Question

The formula for energy \(E\) of a photon gas at temperature \(T\) in a two-dimensional box at equilibrium with \(g_{2d}(\nu)\) denoting the density of states of photons is given below, where symbols \(\nu\), \(h\) and \(k_B\) have their standard meaning. The specific heat (\(C_V\)) of this photon gas obeys
\[ E = \int_0^{\infty} d\nu \, g_{2d}(\nu) \, \frac{h\nu}{\exp\left(\dfrac{h\nu}{k_B T}\right) - 1} \]

• A. \(C_V \propto T\)
• B. \(C_V \propto T^2\)
• C. \(C_V \propto T^3\)
• D. \(C_V \propto T^4\)

Hint:

In \(d\) dimensions the photon density of states goes as \(g_d(\nu)\propto \nu^{d-1}\).
For \(d=2\), working through the integral gives \(E\propto T^3\), so \(C_V=\partial E/\partial T \propto T^2\).

Answer 1

The Correct Option is B

Step 1: Use a scaling argument instead of doing the integral first.
The only place temperature enters the integrand is through the combination $h\nu/(k_BT)$, so the integral is only sensitive to frequencies of size $\nu \sim k_BT/h$. Every mode below this scale contributes about $k_BT$ of energy, and the density of states up to this scale, since $g_{2d}(\nu)\propto \nu$, carries an overall size $\propto (k_BT/h)^2$. Multiplying the number of relevant modes $\propto (k_BT/h)^2$ by the energy per mode $\propto k_BT$ already shows $E \propto T^3$, without touching the exact integral.

Step 2: Differentiate directly under the integral sign for $C_V$, as a check.
\[ C_V = \frac{\partial E}{\partial T} = \int_0^\infty g_{2d}(\nu)\, h\nu \, \frac{\partial}{\partial T}\left[\frac{1}{e^{h\nu/k_BT}-1}\right] d\nu \]
Carrying out the derivative and substituting $x = h\nu/k_BT$ the same way as before turns the integral into a pure number times a power of $T$; tracking the powers of $T$ that come out of $g_{2d}(\nu)\propto\nu \propto T$, the extra $h\nu\propto T$, and one more factor of $T$ from the derivative acting on the exponential term, the net power left over is $T^{1+1+1-1} = T^2$.

Step 3: Cross-check with the general d-dimensional rule.
The same substitution done in general for a $d$-dimensional photon gas, where $g_d(\nu)\propto \nu^{d-1}$, always gives $E \propto T^{d+1}$ and hence $C_V \propto T^d$. For $d=3$ (ordinary blackbody radiation) this correctly reproduces the familiar $C_V\propto T^3$ result, and for $d=2$ it gives $C_V \propto T^2$, matching what both methods above found.

Final Answer:
Two independent routes, the scaling estimate and the general $d$-dimensional rule, both give the same power. \[ \boxed{C_V \propto T^2 \ \text{(Option B)}} \]