# Why are positrons traveling backward in time in Feynman diagrams?

We draw the positron as if it's traveling backward in time. Why? The momentum of the positron is drawn opposite to the time-direction of the diagram. I don't see this having any effect on the calculation. We still arrive at the result $$k=p_1+p_2$$, and not $$k=p_1-p_2$$ like the diagram would indicate. So, if anything, the diagram is misleading? $$p_1$$ and $$p_2$$ are the momenta of the incoming electron and positron, $$k$$ is the momentum of the virtual photon.

• Who are we? Which page? Commented Jul 7, 2022 at 8:06
• @Qmechanic Diagram of type 1 Bhabha process. The incoming positron is drawn to have a momentum opposite to the time direction Commented Jul 7, 2022 at 8:10
• – rob
Commented Jul 7, 2022 at 9:42
• @rob Can we say that the arrows in the Feynman diagrams have no effect on the calculation? They're just there to indicate that electron and positron are similar beasts? Also, if a positron backwards arrow is labelled by a three momentum $\vec{p}$, then I should not conclude that its actual three momentum is $-\vec{p}$, right? Commented Jul 7, 2022 at 9:47
• Both correct. The arrows make it easier to conserve lepton number at each vertex.
– rob
Commented Jul 7, 2022 at 9:52

On the time direction. Real particles never travel backward in time. For virtual particles instead this can happen. For example you can have an initial particles which goes from say $$t_0$$ to $$t_1, then interacting with a photon for example, and propagate from $$t_1$$ to $$t_2>t_0$$. Indeed $$t_1$$ would be an integrated variable, and all the possible values of $$t_1$$ need to be considered. In the convention where time flows from left to right, and for the case $$t_2, there would be a line from center to the left, and a second line going back from the left to the right. In both lines the direction of the arrow would just depend on the particle kind, and not on the time flow.