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Homework Help: Zwiebach page 178

  1. Oct 12, 2007 #1
    1. The problem statement, all variables and given/known data
    Can someone help me figure out where equation 10.73 comes from and why the sentence

    "In momentum space, the gauge transformation relates [tex]\delta A_{\mu}(p)[/tex] to the Fourier transform epsilon(p) of the gauge parameter"

    is true?

    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Oct 12, 2007
  2. jcsd
  3. Oct 12, 2007 #2
    Take the Fourier transform of both sides of equation (10.68)

    This is just a description, in words, of equation (10.73)
     
  4. Oct 12, 2007 #3
    Like this

    [tex] \int \frac{1}{2 \pi} \delta A_{\mu} e^{-i p \cdot x} dx = \int \frac{1}{2 \pi} \partial_{\mu}\epsilon e^{-i p \cdot x} dx [/tex]

    ?
    How does that simplify?
     
  5. Oct 12, 2007 #4
    My bad. I'm not doing so good, my answers to your other question were wrong too. Please review that thread.

    What I should have said is display (10.68) as the Fourier transform of the momentum equation. Like this.
    [tex] \int \frac{d^Dp}{(2 \pi)^D} e^{i p \cdot x} \delta A_{\mu} = \partial_{\mu} \int \frac{d^Dp}{(2 \pi)^D} e^{i p \cdot x} \epsilon (p) [/tex]
     
  6. Oct 12, 2007 #5

    nrqed

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    Like this:

    [tex] \int \frac{1}{2 \pi}~( \delta A_{\mu}(p))~ e^{-i p \cdot x} dp = \int \frac{1}{2 \pi} \partial_{\mu} (\epsilon(p) e^{-i p \cdot x}) dp [/tex]

    (The integrals are over the momentum). Now, in the secondterm, you may apply the derivative to the exponential which leads to the desired result.
     
  7. Oct 12, 2007 #6
    Are you missing a partial in front of the epsilon?
     
  8. Oct 12, 2007 #7

    nrqed

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    You mean a partial derivative? Jimmy moved it in front of the integral! (which he can do since the integral is over th emomentum)
     
  9. Oct 12, 2007 #8
    Interesting. I didn't realize it but we have been assuming that the four-momentum has no explicit x^mu dependence. Is that safe?
     
  10. Oct 12, 2007 #9

    nrqed

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    Yes. Here, the momentum is to be seen as a totally independent variable. what we are doing here is simply a Fourier transform. So think of p as being independent. It's in the same spirit as doing Hamiltonian mechanics with the generalized coordinates and momenta being independent.
     
  11. Oct 12, 2007 #10
    But wait. Even though the integral is over momentum, jimmy moved the partial derivative in front of a term that does have x^mu dependence i.e. the exponential term? That's like saying

    [tex] e^{i p \cdot x} \partial_{\mu} \epsilon (p) = \partial_{\mu} e^{i p \cdot x} \epsilon (p) [/tex]

    ?
     
    Last edited: Oct 12, 2007
  12. Oct 12, 2007 #11

    nrqed

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    No. One possible source of confusion here is that there are two different functions with the same name!! There is the [itex] \epsilon(x) [/itex] and then there is the [itex] \epsilon(p) [/itex] Those are different functions!

    The initial expression was [itex] \partial_\mu \epsilon(x) [/itex]

    Then this epsilon(x) is written as the whole integral over momentum of epsilon(p) times the exponential. The derivative is applied on the whole thing and then can be moved inside and be applied to the exponential (it goes right through the epsilon(ep) )
     
  13. Oct 12, 2007 #12
    I see. So it should really be
    [tex] \int \frac{d^Dp}{(2 \pi)^D} e^{i p \cdot x} \delta A_{\mu}(p) = \partial_{\mu} \int \frac{d^Dp}{(2 \pi)^D} e^{i p \cdot x} \epsilon (p) [/tex]

    where I added a (p) to the delta A.

    Also, was I correct to dot p and x in the exponential because I noticed that Zwiebach does not do that and it seems like he should because the vector potential A has four indices which would imply that x and p would also have four indices?
     
