Mass and velocity of virtual photon in Bhabha Scattering

In summary, to determine the mass of the virtual photon in each of the lowest-order diagrams for Bhabha scattering, one can use the general energy-momentum conservation at the vertex and the relativistic invariant equation E2 - p2c2 = m2c4. For the annihilation diagram, the energy of the virtual photon is equal to the total initial energy, and its mass can be calculated using the equation E2 = m2. After transforming to the lab frame, the mass of the virtual photon can be determined.
  • #1
JamesM86
5
0

Homework Statement



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.)

Homework Equations



For real particles:
E2 - p2c2 = m2c4

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!
 
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  • #2
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
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
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)
 
  • #5


First of all, it is important to clarify that virtual particles are not real particles. They are mathematical constructs used in Feynman diagrams to represent the exchange of energy and momentum between real particles. Therefore, they do not have a well-defined mass or velocity.

In the case of Bhabha scattering, the virtual photon is created and then immediately annihilated, so it does not have enough time to travel a measurable distance and therefore does not have a well-defined velocity. Its mass can also not be determined from the energy and momentum conservation equations, as those equations only apply to real particles.

In order to calculate the mass of the virtual photon, one would need to use the relativistic invariant E2 - p2c2 = m2c4, but this is not a straightforward process and requires a more in-depth understanding of quantum field theory.

In summary, the mass and velocity of a virtual photon in Bhabha scattering cannot be determined in a meaningful way. The virtual photon is simply a mathematical construct used to describe the interaction between real particles.
 

1. What is Bhabha scattering?

Bhabha scattering is a type of scattering process in particle physics where an electron and a positron collide and produce two high-energy photons.

2. How does Bhabha scattering relate to virtual photons?

In Bhabha scattering, the two high-energy photons produced are virtual photons, meaning they exist for a very short period of time before decaying back into their original particles.

3. What is the mass of a virtual photon in Bhabha scattering?

The mass of a virtual photon in Bhabha scattering is not a well-defined quantity, as virtual particles do not follow the same laws of physics as real particles. Therefore, the concept of mass does not apply to them.

4. How is the velocity of a virtual photon in Bhabha scattering determined?

The velocity of a virtual photon in Bhabha scattering is not a measurable quantity, as virtual particles do not actually travel through space like real particles. They are simply mathematical constructs used to describe the interaction between real particles.

5. How is the mass and velocity of a virtual photon in Bhabha scattering calculated?

The mass and velocity of a virtual photon in Bhabha scattering cannot be calculated, as they are not physical quantities. They are simply mathematical concepts used in theoretical calculations to describe the behavior of particles during the scattering process.

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