Energy Transport and the Poynting Vector — Question 10

Question 10

Unequal opposing waves have a mean energy-transport speed below c

Two vacuum plane waves with the same linear polarization propagate in opposite x directions. Their electric amplitudes are A and B. Average over time at a fixed position; the total energy density is the sum of electric and magnetic contributions.

1. Derive the mean energy density and net mean Poynting flux, showing cancellation of interference terms in total energy.

2. Find the ratio of net flux to mean energy density for B=A divided by square root of two. Explain its meaning.

Question 10 - Solution

Unequal opposing waves have a mean energy-transport speed below c

(a) Method and calculation

$$u=\epsilon_0[A^2\cos^2(kx-\omega t)+B^2\cos^2(kx+\omega t)].$$$$\boxed{\langle u\rangle=\tfrac12\epsilon_0(A^2+B^2),\quad\langle S_x\rangle=\tfrac12\epsilon_0c(A^2-B^2)}.$$

(b) Interpretation and further calculation

$$v_E\equiv\frac{\langle S_x\rangle}{\langle u\rangle}=c\frac{A^2-B^2}{A^2+B^2}.$$$$B=A/\sqrt2\Rightarrow\boxed{v_E=c/3}.$$

Physical check

This is net transport divided by stored energy for counterflowing radiation; each individual vacuum wave still propagates at c.