Adding intensities for incoherent waves

[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.))

 

[Doc Al]

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?

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:
$$I \propto A^2/2 + B^2/2$$
$$I_{total} = I_1 + I_2$$

 

[wais]

Hello,

Thank you for your question. The reason for this difference in adding intensities for coherent and incoherent waves lies in the nature of the waves themselves.

Coherent waves have a fixed phase relationship between them, meaning that their crests and troughs align perfectly. This results in constructive and destructive interference, which can be seen in the equation you provided. When adding the amplitudes of two coherent waves, the interference term (2\psi(x)\phi(x)) takes into account this phase relationship and leads to the squared amplitude.

In contrast, incoherent waves have varying phase relationships, as they come from different sources or are scattered in different directions. This means that the interference term will average out to zero, as you correctly pointed out. This is because the waves are not consistently aligned and therefore do not lead to constructive or destructive interference. As a result, the squared amplitudes of the two waves are simply added without any interference term.

Mathematically, this can be shown by considering the average of the product of two incoherent waves:

<\psi(x)\phi(x)> = \int \psi(x)\phi(x) dx

Since the phase relationship between the two waves is random and averages to zero, the average of the product will also be zero. This is why the interference term disappears when adding intensities for incoherent waves.

I hope this helps to clarify the concept. Thank you for providing the references, as they offer additional explanations for this topic. Best of luck with your studies.