Interference of Waves — Question 7

Question 7

Random relative phase removes the coherent cross term on average

Two equal-amplitude waves have instantaneous phase difference $\Delta$. During a long measurement, $\Delta$ samples the interval $[0,2\pi)$ uniformly, slowly compared with the oscillation period. Each wave alone carries mean power $P_0$.

1. Find the long-time averaged total power.

2. Compare with a permanently fixed phase difference $\pi/2$. Can a single averaged power reading distinguish these two situations?

Question 7 - Solution

Random relative phase removes the coherent cross term on average

Method and mathematical working

Average the interference term over the phase distribution. Uniform phase over a complete cycle has zero mean cosine, even though each realization can interfere strongly.$$\begin{aligned}\langle\cos\phi\rangle&=\frac1{2\pi}\int_0^{2\pi}\cos\phi\,d\phi=0,\\\langle\overline P\rangle&=2\overline P_0.\\[4pt]\langle\overline P\rangle&=2P_0[1+\langle\cos\phi\rangle]=2P_0,\\\phi=\pi/2&\Rightarrow\overline P=2P_0\quad\text{in every realization}.\end{aligned}$$

(a) Calculation and conclusion

At fixed relative phase, cycle-averaged power is $2P_0(1+\cos\Delta)$. Uniform phase averaging gives $\langle\cos\Delta\rangle=0$, hence $\langle P\rangle=2P_0$.

(b) Interpretation and limits

At fixed quadrature, $\cos(\pi/2)=0$, so the same power $2P_0$ results. But the coherent quadrature sum has a stable phase and amplitude, whereas the random-phase sum fluctuates. A single mean-power reading cannot establish coherence; phase-sensitive or time-resolved observations are needed.

Physical check

A fixed quadrature phase gives the same mean power but no phase-induced variation from trial to trial. The mean alone cannot distinguish these two statistical situations.