Derivation of number of standing i l i di iwaves per unit volume in radiation cavityy

Cavity surface is FULL with oscillators

Standing wavesNumber of modes?

Number of standing waves y

Cavity is filled with standing waves

b f di i diff

Number of standing waves one can form in a given area (2D case) or volume

L

Number of standing waves is different for different wavelengths

(3D case) with varying orientation

L

Modes in 1Dimentional where nx is integerLnx =2λ

Number of modes between λ and λ+dλ: N(λ) dλ = dλλ2LΔn 2x −=

Number of modes in 2D

BB 2πk2πk)(kiC)f(BB

2λ

yλx

x λ2πk;

λ2πk;ωt)(kxsinCt)f(x, ==−=

k d k d f h kβα

M

AO

2 βα

M

AO

kx and ky are x and y components of the wave vector k.

For forming a standing wave the amplitude of the wave must be zero at x=LAO

2xλ

AO

yx nLnλλ

== cos α2L2Ln

the wave must be zero at x=L

22 yx nLn == cos αλλ

nx

x ==

cos β2L2Ln ==

OM=λ/2

OA=λx/2

OB=λ /2

cos α = λ /λx ;

cos β = λ /λy; cos βλλ

ny

y ==

cos γ2L2Ln

OB=λy/2 y

cos γ = λ /λz ;

cos γλλ

ny

z ==

Number of modes in 2D

BB

2λ

yλ

Condition for formation of “Standing wave” i 2D

λ4Lnn 2

22y

2x =+

2 βα

M

AO

in 2D space 1)βcosαcos( 22 =+Q

2xλ

AO

To know the number of modes of a given wavelength λ in 2D :Consider number space and draw a circle with radius of 2L/ λ.All points lying on the circle will satisfy the condition for standing wave

Number of modes in 3D cavityλ

2Lradiusdλλ

2Lradius+ Each unit volume represents a mode

ny

λ dλλ +

The number of modes between λ and λ+dλ.=The volume of spherical shell

Each unit volume represents a mode.

nnxWe have to consider only +ve integers∴ ONLY of the volume representspossible modes

81

dr) r (4 shell spherical theof Volume 2π modestongcontributiVolume

possible modes

dλλ8L 4π dλ

λ2L

λ2L π4 4

3

2

2

=⎥⎦⎤

⎢⎣⎡= dλ

λL4π dλ

λ8L 4π

81 4

3

4

3

=⎥⎦

⎤⎢⎣

⎡=

⎤⎡Polarization : There are TWO polarizations for each EM wave

dλλ

L4π2 modesofnumberTotal 4

3

⎥⎦

⎤⎢⎣

⎡=

dλλ8π )dN( 4=λλNumber of modes between λ and λ+dλ

per unit volume are:

Number of modes in 2D

B 2πk2πk)(kiC)f(B

2λ

yλx

x λ2πk;

λ2πk;ωt)(kxsinCt)f(x, ==−=

k d k d f h k2 βα

M

AO

kx and ky are x and y components of the wavevector k.

For forming a standing wave the amplitude of the wave must be zero at x=L

2xλ

AO

yx nLnλλ

== cos α2L2Ln

of the wave must be zero at x=L

22 yx nLn == cos αλλ

nx

x ==

cos β2L2Ln ==

OM=λ/2

OA=λx/2

OB=λ /2

cos α = λ /λx ;

cos β = λ /λy; cos βλλ

ny

y ==

cos γ2L2Ln

OB=λy/2 y

cos γ = λ /λz ;

cos γλλ

ny

z ==

Number of modes in 2D cavity

Each dot represents different mode of ny c do ep ese s d e e ode o

different λ or of same λ :Dot 1: λ =2L, wave moving along x-axisDot 2: λ =L, wave moving along x-axis5 6 , g gDot 3: λ =2L/3 wave moving along x-axisDot 4: λ =0.5L wave moving along x-axisDot 5: λ =L wave moving along y-axis

1 2 3 4

7

nx

L5/4=λDots 2 & 5: two modes of λ =LDots 6 & 7: two modes ofDegeneracy

Practical cavities have dimensions of cm or m What happens & We are considering λ in micrometers. ∴ The number of nodes in any direction are : >104.

if the cavity is of micron size

?The number of points will be almost continuously spread in chosen space ⇒All λ will form SW

Number of modes in 2D cavity  

λ2Lradius 2L Each unit area represents a mode.

nyλ

dλλ2Lradius+ The number of modes between λ and λ+dλ.

