# Mandl & Shaw page 297

1. Jul 25, 2013

### Vic Sandler

In the second edition of QFT by Mandl & Shaw, for the unnumbered eqn below eqn (13.120c) on page 297, I get a factor of -1 on the rhs that the book doesn't have. However, in the next two equations, it is clear that the author intends for the eqn to stand as it is. I get:

$$F_{\mu\nu}F^{\mu\nu} = (\partial_{\nu}A_{\mu} - \partial_{\mu}A_{\nu})(\partial^{\nu}A^{\mu} - \partial^{\mu}A^{\nu})$$
$$= \partial_{\nu}A_{\mu}\partial^{\nu}A^{\mu} - \partial_{\mu}A_{\nu}\partial^{\nu}A^{\mu} - \partial_{\nu}A_{\mu}\partial^{\mu}A^{\nu} + \partial_{\mu}A_{\nu}\partial^{\mu}A^{\nu}$$
$$= (\partial_{\nu}A_{\mu})(\partial^{\nu}A^{\mu}) + A_{\mu}\Box A^{\mu} - (\partial_{\mu}A_{\nu})(\partial^{\nu}A^{\mu}) - A_{\nu}\partial_{\mu}\partial^{\nu}A^{\mu} - (\partial_{\nu}A_{\mu})(\partial^{\mu}A^{\nu}) - A_{\mu}\partial_{\nu}\partial^{\mu}A^{\nu} + (\partial_{\mu}A_{\nu})(\partial^{\mu}A^{\nu}) + A_{\nu}\Box A^{\nu}$$
$$= 2A_{\mu}\Box A^{\mu} - 2A^{\mu}\partial_{\mu}\partial_{\nu}A^{\nu} + 2(\partial_{\nu}A_{\mu})F^{\mu\nu}$$

$$\int d^4 x(\partial_{\nu}A_{\mu})F^{\mu\nu} = -\int d^4 x A_{\mu}\partial_{\nu}F^{\mu\nu} = 0$$

using eqn (5.2) on page 74 and s = 0. So I get an extra minus sign on the rhs.

$$-\frac{1}{4}\int d^4x F_{\mu\nu}F^{\mu\nu} = -\frac{1}{2} \int d^4 x A^{\mu}[g_{\mu\nu}\Box - \partial_{\mu}\partial_{\nu}]A^{\nu}$$

What am I doing wrong?

Last edited: Jul 25, 2013
2. Jul 25, 2013

### dextercioby

You started off wrongly

∂νAμ∂νAμ−∂μAν∂νAμ−∂νAμ∂μAν+∂μAν∂μAν = 4div - Aμ □Aμ - 4div + Aν∂μ∂νAμ - 4div + Aμ∂ν∂μAν +4div - Aν□Aν = 4div + 2 Aμgμν□Aν - 2 Aμ∂μ∂νAν

3. Jul 25, 2013

### Vic Sandler

Thanks dextercioby. I am not familiar with the div. How is it defined?

4. Jul 25, 2013

### Avodyne

I can't follow what you do after this.

But by swapping the dummy indices μ and ν, we see that the 4th term above is the same as the 1st, and the 2nd term is the same as the 3rd, and so
$$= 2\partial_{\nu}A_{\mu}\partial^{\nu}A^{\mu} - 2\partial_{\nu}A_{\mu}\partial^{\mu}A^{\nu}$$
Now use integration by parts to move the derivatives off the first A in each term:
$$= -2A_{\mu}\partial_{\nu}\partial^{\nu}A^{\mu} + 2A_{\mu}\partial_{\nu}\partial^{\mu}A^{\nu}$$
In the second term, swap which μ is up and which is down, and commute the derivatives to get
$$= -2A_{\mu}\Box A^{\mu} + 2A^{\mu}\partial_{\mu}\partial_{\nu}A^{\nu}$$

5. Jul 26, 2013

### Vic Sandler

Thanks Avodyne, your solution is simpler for me. I'd still like to know what div in dextercioby's post stands for though.

6. Jul 26, 2013

### Vic Sandler

Wait a minute, are you sure you can do that? I thought

$$\partial_{\nu}A_{\mu}\partial^{\nu}A^{\mu}$$

in the second line of my equation in the OP meant

$$\partial_{\nu}(A_{\mu}\partial^{\nu}A^{\mu})$$

not

$$(\partial_{\nu}A_{\mu})(\partial^{\nu}A^{\mu})$$

7. Jul 26, 2013

### Avodyne

$$\partial_{\nu}A_{\mu}\partial^{\nu}A^{\mu} =(\partial_{\nu}A_{\mu})(\partial^\nu A_\mu)$$
since it comes from the product of the two F's, which have the derivatives acting on their own A's only.

In general, when you have an expression like this, the standard convention is that the derivative acts only on the object to the immediate right, and not on everything to the right.

8. Jul 26, 2013

### dextercioby

A 4-divergence of a tensor of rank 2 is $\partial_{\mu} T^{\mu\nu}$ (for flat space-time). From rank 2 you can define it for any rank with the same pattern.

9. Jul 26, 2013

### Vic Sandler

Thanks to Avodyne and dextercioby for your help.