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I understand that it is the random motion of the molecules that causes them to move from an area of high concentration to an area with a lower concentration and the diffusion will continue until the concentration gradient has been eliminated.

My question is, do light photons behave the same or not?

For example, if I put two light sources of the same power in a vacuum at a finite distance, will I find that it's the same probability of finding a photon when the distance measured from the sources is the same.

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    $\begingroup$ Not likely. For a diffusion motion, the momentum of a particle should be randomiized (scattered) within a reasonable samll legth scale and time scale. It is not likely to be applicable to the photon case. $\endgroup$ – ytlu Mar 19 at 8:39
  • $\begingroup$ Looking at a cloud outside my window. The light I see is photons that have entered the cloud and diffused back out. $\endgroup$ – John Doty Mar 19 at 17:32
  • $\begingroup$ @JohnDoty Should scattering be considered the same as diffusion? $\endgroup$ – BioPhysicist Mar 19 at 17:41
  • $\begingroup$ @BioPhysicist Diffusion is a consequence of scattering. $\endgroup$ – John Doty Mar 19 at 17:45
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No. The light waves (and the associated photons) will just keep on going in a straight line away from each source. In the case of a source emitting equally in all directions, the light intensity falls with distance squared from a given source, so for two such sources in otherwise empty space the net intensity at some point $\bf r$ is $$ P = \frac{P_A}{({\bf r} - {\bf r}_a)^2} + \frac{P_B}{({\bf r} - {\bf r}_b)^2} $$ if the sources are at ${\bf r}_a$ and ${\bf r}_b$ and $P_A$, $P_B$ is the intensity at unit distance from each source respectively.

To get diffusive motion for photons, you need some sort of matter to scatter the photons. For example, this can happen in a plasma such as the hot plasma which makes up a star. In this case the photon diffusion does tend to even out the photon density, but there are other effects as well, such as temperature and gravity, which also influence the distribution. The effect of gravity is to hold the whole star together, and to cause high pressure at the centre, and consequently the temperature is also higher there, which in turn results in more light there, produced by thermal radiation from the electrons as they collide with one another, and there are also further gamma photons associated with the fusion processes amongst the protons.

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  • $\begingroup$ You should clarify that $\frac{1}{r^2}$ is for a point source, or for a source sufficiently distant that it can be treated as such $\endgroup$ – Carl Witthoft Mar 19 at 13:20
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    $\begingroup$ @CarlWitthoft yes I agree; modified accordingly $\endgroup$ – Andrew Steane Mar 19 at 13:34
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Equilibrium can only reach if there is interaction. Molecules interact but photons do not. Therefore a photon gas will not reach equilibrium. The situation changes if the photons are interacting with other matter such as a gas of molecules. Then equilibrium will be reached and the result is black body radiation.

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  • $\begingroup$ THe equlibrium in black body is not diffusive. It is a balance between absorption and emission. $\endgroup$ – ytlu Mar 19 at 16:51
  • $\begingroup$ Or if the photons have sufficient energy for pair production. $\endgroup$ – Lawnmower Man Mar 19 at 19:12

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