# Homework Help: Understanding Lorentz invariance

1. Nov 20, 2007

### jdstokes

On p. 32 of Quantum field theory in a nutshell, Zee tries to derive the propagator for a spin 1 field:

$D_{\nu\lambda} = \frac{-g_{\nu\lambda} + k_\nu k_\lambda /m^2}{k^2 - m^2}$

using the Lorentz invariance of the equation $k^\mu \varepsilon_\mu^{(a)}=0$ where $\varepsilon_\mu^{(a)}$ denotes the three polarization vectors $a = x,y,z$.

I understand that if we have two electromagnetic 4-currents $J_\lambda , J_\nu$, then the amplitude for a particle with momentum $k$ to propagate from one to the other is going to be proportional to

$\sum_{a} \varepsilon_\mu^{(a)}\varepsilon_\lambda^{(a)}$

which is in turn proportional to the propagator $D_{\nu\lambda}$.

According to Zee, Lorentz invariance as well as $k^\mu \varepsilon_\mu^{(a)}=0$ enfornces the condition that

$\sum_{a} {\varepsilon_\nu}^{(a)}{\varepsilon_\lambda}^{(a)}$

is proportional to $g_{\nu\lambda} - k_\nu k_\lambda /m^2$ and that evaluating for k at rest with $\nu = \lambda = 1$ changes the sign.

I'm having trouble understanding each step. I apologise for the lengthiness of this question.

1. Why should $\sum_{a} {\varepsilon_\nu}^{(a)}{\varepsilon_\lambda}^{(a)}$ be Lorentz invariant just because $k^\mu \varepsilon_\mu^{(a)}$ is?
2. Why exactly does Lorentz invariance dictate that it must be a linear combination involving only the metric and $k_\nu k_\lambda$?
3. Why does

$k^\nu{\varepsilon_\lambda}^{(a)} = 0 \implies \sum_{a} \varepsilon_\nu^{(a)}{\varepsilon_\lambda}^{(a)} \propto g_{\nu\lambda} - k_\nu k_\lambda /m^2$

4. How does evaluation of k at rest with $\nu = \lambda = 1$'' change the sign?

Last edited: Nov 20, 2007
2. Nov 21, 2007

### noospace

Since no one has yet replied, let me see if I can get any further withe the first part.

First of all the definition of Lorentz invariance is that the equation remains true in arbitrary intertial reference frames.

The quantity

$A_{\mu\lambda} = \sum_{a} \varepsilon_\mu^{(a)}\varepsilon_\lambda^{(a)}$

is not a scalar, so its components will not be the same in every frame. It is however, the tensor product of two tensors, so in a new reference frame the coordinates will be

$A_{\mu'\lambda'} = \sum_{a} \frac{\partial x^\mu}{\partial x^{\mu'}} \frac{\partial x^\lambda}{\partial x^{\lambda'}}\varepsilon_\mu^{(a)}\varepsilon_\lambda^{(a)}$

Now since $k^\mu_\varepsilon^{(a)}_\mu$ is true in every frame, the polarization is completely determined by the momentum k. Thus we can expect $A_{\mu\lambda}$ being a second rank tensor, to contain a term $k_\nu k_\lambda$. I don't know what the motivation is to include the metric but since we can always set that coefficient to zero we might as well include it also. Thus the most general form for $A_{\mu\lambda}$ is

$A_{\mu\lambda} = c_1 g_{\nu\lambda} + c_2 k_\nu k_\lambda$.

Now to figure out the coefficients $c_1,c_2$. In order for this equation to be dimensionally correct, the coefficient c_2 is going to have units of 1/energy^2. I don't see how $k^\mu \varepsilon_\mu^{(a)}=0$ implies that c_1 and c_2 have opposite sign? Could someone please help me understand this?

3. Nov 21, 2007

### nrqed

Are you sure he does not say Lorentz covariance ?
Lorentz covariance would say that the result must be something that has two spacetime indices (a second rank tensor). By the same argument we had in the other thread, the only two quantities that you have at your disposal of this form are $$k_\nu k_\lambda$$ and $$g_{\mu \lambda}$$
Like in the other thread, write the result as the general expression $$A(k) k_\nu k_\lambda + B(k) g_{\mu \lambda}$$, tne impose that dotting this with k^mu or k^lambda gives zero and you will get the form given (modulo an overall arbitrary constant)
I don't know that that means, I will have to look at the book. What page is that?

4. Nov 22, 2007

### jdstokes

He says invariance on the 3rd line from the bottom on page 31. I think he means covariance, however.

What exactly is covariance anyway? Am I correct in saying that a quantity in Lorentz covariant if and only if it is a tensor (in the relativistic sense)?

Your explanation of the coefficients makes perfect sense. What he means is that if you evaluate $\sum_{a}\epsilon^{(a)}_\nu\epsilon^{((a)}_\mu$ at \nu = \lambda = 1 you get 1, whereas if you evaluate the RHS you get minus 1, this fixes the overall sign of minus 1.

5. Nov 22, 2007

### nrqed

I take back what I said. Now it comes back to me that people use "covariant" when referring to an equation (like "Dirac equation is covariant").
Ah, ok. Yes, that makes sense.

Last edited: Nov 22, 2007