3Quantum gases
II Statistical Physics
3.2 Black-body radiation
Unfortunately, in this chapter, we do not have much opportunity to deal with
genuine gases. Instead, we are going to take some system that is not a gas, and
pretend it is a gas. Of course, nothing in our previous derivations relied on the
system actually being a gas. It just relied on the fact that we had a lot of things.
So the results should still hold.
In this chapter, we suppose have a gas of photons (which is just a fancy way
to say “photons”) in a box with opaque walls.
We suppose this box of photon is at equilibrium with temperature
T
. We will
use the box of photons as a model of a “black body”, i.e. a perfectly absorbing
object. The gas of photons inside is called black-body radiation. Later on, we
will argue that the photons emitted by a black body must indeed follow the
same distribution as the photons inside our box.
We begin by reminding ourselves of some basic properties of photons: they
are massless and have energy
E = ~ω, ω =
2πc
λ
,
and λ is the (genuine) wavelength of the photon.
Interactions between photons are negligible, so we can treat them as noble
gas.
Photons have two polarization states, since given any direction of travel
k
,
the electric and magnetic fields have to obey
E · k = B · k = B · E = 0,
and there are two independent choices for
E
and
B
. This has implications for
our counting, as we need an extra factor of 2 in our density of states. So the
density of states is given by
g(E) dE =
V E
2
π
2
~
3
c
3
dE.
Using the fact that
E = ~ω,
it is often convenient to instead write this as
g(ω) dω =
V ω
2
π
2
c
3
dω.
This is an abuse of notation, as the two
g
’s we wrote down are different functions,
but this is a physics course.
The final important property of the photons is that photon numbers are not
conserved. Indeed, they are absorbed and emitted by walls of the box. So we
must sum over all possible photon numbers, even though we are in the canonical
ensemble. In practice, what we get is the same as the grand canonical ensemble,
but with µ = 0.
We can begin. While we have figured out the density of states, we will not
use that immediately. Instead, we still begin by assuming that we have a discrete
set of states. Only after doing all the manipulations, we replace all remaining
occurrences of sums with integrals so that we can actually evaluate it.
We can work with this in more generality. Consider a system of non-
interacting particles with one-particle states
|ii
of energies
E
i
. Then the general
accessible state can be labelled by
{n
1
, n
2
, ···},
where
n
i
is the number of particles in
|ii
. Since particle numbers are not
conserved, we do not impose any restrictions on the possible values of the
n
i
.
The energy of such a state is just
X
i
n
i
E
i
.
As before, in the canonical ensemble, the probability of being in such a state is
p({n
i
}) =
1
Z
e
−β
P
n
j
E
j
,
and
Z =
X
{n
k
}
e
−β
P
n
j
E
j
=
∞
X
n
1
=0
e
−βn
1
E
1
∞
X
n=2
e
−βn
2
E
2
··· =
Y
i
1
1 − e
−βE
i
.
So we find that
log Z = −
X
i
log(1 − e
−βE
i
).
Using this, we have
hn
i
i =
X
{n
k
}
n
i
p({n
k
})
=
X
{n
k
}
n
i
e
−β
P
j
n
j
E
j
Z
= −
1
β
∂
∂E
i
log Z.
Using the formula of log Z we had, we find
hn
i
i =
1
e
βE
i
− 1
.
Applying this to photons, we replace the sum with an integral, and
E
i
=
~ω
.
Using our density of states, we have
log Z = −
V
π
3
c
3
Z
∞
0
dω ω
2
log(1 − e
−β~ω
).
Similarly, the average number of photons with frequency between ω and dω is
n(ω) dω = g(ω) dω ·
1
e
β~ω
− 1
=
V ω
2
dω
π
2
c
3
(e
β~ω
− 1)
.
Thus, the total energy in this range is
E(ω) dω = ~ωn(ω) dω =
V ~
π
2
c
3
ω
3
e
β~ω
− 1
dω.
This is the Planck distribution.
Let’s check that this makes sense. Let’s try to compute the total energy of
all photons. We have
E = −
∂ log Z
∂β
V
=
V ~
π
2
c
3
Z
∞
0
dω
ω
3
e
β~ω
− 1
=
Z
∞
0
E(ω) dω,
as we would expect.
Putting
ω
=
2πc
λ
, we find that the average energy with wavelength between
λ and λ + dλ is given by the horrendous formula
˜
E(λ) dλ =
V ~(2πc)
4
π
2
c
3
λ
5
dλ
e
β~2πc/λ
− 1
.
We can plot this for some values of T :
λ
E(λ)
Note that the maximum shifts to the right as temperature lowers, and the total
energy released also decreases. For the maximum to be near the visible range,
we need T ∼ 6000 K, which is quite a lot.
In general, if we want to figure out when E(ω) is maximized, then we set
dE
dω
= 0.
This gives us
ω
max
= ζ
kT
~
,
where ζ ≈ 2.822 is the solution to
3 − ζ = 3e
−ζ
.