  14. Oct 12, 2007 #13
    Yes, I left off the (p) in error. You are correct that the dot is missing from the exponent in equation (10.71), but present in (10.21) on page 170. It should always be there, but there is no fear of confusion if it is missing.
     
  15. Oct 12, 2007 #14
    So [tex] \int \frac{d^Dp}{(2 \pi)^D} e^{i p \cdot x} \delta A_{\mu}(p) = \partial_{\mu} \int \frac{d^Dp}{(2 \pi)^D} e^{i p \cdot x} \epsilon (p) [/tex]

    [tex] \int \frac{d^Dp}{(2 \pi)^D} e^{i p \cdot x} \delta A_{\mu}(p) = {\mu} \int \frac{d^Dp}{(2 \pi)^D} ip_{\mu} e^{i p \cdot x} \epsilon (p) [/tex]

    So, it seems like you get equation 10.73 by setting the integrands equal to each other? Is that allowed?

    Does [tex] \int_0^{\pi} \sin (x) dx = \int_0^{\pi} 2/\pi dx [/tex] imply that sin(x) = 2/pi ?

    EDIT: the mu in front of the last equation is a mistake
     
    Last edited: Oct 13, 2007
  16. Oct 13, 2007 #15
    In general no, but for Fourier transforms, yes. The operator F that takes a function to its Fourier transform is invertable. Therefor
    [itex]F(g) = F(h)[/itex] implies [itex]g = F^{-1}F(g) = F^{-1}F(h) = h[/itex]
    In the present case, g is [itex]\delta A_{\mu}(p)[/itex] and h is [itex]ip_{\mu}\epsilon (p) [/itex]
     
  17. Oct 13, 2007 #16
    Could you also just say that

    [tex] \int \frac{d^Dp}{(2 \pi)^D} e^{i p \cdot x} \delta A_{\mu}(p) =\int \frac{d^Dp}{(2 \pi)^D} ip_{\mu} e^{i p \cdot x} \epsilon (p) [/tex]

    implies

    [tex] \int \frac{d^Dp}{(2 \pi)^D} e^{i p \cdot x} \delta A_{\mu}(p) = ip_{\mu} \int \frac{d^Dp}{(2 \pi)^D} e^{i p \cdot x} \epsilon (p) [/tex]

    implies

    [tex] \delta A_{\mu}(p) = ip_{\mu} \epsilon (p) [/tex]

    implies


    [tex] \delta A_{\mu}(x) = ip_{\mu} \epsilon (x) [/tex]

    ?
     
  18. Oct 13, 2007 #17

    nrqed

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    You can't move the p_mu out of the integral since it is what is being integrated over
     
  19. Oct 13, 2007 #18
    OK. So like jimmy said we need to take an inverse Fourier transform:

    [tex] \int \frac{d^Dp}{(2 \pi)^D} e^{i p \cdot x} \delta A_{\mu}(p) =\int \frac{d^Dp}{(2 \pi)^D} ip_{\mu} e^{i p \cdot x} \epsilon (p) [/tex]

    implies

    [tex]\int \frac{d^Dx}{(2 \pi)^D} e^{-i p \cdot x} \int \frac{d^Dp}{(2 \pi)^D} e^{i p \cdot x} \delta A_{\mu}(p) =\int \frac{d^Dx}{(2 \pi)^D} e^{-i p \cdot x} \int \frac{d^Dp}{(2 \pi)^D} ip_{\mu} e^{i p \cdot x} \epsilon (p) [/tex]

    implies

    [tex]\int \int \frac{d^Dx}{(2 \pi)^D} \frac{d^Dp}{(2 \pi)^D}\delta A_{\mu}(p) = \int \int \frac{d^Dx}{(2 \pi)^D} \frac{d^Dp}{(2 \pi)^D} ip_{\mu}
    \epsilon (p) [/tex]

    and that is supposed to imply that the integrands are equal somehow?
     
