What happens when light hits light?


by junguo93
Tags: hits, light
andrien
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#19
Dec31-12, 04:27 AM
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it depends on when one has read that two photon sources can not interfere.It is mentioned in older books that two distinct photon sources can not interfere because of random phase relationship.But this statement is quite wrong now because it is possible to make sources which have constant phase relationship over a long time.
DaleSpam
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#20
Dec31-12, 06:21 AM
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There is also the photon-photon interaction where two highly energetic photons can interact with each other to produce e.g. an electron and a positron.

http://en.wikipedia.org/wiki/Two-photon_physics
andrien
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#21
Dec31-12, 08:33 AM
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it is related to scattering of light by light.it is called delbruck scattering.
http://en.wikipedia.org/wiki/Delbruck_scattering
sophiecentaur
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#22
Jan1-13, 05:45 PM
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Quote Quote by andrien View Post
it depends on when one has read that two photon sources can not interfere.It is mentioned in older books that two distinct photon sources can not interfere because of random phase relationship.But this statement is quite wrong now because it is possible to make sources which have constant phase relationship over a long time.
I think you are making assumptions about the nature of the photon. Just because two sources are in phase doesn't imply that photons have to come from both sources 'at once' to produce interference. We are quite happy to bandy about the notion of virtual photons when talking about charges interacting so why not regard the interference effect as being virtual too? Whenever I come across discussions which involve 'shoehorning' particles into explanations of phenomena that are essentially wavelike, it seems that the explanations get more and more convoluted in order to accommodate them. I know there is overwhelming evidence of the Quantisation of EM but, beyond that, the actual nature of photons is pretty unclear. Calling them particles doesn't actually seem to help - I know that Feynman insisted that they are but I never heard of him justifying that statement where RF is concerned. His famous diagrams were never intended as pictures of an actual event, I'm sure; rather, they were two dimensional functional diagrams. I thought that the whole point of QM is that it's not actually 'like' anything else. I'll bet that it is only in the minds of a very few, really well informed Physicists that, when the word 'photon' comes up, the little bullet model doesn't lurk somewhere in there.

It has been possible since the invention of the radio valve! Like I said before, people seem to regard the Laser phenomenon as being something special and needing special terms to explain it with. The equivalent has been around at lower frequencies for ages and needs, just as much, to be explained in common terms, surely.

The fact that photons with high enough energies can interact to produce massive particles is an interesting facet but a bit difficult to reconcile with a 'linear space'. It seems that there are no particles with low enough mass for visible light photons to create. Does this, somehow imply a limit to the linearity of space as a medium? A sort of gear change, which occurs when the energies correspond to a lepton (or possiby a quark?) mass.
andrien
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#23
Jan2-13, 12:40 AM
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well,the statement I have just written comes from the book of feynman lectures itself.I don't know what you mean by linear space here.Some vector space,minkowski space,affine space?
sophiecentaur
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#24
Jan2-13, 03:56 AM
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Quote Quote by andrien View Post
well,the statement I have just written comes from the book of feynman lectures itself.I don't know what you mean by linear space here.Some vector space,minkowski space,affine space?
I guess I mean minkowski space, in as far as SR doesn't imply quantum effects. I am talking in terms of Maxwellian (?) space, in which E = E1 + E2 for all values of the Es.

The point I am making is that two optical quanta will not (afaik) mutually annihilate to produce a particle but they will have no effect on each other in the same way that two AC signals on a linear transmission line will not.

Of course, people quote Feynman all over the place because his lectures, interviews and books are highly respected. I am not arguing against what I know of what he's said but did he ever discuss my point publicly? I do know that his 'diagrams' were not intended to be taken literally, any more than Equations are intended to be a physical representation. I have heard him assert that photons 'are' particles but do we all mean the same thing as he did when he used the word particle? Is there any record of what he had to say about low frequency photons, for instance?

