Photon interactions

Edi

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

malawi_glenn

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.

Bob S

We are discussing photon scattering off of matter without electrons. Your diagram is for electron-electron scattering with a photon mediator.

Bob S

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.

kanato

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.

Vanadium 50

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.

Bob_for_short

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.

Bob S

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).

Edi

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.

Bob_for_short

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.