# Resultant intensity of two waves

For long I've been thinking about this issue, but ended up with nothing :

Suppose you have a point source $S_1$ sending a spherical wave in space of equation
$$\phi=\dfrac{A_0}{r}\sin(kr-\omega t)$$ and intensity $$I=h\left(\dfrac{A_0}{r}\right)^2=\dfrac{I_0}{r^2}.$$ Now suppose we place at the same position of $S_1$ another source $S_2$ sending just the same wave $$\phi=\dfrac{A_0}{r}\sin(kr-\omega t).$$ What is the resultant intensity?

One may say that two sources will produce double the power, so double the intensity.

Another may say that the two waves, superposing in phase, will produce double the amplitude so four times the initial intensity.

Which one is correct? It is said that the second answer is the right one in the case of interference and that the extra energy comes from dark fringes, but here there are no dark or bright fringes; the two waves superpose constructively everywhere.

Please help.

• The space in which the wave propagates has an impedance. If it can be excited with twice the amplitude then that takes 4 times the power to do so. Think of Ohm's law, $V=IR$ but $P=V^2/R$. – hyportnex Apr 4 '16 at 12:57

## 6 Answers

Fairly easy, just add the two waves using superposition.

$\phi^\prime = 2 \phi = 2 \frac{A_0}{r} sin(kr - \omega t)$ for the resulting wave (the prime denotes superposition)

Using $I = h(A_0/r)^2$ we see that $I^\prime = I \times (2A_0/A_0)^2 = 4I$.

Therefore the intensity has become four times larger.

This is an interesting observation, thank you!

Well, of course the second answer is correct, the intensity is 4 times as big -- but where does the energy come from?

Think about where the energy of a single wave comes from. The moving charges, that produce the wave, are obviously doing work - against the force of the field coming from their own radiation.

(Wow, that's by the way a nice way one can see, that the intensity has to be quadratic in the frequency: the velocity is increased and the field is increased, so the power $F\cdot v$ is quadratic!)

So in this case the radiation fields will superpose, and every of the charges will do double work (same velocity against a double field) per period. There you have the energy.

In the usual system of distinct sources you can (to a certain approximation) neglect the action of one on the other, so the net energy production is just (approximately, as I see now) the sum of the energies produced by each of the sources alone.
This is consistent to the other, rather qualitative, reasoning: if you consider a shell around both sources, there will be some areas where the waves interfere positively (there you have 4 times the single intensity) and others where they interfere destructively and leave zero intensity, so when you take the mean you get approximately the double intensity of one source.

Very nice.

• The case of distinct sources was discussed here, see there is no summation for intensities. – Tofi Apr 4 '16 at 14:24
• you are right, I used the wrong words. In any point the amplitudes get added and the intensity is the square of this sum. I meant, that when you take the mean over the intensities (which are therefore distributed from zero to four) you get two times the single intensity. – Ilja Apr 4 '16 at 14:38

I would say neither answer considered in your question is correct in all situations.The two sources are in the near field (https://en.wikipedia.org/wiki/Near_and_far_field) of each other, so they significantly influence each other (they present loads for each other). So, theoretically, the two sources can radiate four times more power than each of them separately. In that case, the extra power comes from the "power sources" of the two sources. For example, if the "power source" for both sources is the electric grid (mains), the two sources will be a four times greater load for the grid than one source. It is possible, however, that the "power sources" can only provide limited power to the sources (or power radiated by the sources is limited otherwise, say, by some nonlinear mechanisms). Then the sources will radiate less than four times the power radiated by one of them separately.

Consider a mechanical analog: a wave on a string? No, too complex--how about a vibrating spring with a mass $m$? Still too much--let's do just a static spring with constant $k$. If you compress it L, then the energy is:

$E = \frac{1}{2}kL^2$

Now someone else comes along and compresses it $L$ again:

$E' = \frac{1}{2}k(L+L)^2 = 4E$.

This is pretty much what the earlier answers said--the second source is working against an non-zero medium, and hence requires more energy than the 1st one.

The total power is twice that of one source.

Where the two add constructively, the field is twice as big, so the energy density is four times as big.

Where they add destructively, the result is zero.

You average “four in some places and zero in others” over all space and get an average energy density of “twice”, just what the sources are putting out.

How would one perform an experiment to see what happens? Remember, physics is a science and that requires that one can tests one's understanding somehow by experimental observations. The only way I can think of to put two sources at the same point is to have one point source with a dial so that you can change the amplitude or power of this source. By changing the dial from 1 to 2 one gets the effect of having two sources instead of one.

Now the question is, in what units does the dial vary? Is it the magnitude of the electric field? Then it changes the amplitude of the source, which means dailing it from 1 to 2, one changes the power by a factor 4. Is it the power of the source, dailing it from 1 to 2, one changes the power by a factor of 2, but the amplitude only changes by a factor of $\sqrt{2}$.

If you can come up with a physical experiment that would have two distinct sources instead of just one, then one can discuss that case.