Radiation Pressure — Question 9

Question 9

An isotropic radiation field has pressure one third of energy density

A cavity contains isotropic radiation with total energy density u. Its energy density per solid angle is u divided by 4 pi. The wall perfectly reflects. A ray at angle theta from the wall normal has incident energy flux c cos(theta) times its directional energy density.

1. Write the angular integral of normal momentum transfer at the wall.

2. Evaluate it and compare with a single normally incident beam having energy density u.

Question 9 - Solution

An isotropic radiation field has pressure one third of energy density

(a) Method and calculation

$$dp=\frac{2\cos\theta}{c}\left(\frac{uc\cos\theta}{4\pi}d\Omega\right).$$$$p=\frac{u}{2\pi}\int_0^{2\pi}d\phi\int_0^{\pi/2}\cos^2\theta\sin\theta\,d\theta.$$

(b) Interpretation and further calculation

$$\int_0^{\pi/2}\cos^2\theta\sin\theta\,d\theta=1/3\Rightarrow\boxed{p=u/3}.$$$$p_{\mathrm{beam}}=2I/c=2u\Rightarrow\boxed{p/p_{\mathrm{beam}}=1/6}.$$

Physical check

The cavity u includes all directions, including rays moving away from this wall; the comparison beam u is entirely incident normally.