D.27 Radiation from a hole

To find how much blackbody radiation is emitted from a small hole in a box, first imagine that all photons move in the direction normal to the hole with the speed of light $c$. In that case, in a time interval ${\rm d}{t}$, a cylinder of photons of volume $Ac{\rm d}{t}$ would leave through the hole, where $A$ is the hole area. To get the electromagnetic energy in that cylinder, simply multiply by Planck’s blackbody spectrum $\rho$. That gives the surface radiation formula except for an additional factor ${\textstyle\frac{1}{4}}$. Half of that factor is due to the fact that on average only half of the photons will have a velocity component in the direction normal to the hole that is towards the hole. The other half will have a velocity component in that direction that is away from the hole. In addition, because the photons move in all directions, the average velocity component of the photons that move towards the hole is only half the speed of light.

More rigorously, assume that the hole is large compared to $c{\rm d}{t}$. The fraction of photons with velocity directions within in a spherical element $\sin\theta{\rm d}\theta{\rm d}\phi$ will be $\sin\theta{\rm d}\theta{\rm d}\phi$$\raisebox{.5pt}{$/$}$$4\pi$. The amount of these photons that exits will be those in a skewed cylinder of volume $Ac\cos\theta{\rm d}{t}$. To get the energy involved multiply by $\rho$. So the energy leaving in this small range of velocity directions is

\begin{displaymath}
\rho Ac {\rm d}t \cos\theta \frac{\sin\theta{\rm d}\theta{\rm d}\phi}{4\pi}
\end{displaymath}

Integrate over all $\phi$ and $\theta$ up to 90 degrees to get $\frac14{\rho}Ac{\rm d}{t}$ for the total energy that exits.

Note also from the above expression that the amount of energy leaving per unit time, unit area, and unit solid angle is

\begin{displaymath}
\frac{\rho c}{4\pi} \cos\theta
\end{displaymath}

where $\theta$ is the angle from the normal to the hole.