# Advice on a derivation from Guage Fixing Lagrangian to Equation or motion?

1. Feb 1, 2010

### neorich

Hi All,

I am trying to derive the equation of motion from a Lagrangian with a gauge fixing term, and I think get quite close to the result, but am missing some steps somewhere. These indices and notation is new to me, so please feel free to bring any mistakes to my attention.

So the Lagrangian is:

$${\cal L} = -\,\frac{1}{4}\,F_{\mu\nu}F^{\mu\nu} +\, j_{\mu}A^{\mu} -\,\frac{1}{2\xi}\,(\partial_{\mu}A^{\mu})^2$$

And the equation of motion I have to derive is:
$$\left(\Box g^{\mu\nu} - \Big(1-\frac{1}{\xi}\Big) \partial^\mu \partial^\nu \right) A_\nu = j^{\mu}$$

using the Euler-Lagrange Equation:

$$\partial_{\mu}\,(\frac{\partial{\cal L}}{\partial\,(\partial_{\mu}A_{\nu})}) -\, \frac{\partial{\cal L}}{\partial\,A_{\nu}}=0$$

So the first term in the E-L equation will only act on the derivatives in the Lagrangian, and the second term will only act in the non-derivatives, so I get:
$$\frac{\partial{(-\,\frac{1}{4}\,F_{\mu\nu}F^{\mu\nu})}}{\partial\,(\partial_{\mu}A_{\nu})}=-F_{\mu\nu}=\partial_{\nu}A_{\mu}-\partial_{\mu}A_{\nu}$$

$$\frac{\partial{(-\,\frac{1}{2\xi}\,(\partial_{\mu}A^{\mu})^2)}}{\partial\,(\partial_{\mu}A_{\nu})}=-\,\frac{1}{\xi}\,g^{\mu\nu}(\partial_{\mu}A_{\nu})$$

$$\frac{\partial{(j_{\mu}A^{\mu}})}{\partial\,A_{\nu}}=\frac{\partial{(j_{\mu}g^{\mu\nu}A_{\nu}})}{\partial\,A_{\nu}}=g^{\mu\nu}j_{\mu}=j^{\nu}$$

Now, I am not entirely sure about my use of gmunu's here. Anyway, combining these, I get:

$$\partial_{\mu}(\partial_{\nu}A_{\mu}-\partial_{\mu}A_{\nu})-\frac{1}{\xi}g^{\mu\nu}\partial_{\mu}\partial_{\mu}A_{\nu}-j^{\nu}=0$$

Can you please tell me if this is correct so far, and if so, can you please direct me as to how to obtain the box times the metric, and the two derivatives with upper indices on the right of the left hand side on the equation of motion?

Regards

neorich.

2. Feb 1, 2010

### torquil

You index positions are incorrect. E.g. in the first term, you have $$\mu$$ as a lower index in two places. In the second term there is also a problem. Be careful about those indices, e.g. which are up and which are down, and which are being summed!

And e.g. when you calculate the derivative of F^2 you are using mu,nu in too many places, although in that expression you probably calculated correct because you already knew that it was only the mu, nu in the F^2 contraction that were to be summed over.

You also use mu too many places in the calculation of the derivative of the gauge fixing "(dA)^2" term.

In the calculation of the derivative of F^2 you should end up with both indices as upper indices.

In the derivative of the gauge fixing term (dA)^2 you have two free indices on the left, but no free indices on the right, so that is not correct.

Torquil

3. Feb 1, 2010

### neorich

So for the F^2, I should have upper indices, OK then, so the equation becomes:

$$\partial_{\mu}(\partial^{\nu}A^{\mu}-\partial^{\mu}A^{\nu})-\frac{1}{\xi}g^{\mu\nu}\partial_{\mu}\partial_{\mu}A_{\nu}-j^{\nu}=0$$

Quote: "You also use mu too many places in the calculation of the derivative of the gauge fixing "(dA)^2" term."

Could you explain this a little further please, as I don't follow. In the equation below, I must use $$(\partial_{\mu}A^{\mu})^2$$ as the function being differentiated (as this is given to me in the Lagrangian), and I must use $$\partial_{\mu}A_{\nu}$$ as the variable I am differentiating with respect to (as this is also given to me in the E-L equation). Is this correct?

Which gives me the LHS

$$\frac{\partial{(-\,\frac{1}{2\xi}\,(\partial_{\mu}A^{\mu})^2)}}{\partial\,(\partial_{\mu}A_{\nu})}$$

So are you saying I have too many mu's in the LHS or the RHS?

If it's the LHS, what should the indices be? If it's the RHS, then what's wrong with the derivation below?

$$\frac{\partial{(-\,\frac{1}{2\xi}\,(\partial_{\mu}A^{\mu})^2)}}{\partial\,(\partial_{\mu}A_{\nu})}=\frac{\partial{(-\,\frac{1}{2\xi}\,(g^{\mu\nu}\partial_{\mu}A_{\nu})^2)}}{\partial\,(\partial_{\mu}A_{\nu})}=-\,\frac{1}{\xi}\,g^{\mu\nu}(\partial_{\mu}A_{\nu})$$

I know you already mentioned having more free indices on the LHS than RHS, but I am not sure how to figure out which indices are free, summed over etc.

