# Adding intensities for incoherent waves

1. Dec 28, 2004

### fliptomato

Greetings all--I have a relatively basic question regarding waves:

For coherent waves, the amplitude of the two waves add ($$\psi(x) + \phi(x)$$) and the resulting intensity is the square of this: $$(\psi(x)+\phi(x))^2$$, or $$\psi(x)^2 +2\psi(x)\phi(x)+\phi(x)^2$$

However, if youre superimposing incoherent waves, then textbooks tell me that you add intensities , i.e. the resulting intensity is $$\psi(x)^2 +\phi(x)^2$$ ... i.e. there's no interference term.

Why is this? It seems a little hokey just to say that the average observed intensity becomes the average of the two component intensities without any motivation for this. I.e. is there a way to show this (mathematically) from the addition of amplitudes?

Thus, if I have waves $$\psi(r(t)\cdot x)$$ and $$\phi(s(t)\cdot x)$$ where $$r(t)$$ and $$s(t)$$ are time-dependent "random" functions that average to zero. When I add these waves, the result is that the term $$2\psi(r(t)\cdot x)\phi(r(t)\cdot x)$$ averages to zero.

References:
Tipler, Paul. Physics for Scientists and Engineers, 4th ed. p. 487
Aitchison and Hey. Gauge theories in Particle Physics, Vol I, 3rd ed. P. 212 (middle paragraph)

((In case you were curious about the odd second reference, I originally was startled by this because I was trying to understand the "spin averaging" in the amplitude for spin-dependent scattering in, say, QED.))

2. Dec 29, 2004

### Staff: Mentor

Sure. Try this: Let $\psi = A cos\theta_1$ and $\phi = B cos\theta_2$. Then:
$$I \propto (\psi + \phi)^2 = (A cos\theta_1 + B cos\theta_2)^2$$
$$I \propto A^2 cos^2\theta_1 + B^2 cos^2\theta_2 + 2AB cos\theta_1 cos\theta_2$$

The average value of $cos^2$ is 1/2; using a bit of trig, you should be able to convince yourself that the cross term averages to zero if the phases are random. Thus:
$$I \propto A^2/2 + B^2/2$$
$$I_{total} = I_1 + I_2$$

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