# Why do spherical waves diminish as 1/$r^2$?

Seeing the derivation of a plane wave in Giancoli, I can't understand why a spherical wave would diminish in amplitude. Shouldn't the peaks still be mutually induced, and therefore nondiminishing?

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Similar to this: physics.stackexchange.com/questions/70285 –  Will Jul 15 '13 at 3:25
If the link doesn't help, think about conservation of energy. –  Will Jul 15 '13 at 3:27
Wait! You asked that question. You're still confused? What's confusing you? –  Will Jul 15 '13 at 3:30
–  dmckee Jul 15 '13 at 3:31
No no no. Do you have background in Maxwell's equations and the associated mathematics? I would guess not based on your confusion. Think about ripples in a pond when a small pebble is dropped into it. It's analogous. The idea is the same. Energy is propagating out, the energy is evenly distributed throughout each ring. As each ring gets bigger, the energy becomes more dispersed (the energy of the entire ring is constant), and therefore the amplitude must decrease. –  Will Jul 15 '13 at 5:15

I'll try to capture what I think is the key difference, which has been mentioned in the comments. Although the question is principally about electromagnetism, I'll use sound waves as an illustration.

A plane wave is an idealization - its the simplest possible wave travelling in a given direction. Being an idealization, you might expect that it would be rather hard to produce, and indeed it is!. The source (antenna for EM, vibrating surface for sound), for an ideal plane wave has to cover an infinite two dimensional plane. Yes, for a sound wave that's an infinite flat loudspeaker surface! The wave then propagates in the direction perpendicular to that plane. If the propagation medium is ideal and it doesn't dissipate the wave due to its mechanical properties, it goes on for ever, and there is no reason why the amplitude of the oscillations changes in the z direction (assume the plane is in the x and y directions).

To make a spherical wave, however, we don't have an infinite sized source. We just need the surface of a sphere. Suppose we now switch on this spherical source and let it run for a second, putting out a Joule of energy (maybe think of a pulsating spherical membrane producing a spherically symmetric sound wave). This spherically symmetric packet then spreads out over an ever increasing radius, so the Joule is distributed over a bigger and bigger area as time progresses. Suppose I put a detector at the point, say $\theta=0, \phi=0, r=1$ and measure the energy. If I then repeat the experiment at $\theta=0, \phi=0, r=1000$, i.e same direction, but further away, then the energy at my detector will be a LOT smaller just due to the spreading.

You can get additional insight into the difference by considering Huygen's principle. This states that, during wave propagation, you can take your wavefront and consider each point on it as a source of spherical waves. You then add up all these spherical waves to get the next wavefront. If you do this for an infinite plane, then all the radiating point sources add up give another identical infinite plane - the energy density is unchanged. If you do it for a spherical surface, the sources add up to give you another spherical surface, but at a larger radius, so the energy density has been reduced in proportion to the square of the radial distance. This dissipation is purely due to the geometry.

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A nice answer, but altogether unnecessary. Not only was this covered in the comments, resulting with a happy @Tony, but it was explained for EM in the link. –  Will Jul 15 '13 at 13:35
@Will: so we should answer questions in the comments now? Pretty sure the whole point IS to get a nice answer (posted under "answer", not "comment"). –  Kyle Oman Jul 15 '13 at 13:59
@Kyle The reason I didn't post under the answer section was that it was already covered in a previous question asked (by the same user). In trying to discover the confusion, I answered the question in comments. Sure I could add it to an answer... –  Will Jul 15 '13 at 14:27

@Tony asked a very similar question in: If fields die off proportional to $R^2$, why does light keep going?

I gave an explanation in terms of solutions to the source-free Maxwell's equations, that is the equations governing EM waves. I also talked about plane waves there. In the comments above, it is mentioned that @Tony doesn't have much background in Maxwell's equations, so I will add comments here that make the results more accessible to others in the same situation as @Tony, following my explanation given in the comments.

Conservation of energy is a very useful tool, which can be used here to understand the $\frac{1}{r^2}$ decrease in energy density with distance. All we need to know is that the energy density per unit time of an EM wave is proportional to $|\vec{E}|^2$, where $|\vec{E}|$ is the amplitude of the electric field in the wave, which I will simply denote by $E$.

My argument is the following: take a spherically symmetric source radiating a fixed amount of energy per second. This produces spherical shell waves propagating out from the surface. Each infinitesimal shell (this represents some release of energy from the source in infinitesimal time $dt$) has an energy density $U_0$ (energy per unit distance, because we are dealing with an infinitesimal shell). By energy conservation, as this shell expands, it has the same total energy density (energy per unit distance), but it is dispersed over the entire shell, of area $4\pi r^2$. Thus the energy density (energy per unit volume) at any distance $r$ is $$u_0 = \frac{U_0}{4\pi r^2}\propto \frac{1}{r^2}$$ as expected. Now, because $u_0\propto E^2$, this tells us that $$E\propto\frac{1}{r}$$ as explained in: If fields die off proportional to $R^2$, why does light keep going?

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