## Photon interactions

You serious or you just making fun of me? One can bend photons with powerful enough electro magnet? Of course, to do it on macro scale would require extreme amounts of energy and it would be kinda hard to make the magnet, but still - photons can be bent by by a powerful enough electro magnet??

wow, thats something completely new. I must have died and changed universe i live/d in... :noface:
Of course i knew taht photon has magnetic and electric field, but... yeah
 Blog Entries: 9 Recognitions: Homework Help Science Advisor Well, the photon electric field is oscillating, so it would just be averaged out. Since you are asking this in the particle physics thread, the EM field's quanta are photons, and photons do not couple directly to photons. To lowest order, the pair prodcution Feynman diagrams are: http://musr.physics.ubc.ca/~jess/p200/emc2/img48.gif (Z = the nucleus with atom number Z) Now recall that this is not what happens in reality, Feynman diagrams are just mathematical tools and representations of terms in an perturbative expansion of the interaction.

 Quote by mheslep I think the basic Feynman diagram describes the situation. The electron shown is of course charged matter. The photon (green gamma line in the diagram) can either be absorbed or emitted. If a photon detector is inserted so that it views the emitted photon, then the electron is 'seen'. Or conversely if the photon has an external source and is absorbed, the lack of its detection also is evidence of the electron, i.e. it is 'seen'. I suppose the same holds true for nuclei. http://media-2.web.britannica.com//e...4-4247634B.gif
We are discussing photon scattering off of matter without electrons. Your diagram is for electron-electron scattering with a photon mediator.
 Here is the Feynman diagram for electron pair photoproduction off a nucleus or proton. See http://www.irs.inms.nrc.ca/EGSnrc/pirs701/node22.html This is independent on whether or not any electrons are present, because the primary pair production occurs off a nucleus. In aluminum, the pair production cross section is about 350 millibarns per atom at 20 MeV. For pair production off a proton, it is about 2 millibarns.
 What about something like Bragg diffraction? The picture often given in introductory solid state texts is that xrays reflect off of planes of nuclei and constructively interfere at the correct angles. It seems like that might be relevant to this situation but I am not sure that the simplified picture is correct.

Mentor
 Quote by Bob S The three ways photons interact with ordinary matter are the photoelectric effect (including deep core photoejection), Compton scattering, and pair production.
True. But irrelevant.

The OP was talking about visible light photons. Your reply is only relevant for photons of much higher energy. And now the thread is hopelessly derailed.

You're a smart guy, Bob. People recognize that. But not every question requires a PhD-level answer.
 There are two Bobs here. I, Bob_for_short, say that when the photon wave-length is long, etc., the positive charge matter can be reflective thus visible. No nuclear reactions, no pair production, just a conductor in an external EMW. Vladimir Kalitvianski.
 I am the other Bob. When the photon wavelength is long (e.g., visible light), it can scatter off of free electrons. See the Thomson cross section: http://en.wikipedia.org/wiki/Thomson_scattering. When the electons are absent, these visible photons can scatter off of free protons by the same process, but the yield of scattered photons is lower by a factor of the mass-ratio squared (18372 = 3,374,000).
 So, the photon does interact with the electro magnetic field (around the nuclei). On macro scale one cant really bend light using a magnet, because photons field oscillates thus averring out and not bending in one direction or another. On particle scale the frequency of relatively long waves isn't enough to average out, so it interacts via the field around nuclei, but instead of the photon changing its path, the nuclei moves with the changing field, thus receiving energy from the photon and later emitting it via another photon in different direction.

 Quote by Edi So, the photon does interact with the electro magnetic field.
Wrong, on the contrary. The electromagnetic field appears in the charge equations of motion as a superposition of fields from different sources. Superposition means non interaction.

The field equations themselves contain charges as a source, not the other fields.

Bob_for_short.

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