Dimensional crossover of thermal cavity radiation ($U_{2D}=\frac{\zeta (3)}{\pi }\frac{A(k_{B}T{)}^3}{{\hslash }^2{c}^2}$)

In a thin rectangular electromagnetic cavity, the window $\hslash c\pi /L_x,\hslash c\pi /L_y\ll k_{B}T\ll \hslash c\pi /L_z$ freezes the nonzero thickness indices while retaining many in-plane normal modes. The zero-thickness-index family has one polarization of an electromagnetic wave and density of states $dN=Akdk/(2\pi )=2\pi A\nu d\nu /c^2$. Multiplying by $h\nu /(e^{h\nu /k_{B}T}-1)$ and integrating gives $U_{2D}=4\pi \zeta (3)A(k_{B}T{)}^3/(h^2{c}^2)$. Thus thermal energy scales as area times $T^3$. The Riemann zeta function appears through ${\int }_0^{\infty }{x}^2/(e^x-1)dx=2\zeta (3)$.

This is an intermediate low-temperature regime of an anisotropic finite cavity. When $k_{B}T$ also falls below the first in-plane excitation, thermal energy becomes exponentially small. If a different model excludes every zero thickness index, its first transverse mode has a finite gap and the proposed $T^3$ regime does not occur. In the three-dimensional continuum regime the corresponding density of states is proportional to volume times ${\nu }^2$ and thermal energy is proportional to volume times $T^4$.