# Physical interpretation of Mandelstam variables

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?

Related Quantum Interpretations and Foundations News on Phys.org
Orodruin
Staff Emeritus
Homework Helper
Gold Member
Did you try answering these questions yourself based on how the variables are defined?

I can see why ##t## is called the four-momentum transfer since it is the square of the difference between one of the initial momenta and one of the final momenta.

Still, the terminology four-momentum transfer does not seem to precisely interpret the variable ##t## since the initial momentum and the final momentum appear to chosen arbitrarily.

To illustrate, why not define ##t## as ##t=(p_{1}-p_{4})^{2}=(p_{2}-p_{3})^{2}##?

But then ##u## is defined using ##u=(p_{1}-p_{4})^{2}=(p_{2}-p_{3})^{2}##, so both ##t## and ##u## together appear to describe the different possibilities of what can be meant by four-momentum transfer.

Is the use of the terminology four-momentum transfer a bit of a hand-waving?

Orodruin
Staff Emeritus
Homework Helper
Gold Member
Yes, both ##t## and ##u## are 4-momentum transfers squared. They depend on which particle you assign as 3 and 4, although many times there are "natural" choices.

The entire point is that ##s## is the square of the 4-momentum of the propagator in an ##s##-channel diagram whereas ##t## and ##u## are the squares of the 4-momenta of the propagators in ##t## and ##u## channel diagrams, respectively.

vanhees71
Yes, both ##t## and ##u## are 4-momentum transfers squared. They depend on which particle you assign as 3 and 4, although many times there are "natural" choices.
How do you decide on a natural choice?

Also, I was wondering if the s-channel, t-channel and u-channel the only possible Feynman diagrams for ##2-2## scattering?

Orodruin
Staff Emeritus
Homework Helper
Gold Member
How do you decide on a natural choice?
If you have an elastic scattering, for example ##e^+ \mu^+ \to e^+ \mu^+##, the "natural" choice would be to assign the ##t## variable to the difference squared of the positron 4-momenta (or equivalently, muon 4-momenta). What is referred to as ##t## is usually taken where the particles considered in the in and out states are the most alike.

Also, I was wondering if the s-channel, t-channel and u-channel the only possible Feynman diagrams for 2−22−22-2 scattering?
This depends on the theory. In QED, they are the only possibilities at tree level. Of course, there are many more options at loop level.

vanhees71
If you have an elastic scattering, for example ##e^+ \mu^+ \to e^+ \mu^+##, the "natural" choice would be to assign the ##t## variable to the difference squared of the positron 4-momenta (or equivalently, muon 4-momenta). What is referred to as ##t## is usually taken where the particles considered in the in and out states are the most alike.
Is there a reason for why this is so?

This depends on the theory. In QED, they are the only possibilities at tree level. Of course, there are many more options at loop level.
Ok, so I presume that the Mandelstam variables are defined not only for scalar theories, but also for fermions and gauge bosons.

The diagrams for Mandelstam variables seem to give the impression that the Mandelstam variables are valid only for Feynman diagrams of ##\phi^{3}## theory with two vertices (with the exchange of one intermediate particle), or can the dotted lines have an arbitrary number of vertices for an arbitrary ##\phi^{n}## theory?

Orodruin
Staff Emeritus
Homework Helper
Gold Member
Is there a reason for why this is so?
Convention.

The Mandelstam variables a priori have nothing to do with what diagrams you are using. They appear in the propagator if you have an s-, t-, or u-channel diagram, but that is another matter. It is just a convenient way of parametrising the kinematic setup you are considering.

Convention.

The Mandelstam variables a priori have nothing to do with what diagrams you are using. They appear in the propagator if you have an s-, t-, or u-channel diagram, but that is another matter. It is just a convenient way of parametrising the kinematic setup you are considering.
So you mean that the s-channel, for example, could be part of a larger. more complicated Feynman diagram?

Orodruin
Staff Emeritus