Griffiths Ex. 2.4 in his book "Elementary particles" says:
"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)"
I know that the mass in the Annihilation Diagram is $m=2m_e$ using the Center of Momentum system and the formula $E^2 – p^2c^2 = m^2c^4$. I'm considering as three different steps
($e^-e^+,\gamma,e^-e^+$),
and conservation over each of them (each vertex before and after).
(Time goes horizontally).
But what about the Scattering diagram
The virtual photon is it part of the "before interaction" or "after interaction"? Is it part of both? that would make $m=0$. If not it would be $m=2m_e$ considering the virtual photon as a middle step.
Best Answer
(2nd diagram) Since the photon is virtual, its exchange is the process that happens during interaction, neither before nor after.
Now, energy and 3-momentum is conserved at each vertex. Also keep in mind that there is a total [initial(i) or final(f)] energy-momentum 4-vector conservation delta function in any scattering process: $\delta(\sum p_i^\mu - \sum p_f^\mu)$.
Using these, in the 2nd diagram, for the topmost vertex (with photon energy/momentum flowing into the vertex), energy conservation gives $E_{e^+(in)} + E_\gamma = E_{e^+(out)}$. Since $E_{e^+} = m$ for both, $E_\gamma=0$. Similarly, $\vec{p}_{e^+(in)} + \vec{p}_\gamma = \vec{p}_{e^+(out)}$ gives $\vec{p}_\gamma = 0$, and therefore, photon velocity $=0$. Now, using the energy-momentum dispersion relation for these values of $E_\gamma$ and $\vec{p}_\gamma$, we find that $m_\gamma=0$.
As Griffiths points out, this is not possible for real photons: they can never have zero velocity.