4
$\begingroup$

In $2-2$ scattering, the Mandelstam variables $s$, $t$ and $u$ encode the energy, momentum, and angles of particles in a scattering process in a Lorentz-invariant fashion.

$$s=(p_{1}+p_{2})^{2}=(p_{3}+p_{4})^{2}$$ $$t=(p_{1}-p_{3})^{2}=(p_{2}-p_{4})^{2}$$ $$u=(p_{1}-p_{4})^{2}=(p_{2}-p_{3})^{2}$$

where $p_1$ and $p_2$ are the four-momenta of the incoming particles and $p_3$ and $p_4$ are the four-momenta of the outgoing particles.


How is $s$ is the square of the center-of-mass energy?

How is $t$ the square of the four-momentum transfer?

What is the physical interpretation of $u$?

Are $s$, $t$ and $u$ related to the $s$-channel, $t$-channel and $u$-channel respectively?

$\endgroup$
5
$\begingroup$

Yes, they are related to the $s,t$ and $u$ channels as you say. Consider a scattering process where the incoming particles have four-momenta $p_1$ and $p_2$ while the particles that go out have momenta $p_3$ and $p_4$. Then, diagrammatically, there are three possible way the process can take place, described by the figure below.

enter image description here

  • $s$-channel: The two particles "merge" into a virtual intermediate particle that finally splits into the two final particles. Note that since four-momentum is conserved at each vertex we have for the momentum $q$ of the intermediate particle $$q^2=(p_1+p_2)^2=s$$ Besides, since the four-momenta are $p_i=(E_i,\vec{p}_i)$ and the center of mass frame is defined by the constraint $\vec{p}_1=-\vec{p}_2$ one has $$s=(p_1+p_2)^2=|(E_1,\vec{p}_1)+(E_2,\vec{p}_2)|^2=|(E_1,\vec{p}_1)+(E_2,-\vec{p}_1)|^2=|(E_1+E_2,\vec{0})|^2=E^2$$ where $E$ is the total energy of the two particles in the c.o.m. frame.

  • $t$-channel: The particle $1$ emits a virtual particle and in doing so it turns into particle $3$. The virtual particle is absorbed by the particle $2$, that as a consequence of this interaction turns into particle $4$. Since the first particle had momentum $p_1$ before the emission and particle $3$ has momentum $p_3$, the difference of these must have gone into the emitted virtual particle, that thus has (squared) momentum $$q^2=(p_1-p_3)^2=t$$ Note that by reasoning in the same way about the absorption by the particle $2$ we find also that $$q^2=(p_4-p_2)^2=t$$ We can say then that the particle $1$ has "lost" the momentum $q$, that has been transferred to the particle $2$.

  • $u$-channel: As you can see from the image, the $u$ channel is analogous to the $t$ channel with the roles of $p_3$ and $p_4$ interchanged. Therefore we can still interpret $u$ as the momentum transfer from particle $1$ to particle $2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.