Has nothing, since his time, ever successfully challenged his ideas? I am not challenging them (far too humble) but I do wonder, sometimes, whether people follow more what they thought he said rather than what he meant. There is so much 'interpretation' involved in this subject (like the Koran and Bible??) and I don't think my question actually goes against the Copenhagen Interpretation.
andrien
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#25
Jan2-13, 05:37 AM
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Is there any record of what he had to say about low frequency photons, for instance?
yes,I know what he has said.He said to give photon a mass in low energy limit to avoid infrared divergence and then combining it with cut-off provided by bethe in non-relativistic limit gives a fully divergenceless result i.e. no infrared divergence!and take mass zero in the end.mass is invariant.
sophiecentaur
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#26
Jan2-13, 10:53 AM
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Quote Quote by andrien View Post
yes,I know what he has said.He said to give photon a mass in low energy limit to avoid infrared divergence and then combining it with cut-off provided by bethe in non-relativistic limit gives a fully divergenceless result i.e. no infrared divergence!and take mass zero in the end.mass is invariant.
I don't really understand that except that it confirms that photons do not exhibit mass. (I'll take "infra red" to include LF radio too.) When I asked what he said about low frequency photons, I was wondering about the situation with interference from two independent synchronous, RF sources. This is such an easily produced phenomenon at RF and not that difficult with two phase locked lasers, apparently.
If, indeed, it were really necessary for actual photons from each RF source to interfere with each other in order to produce an interference pattern then surely we would expect an awful lot of photons from each source not to interfere ('cancel' in one direction) - producing a very diluted interference pattern. In fact, very deep, stable nulls (0.1% amplitude) can be formed from two independent RF sources. If the sources are actually not phase locked at all but just very stable and slowly drifting past each other, a very clean interference pattern can result. (All that is necessary is that the amplitudes are equal, of course)
So the choice is between saying that, magically, pairs of matched photons are always present in the two 'beams' so that they can combine into a perfect pattern or that the interference is nothing to do with photons at all and the photons only turn up when they're actually detected (which is in line with the Copenhagen interpretation, I think).

I wish someone could comment on this and find the flaw - if there is one.
sophiecentaur
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#27
Jan2-13, 10:59 AM
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Quote Quote by Cthugha View Post
No, two-photon interference is exactly not that. It is very different. This is why you need to be careful. An interesting discussion on TPI can be founs in "Can Two-Photon Interference be Considered the Interference of Two Photons?", Phys. Rev. Lett. 77, 19171920 (1996). You can also find free copies of it all over the web.

In a nutshell, TPI arises not due to superposition of waves, but due to superposition of indistinguishable probability amplitudes associated with the simultaneous detection of two photons. I do not know your background, so it is somewhat hard to tell, whether it is sensible to go into details or not. If you are not interested in very special stuff, that does not occur in eveyday life, "different light sources do not interfere" is rather correct.

Typically you need indistinguishable photons for two-photon interference. That also means you need the same polarization. Interference terms will also cancel out on average, if you do not have a fixed phase relationship between the two fields of interest. To maintain a fixed phase relationship over a longer timescale, you need the fields to be as similar as possible. Just taking two arbitrary light beams will therefore not create interference. Two-photon interference is therefore a rather rare thing happening only under lab conditions.
Is there a real difference between these two things? Is it not just two ways of saying the same thing?
Using RF sources tends to take care of the polarisation issue.
I looked for that article but could only find sources that charge for it.
andrien
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#28
Jan3-13, 05:27 AM
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you can not count about every photon,in most precise definition a photon interferes with itself.All those classical ideas fails,there i no intuition.
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Jan3-13, 06:56 AM
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Quote Quote by andrien View Post
you can not count about every photon,in most precise definition a photon interferes with itself.All those classical ideas fails,there i no intuition.
You can count individual gamma photons by the clicks they give on a GM tube.
You are right enough about intuition failing. I think this is the general problem that people have when talking of photons. When Feynman asserted that they were particles, I think he did no one any favours because the particle word that he used was not the particle that comes to most people's minds. There are so many apparent paradoxes involved when we compare what happens with photons of different energies and those paradoxes should be taken as a strong message that photons are nothing like most people think.

