# Book by Srednicki

Could someone help me how in this book by Srednicki I get from eq. 54.23 to 54.24?

thank you

(The book is free on the net, I'm not allowed yet to post the link on this forum, maybe some other can do.)

It's just simple math, there are no sneaky tricks there. Start with 24 and plug in the formula for K. Excepting two terms you should be able to cancel all of the terms in the first group with the terms in div K, leaving you with eqn 23.

thanks David, but I still can't see it. Where does the first term in 5.24 come from? I just see that the last two terms in 5.24 are 5.23, but the first term in 5.24??

When you use the product rule on the first term on div K you will get the first term in 5.24, but with opposite sign and they will cancel.

The thing is that $$\partial_\mu \partial^\mu = \partial^2$$ the d'Alembertian operator, and $$A^\nu=g^{\mu\nu}A_{\mu}$$ where $$\mu$$ is a different index! The first one was already summed over.

When you use the product rule on the first term on div K you will get the first term in 5.24, but with opposite sign and they will cancel.
OK, got that.

The thing is that the d'Alembertian operator, and where is a different index!
I also understand what you are saying here. But still I can't see, even after starring another 30 minutes on the equation, how that get me to that first term in 54.24.

nrqed
Homework Helper
Gold Member
OK, got that.

I also understand what you are saying here. But still I can't see, even after starring another 30 minutes on the equation, how that get me to that first term in 54.24.
You have to add that term in order to be able to subtract $$\partial^\mu K_\mu$$ at the end.

Here is the way to do it:

Take 54.23. Now add zero in the form $$\partial^\mu K_\mu - \partial^\mu K_\mu$$.

Now expand out the positive$$\partial^\mu K_\mu$$ , leaving the subtracted one in the form $$- \partial^\mu K_\mu$$ . You should recover 54.24

But I don't get the indices match, I don't get them up and down like they supposed to in 5.24. The two upper indice partial derivative in 5.24, where does that come from?

On the same page on the second to last line, the author does this trick again, but I can't see how.

malawi_glenn
Homework Helper
are we talking about eq 5.24 or 54.24?

nrqed
Homework Helper
Gold Member
But I don't get the indices match, I don't get them up and down like they supposed to in 5.24. The two upper indice partial derivative in 5.24, where does that come from?

On the same page on the second to last line, the author does this trick again, but I can't see how.

Consider the expression

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

First, you can relabel mu -> nu and nu -> mu (they are summed over so you may rename those indices) This gives

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

The next trick is to use the fact that you may lower the index mu on the gauge field if you move up the corresponding index on the partial derivative, which gives

$$A_\mu \partial^\nu \partial^\mu A_\nu$$

Using this and the approach I mentioned in my previous post, you should get the result.

Now I see!! thanks everybody

After I took up so much of everybody's time over such a trival matter, has someone the nerve to look at 55.4? Is there a quick way to see that this gives the Coulomb gauge? Just a hint would be fine.

thank you

nrqed
Homework Helper
Gold Member
Now I see!! thanks everybody

After I took up so much of everybody's time over such a trival matter, has someone the nerve to look at 55.4? Is there a quick way to see that this gives the Coulomb gauge? Just a hint would be fine.

thank you

the idea is this: Consider a totally arbitrary A_j. Now, consider the expression on the rhs of 55.4 (where A_j is still completely arbitrary). Let's call this $$P_{ij} A_j$$. It's clear that this expression obeys

$$\nabla_j (P_{ij} A_j) = 0$$

Therefore the quantity $$P_{ij} A_j$$ can be used to represent a gauge field in the Coulomb gauge (where, again, the A_j that is there is arbitrary)

I think the simplest way to see that you are now in the Coulomb gauge is to do as Srednicki says and look at div A in k-space. You can see that $$k^i \tilde{A_i}=0$$ which implies that in x-space that $$\nabla^i A_i = 0$$.

thanks David!

or k_i(d_ij - k_i *k_j/k^2)= k_j - k_j = 0

I hope that's readable, not good at that latex thing (d is Kronecker delta)

Exactly! Bingo.

Oh and also in k-space it easily looks like the classic projection operator that we're all used to from vector algebra/calc. I'm uncomfortable with the differential operator version because of that inverse Laplacian operator, very funky.

Again thanks nrqed, thanks David.