Finally was my last term $$j^{\nu}$$ correct?

Thanks

Regards

neorich.

4. Feb 1, 2010

### torquil

Yes, it should look like this:
$$\frac{\partial{(-\,\frac{1}{2\xi}\,(\partial_{\rho}A^{\rho})^2)}}{\partial\,(\partial_{\mu}A_{\nu})}$$

Otherwise the reader doesn't know if you are summing the two upper \mu's, or one upper \mu with one of the \mu's in the denominator. This "objection" was not really a problem for you, because you knew already that it was the two upper indices that were supposed to be summed. But if you want to write it correctly, you shouldn use each index letter either only once, or twice (one down and one up)

I'll answer the other question in my next post.

Torquil

5. Feb 1, 2010

### torquil

$$\frac{\partial(\partial_\alpha A^\alpha)^2}{\partial(\partial_\mu A_\nu)} = \frac{\partial(g^{\alpha\beta}\partial_\alpha A_\beta)^2}{\partial(\partial_\mu A_\nu)} = 2 (g^{\alpha\beta} \partial_\alpha A_\beta) g^{\mu\nu} = 2(\partial_\alpha A^\alpha)g^{\mu\nu}$$

where I have used that
$$\frac{\partial(\partial_\alpha A_\beta)}{\partial(\partial_\mu A_\nu)} = \delta^\mu_\alpha \delta^\nu_\beta$$

Torquil

6. Feb 1, 2010

### torquil

As with the others, you need to rename the contracted indices before you start differentiating with respect to a quantity that uses the same index name:

$$\frac{\partial{(j_{\alpha}A^{\alpha}})}{\partial\,A_{\nu}}= \frac{\partial{(j^{\alpha}A_{\alpha}})}{\partial\,A_{\nu}}=j^\alpha \delta^\nu_\alpha = j^\nu$$

This time I just switched the up and down positions of the contracted indices. This is allowed, and you can prove it by doing it step by step, and including metrix tensors. I also used:

$$\frac{\partial A_\alpha}{\partial A_\beta} = \delta^\beta_\alpha$$

So if you follow this precedure, you'll be able to also get the correct indices on the F^2 differentiation, and you should get the correct equation in the end.

Torquil

7. Feb 1, 2010

### neorich

Torquil, Thanks alot,

So I follow the derivation of $$j^{\nu}$$ ok.

For the (dA)^2 term, I have added another step to see if I follow correctly,

$$\frac{\partial(\partial_\alpha A^\alpha)^2}{\partial(\partial_\mu A_\nu)} = \frac{\partial(g^{\alpha\beta}\partial_\alpha A_\beta)^2}{\partial(\partial_\mu A_\nu)} = 2 (g^{\alpha\beta} \partial_\alpha A_\beta) \delta^\mu_\alpha \delta^\nu_\beta = 2 (g^{\alpha\beta} \partial_\alpha A_\beta) g^{\mu\nu}=2(\partial_\alpha A^\alpha)g^{\mu\nu}$$

Thus $$\delta^{\mu}_{\alpha}\delta^{\nu}_{\beta} = g^{\mu\nu}$$ Correct? If so why do we pick mu and nu, rather than $$g_{\alpha\beta}$$ ?

Anyway, so working through I now get this equation:

$$\partial_{\mu}((\partial^{\nu}A^{\mu}-\partial^{\mu}A^{\nu})-\frac{1}{\xi}(\partial_{\alpha}A^{\alpha})g^{\mu\nu})-j^{\nu}=0$$

Firstly, is this now correct? Secondly, how do I get from alpha's to nu's as required in the question for the second term, and thirdly how do I get $$\Box g^{\mu\nu} A_{\nu}$$?

Regards

neorich.

8. Feb 1, 2010

### torquil

No, this intermediate step that you added is not correct, because your have alpha both over and under in the parenthesis (ie. they are summed over), at the same time as alpha and beta in the postmultiplied factor.

No this is not correct. On the left side you have four free indices, and on the right only two.

The correct expression in your intermediate step is

$$2 (g^{\gamma\delta} \partial_\gamma A_\delta) g^{\alpha\beta}\delta^\mu_\alpha \delta^\nu_\beta$$

Here I had to use different names than alpha,beta for the contraction within the parenthesis, because I had to use the alpha,beta on the metric g that appeared from the differentiation.

Sorry, I don't have time to check your final equation. You should check the rules for doing these tensor calculations, maybe in a book about gauge theory or einstein gravity.

Good luck!

Torquil

9. Feb 8, 2010

### neorich

Torquil,

Thanks for your help, that was useful.

neorich