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In tree-level electron-positron scattering one has two possible channels corresponding to Mandelstam variables $s$ and $t$. The $s$-channel ist fine, there $\sqrt s$ is just the center of mass energy available for the gauge boson that the two fermions annihilate into. A photon with four-momentum $(\sqrt s, 0, 0, 0)$ would be off-shell but that is fine. Its rest-mass would also be $\sqrt s$ which makes sense as it needs to carry all that energy.

Here I depict the two particles annihilating in the CMS frame into a resting photon with mass $\sqrt s$. After a given time it will decay into two new particles. As the photon is a virtual particle it does not bother me that it has a rest-mass and is not moving.

The $t$-channel corresponds to the electron and positron doing back-to-back scattering. It is probably better if one takes two electrons such that they actually repel. Nevertheless, I would have momenta $(E, \vec p)$ and $(E, -\vec p)$ for the two particles. Then $t = - 4 \vec p^2$. I do not quite understand what that is supposed do mean. Sure, $p_1 - p_1'$ is just $(0, 2 \vec p)$ and then $t$ is the Minkowski-norm of that state. The math is fine.

But what does $t = - 4 \vec p^2$ mean for the exchanged photon? I would interpret this as a photon with a negative rest mass (arising from space-like propagation). But there is no mass transferred from one lepton to the other, only the direction of the three-momentum $\vec p$ has changed for both of the leptons.

On the other hand, if $t = 0$ as I would expect for an on-shell photon, then one would have $(|\vec k|, \vec k)$ as four-momentum of the photon. The photon would then not only change the momentum of the particles but also transfering energy from one to the other.

What would be a good way to think about $t$?


Perhaps the main source of confusion is the interpretation of the invariant momentum transfer $Q^2$. So if we have $Q^2 = s$ we have a time-like momentum transfer. That is mostly an energy transfer. If we have $Q^2 = t$ it is usually negative and is a space-like momentum transfer.

If one looks at the four-vectors, a time-like momentum transfer obviously is mostly energy transfer and a space-like transfer mostly three-momentum.

When one does quantum field theory and refers to an energy scale, one means this $Q^2$. But one can still have some strong effects for $Q^2 \approx 0$ if one exchanges a lot of on-shell photons. So what it the merit of that $Q^2$ in those cases?

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  • $\begingroup$ It would probably help clarify your doubts and questions if you considered a generic case, not the particular case of $e^- e^+ \to e^- e^+$ scattering. $\endgroup$ – xi45 May 16 '16 at 13:29
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Mandelstam $t = (p_1 -p_1')^2$ corresponds to the square of the momentum transfered between the two scattering particles, in an elastic scattering process.

If you look at the scattering in the centre of mass frame, like you suggest, then clearly they cannot transfer any energy between them, since that would vilolate conservation of momentum.

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  • $\begingroup$ Energy transfer is forbidden when the particles have the same mass. That might not be the case when the particles are not of the same mass. I also thought about another angle to the question, I'll update my initial post. $\endgroup$ – Martin Ueding May 16 '16 at 11:03
  • $\begingroup$ In the COM frame, energy transfer is forbidden even between different mass particles. For elastic scattering that is. $\endgroup$ – Ihle May 16 '16 at 12:20

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