= The area between two rings

ac u e ep ese s a ode.

nx

We have to consider only +ve integers∴¼ area ONLY represents possible modes

dr)r(2ringsbetwen twoArea π modestongcontributiArea

∴¼ area ONLY represents possible modes

dλλ

4L 2π dλλ2L

λ2L π2

dr)r (2rings betwen two Area

3

2

2 ==

π

dλλ

L2π dλλ

4L 2π41

modestongcontributiArea

3

2

3

2

=⎥⎦

⎤⎢⎣

⎡=

λλλ

Polarization : There are TWO polarizations for each EM wave∴ Number of modes get doubled

dλλ4π )dN(: 3=λλNumber of modes between λ and λ+dλ

per unit area are:

Number of modes in 3D cavity

λ2Lradius

dλλ2Lradius+

Each unit volume represents a mode.ny

λ dλλ +

The number of modes between λ and λ+dλ.=The volume of spherical shell

nnxWe have to consider only +ve integers∴ ONLY of the volume representspossible modes

81

dr) r (4 shell spherical theof Volume 2πL48L1

modestongcontributiVolume33 ⎤⎡

possible modes

dλλ8L 4π dλ

λ2L

λ2L π4 4

3

2

2

=⎥⎦⎤

⎢⎣⎡= dλ

λL4π dλ

λ8L 4π

81 4

3

4

3

=⎥⎦

⎤⎢⎣

⎡=

⎤⎡Polarization : There are TWO polarizations for each EM wave

dλλ

L4π2 modesofnumberTotal 4

3

⎥⎦

⎤⎢⎣

⎡=

dλλ8π )dN( 4=λλNumber of modes between λ and λ+dλ

per unit volume are:

Blackbody radiation: Ultraviolet catastrophe

dλλ8π :are case 3Din eunit volumper modes ofnumber Total 4

kTdλλ8π dλ)λ(ρ 4 ⎥⎦

⎤⎢⎣⎡=

Average energy of the mode : kT

Energy density inside the 3D cavity: λ ⎦⎣

Why radiation is less at longer wavelengths?: there just isn’t enough room in the box to fit

gy y y

T2

Ultraviolet catastrophej g

many standing wave of longer wavelengths

There is no limit to the number of short l th d th t i t i th b d

T2

T2> T1

length modes that can exist in the box, and so we predict the “Ultraviolet catastrophe”

At short wavelengths the classical curve is T2At short wavelengths the classical curve is disagrees completely with experiment.

OUT OF LINE Solution by Planck

Oscillator energies E = nhν n = 0 1 2 ;Edλ

λ8π osc4 ⎥⎦

⎤⎢⎣⎡=

Oscillator energies E = nhν, n = 0,1,2…;h (Planck’s constant) = 6.62x10-34 Js

Oscillator’s energy can only change by discrete

1eλhcE kThc/osc −

= λ

Oscillator s energy can only change by discrete amounts, (Oscillators can absorb or emit energy in small packets – “quanta”; Equantum = hν ONLYq

Average energy of oscillators is derived assuming that oscillator energies can vary only specific values E = 0, hν, 2hν, 3hν, & using g y y p , , , , g“Boltzmann factor” n(E) = e-E/kT

When Planck was deriving the formula even he did notWhen Planck was deriving the formula even he did not know what to make of his assumption, but it was the

first hint of “quanta” and Universal constant hfirst hint of quanta and Universal constant h

Planck called his theory “an act of desperation”.

Number of modes or Density of states between λ and λ+dλ

1D case per unit length

2

2D case perunit area

4π

3D case per unit volume

8πdλλ2

2 dλλ4π

3 dλλ8π

4

ny

Density of states in term of frequencyy

2D Case: Density of states ν and ν+dνnx

π 2Lυ2Lrwheredr)r(2ringsbetwen twoArea ==

per unit area are:

