# inverse square law and far-field radiation

This question is related to Feynman on inverse square law of EM radiation.

It is basicly the same question except that I don't see that the question was ever answered, and I hope someone will answer it.

First off, inverse square law. You can expect it to apply to anything that spreads out across the same proportional area as it covers the same distance. The area increases with square of distance. It should apply to gravity, electromagnetic force, radiation, to light, etc.

Wikipedia says it applies to gravity, to EM force, to light, to sound in an unconstrained gas, etc.

However, the equations that describe EM force put the static (and constant velocity) force at inverse square law of distance, and radiation (from acceleration in some directions) at simple inverse of distance. Everything in EM force except radiation falls off faster than the radiation, so at near distance there's a component that depends on those, and farther away they are too small to bother with and radiation is all that matters.

My obvious question is what is going on with light that keeps it from spreading out across an area, like everything else. The inverse square law is simple geometry. Why doesn't the geometry work with light? Why is it that when you double the distance, the force from light is cut in half instead of by four?

The previous question had two answers. One of them stated that the inverse square law is true for radiation field intensity. But since intensity of an EM wave is proportional to the square of amplitude, the result is that it falls only linearly. This explanation made no sense to me.

The second answer was that the inverse square law applies to static sources. But when you shake an electric charge then the law doesn't apply and the force drops linearly with distance. This was only a restatement of the question, without the question mark.

How can the force fall only linearly when it spreads through an area, like all the other forces? What allows that to happen?

The radiative field can be visualized, at least in simple cases, as a “kink” in the field lines. (See Figure 2 here.) The field lines apart from the kink do spread out with a $$1/r^2$$ decrease in the field strength, but the kink itself, which is what transports energy to infinity, has a field strength that falls off as $$1/r$$. The kink is tangential, not radial, so the usual “same flux of the field over any sphere” argument from Gauss’ Law doesn’t apply. Instead, “same flux of energy over any sphere” applies, by energy conservation.

• Thank you! I still don't understand, but it sounds like you do. I'll try to say it back in my own words. Electric force from acceleration creates a rotation, and the angle of rotation stays the same independent of distance. So the force doesn't spread out in that direction, but only in the direction of axis of rotation. I can imagine that happening, but as the sphere gets bigger, the surface of the sphere still increases by the square while the energy stays the same. So I'm missing something. Does the force from light actually decrease by 1/r over distance? – J Thomas Feb 17 '19 at 15:43

My answer applies to a thermic source with equal distribution in all directions. A good old light bulb is such source and the accelerated electrons in the wire emit a huge number of photons. Photons are indivisible units from their emission to their absorption and not loosing energy during their life.

Since the surface of a sphere with doubled radius is four times bigger, it is obvious that the number of photons follows the inverse square law.

But there is a moment when the inverse quadratic law is no longer applicable. The number of photons emitted for a given time is huge, but not unlimited. At sufficiently large distances, the number of photons for a given surface is simply less than one.

• This doesn't relate to the question, which is a classical question about the field. – user4552 Feb 17 '19 at 15:18
• @Ben Want to discuss this with an astronomer which gets the light from a star 13 billion light years away in photons over days of observation? – HolgerFiedler Feb 17 '19 at 15:33
• @Ben The explanation with EM field should be the same as with a stream of photons. Both have to be equivalent. – HolgerFiedler Feb 17 '19 at 15:34
• It looks to me like at large distances, the probability that a photon will arrive within the aperture of a telescope should decrease by the inverse square. So the inverse square relation would still hold. – J Thomas Feb 17 '19 at 15:46
• @JThomas Over a short time observation you get a fluctuation of single arriving photons. Over long time observation it is a constant number for some time period and of course you may apply the inverse square law. – HolgerFiedler Feb 17 '19 at 15:56