Why is u-channel diagram kinked in such way?

Asked Oct 31 '19 at 23:37

Active Oct 31 '19 at 23:37

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Let us consider Bhabha scattering as an example. Using Mandelstam variables, we can think about u-channel Feynman diagrams (see below). Notice the kink at the first two vertices where virtual particle photon is being exchanged. I suspect that the kinks are due to momentum transfer (i.e. momentum conservation), however, I don't understand them why. It seems like they should be kinked the other way around like the kink Bremsstrahlung radiation (see second picture).

It would be appreciated if someone could shed some light on this.

asked Oct 31 '19 at 23:37

Nugi

The shape has nothing to do with momentum. It’s just that if you don’t draw a kink, you can’t tell it apart from a t-channel diagram. – knzhou Nov 01 '19 at 00:04
@knzhou Why can't you tell it apart from a t-channel? You can identify it by where the virtual particle propagates. – Nugi Nov 01 '19 at 00:28
None of the electrons in the u-channel diagram you drew are virtual. – knzhou Nov 01 '19 at 00:53
But in t-channel are they virtual? – Nugi Nov 01 '19 at 01:32
No. Only the photon is virtual. – knzhou Nov 01 '19 at 01:33
I meant the photon in the t-channel. – Nugi Nov 01 '19 at 01:34
In both the t-channel and the u-channel the photon is in general virtual. – knzhou Nov 01 '19 at 01:35
I see. Can you explain why the electrons in the u-channel are virtual, and in t-channel real? – Nugi Nov 01 '19 at 01:50
I can't explain that because it is not true. None of the electrons are virtual. – knzhou Nov 01 '19 at 01:51
I said wrong things. Let me rephrase what I'm going to say. Can you elaborate why none of the electrons in the u-channel are virtual compared to t-channel? – Nugi Nov 01 '19 at 01:53

Why is u-channel diagram kinked in such way?

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