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
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.