Talk:Mandelstam variables

Call me stupid but isn't the u channel a twist of the s channel, not the t channel

I'm just learning I'll leave it to the true scholars (ie those get paid rather than pay) amongst us to make the edit

• No, the u-channel is just like the t-channel with the final-state particles' roles exchanged. You may be referring to the fact that on the graphic in this article the u-channel looks like the s-channel, but the relevant bit is whether the two initial-state particles annihilate or not, which they don't in the u-channel. The graphic should be changed to better reflect this, though; in particular, the intermediate particle in the s-channel has a timelike momentum, whereas it is spacelike in the t- and u-channels, and it is pedagogically useful for the direction of the lines in the diagrams to reflect this. 142.3.164.195 23:51, 10 May 2006 (UTC)

• Wrong sign: if it is p1 + p2 = - p3 - p4 then it should be t = (p1 + p3)^2 = (p2 + p4)^2, u = (p1 + p4)^2 = (p2 + p3)^2.

wrong sign

I think the conservation of 4-momentum should be $p_{1}+p_{2}=p_{3}+p_{4}$ instead of $p_{1}+p_{2}=-p_{3}-p_{4}$ ... -- Cheesus 14:47, 10 November 2006 (UTC)

s-channel discovery

This statement needs a citation or needs to be removed:

the s-channel is the only way that resonances and new unstable particles may be discovered provided their lifetimes are long enough that they are directly detectable.

It is a statement depending on what kind of collider is used, it is for example a possibility that a process does not proceed through the s-channel. See for example Møller scattering with a heavy photon or a Z' like particle. — Preceding unsigned comment added by 87.174.241.156 (talk) 20:16, 28 July 2014 (UTC)

Feynman diagrams

The equation claiming equality of the inner products p1.p2 = p3.p4 is not correct. If m3^2+m4^2 is not equal to m1^2+m2^2, the inner products must also differ for the square of the sum to be equal (p1+p2)^2 = (p3+p4)^2.Matti Heikinheimo (talk) 14:00, 31 May 2022 (UTC)