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Bookmark or Share- What's the difference between $t$-channel and $s$-channel in particle ...Then if the collision happens at energy of about Z boson mass, i.e. sqrt(s) ~= m(Z), the s-channel dominates. But depending on the kinematic phase space of the interaction, you might have a situation where the t-channel is dominant. These two cases correspond to the answer by @JamalS: M ~= 1/(s-m^2) @ s~=m^2 VS M ~= 1/(t-m^2) @ t~=m^2. $\endgroup$
- Mandelstam variable equality derivation for $s$-channel processThis equation should hold in any inertial frame, so let's boost into the COM frame. In the center of mass frame (by definition), $\vec p_1 + \vec p_2 =0$ (since each are the regular 3-momenta of the particles), so
- Compton Scattering Lowest Order Feyman DiagramHere is a clearer sum of the two lowest order diagrams:. It shows the two geometric ways energy and momentum can be exchanged between the two incoming particles to produce the two outgoing, to first order in a series expansion.
- Equivalence between $t$ and $u$ channels - Physics Stack ExchangeFor real particles it's the rest mass, for virtual particles it also looks like a squared-mass, except that is negative in the t-channel. $\endgroup$ – JEB Commented Dec 4, 2018 at 20:38
- How do I know which Feynman diagram is right? ($s,t,u$-channel)In the specific case of four-body processes (like 2 particles scattering off each other into a 2 particle state) you may find that some of the diagrams have a term like $1/(s-m^2)$ or $1/(t-m^2)$. We call those diagrams the "s-channel" or "t-channel" diagrams.
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