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: Understanding the Concept:
This is a blackbody-radiation-style problem in a lower dimension. For a photon gas confined in a box of dimension \(d\), the number of modes per unit frequency, the density of states \(g_d(\nu)\), scales as \(\nu^{d-1}\), because it comes from counting standing wave modes inside a region of radius proportional to \(\nu\) in \(d\)-dimensional wavevector space, and the surface of that region scales as (radius)\(^{d-1}\). Here the box is two-dimensional, so \(d=2\) and \(g_{2d}(\nu) \propto \nu\).

Step 2: Key Formula or Approach:
Put \(g_{2d}(\nu) = A\nu\) into the given energy integral and change to the dimensionless variable \(x = h\nu/(k_BT)\):
\[ E = \int_0^\infty A\nu \cdot \frac{h\nu}{\exp(h\nu/k_BT)-1}\, d\nu = A\int_0^\infty \frac{h\nu^2}{\exp(h\nu/k_BT)-1}\, d\nu \]

Step 3: Detailed Explanation:
With \(\nu = \dfrac{k_BT}{h}x\) and \(d\nu = \dfrac{k_BT}{h}dx\), every factor of \(\nu\) pulls out a factor of \(T\):
\[ E = A\int_0^\infty h\left(\frac{k_BT}{h}x\right)^2 \frac{1}{e^x - 1}\cdot\frac{k_BT}{h}\,dx = A\,\frac{(k_BT)^3}{h^2}\int_0^\infty \frac{x^2}{e^x-1}\,dx \]
The remaining integral over \(x\) is just a fixed pure number, it does not depend on \(T\) at all, so all the \(T\)-dependence of \(E\) sits in the prefactor:
\[ E = a\,T^3, \qquad a = A\,\frac{k_B^3}{h^2}\int_0^\infty \frac{x^2}{e^x-1}\,dx = \text{constant} \]

Step 4: Get the specific heat and rule out the other options.
Differentiating with respect to temperature at constant volume gives \(C_V = \dfrac{\partial E}{\partial T} = 3aT^2\), so \(C_V \propto T^2\). Option (A), \(C_V\propto T\), would be the result for a one-dimensional photon gas, where \(g_{1d}(\nu)\) is constant, not linear in \(\nu\). Option (C), \(C_V\propto T^3\), is the familiar three-dimensional blackbody result, it is a trap for anyone who reflexively uses the 3D scaling without checking the actual dimensionality stated in the question. Option (D), \(C_V\propto T^4\), does not correspond to any of these standard photon-gas dimensionalities.

Final Answer:
In a 2D photon gas, \(E\propto T^3\), so \(C_V\propto T^2\), option (B). \[ \boxed{C_V \propto T^2} \]