# Intuitive picture for Mandelstam $t$

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?

• 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. – xi45 May 16 '16 at 13:29

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.