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| From https://meanjin.com.au/blog/the-slow-light-of-truth/ |
(Note -- WEDNESDAY)
Mohammad Sayeh
Slow light (photons calmed)
[recording]
Two papers (click):
And the book suggested by Mohammad: "Fast Light, Slow Light and Left-Handed Light" by P W Milonni, 2005. [Amazon, PDF]
In comments use dollar signs and latex to get math formating. For instance $\frac{7}{3}=\det\begin{bmatrix}a&b\\ c&d\end{bmatrix}$.
I’m wondering if a basic question is to guarantee that the exiting photons are not simply emitted by the Na atoms, and that, in fact, the initial photon stream have/had all been absorbed. Excuse me if this is “obvious”.
ReplyDeleteDid you think one of the ideas of the slow light is to develop an invisibility cloak [shield]? I think I had seen something about this for real.
ReplyDeleteYou are correct! They are articles on that ideas.
DeleteNice talk, Mohammad. I think the problem posed by Mohammad, in the end, is intriguing.
ReplyDeleteThe energy-momentum four-vector for light in vacuum is a light-like (null) vector. That is, $E=pc$. Thus, using the formula $E^2=p^2c^2+m^2c^4$ we conclude that $m=0$ and $E=pc$. This is the characteristic of light in vacuum. In a medium speed of light is slower than the speed of light in vacuum $c$. Thus, it would seem that, there exists a frame in which momentum $p$ is zero and thus using $E=mc^2$ in this frame we can deduce an effective mass for light in this medium. This, in my interpretation, is what Mohammad is suggesting.
I have been able to show, after the lecture yesterday, that the energy-momentum four-vector is a light-like (null) vector in a non-dispersive (frequency independent) medium. So, light is massless even in such a hypothetical medium. Using straight forward analysis, I can demonstrate this for a monochromatic wave in vacuum and repeat the calculation for a monochromatic wave in a non-dispersive medium of infinite extent.
However, real media, and especially the discussion of yesterdays talk, was in the context of dispersive media. So, not surprisingly, it is a harder problem.
The above are my two cents.
I have the same (basically) comment as KVS. This formula can also be written as a Lorentzian square of 4-momentum equal to m^2c^2, g_{mu nu}p^{mu}p^{nu}=m^2c^2, where p^{mu}=(E/c,\vec{p}) (one then multiplies both sides by c^2 and moves p^2c^2 to the other side.) Photons always move and "rest mass" for them basically does not make sense, so we take it as zero. All energy of a photon is in its momentum, as the formula E=pc suggests. Next, the motion of a photon in a medium is difficult to follow, but it is probably a chain of absorption and emission events, as well as scatterings on electrons, so that the photon we see at the exit is different than the one at the entrance. I am not sure if anybody could prove it experimentally, but it is not excluded that in between these (very frequent in even moderately dense matter) events photons still move with the speed of light as they should. Only the effective speed is lower. Moreover, of we deal with a beam of light, it consists of huge number of photons, and each of them underlies its history of events, before we see the light at the exit from the medium.
DeleteWhat if photons really do not exist! Some electrons lose energy and then some electrons gain energy.
DeleteThanks KVS. Let's think about the dispersive case.
ReplyDelete