Mandelstam Variables and Veneziano Amplitudes

Quantum mechanics is a probabilistic theory. The probabilities are the squared magnitudes of (probability) amplitudes. Amplitudes are computed through perturbation theory, or whatever method one can find. The state of the art method is to use Feynman diagrams to find the amplitudes.

When one considers interactions between two particles, there are two Feynman diagrams that are of particular interest, the “t-channel” and the “s-channel”. The “s” and “t” are Mandelstam variables. These variables define the interaction in a way that automatically preserves Lorentz invariance.

Special Relativity Momenta

This is a relativistic problem, so all momenta are 4-momenta, that is, they are vectors with four components. The time or zero portion of the vector gives the negative energy of the particle, while the space or 1-3 portion of the vector gives its momentum. A 2-particle interaction involves 4 momenta, two for the incoming particles, and two more for the outgoing particles. That’s a total of 4×4 = 16 variables.

Let p_1, p_2, p_3, p_4 be the four momenta. The first two are for the incoming particles, that is, the ones before the interaction, the 3rd and 4th are for the outgoing particles:
Definition of momenta for four particles
Now each of these p_i are a 4-vector. I will write them as p_i^\mu where p_i^0 is the energy of the particle, p_i^1 is the momentum in the x-direction, etc.

The Minkowski norm of a 4-vector is written as a square. It is defined by computing the difference between the square of the energy/time portion of the 4-momenta and the squared magnitude of the 3-momentum part of the 4-momenta:
(p_i)^2 = -(p_i^0)^2/c^2 + (p_i^1)^2 + (p_i^2)^2 + (p_i^3)^2.

The Minkowski norm of a particle does not depend on the choice of reference frame. What attribute of a particle doesn’t depend on the reference frame? Yes, the proper mass does not depend on reference frame. This means that we can get the mass of the particle by squaring its 4-momentum. The square of the speed of light provides a constant of proportionality, and, by convention, the difference is taken so that the square is negative:
(p_i)^2 = -m^2\;c^2.

Mandelstam Variables

So 4-momenta do not depend on choice of reference frame. In addition, we can add 4-momenta. This means that given two particles, we can sum their 4-momenta, and then compute its square. While this is not the mass of any particle, it is a conserved quantity and it does not depend on the choice of reference frame.

The principle of conservation of 4-momenta for this 2-body problem says that the total 4-momenta is the same before and after the interaction, that is:
p_1+p_2 = p_3 + p_4.
We can square the two sides of the above equation. The result is the “s” Mandelstam variable:
s = (p_1+p_2)^2 = (p_3 + p_4)^2.
The other Mandelstam variables come from exchanging one or the other of the 4-momenta to the opposite of the equation. We have (using the fact that the square of the negative of a 4-momenta is the same as the square of the 4-momenta by definition):
p_1-p_3 = p_4 - p_2,\;\;\textrm{so}\;\;t=(p_1-p_3)^2 = (p_2-p_4)^2,
p_1-p_4 = p_3 - p_2,\;\;\textrm{so}\;\;u=(p_1-p_4)^2 = (p_2-p_3)^2.

You can see that the “u” and “t” variables are swapped under the exchange of 3 and 4. So physicists don’t talk much about “u-channel”; they instead talk about the “s-channel” and “t-channel”. S is the square of the center of mass-energy or the square of the invariant mass. This is a measure of how much energy is available for the interaction. It tells you what can be created in the interaction. T is the square of the momentum transfer. This tells how much energy was moved from one particle to the other.

In a “t-channel” interaction, two particles interact with the exchange of a presumably small amount of momentum. This is the kind of picture we associate with two particles that do not get near each other. A Feynman diagram for the t-channel might look like this:
T channel process, boson exchange particle

In an “s-channel” interaction, one thinks of the two particles as being a combined entity; hence the center of mass energy or invariant mass. A typical s-channel interaction could be much more complicated than the above t-channel interaction. For a resonance, the two particles might interact for quite some time before parting ways and this gives time for them to exchange many particles:
Particle resonance, an s-channel interaction

The “1” incoming particle has 4 degrees of freedom, but given that the square of its 4-momenta gives its mass, only 3 of those degrees of freedom are real. The same can be said about the other three particles. So the 4-momenta of the particles themselves has 4×3 = 12 real degrees of freedom which we can take to be the 3-momenta. But 3-momenta is conserved, so given three of the particles we can compute the 3-momenta of the 4th. This gets us down to just 9 real degrees of freedom.

Lorentz symmetry requires that the interaction not depend on choice of reference frame. We can boost in any of 3 directions, and we can also rotate. Together, these give 6 degrees of freedom that do not need to be included to describe the physical character of the interaction. What’s left is 3 real degrees of freedom. These are given by the s, t, and u variables. For any given interaction, one of these might be a lot larger than the others. For examle, if s >>t,u, then we have relatively heavy particles that are on parallel tracks. This would be the natural case for two particles bound together.

