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I know that light (or electromagnetic radiation in general) attenuates in intensity as the square of the distance it travels.

  • Why does it attenuate?
  • Are the photons being scattered by the medium they pass through?

I also know that the energy carried by light quanta is a function of their frequency only.

  • If the answer to B is yes, then can we possibly transmit light energy (and information) over infinite distances through a perfect vacuum?

I may be wrong on some assumptions, so please enlighten me.

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3 Answers

up vote 8 down vote accepted

A. All light sources (even lasers) are subject to a diffraction limit, so any light beam will eventually diverge with an angle $\theta$ given by

$$\theta \approx \frac{\lambda}{A_T}$$

where $\lambda$ is the wavelength of the light and $A_T$ is the aperture of the light beam source (and "eventually" means for distances much greater than $A_T$).

Any beam diverging with a constant angle will have an intensity following an inverse-square law, though the total beam power will be unaffected (if we can neglect light absorption and scattering).

The diffraction limit can be seen as a consequence of the Heisenberg uncertainty principle: calling one transverse coordinate $x$ and applying the position-momentum uncertainty relation at the source (assuming that the inequality is saturated and the position uncertainty is equal to the aperture size), we get

$$\Delta p_x^{source} \Delta x^{source} \approx \hbar$$

$$\Delta p_x^{source} \approx \frac{\hbar}{A_T}$$

Far from the source, at a distance $R \gg A_T$, the transverse position uncertainty will be dominated by the transverse momentum uncertainty at the source, giving

$$\theta \approx \frac{\Delta x^{far}}{R} \approx \frac{R \Delta p_x^{source}}{p}\frac{1}{R} \approx \frac{\hbar\lambda}{A_T \hbar 2\pi} = \frac{1}{2\pi}\frac{\lambda}{A_T}$$

that differs from the previously given result by a constant, due to the approximations involved and the imprecise nature of the "deltas". A more precise treatment shows that $\theta = \lambda/A_T$ is a better approximation.

B. Photons are not scattered in a perfect vacuum. And the intergalactic space, while not a perfect vacuum, is so empty that even photons originated in galaxies at billions of light years can be received.

C. Yes, but you will need a receiver with a big aperture to receive this light. In more precise terms, you will need a receiver with an aperture $A_R$ given by

$$A_R \approx \frac{\lambda}{A_T}R$$

where $\lambda$ is the wavelength, $A_T$ is the aperture of the transmitter and $R$ is the distance from the transmitter to the receiver.

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You might add that the above diffraction limit can be thought of as an application of Heisenberg's uncertainty principle (with regard to the transverse position vs momentum of the light). –  Carl Brannen Feb 24 '11 at 1:02
    
@Carl Brannen Thanks for the suggestion: I've added a derivation along these lines. Feel free to improve it, as I'm not an expert by any means. :-D –  mmc Feb 24 '11 at 12:43
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A. Because the total energy is divided over the whole sphere $4\pi R^2$ of radius $R$ around the source of light which means that the energy density over unit area goes like $1/R^2$. However, this is only true for unfocused light - with chaotic directions. Lasers may create coherent light whose intensity per unit area doesn't drop. The beam may move for kilometers without changing its shape and intensity.

B. It depends on the environment: photons don't scatter in the vacuum.

C. No, because the vacuum doesn't interfere with the photons' motion at all, we can transfer energy by photons over any distance. In particular, there's no exponential decrease of the intensity in the vacuum. In this sense, the vacuum is not a "medium" - there is no "luminiferous aether" that fills it.

When the intensity of energy goes down like $1/R^2$, it's the number of photons per unit area that decreases. The energy of each photon is unchanged.

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Your explained answer to C seems to imply "Yes", however, you say "No". Am I misreading? –  Justin L. Feb 23 '11 at 9:11
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Even coherent laser beams have some degree of dispersion, i.e. the photons are not on exactly parallel paths. So you will still end up with the 1/R squared falloff. see mmc below. –  Omega Centauri Feb 23 '11 at 23:50
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Well, laser beam will eventually diverge but it is extremely far and I still think that it may be kilometers without changing the width by more than a percent or so. –  Luboš Motl Feb 24 '11 at 6:29
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Any beam has at least the divergence by the diffraction at the "Pupilla". High power lasers can do better, but this comes from autofocus by nonlinear optics in air. –  Georg Feb 24 '11 at 9:49
    
Assume 1m wide laser aperture. 1% dispersion at 1km gives theta = 1.0E-5 So wavelength of laser no greater than 1.0E-5m = 10,000 nm, which is deep in the microwave region (visible is around 600nm). So Lubos' estimate is approximately correct. –  Carl Brannen Feb 25 '11 at 0:20
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The diminishing of intensity is best described if you use the wave model of light.

We make one basic assumption:

  • Total light energy is conserved over time

Let's say you have a light bulb in a vacuum that makes a flash. Let's say we enclose that light bulb in a ball, and measure the energy of the light that hits the ball's inside walls.

Then we try it again, but this time use a larger ball, with a bigger radius.

Because it's the same "amount" of light that is flashed, both balls, with different radii, should measure the same total energy.

However, Intensity is energy per area (per second).

It should be clear that with the larger ball, there is less energy per area than with the smaller ball. The energy is more concentrated in the smaller ball.

From this it can be seen that because the surface area of a sphere increases with $r^2$, the intensity of light (energy per area per second) decreases with $r^2$.

Light's fading intensity, therefore, is a direct consequence of the conservation of energy.

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