# Homework Help: Need help with gamma dirac matrices

1. Dec 17, 2005

### goulio

Hello,

This problem is related to the beta decay of a neutron in a proton an electron and a anti-neutrino. I need to prove that, in the limit where the mass of the neutron and the proton goes to infinity, $m_P, m_N \to \infty$, we have
$$\bar{u}\gamma^\mu(1-\alpha \gamma_5)(\gamma^\alpha k_\alpha + m_P)\gamma^\nu(1-\alpha \gamma_5)u = 4m_P^2(c^\mu g^{\mu \nu} - \alpha(\delta^\mu_0 \delta^\nu_3+\delta^\mu_3 \delta^\nu_0)-i\alpha \epsilon^{0 \mu \nu 3})$$
where $k$ is the momentum 4-vector of the proton and
$$u=\sqrt{m_N}(1,0,1,0)$$
is the spinor of the neutron, which is at rest, aligned with positive z-axis and $c^0=1$, $c^i=-\alpha^2$ for $i=1,2,3$ and $\alpha = 1.22$.

I really can't figure out how to do this...

Any help greatly appreciated

Last edited: Dec 17, 2005
2. Dec 20, 2005

### CarlB

I'm not completely sure what you're asking. I'm going to assume that you're given the LHS and you want to derive the RHS under the assumption that the masses go to infinity.

I would first try to take the m-> infinity limit on the RHS. That should simplify the proton propagator term in the center as the proton will no longer be moving, I would think. That should give you one factor of m_P. I guess the other will come from the spinor after you take a trace or get rid of all the intervening gammas between the two spinors.

To deal with the gamma_5s, use the commutation / anticommutation rules to move them all together into one place where you can square them into +/-1 (I forget which, and if I recall, it depends on your signature assumption. I work in Clifford algebras and would be inclined to rewrite gamma_5 in terms of the other matrices.)

Some of the most useful relations (for this sort of problem) when dealing with gamma matrices is stuff of the form:

$$\gamma^0 (A\gamma^0 + B\gamma^1 + C \gamma^2\gamma^3) = (+ A\gamma^0 - B\gamma^1 + C\gamma^2\gamma^3) \gamma^0$$

That is, you need to figure out how moving a gamma matrix around a complicated sum of gamma matrices changes that complicated sum. The general rule is that like gamma matrices commute and unlike gamma matrices anticommute, as in the example above. In the example above, the 0 matrix commutes with itself, and with any even product of non-zero (i.e. spatial) matrices. It anticommutes with odd products of spatial gamma matrices.

If you generalize this relation to one that allows you to commute the gamma-5s, you will be able to get them together into one product and get rid of them.

It's not stressed in the books, but if you understand what I've written above, it will save you great amounts of algebraic headache. You can sometimes make calculations of great beauty this way.

That is, with this method, I think you're still going to have to multiply out the LHS to get the product terms, but then cancelling out the gamma 5s will be a lot easier.

When you commute the gamma5 around the gamma^mu, it will do different things according to mu and that is where all those delta functions are going to pop up. That is, the basic structure of this problem is the fact that spinors are normalized, so you need to get those u's together with the simplest possible thing between them.

Tell us if this helps.

Oh, and thanks for putting the problem up on the thread. My regular job involves too much things like spud wrenches. I'm going to guess that, efficiently done, your answer should fit on one side of piece of paper.

Carl

Last edited: Dec 20, 2005