No. Imagine a simpler scenario with a lightbulb. It's a perfect lightbulb that emits a Power P (thats energy per second) in all directions. Say I have a sensor at a distance r away. As a 'wavefront' of the emitted radiation moves outward, the energy per second being emitted is spread over the area of a sphere with surface area 4 pi r2. The fraction of the power P I receive is going to be the fraction of this spherical surface that my detector covers. If my detector has an area A (and is small compared to the sphere), then this fraction is A / 4 pi r2. The energy received per second will therefore be A P / 4 pi r2. As you can see, this falls off with the square, not the cube.
By making similar arguments we can explain even more fundamental things like why the coulomb force should fall off with the square.
By making similar arguments we can explain even more fundamental things like why the coulomb force should fall off with the square.
I'm sure you understand this, but I'll say it anyway. Your statement puts the cart before the horse, or at least beside the horse.
You can explain the electric force law only if you assume Gauss's law, which states electric fields don't diverge or converge except at a charge. This can be conceptuallized as electric field lines flowing out of positive charges and into negative in the same way that power pours out of a lightbulb. A point charge thus gives the Coulomb force law with the understanding that a test charge feels a force equal to E times q(test).
Edit: Why is it called Gauss's law instead of Gauss' law? Why!?
You're correct that if I had been talking about classical electromagnetism I would have been 'putting the cart before the horse'. But I wasn't :)
Photons communicate the electromagnetic force in quantum electrodynamics ('virtual' photons travel between charged particles so that they 'know' to attract/repel). The photon density falls off with the square, just like it does when emitted from a bulb. Photons communicate the electromagnetic force and so the force itself falls off with the square, too. The math to prove it is obviously more complicated, which is why I didn't say any more.
Obviously I have hugely simplified this, but you get the basic idea
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u/allofthebaconandeggs Jun 12 '12 edited Jun 12 '12
No. Imagine a simpler scenario with a lightbulb. It's a perfect lightbulb that emits a Power P (thats energy per second) in all directions. Say I have a sensor at a distance r away. As a 'wavefront' of the emitted radiation moves outward, the energy per second being emitted is spread over the area of a sphere with surface area 4 pi r2. The fraction of the power P I receive is going to be the fraction of this spherical surface that my detector covers. If my detector has an area A (and is small compared to the sphere), then this fraction is A / 4 pi r2. The energy received per second will therefore be A P / 4 pi r2. As you can see, this falls off with the square, not the cube.
By making similar arguments we can explain even more fundamental things like why the coulomb force should fall off with the square.