This is known as Wien’s displacement law. In particular, we see that this is
linear in T .
Recall that we didn’t actually find the value of
E
. To do so, we have to do
the integral. We perform the change of variables x = β~ω to get
E =
V (kT )
4
π
2
c
3
~
3
Z
∞
0
x
3
dx
e
x
− 1
.
The remaining integral is just some constant, which isn’t really that important,
but it happens that we can actually evaluate it, and find the value to be
Z
∞
0
x
3
dx
e
x
− 1
= Γ(4)ζ(4) =
π
4
15
,
where ζ is the Riemann zeta function.
The energy density of the box is thus
E =
E
V
=
π
2
k
4
15~
3
c
3
T
4
∝ T
4
.
Now if we are experimentalists, then we have no way to figure out if the above
result is correct, because the box is all closed. To look into the box, we cut a
small hole in the box:
Let’s say this hole has area
A
, which is sufficiently small that it doesn’t mess
with what is going on in the box. How much energy do we expect the box to
leak out of the hole?
For the purposes of this analysis, for a wave vector
k
, we let
S
(
k
) be the set
of all photons with wave-vector within d
3
k
of
k
. We further suppose
k
makes
an angle θ with the normal to the hole.
k
θ
What we want to know is what portion of
S
(
k
) manages to get out of the hole,
say in a time period dt. To do so, we look at the following volume:
The volume of this box is
Ac cos θ
d
t
. So the proportion of energy that gets out
is simply
Ac cos θ
V
dt.
We now write
E(|k|) d
3
k = total energy in S(k).
Then the total energy leaving the hole in time dt as
Z
θ∈[0,π/2]
d
3
k
Ac cos θ
V
dt E(|k|).
To evaluate this integrate, we simply have to introduce polar coordinates, and
we get
cA dt
V
Z
2π
0
dϕ
Z
π/2
0
dθ sin θ cos θ
Z
∞
0
d|k| |k|
2
E(|k|).
The first two integrals give 2
π
and
1
2
respectively, and we can rewrite the last
integral back into Cartesian coordinates, and it is
1
4π
Z
d
3
k E(|k|) =
E
4π
.
Putting these all together, we find that the rate of energy emission is
1
4
cAE dt.
If we define the energy flux as energy per unit time per unit area leaving the
hole, then the energy flux is given by
c
4
E = σT
4
,
where
σ =
π
2
k
4
60~
3
c
2
≈ 5.67 × 10
−8
J s
−1
m
−2
K
4
.
What was the point of computing this? We don’t really care that much about
the experimentalists. Suppose we had any black body, and we want to figure
out how much radiation it is emitting. We imagine we put it inside that box.
Then to the box, the surface of the black body is just like the hole we drilled
through the box. So we know that the black body is absorbing
σT
4
energy per
unit time per unit area.
But the system is in equilibrium, so the black body must emit the exact same
amount of radiation out. So what we have derived is in fact how black bodies
behave!
The best example of a black body we know of is the cosmic background
microwave radiation of the universe. This is a black body radiation to incredibly
high accuracy, with a temperature of
T
=
2.7 K
. This is the temperature of
space.
Let’s quickly calculate some other thermodynamic quantities we might be
interested in. We have
F = −kT log Z
= −
V kT
π
2
c
3
Z
∞
0
dω ω
2
log(1 − e
−β~ω
)
Integrating by parts, we obtain
= −
V ~
3π
2
c
3
Z
∞
0
dω
ω
3
e
−β~ω
1 − e
−β~ω
= −
V ~
3π
2
c
3
1
β
4
~
4
Z
∞
0
x
3
dx
e
x
− 1
= −
V π
2
k
4
45~
3
c
3
T
4
.
The free energy is useful, because we can differentiate it to get pressure:
p = −
∂F
∂V
T
=
E
3V
=
1
3
E =
4σ
3c
T
4
.
This is radiation pressure. Since
c
is a big number, we see that radiation pressure
is small.
We can also get the entropy from this. We compute
S = −
∂F
∂T
V
=
16V σ
3c
T
3
.
Another quantity we can be interested in is the heat capacity, which is
C
V
=
∂E
∂T
V
=
16V σ
c
T
3
.
The classical/high temperature limit of black body radiation happens when
~ω kT . In this case, we have
1
e
β~ω
− 1
≈
1
β~ω
.
So we have
E(ω) dω ≈
V ω
2
π
2
c
3
kT dω ≡ E
classical
(ω) dω.
This is known as the Rayleigh-Jeans law. Note that there is no
~
in it. So it is
indeed a classical result. It also agrees with equipartition of energy, as it does
give
kT
per normal mode of the electromagnetic field, with each mode viewed
as a harmonic oscillator. We get kT because here we have a potential.
However, this has the obvious problem that
E
(
ω
) is unbounded as
ω → ∞
.
So if we tried to compute the total energy, then we get infinity. This is called
the ultraviolet catastrophy. This showed that classical reasoning doesn’t actually
work at high energies, and this is what eventually led Planck to come up with
Planck’s constant.