Solved thermodynamics exercise

The photon gas in the Universe

Exercise 39 · Lesson 7 — Mathematical Structure: The Fundamental Relations

photon gas
radiation
cosmic microwave background
fundamental relation
adiabatic expansion
homogeneous function
zero chemical potential

Statement

In the early Universe, matter is a hot plasma of nuclei, electrons and photons. At around $T_{i} \approx 3000 K$ , the first atoms form and the photons become free. We assume that at this moment the photon gas is in equilibrium at temperature $T_{i}$ , and that it then undergoes a reversible adiabatic expansion as the Universe expands. We take as given its fundamental relation

S (U, V) = \frac{4}{3} a V^{1/4} U^{3/4},

where $a$ is a positive dimensional constant. The entropy does not depend on the number $N$ of photons: this number is not conserved.

Recall the definition of a homogeneous function of degree one and show that $S$ is one.
Sketch $S$ as a function of $U$ at fixed $V$ . Is this function concave? Comment.
Determine $T$ , $P$ and $μ$ from the derivatives of $S$ . Show in particular that $U = a^{4} V T^{4}$ and that $P = U / (3 V)$ , then comment on the value of $μ$ .
Write the first law for this gas and specialize it to a reversible adiabatic expansion. Deduce a relation between $U$ and $V$ along this transformation.
Deduce that $V T^{3}$ remains constant during the expansion.
Recover this result directly, from the entropy expressed as a function of $T$ and $V$ .
The photon gas is observed today at $T_{f} \approx 3 K$ : this is the cosmic microwave background. By what factor has the volume of the Universe increased since the emission?

Hint

In the entropy representation,

1/ T = (\partial S / \partial U)_{V}

P / T = (\partial S / \partial V)_{U}

and

μ / T = - (\partial S / \partial N)_{U, V}

. For Question 4, a reversible adiabatic transformation satisfies

d S = 0

, so

d U = - P d V

: separate the variables.

Detailed solution

Solution

Question 1. A function

f

of several variables is homogeneous of degree one if

f (λ x, λ y) = λ f (x, y)

for all

λ > 0

. Here,

S (λ U, λV) = \frac{4}{3} a (λV)^{1/4} (λ U)^{3/4} = \frac{4}{3} a λ^{1/4 + 3/4} V^{1/4} U^{3/4} = λ S (U, V) .

The entropy of the photon gas is therefore extensive.

Question 2. At fixed

V

S

is proportional to

U^{3/4}

. Its first derivative,

(\frac{\partial S}{\partial U})_{V} = a V^{1/4} U^{- 1/4} > 0,

is positive: $S$ increases with $U$ , and the temperature is positive. Its second derivative,

(\frac{\partial ^{2} S}{\partial U ^{2}})_{V} = - \frac{a}{4} V^{1/4} U^{- 5/4} < 0,

is negative: $S$ is concave in $U$ . The curve starts from the origin with a vertical tangent, then bends over while remaining above its chords. This concavity reflects the stability of the equilibrium of the photon gas.

Question 3. From Question 2,

\frac{1}{T} = a (\frac{V}{U})^{1/4} ⟹ (\frac{U}{V})^{1/4} = a T ⟹ U = a^{4} V T^{4} .

The derivative with respect to $V$ at fixed $U$ is

\frac{P}{T} = (\frac{\partial S}{\partial V})_{U} = \frac{4}{3} a \cdot \frac{1}{4} V^{- 3/4} U^{3/4} = \frac{a}{3} (\frac{U}{V})^{3/4} .

Now $(U / V)^{3/4} = ((U / V)^{1/4})^{3} = a^{3} T^{3}$ , hence

P = T \cdot \frac{a}{3} \cdot a^{3} T^{3} = \frac{a ^{4} T ^{4}}{3} = \frac{U}{3 V} .

Finally, $S$ does not depend on $N$ , so $μ = - T (\partial S / \partial N)_{U, V} = 0$ . The chemical potential of photons is zero: since their number is not conserved, it adjusts freely at equilibrium, and adding one costs nothing.

Question 4. Since the term

μ d N

is zero, the differential fundamental relation reads

d U = T d S - P d V

. For a reversible adiabatic expansion,

d S = 0

, so, with

P = U / (3 V)

d U = - P d V = - \frac{U}{3 V} d V ⟹ \frac{d U}{U} = - \frac{1}{3} \frac{d V}{V} .

Integrating between the initial state and any state during the expansion, $ln (U / U_{i}) = - \frac{1}{3} ln (V / V_{i})$ , that is,

U V^{1/3} = const .

Question 5. Replace $U$ by $a^{4} V T^{4}$ :

a^{4} V T^{4} \cdot V^{1/3} = a^{4} V^{4/3} T^{4} = const .

Since $a$ is constant, $V^{4/3} T^{4} = (V^{1/3} T)^{4}$ is constant, so $V^{1/3} T$ is too; cubing, $V T^{3} = const$ .

Question 6. Express

S

in terms of

T

and

V

by replacing

U

with

a^{4} V T^{4}

S = \frac{4}{3} a V^{1/4} (a^{4} V T^{4})^{3/4} = \frac{4}{3} a V^{1/4} a^{3} V^{3/4} T^{3} = \frac{4}{3} a^{4} V T^{3} .

A reversible adiabatic expansion is isentropic: $S$ is constant, and so is $V T^{3}$ . This directly recovers the result of Question 5.

Question 7. Since

V T^{3}

is conserved,

\frac{V _{f}}{V _{i}} = (\frac{T _{i}}{T _{f}})^{3} = (\frac{3000}{3})^{3} = (1 0^{3})^{3} = 1 0^{9} .

The volume of the Universe has increased by a factor of a billion since the emission, which corresponds to a factor of $1 0^{3}$ in distances. The radiation emitted at about $3000 K$ , cooled by the expansion, is the cosmic microwave background measured today at $2, 7 K$ .