Poles in Transition Amplitudes

In QFT, a Feynman diagram stands for the interactions that are topologically equivalent. One must count up the contributions by all these interactions. One typically does this by integrating over the possibilities. All this is generally done in momentum space for the simple reason that in momentum space, momentum is conserved. So any time we have a diagram with no loops, that is, a tree diagram, (like the example t-diagram shown above), it is possible to compute the momenta exactly on each of the lines of the diagram as functions of the momenta of the incoming and outgoing particles.

The loops are another story. A Feynman diagram with a loop needs to be integrated over all possible momenta. From the incoming and outgoing momenta one can compute the differences between the momenta on different legs of the loops, but there is always one free 4-momenta left over and this is integrated over.

Now the problem with this process is that when you integrate over the loop momenta you get infinity. This is called a loop divergence. LOL! For engineers this would be an issue (we would trash the theory and get something that worked) but for physicists it is no big deal. They cancel the infinities off against each other, or ignore them one way or another.

The infinities arrive in the form of poles, that is, singularities in a complex function that act like 1/z^n near z=0. For the poles created by the loop momenta, the function with the pole is the probability amplitude. The probability amplitude depends on the Mandelstam variables s, t, and u. The Mandelstam variables are real variables; to make them into complex variables they are analytically continued.

The s in the s-channel is the square of the center of mass energy. When the transition amplitude has a pole in the s-channel we interpret this as a particle, a resonance: two particles arrive and temporarily combine into a composite particle. This particle eventually decays. This is an s-channel resonance. And since 4-momenta is conserved, we know that the mass of the resonance, that is, the mass of the composite particle, is given by \sqrt{s} . When physicists talk about meson resonances, such as are described by the Particle Data Group, they are talking about these poles. In short, a pole is a particle.

Veneziano Amplitudes

In 1968, Gabriele Veneziano noticed that two quantum interactions between 2 particles, the “t-channel” and “s-channel”, though very much different physically in terms of interpretation, nevertheless must produce the same amplitudes. They are, after all, just different ways of describing the same physical interaction; particles 1 and 2 interact and produce particles 3 and 4. In either case, there had to be the same set of poles in the transition amplitude. In terms of writing the probability amplitude T in terms of s, t, and u, this means that T(s,t,u) = T(t, u, s) = T(s,u,t) = etc. Since this is two ways of looking at the same resonances, he called this the dual resonance model.

Veneziano’s proposal for the probability amplitude, as a function of s, t, and u, is given by
T = A(s,t) + A(s,u) + A(t,u) where
A(s,t) = \Gamma(-\alpha(s))\Gamma(-\alpha(t))/\Gamma(-\alpha(s)-\alpha(t))
where \alpha(s) is a linear function of s. The \Gamma or Gamma function is the analytic continuation of the factorial function, (n-1)!. As you can see, the above amplitude is symmetric with respect to s, t, and u. That is, it treats these variables equally so there will be poles in one variable where there are poles in the others.

The Gamma function has poles at the non positive integers. As noted above, these will be interpreted as particles or resonances. Since \alpha(s) is a linear function of s, its presence in a Gamma function will cause a sequence of resonances to appear, a Regge trajectory, that is characterized by a linear relationship in the square of the mass. For instance, if the Gamma function causes poles to occur for s = 3,5,7…, then the corresponding particles will have masses of \sqrt{3}, \sqrt{5}, \sqrt{7} and so on. For more on Regge trajectories, including their interpretation in string theory, see the class notes of Stephen D. Ellis.


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5 responses to “Mandelstam Variables and Veneziano Amplitudes

  1. Kea

    Wonderful introduction! Can’t wait to see further developments.

  2. carlbrannen

    Kea, the thing I’m doing is similar except that I’m in qubit space. So I don’t have any s, t, or u variables, but I do have the equivalence between the poles in the s and t-channels but I do use Feynman diagrams and you can classify them as s-channel like or t-channel like.

    What that all amounts to is that the preon interaction, which is a complicated s-channel interaction, can be modeled as a simple t-channel interaction. This is theoretical support for the computation for “that damned number.”

    That is, the calculation for tdn will be in a t-channel form but it will also be appropriate for the s-channel Feynman diagrams.

    By the way, I’m slowly typing this up for submission to Foundations of Physics.

  3. Very snazzy, although a good chunk is above my head. Within a year or so I should have had the requisite classes. It’s interesting stuff, and I’ll keep reading to learn more. Cool to see perturbation theory mentioned as well, since I’ve got a post demonstrating a (very elementary undergrad!) example from the GRE in the works.

    Also, I think I’ve caught a typo – you’ve listed the Gamma function as the analytic continuation of the exponential function, when I believe it’s the continuation of the factorial function. The exponential function is thankfully analytic over the entire complex plane.

    [Fixed! Thank you. Even little typos make stuff a lot harder to read.]

  4. Ark

    Please take a look at
    in connection with your posting on Veneziano amplitudes

  5. Bob

    A good introduction, though i think you may have misrepresented the consistent process of renormalization, mr engineer.

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