υυπυυ

π

dcL4 d

c2L

c2L π2

41(2)

cλrwhere,dr)r (2 rings betwen two Area

2

2

=⎥⎦⎤

⎢⎣⎡=

Polarization factor Only +ve integers contribute

Total number of oscillators with natural frequency ν are No

State Energy Number of oscillators Energy Average energyin the given state total

Ground E0=0 E0N1

1st excited E =hν E N/kT)Eexp(AN =/kT)Eexp(A N 01 −= hE /kThosc = υ

υ1st excited E1=hν E1N2

2nd excited E2=2hν E2N3

3rd excited E3=3hν E3N4

/kT)Eexp(A N 23 −=/kT)Eexp(A N 12 −=

/kT)Eexp(A N 34 −=

1e /kThosc −υ

1λehcE OR kThc/λosc −

=34

hcdλ8π)( ⎥⎤

⎢⎡=λλρ d

1eλdλ

λ )( kThc/4 −⎥⎦⎢⎣= λλλρ d

Number of modes or Density of states between λ and λ+dλ

1D case per unit length

2

2D case perunit area

4π

3D case per unit volume

8πdλλ2

2 dλλ4π

3 dλλ8π

4

ny

Density of states in term of frequencyy

2D Case: Density of states ν and ν+dνnx

π 2Lυ2Lrwheredr)r(2ringsbetwen twoArea ==

per unit area are:

υυπυυ

π

dcL4 d

c2L

c2L π2

41(2)

cλrwhere,dr)r (2 rings betwen two Area

2

2

=⎥⎦⎤

⎢⎣⎡=

Polarization factor Only +ve integers contribute

Density of states between ν and ν+dν : in 3D case

ny

π

L82L2L1

c2Lυ

λ2Lr where, dr) r (4 spheres betwen two shell theof Volume

232

2

⎤⎡ ⎞⎛

==

nx

υυπυυ dcL8 d

c2L

c2L π4

81(2) 3

232

=⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

Polarization factor Only +ve integers contributePolarization factor Only +ve integers contribute

1D case per unit length

2D case perunit area

3D case per unit volume

dυc2

dυcυ4π2 dυ

cυ8π3

2

Number of modes or Density of states between k and k+dknyCny

22Lk

λ2Lr where, dr)r (2 rings betwen two Area ==

ππ

2D Case:

nxdkπ

kL dkπL

πLk π2

41(2)

2λ2

=⎥⎦⎤

⎢⎣⎡=

π

Density of states between k and k+dk : in 3D case

Polarization factor Only +ve integers contribute

dkkLLdkLkπ41(2)

22Lk

λ2Lr where, dr) r (4 spheres betwen two shell theof Volume

232

2

⎥⎤

⎢⎡

⎟⎞

⎜⎛

==π

π

dkπ

ππ

π48

(2) 2=⎥⎥⎦⎢

⎢⎣

⎟⎠

⎜⎝

=

1D :unit length 2D : unit area 3D :unit volume

dkπ1

dkπ

kdk

πk

2

2

Number of modes or Density of states between p and p+dpnyCny

h2Lp

λ2Lr where, dr)r (2 rings betwen two Area ==π

2D Case:

nxdph

pπL4 dph

2Lh

2Lp π241(2) 2

2

=⎥⎦⎤

⎢⎣⎡=

Density of states between p and p+dp : in 3D case

Polarization factor Only +ve integers contribute

dppL8π2Ldp2Lpπ41(2)

h2Lp

λ2Lr where, dr) r (4 spheres betwen two shell theof Volume

232

2

⎥⎤

⎢⎡

⎟⎞

⎜⎛

==π

dph

hh

π48

(2) 2=⎥⎥⎦⎢

⎢⎣

⎟⎠

⎜⎝

=

1D :unit length 2D : unit area 3D :unit volume

dph2 dp

hp4π2= dp

hp8π3

2

=

Number of modes or Density of states between E and E+dEnyCny

h2mE2L

λ2Lr where, dr)r (2 rings betwen two Area ==π

2D Case:

nxdEh

mπL4 dEE

2m21

h2L

h2mE2L π2

41(2) 2

2

=⎥⎦

⎤⎢⎣

⎡=

Density of states between E and E+dE : in 3D case

Polarization factor Only +ve integers contribute

3232

2

dEELπm 8dE2m12L2mE2Lπ41(2)

h2mE2L

λ2Lr where, dr) r (4 spheres betwen two shell theof Volume

⎥⎤

⎢⎡

⎟⎟⎞

⎜⎜⎛

==π

3h dE

E2hh π4

8(2) =

⎥⎥⎦⎢

⎢⎣

⎟⎟⎠

⎜⎜⎝

=

1D :unit length 2D : unit area 3D :unit volume

dEE

2m dEhm4π2= dE

hEπm28

3

23

=