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In Griffith's Elementary Particles we are asked to draw all four vertex Feynman Diagrams for Compton Scattering. Most are straightforward but one puzzles me.

Four vertex

As has been said here: https://www.physicsforums.com/threads/special-fourth-order-feynman-diagram-compton-scattering-why-is-it-allowed.805039/

But my question is, does this not represent pair annihilation instead of Compton scattering? All other fourth-order Feyman Diagrams feature electron and photon going in and electron and photon going out. This one however an electron-positron pair going in and two photons going out. I understand that the two processes are related by crossing symmetry, but are they considered to be the same process or am i missing something important?

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  • $\begingroup$ If time "goes from left to right", what makes you think you have ee+ going in and two gammas coming out ?? $\endgroup$ – DanielC Aug 23 '17 at 7:48
  • $\begingroup$ Every other diagram in the book uses the time from bottom to top convention, therefore i read the diagram that way. $\endgroup$ – Zarathustra Aug 23 '17 at 8:00
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Time in this diagram (as the arrow shows) goes from left to right instead of bottom to top. Thus the particles on the left (an electron and photon) are the initial particles and those on the right (an electron and photon) are the final particles. Hence this is Compton scattering instead of pair production.

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  • $\begingroup$ This seems like a bizarre inconsistency, since every other diagram in the book uses the time flows from bottom to top convention. Thank you for the answer. $\endgroup$ – Zarathustra Aug 23 '17 at 7:59
  • $\begingroup$ @Zarathustra ye I know, its odd but not all that uncommon. When looking at Feynman diagrams it is at least worth checking which direction time is - just in case. $\endgroup$ – Quantum spaghettification Aug 23 '17 at 8:01

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