{title2=$s$}
{wiki}

For a homogeneous equilibrium system with <energy density> $\rho$, <pressure> $P$, <chemical potential> $\mu$, and <number density> $n$, the entropy density is
$$
s=\frac{\rho+P-\mu n}{T}.
$$

Entropy density

For an <ultrarelativistic particle>, $E\simeq|\mathbf p|$. Setting $\mu=0$ and writing the three-dimensional momentum integral radially gives
$$
n=\frac{gT^3}{2\pi^2}\int_0^\infty\frac{x^2\,dx}{e^x\mp1},
\qquad
\rho=\frac{gT^4}{2\pi^2}\int_0^\infty\frac{x^3\,dx}{e^x\mp1},
\qquad
P=\frac{\rho}{3}.
$$
The standard Bose integrals yield
$$
n_b=\frac{g\zeta(3)}{\pi^2}T^3,
\qquad
\rho_b=\frac{g\pi^2}{30}T^4,
\qquad
P_b=\frac{\rho_b}{3}.
$$
For <fermions>, the corresponding integrals differ by the familiar factors
$$
n_f=\frac34\frac{g\zeta(3)}{\pi^2}T^3,
\qquad
\rho_f=\frac78\frac{g\pi^2}{30}T^4,
\qquad
P_f=\frac{\rho_f}{3}.
$$
Thus the <number density> scales as $T^3$, while the <energy density> and <pressure> scale as $T^4$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 310 / 1 / a / iii / Solution

Summing over the $g$ internal states, the <number density>, <energy density>, and isotropic <pressure> are
$$
n=g\int\frac{d^3p}{(2\pi)^3}f(\mathbf p),
\qquad
\rho=g\int\frac{d^3p}{(2\pi)^3}E(\mathbf p)f(\mathbf p),
$$
and
$$
P=g\int\frac{d^3p}{(2\pi)^3}
\frac{\mathbf p^2}{3E(\mathbf p)}f(\mathbf p).
$$
The factor $\mathbf p^2/(3E)$ in the <kinetic pressure of an isotropic gas> is the angular average of one diagonal component of the momentum flux.

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 310 / 1 / a / ii / Solution

{title2=$\rho$}
{wiki}

The energy density is energy per unit volume. A homogeneous particle species with one-particle energy $E(\mathbf p)$ has
$$
\rho=g\int\frac{d^3p}{(2\pi)^3}E(\mathbf p)f(\mathbf p).
$$


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