Class 11: Thermal radiation

Class 11: Thermal radiation
By analyzing the results from a number of experiments, Planck found the energy density of the radiation
emitted by a ‘black body’ in wavelength interval ( λ , λ + d λ ) was well described by the formula
u (λ ) dλ =
8π hc
λ
5
1
e
hc ( λ kT )
−1
d λ,
(10.1)
where T is the temperature of the black body and h and k are constants. Planck, a short time later,
developed a theory that explained this result. Before considering Planck’s theory, it is useful to see what
classical physics predicts for black body radiation. As a preliminary we need a result from statistical
mechanics due to Boltzmann.
The Boltzmann factor
Consider a physical system that can be in a number of states. Associated with each state is an energy.
Boltzmann showed that when the system is thermal equilibrium with a heat reservoir of temperature T,
the probability, p, that the system is in a particular state of energy, E, is
p = Ae − E ( kT ) ,
(10.2)
where k is a constant, now known as the Boltzmann constant, and A is a constant normalization factor
that is determined by the requirement that the probabilities sum to unity. The term e
Boltzmann factor.
− E ( kT )
is called the
As an example of application of the Boltzmann factor consider a system that has two states: a ground
state of energy 0 and an excited state of energy E. The probabilities that the system is one of the two
states are A for the ground state and Ae
− E ( kT )
for the excited state. By summing the probabilities, we find
A=
1
1 + e − E ( kT )
.
(10.3)
We see that at low temperature ( kT ≪ E ) , the probability that the system is in its ground state is close
to 1, and the excited state has a very low probability. At high temperatures ( kT ≫ E ) , the Boltzmann
factors are equal and the two states have equal probability. The average energy of the system is
0 + Ee − E ( kT )
E
E=
= E ( kT ) .
− E ( kT )
1+ e
e
+1
1
(10.4)
For a system with many states, it is likely that some states have the same energy. The number of states
with a particular energy is called the statistical weight if the energies are discrete. For a continuous
energy distribution, the corresponding quantity is called the density of states.
The Rayleigh-Jeans law
To develop an expression for the energy density of black body radiation, Rayleigh considered the
radiation emitted through a small hole in a cubical cavity with walls kept at temperature, T. The walls
contain classical harmonic oscillators that can absorb and emit radiation. This provides the coupling
between the system, i.e. the radiation in the cavity, and the heat reservoir, i.e. the cavity walls, needed for
the system to be in thermal equilibrium, so that we can use the Boltzmann factor.
The first step in the calculation is determine the density of states. To do this it is assumed that walls of the
cavity are perfect electrical conductors. This means that the electrical field associated with the
electromagnetic waves in the cavity must be zero on the cavity walls to avoid generating currents of
infinite magnitude. The electromagnetic waves are then standing waves with nodes at the cavity walls. A
1D analogy is standing waves on a string of length L with fixed ends. The allowed wavelengths for these
waves are
2L
,
n
λn =
(10.5)
where n is a positive integer. Treating n as a continuous variable, the number of waves with wavelengths
in the interval ( λ , λ + d λ ) is
dn =
2L
λ2
d λ.
(10.6)
Note that dn is taken to be positive. We see that the density of states is higher at shorter wavelengths.
The generalization of equation (10.5) to three dimensions is
( 2L )
2
λ =
(n
2
x
2
+ n y 2 + nz 2 )
,
(10.7)
where nx, ny, and nz are independent positive integers. To determine the density of states, we treat the
integers as continuous variable and note that once we restrict to positive integers the volume element in n-
(
)
space is π n 2 2 dn. Because there are two independent polarizations of light, the number of waves in
this volume element is
2×
π
2
( 2L )
n dn = π
2
2
λ
2
where V is the volume of the cavity.
2
2L
λ
2
dλ =
8π V
λ4
d λ,
(10.8)
The second step is to consider the mean energy of a wave of wavelength λ. In classical physics, the
energy of the wave is proportional to the square of the amplitude of the wave, and there are no other state
variables, i.e. the statistical weight is 1. The mean energy can be found by employing the Boltzmann
factor:
∞
∫ Ee
E=
0
∞
∫e
−
E
kT
−
E
kT
dE
= kT .
(10.9)
dE
0
Combining these results, we find that the energy density per unit wavelength is
u=
8π
λ4
kT .
(10.10)
This is the Rayleigh-Jeans law. Integration over all wavelengths gives the troubling result that the cavity
should contain an infinity amount of energy! Since the expression for the energy density diverges as the
wavelength becomes shorter, this result was called the ultraviolet catastrophe.
Note that at long wavelengths (low frequencies), the Planck formula gives
u (λ ) =
8π hc
λ
5
1
e
hc ( λ kT )
−1
=
8π hc
5
λ
1
hc


