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Standing waves derivation
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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
=