal equilibrium at temperature T,

the average kinetic energy of a molecules per degree of freedom is kT/2,

is Boltzmann constant.Kjoulek o/1038.1 23

average total energy of each standing wave : KTKT 2/2

the energy density between ν and ν+dν:

kTdc

dT 3

28)( Rayleigh-Jeans blackbody radiation

ultraviolet catastrophe


1.4 Planck’s theory of cavity radiation

),( T Planck’s assumption: and 0,0

kT the origin of equipartition of energy:

Boltzmann distribution kTeP kT /)( /

:)( dP probability of finding a system with energy between ε and ε+dε

kT

kTekTekTkT

dkT

edP

ekTkT

dkT

edP

dP

dP

kTkT

kT

kTkT

])(|)([1

)(

1|)(1

)(

)(

)(

0

/0

/

0 0

/

0/

0

/

0

0

0


Planck’s assumption: ..............4,3,2,,0 kTkT ,

kTkT ,

kTkT ,

kT0 (1) small ν

(2) large large ν0

sjoulh

h

341063.6

Planck constant

Using Planck’s discrete energy to find

kTh

e

enkT

ekT

ekTnh

P

p

nnh

n

n

n

n

n

kTnh

n

kTnh

n

n

/

1)(

)(

......3,2,1,0,

0

0

0

/

0

/

0

0


0

0

0

0

0

0

0

ln

n

n

n

n

n

n

n

n

n

n

n

n

n

n

e

en

e

edd

e

edd

ed

d

00

ln]ln[n

n

n

n ed

dhe

d

dkT

1132

32

0

)1()1(.......1

.....1

eXXXX

eeee

eX

n

n

11)

1

1(

)]1ln([)()1ln(

/

1

kThe

h

e

he

eh

ed

dhe

d

dh

01

/1/

/

hekTh

kTkThekThkTh

kTh


energy density between ν and ν+dν: 1

8)( /3

2

kThT e

h

c

1

18)()()(

)()(

/52

kThcTTT

TT

e

hcc

d

d

dd

Ex: Show )()/4()( TT Rc

dA

dV

r22 4

cos

4

ˆ

r

dA

r

rAd

solid angle expanded by dA is

spectral radiancy:

)(4

sin4

cos)(

)/()4

cos()()(

2220

2/

0

2

0

2

T

tc

T

TT

c

drrtr

dd

tdAr

dAdVR


Ex: Use the relation between spectral radiancy

and energy density, together with Planck’s radiation law, to derive

Stefan’s law

dcdR TT )()/4()(

32454 15/2, hckTRT

44

3

4

2

0

3

3

4

2

0 /

3

200

15

)(2

1

)(2

1

2)(

4)(

Th

kT

c

dxe

x

h

kT

c

de

h

cd

cdRR

x

kThTTT

15/)1/(

/

4

0

3

dxex

kThx

x

32

45

15

2

hc

k


Ex: Show that 15/)1( 41

0

3

dxex x

dyeyn

dxexdxeexI

eeee

dxeexdxexI

y

nn

xn

n

nxx

n

nxxxx

xxx

0

3

04

00

)1(3

00

3

0

21

1

0

31

0

3

)1(

1

.....1)1(

)1()1(

Set yxn eenyxndydxxny )1(33 ,)1/()1/()1(

14

04

0

3

16

)1(

16

6

nn

y

nnI

dyey by consecutive partial integration

?1

14

n n

90

1148

18

5)(

6

1)(

4

14

14

12

24

44

2

12

2

nnn

x

n

x

nnnxF

nxF :F Fourier series expansion


Ex: Derive the Wien displacement law ( ), Tmax ./2014.0max khcT

15

0)1(

50

)(

1

8)(

2/

/

/

/5

x

kThc

kThc

kThcT

kThcT

ex

e

e

kT

hc

ed

d

e

hc

kThcx /

xeyx

y 21 ,5

1

Solve by plotting: find the intersection point for two functions

5/11 xy

xey 2

Tmax

5

Y

X

intersection points:965.4,0 xx

khcT /2014.0max


1.5 The use of Planck’s radiation law in thermometry

(1) For monochromatic radiation of wave length λ the ratio of the spectral

intensities emitted by sources at and is given byKT o1 KT o

2

1

12

1

/

/

kThc

kThc

e

e

:

:

2

1

T

T standard temperature ( Au )

unknown temperature

CT omelting 1068

(2) blackbody radiation supports the big-bang theory. Ko3

optical pyrometer


1.6 Planck’s Postulate and its implication

Planck’s postulate: Any physical entity with one degree of freedom whose

“coordinate” is a sinusoidal function of time

(i.e., simple harmonic oscillation can posses

only total energy nh

Ex: Find the discrete energy for a pendulum of mass 0.01 Kg suspended

by a string 0.01 m in length and extreme position at an angle 0.1 rad.

295

333334

5

102105

10)(106.11063.6

)(105)1.0cos1(1.08.901.0)cos1(

sec)/1(6.11.0

8.9

2

1

2

1

E

EJhE

Jmgmgh

l

g

The discreteness in the energy is not so valid.

chapter 1 thermal radiation and planck’s postulate 1.2 thermal radiation thermal radiation: the...

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