+… −1
1 +
 λ kT

≈
8π
λ4
kT ,
(10.11)
which agrees with the Rayleigh-Jeans result.
Planck’s theory
Planck’s approach to avoiding the ultraviolet catastrophe was to find a way to make the average energy of
a high frequency oscillator less than kT, yet leave the energy of a low frequency oscillator close to kT to
get correct low frequency limit. He postulated that the oscillators could emit or absorb energy in discrete
packets, with the energy in a packet being an integer multiple of a constant times the frequency
En = nhf ,
(10.12)
where h is Planck’s constant, and n = 0, 1, 2, …. The mean energy of a wave is now
∞
∑ nhfe
E=
− nhf kT
n=0
∞
∑e
.
− nhf kT
n=0
To perform the summations, let x = hf kT . The denominator is
3
(10.13)
∞
Z = ∑e
∞
− nhf kT
= ∑e
n=0
∞
− nx
= ∑ ( e− x ) =
n =0
n
n=0
1
.
1 − e− x
The numerator is
∞
∑ nhfe−nhf
∞
kT
n =0
= hf ∑ ne − nx = − hf
n=0
d ∞ − nx
d
1
e− x
e
hf
hf
=
−
=
.
∑
−x 2
dx n = 0
dx 1 − e − x
1
e
−
(
)
Hence
E
e− x
x
=x
= x .
−x
kT
1− e
e −1
(10.14)
We see that when x = hf kT ≪ 1, the mean energy is close to kT, but when x ≫ 1, the exponential term
in the denominator dominates and the mean energy is much less than kT. Also note that the classical result
is recovered if Planck’s constant is replaced by 0.
Using equation (10.14) in place of equation (10.9), we find that the energy density per unit wavelength is
u (λ ) =
8π
hf
4
hf kT
λ e
−1
=
8π
hc
5
hc λ kT
λ e
−1
,
(10.15)
which is the Planck result.
Planck’s postulate that the oscillator could absorb or emit only in discrete packets of energy was
revolutionary, and marks the beginning of quantum physics.
Radiation intensity
The radiation from a small hole in a cavity wall is emitted not in a single direction but over one
hemisphere of directions (i.e. into solid angle 2π steradians). By averaging over all possible directions, it
is found that ¼ of the radiation flux is directed outward perpendicular to the wall. The radiation intensity
(or flux) per unit wavelength is then
I (λ ) =
1
cu ( λ ) .
4
(10.16)
Integrating over all wavelengths gives the total flux
∞
∞
1
1 8π
hc
F = c ∫ u ( λ ) d λ = c ∫ 5 hc λ kT
d λ.
4 0
4 0λ e
−1
Making the substitution x = hc ( λ kT ) , we have
4
(10.17)
F =c
2π ( kT )
( hc )
3
4 ∞
x3
∫0 e x − 1 dx.
(10.18)
Hence the radiative flux is proportional to the 4th power of the temperature of the black body, in
agreement with Stefan’s law (also called the Stefan-Boltzmann law):
F = σT 4.
(10.19)
On evaluating the integral, we find that the Stefan-Boltzmann constant is
σ=
2π 5 k 4
.
15 c 2 h3
(10.20)
The Wien displacement law
The figure below shows the Planck function plotted against wavelength for a temperature equal to the
effective temperature of the Sun.
3.0x10
14
5777 K blackbody
Bλ
(erg cm-3 s-1 sr-1)
2.5x1014
2.0x1014
Rayleigh - Jeans law
1.5x1014
1.0x10
14
5.0x10
13
UV Visible Infra-red
0.0
0
200
400
600
800 1000 1200 1400 1600 1800 2000
λ (nm)
We see that, by no coincidence, the peak of the black body radiation lies in the visible part of the
electromagnetic spectrum, near a wavelength of 500 nm.
The figure below shows the Planck function for three temperatures. We see the wavelength of the peak
depends inversely on the temperature of the black body. This is Wien’s displacement law, which can also
5
10
T⊙/2
T⊙
251 nm
15
T⊙×2
1014
502 nm
1013
Bλ
(erg cm-3 s-1 sr-1)
1016
1004 nm
10
12
1011
0
200
400
600
800
1000 1200 1400 1600 1800 2000
λ (nm)
be found by differentiation of the energy density given in equation (10.15). Again let x = hc ( λ kT ) , so
that
u (λ ) =
8π ( kT )
( hc )
4
5
H ( x) ,
(10.21)
where
H ( x) =
x5
.
ex −1
(10.22)
The function H(x) has a maximum at x = 4.9651, and so the peak occurs at wavelength
λ peak = 0.2014
hc 2.898 × 106
=
nm ⋅ K.
kT
T
6
(10.23)

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