- #1

Icosahedron

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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.)

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- Thread starter Icosahedron
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- #1

Icosahedron

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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.)

- #2

DavidWhitbeck

- 351

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- #3

Icosahedron

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- #4

DavidWhitbeck

- 351

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

- #5

Icosahedron

- 54

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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 staring another 30 minutes on the equation, how that get me to that first term in 54.24.

- #6

nrqed

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OK, got that.

I also understand what you are saying here. But still I can't see, even after staring 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 [tex] \partial^\mu K_\mu [/tex] at the end.

Here is the way to do it:

Take 54.23. Now add zero in the form [tex] \partial^\mu K_\mu - \partial^\mu K_\mu [/tex].

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

- #7

Icosahedron

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

- #8

malawi_glenn

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are we talking about eq 5.24 or 54.24?

- #9

nrqed

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

Ah, I see your problem.

Consider the expression

[tex] A^\nu \partial^\mu \partial_\nu A_\mu [/tex]

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

[tex] A^\mu \partial^\nu \partial_\mu A_\nu [/tex]

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

[tex] A_\mu \partial^\nu \partial^\mu A_\nu [/tex]

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

- #10

Icosahedron

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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

- #11

nrqed

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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 [tex] P_{ij} A_j [/tex]. It's clear that this expression obeys

[tex] \nabla_j (P_{ij} A_j) = 0 [/tex]

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

- #12

DavidWhitbeck

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- #13

Icosahedron

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or k_i(d_ij - k_i *k_j/

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

- #14

DavidWhitbeck

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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.

- #15

Icosahedron

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I will now read on but I'm afraid I will have to return to this thread sooner or later.

But so far, thank you guys!

- #16

DavidWhitbeck

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