Mass and velocity of virtual photon in Bhabha Scattering

1. Jan 31, 2013

JamesM86

1. The problem statement, all variables and given/known data

Determine the mass of the virtual photon in each of the lowest-order diagrams for Bhabha scattering (assume the electron and positron are at rest). What is its velocity? (Note that these answers would be impossible for real photons)

Note: You can "just write down" the answers or provide a more rigorous solution. The rigorous solution requires use of the relativistic invariant E2 - p2c2 = m2c4.)

2. Relevant equations

For real particles:
E2 - p2c2 = m2c4

3. The attempt at a solution

Two two feynman diagrams which I come up with are the same ones that are on wikipedia:
http://en.wikipedia.org/wiki/Bhabha_scattering

I am EXTREMELY confused about the fundamental nature of this problem. I've been told that "energy and momentum" are conserved at each vertex. If that is the case, then for the annihilation diagram, I see no reason why the photon isn't real. For the "scattering" diagram, I suppose that the energy of the photon must equal the initial (and final) energy of the leptons. But even if you know that energy, how do you get a mass of the photon from it, if the above equation only applies to real particles?

Furthermore, in "4-momentum" (which I barely understand), if I somehow get the momentum of the virtual particle, how do I get its mass? Do I simply divide by c?

I've been racking my brain about this problem for days, and even spoke to the professor about it in person, and his explanation was wonky and not terribly helpful. And apparently nobody on the internet has ever tried to solve this type of problem, because I can't find anything anywhere on how to figure this out. Any help here would be amazing and extremely appreciated.

Thank you!

2. Jan 31, 2013

andrien

the annihilation diagram of bhabha scattering comprise of a time like photon while the simple one is having a spacelike part.use the general energy momentum conservation at vertex and use einstein formula to determine mass.for a real photon,the m is zero but not for virtual one.Also the diagram drawn on wiki page is not the usual way and in a awkward way it is wrong but it is still right if you don't follow the feynman usual way.

3. Jan 31, 2013

JamesM86

I know that the wikipedia diagrams aren't really correct because they label the antiparticles and have the arrows the wrong way, which isn't convention. But they're correct other than that, no?

So you're saying that, for the annihilation case, the energy of the virtual photon is equal to the total initial energy:

Ephoton = 2mec2

And since

E=mc2

The mass of the photon is therefore 2me

Is that right?

4. Feb 1, 2013

andrien

I am saying to use
E2=p2+m2
For annihilation diagram,you can best convert to CM system in which electrons and positrons move in opposite direction and resulting photon's momentum is zero.so in this frame
for photon,
E2=m2
and E=2(√P2+M2),WHERE P is CM system momentum of any electron or positron.now you can calculate photon mass.You can transform to lab frame after that in terms of original momentas.(mass is invariant of frame)