I don't think it helps that most specialists on photons seem to be concerned with optical photons which are only a small sub-set of the beasts.
jim hardy
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#30
Jan3-13, 08:08 AM
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Interesting thoughts there. Thanks i learned quite a bit

From my experiments as a teenager installing aircraft landing lights in our automobiles-
superposition makes the best sense.
When you place two of them face to face and energize both , one of the filaments soon melts.


Ohh, nostalgia!
andrien
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#31
Jan3-13, 08:44 AM
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Quote Quote by sophiecentaur View Post
You can count individual gamma photons by the clicks they give on a GM tube.
I meant there that you can not follow a single photon's path.Photon is a result of quantization of electromagnetic field.It is just a quantum of EM field.In large occupation number limit,you can treat photons as light.Photons and light are same thing.
DaleSpam
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#32
Jan3-13, 10:06 AM
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From a practical standpoint, I don't get the obsession with throwing the photon concept into situations which are perfectly adequately described classically. Unless the light the OP is beaming is such high frequency that you get photon-photon interaction or such low amplitude that you get single quanta, then just use Maxwell's equations and superposition.
sophiecentaur
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#33
Jan3-13, 10:42 AM
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Quote Quote by andrien View Post
I meant there that you can not follow a single photon's path.Photon is a result of quantization of electromagnetic field.It is just a quantum of EM field.In large occupation number limit,you can treat photons as light.Photons and light are same thing.
That's right, more than that starts to become a strreeeettch in thinking. The only actual evidence for photons is when they are are formed or detected. What goes on in between is a total mystery. To describe the nature of a photon whist energy is being transferred (in the wave) is, to my mind, a bit glib. And I think this applies however low the flux happens to be.

At the high frequency end, where photons interact to produce matter, the situation can still obtain. I would like to know just what is the minimum frequency for this to happen, though, and what particle is involved. It seems here must be a major change in the Physics of EM at that point. Is there some kind of breakdown in the way 'space works' then or could it be looked upon as some sort of minimum quantum EM energy for a change of 'mass state'?
DaleSpam
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Jan3-13, 12:22 PM
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Quote Quote by sophiecentaur View Post
At the high frequency end, where photons interact to produce matter, the situation can still obtain. I would like to know just what is the minimum frequency for this to happen, though, and what particle is involved. It seems here must be a major change in the Physics of EM at that point. Is there some kind of breakdown in the way 'space works' then or could it be looked upon as some sort of minimum quantum EM energy for a change of 'mass state'?
Two oppositely-travelling 511 keV photons could interact to produce an electron-positron pair. If the photons were just barely 511 keV then the resulting electron and positron would have very little KE and so they would attract each other, anhilate, and produce two 511 keV photons. The net result would be scattering of the photons.
sophiecentaur
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Jan3-13, 05:18 PM
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Quote Quote by DaleSpam View Post
Two oppositely-travelling 511 keV photons could interact to produce an electron-positron pair. If the photons were just barely 511 keV then the resulting electron and positron would have very little KE and so they would attract each other, anhilate, and produce two 511 keV photons. The net result would be scattering of the photons.
So, is 511keV the minimum? This would make 511keV a very significant energy quantity, wouldn't it? It would seem to be some sort of threshold value for the production of 'free mass', rather than just 'mass defect'.
DaleSpam
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#36
Jan3-13, 05:21 PM
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Quote Quote by sophiecentaur View Post
So, is 511keV the minimum? This would make 511keV a very significant energy quantity, wouldn't it?
It is pretty significant, it is the mass of an electron.

EDIT: actually, I guess this could happen for neutrinos also, at much lower energies.


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