  20. Oct 13, 2007 #19

    nrqed

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    In your first step, when you do the inverse Fourier transform, you must do it with respect to a different momentum, say p'. Then you must use that the integral over x of the product of the two exponentials will give a Dirac delta function [itex] \delta(p - p') [/itex] (you have to watch out for factors of 2 pi and so on). Then you will be left at the end with [itex] \deltaA_\mu(p;) = i p'_\mu \epsilon(p') [/itex]
     
  21. Oct 13, 2007 #20
    Ehrenfest, if you look on page 316, there is another discussion of equation (10.68). Here it is labeled (15.41). In this discussion, he adds a new twist, the condition that epsilon vanishes sufficiently rapidly at infinity. With this extra condition, you can get equation (10.73) directly from (10.68) by Fourier transforming both sides. I don't have time to show you how right now, but I think you can do it on your own anyway.
     
  22. Oct 13, 2007 #21
    I see. So it is:

    [tex]\int \int \frac{d^Dx}{(2 \pi)^D} \frac{d^Dp}{(2 \pi)^D}\delta A_{\mu}(p)\delta)(p'-p) = \int \int \frac{d^Dx}{(2 \pi)^D} \frac{d^Dp}{(2 \pi)^D} ip_{\mu} \epsilon (p) \delta(p'-p)[/tex]

    implies

    [tex]\int \frac{d^Dx}{(2 \pi)^D} \frac{1}{(2 \pi)^D}\delta A_{\mu}(p')= \int \frac{d^Dx}{(2 \pi)^D} \frac{1}{(2 \pi)^D} ip_{\mu} \epsilon (p') [/tex]

    and that gives the result we want.

    Was I correct to leave that 1/(2 pi)^D in there when I applied dirac delta to the momentum integral?
     
  23. Oct 13, 2007 #22

    nrqed

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    Not quite.

    It's only after you have done the x integration that you have a Dirac delta function . Do you see what I mean? It's the integral [itex] \int dx^D e^{i(p-p')x} [/itex] which is proportional to [itex] \delta(p-p') [/itex] (I don't have the time to check the normalization used by Zwiebach right now).

    So once you have the dirac delta, the integral over x is gone completely. Then you use the delta function to do the integral over p. So at the end you have no integrals left at all.
     
  24. Oct 13, 2007 #23
    You're right. So it should be:

    [tex]\int \frac{d^Dp}{(2 \pi)^D}\delta A_{\mu}(p) \int \frac{d^Dx}{(2 \pi)^D} e^{i (p-p') \cdot x} = \int \frac{d^Dp}{(2 \pi)^D} ip_{\mu} \epsilon (p) \int \frac{d^Dx}{(2 \pi)^D} e^{i (p-p') \cdot x} [/tex]

    So, I guess we will assume Zwiebach normalized his Fourier transforms correctly:

    [tex]\int \frac{d^Dp}{(2 \pi)^D}\delta A_{\mu}(p) \delta(p-p') = \int \frac{d^Dp}{(2 \pi)^D} ip_{\mu} \epsilon (p) \delta(p-p') [/tex]

    implies

    [tex] \frac{1}{(2 \pi)^D}\delta A_{\mu}(p') = \frac{1}{(2 \pi)^D} ip_{\mu} \epsilon (p) \delta(p-p') [/tex]

    Was I correct to keep the 1/(2 pi)^D in the last step given that we are assuming that Zwiebach normalized his Fourier transforms correctly?
     
  25. Oct 14, 2007 #24
    That should read:

    [tex]\frac{1}{(2 \pi)^D}\delta A_{\mu}(p') = \frac{1}{(2 \pi)^D}ip_{\mu}\epsilon (p')[/tex]
    I think that the factor of 1/2pi is wrong, it was supposed to fall out in one of the integrations. But in this case it is irrelevant since you can cancel it out anyway. You can find a proof that the Fourier transform is invertible in a book on Fourier analysis, or perhaps in a book on mathematical physics. As I recall, the proof is more complicated than the one given here, but I don't remember what the issue is.
     
  26. Oct 14, 2007 #25
    I see. Thanks.
     
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