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Anyone prove Fourier representation of the Coulomb potential!

  1. May 30, 2006 #1
    I've seen the Fourier representation of the Coulomb potential is [tex]-\frac {Ze} {|\mathbf{x}|} = -Ze 4\pi \int \frac {d^3q} {(2\pi)^3} \frac {1} { |\mathbf{q}|^2} e^{i\mathbf{q}\cdot\mathbf{x}}[/tex]

    Will anyone show me how to prove it? (yes, it's the Coulomb potential around an atomic nucleus.)

    Thanks in advance!
     
    Last edited: May 30, 2006
  2. jcsd
  3. May 30, 2006 #2

    Hurkyl

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    Well, if you take the inverse Fourier transform of
    [tex]-\frac {Ze} {|\mathbf{x}|}[/tex]
    then, the Fourier transform of the result will be your potential.

    Or, maybe you could take the Fourier transform of both sides, and find some way to justify reversing the order of integration?


    That being said, why not simply integrate? Won't that integral be pretty easy in spherical coordinates? And even easier if you choose your spherical coordinates to be relative to a clever choice of axes?
     
  4. Jun 20, 2006 #3
    The problem is that the fourier transform of the coulomb potential does not converge directly. you must use the yukawa potential Exp(-u r)/r and after the fourier calculation take the limit u ->0

    salu2
    K
     
  5. Jun 21, 2006 #4
    Just fourier transform Laplace equation with a delta source, and you get the right side of that
     
  6. Jun 21, 2006 #5
    by the way, how do you post equations here?
     
  7. Jun 21, 2006 #6
    The forum uses LaTex. There some info on the syntax somewhere on this site. To write an equation you might write it in LaTex as

    -\frac {Ze} {|\mathbf{x}|} = -Ze 4\pi \int \frac {d^3q} {(2\pi)^3} \frac {1} { |\mathbf{q}|^2} e^{i\mathbf{q}\cdot\mathbf{x}}

    For the server to recognize it as LaTex, you must place it between the tags tex tags like so:

    [tex]-\frac {Ze} {|\mathbf{x}|} = -Ze 4\pi \int \frac {d^3q} {(2\pi)^3} \frac {1} { |\mathbf{q}|^2} e^{i\mathbf{q}\cdot\mathbf{x}}[/tex]

    Click on it to see the tags.
     
  8. Jul 28, 2007 #7
    Use spherical polar coordinate (z-axis along the [tex]\vec{q}[/tex] direction)
    The integrand becomes:
    [tex]\frac{e^{i q r cos(\theta)}}{r}[/tex]

    and the differential:
    [tex]d^3r = r^2 sin(\theta) d\phi d\theta dr[/tex]

    the integral is then quite trivial.
     
  9. Nov 14, 2008 #8
    It is not. As karenas73 pointed out, it does not converge. I know this was an old discussion, but I am having the same problem right now. How does one justify the trick with the Yukawa potential?
     
  10. Feb 24, 2009 #9
    It just took me a while to do this "trivial" integration correct, so I thought it might be of general interest. I don't know about the mathematical justification for the Yukawa trick. Most physicists simply accept this kind of swapping of limits and integrations without thinking about it too hard...

    [tex]
    \begin{array}{lll}
    \int \mathrm{d}^3 x \frac{1}{|\mathbf{x}|} e^{- \mathrm{i} \mathbf{q} \cdot
    \mathbf{x}} & = & \lim_{u \rightarrow 0} \int \mathrm{d}^3 x \frac{\exp \left(
    - u \left| \mathbf{x} \right| \right)}{\left| \mathbf{x} \right|} e^{-
    \mathrm{i} \mathbf{q} \cdot \mathbf{x}}\\
    & = & \lim_{u \rightarrow 0} \int \mathrm{d} r \mathrm{d} \cos \vartheta 2 \pi r^2
    \frac{\exp \left( - ur \right)}{r} e^{- \mathrm{i} qr \cos \vartheta}\\
    & = & 2 \pi \lim_{u \rightarrow 0} \int \mathrm{d} rr \int \mathrm{d} \cos
    \vartheta \exp \left( - \left( u + \mathrm{i} q \cos \vartheta \right) r
    \right)\\
    & = & 2 \pi \lim_{u \rightarrow 0} \int \mathrm{d} r \frac{r}{- \mathrm{i} q}
    \left[ \exp \left( - \left( u + \mathrm{i} q \right) r \right) - \exp \left( -
    \left( u - \mathrm{i} q \right) r \right) \right]\\
    & = & \frac{2 \pi}{- \mathrm{i} q} \lim_{u \rightarrow 0} \int \mathrm{d} r \left[
    r \exp \left( - \left( u + \mathrm{i} q \right) r \right) - r \exp \left( -
    \left( u - \mathrm{i} q \right) r \right) \right]\\
    & & \left[ \begin{array}{l}
    \int_0^{\infty} \mathrm{d} rr \exp \left( - ar \right) = \left[ \frac{1}{- a}
    \exp \left( - \tmop{ar} \right) \right]_0^{\infty} = \frac{1}{a}
    \end{array} \right.\\
    & = & \frac{2 \pi}{- \mathrm{i} q} \lim_{u \rightarrow 0} \left[ \frac{1}{u +
    \mathrm{i} q} - \frac{1}{u - \mathrm{i} q} \right]\\
    & = & \frac{2 \pi}{- \mathrm{i} q} \lim_{u \rightarrow 0} \left[ \frac{- 2
    \mathrm{i} q}{u^2 + q^2 } \right]\\
    & = & \lim_{u \rightarrow 0} \left[ \frac{4 \pi}{u^2 + q^2 } \right]\\
    & = & \frac{4 \pi}{q^2 }
    \end{array}
    [/tex]
     
  11. Mar 10, 2009 #10
    Step 3 (integration over cos(phi)) can be written more simply by defining cos(phi) = s, and needs the r as an integration factor. Its nice to be able to look up an answer to your problems online...

    [tex]

    \begin{array}{lll}
    \int \mathrm{d}^3 x \frac{1}{|\mathbf{x}|} e^{- \mathrm{i} \mathbf{q} \cdot
    \mathbf{x}} & = & \lim_{u \rightarrow 0} \int \mathrm{d}^3 x \frac{\exp \left(
    - u \left| \mathbf{x} \right| \right)}{\left| \mathbf{x} \right|} e^{-
    \mathrm{i} \mathbf{q} \cdot \mathbf{x}}\\
    & = & \lim_{u \rightarrow 0} \int_0^\infty \mathrm{d} r \int_{-1}^1 \mathrm{d} \cos \vartheta 2 \pi r^2
    \frac{\exp \left( - ur \right)}{r} e^{- \mathrm{i} q r \cos \vartheta}\\
    & = & 2 \pi \lim_{u \rightarrow 0} \int_0^\infty \mathrm{d} r \int_{-1}^1 \mathrm{d}s r \exp \left( - r \left( u + \mathrm{i} q s \right) \right)\\
    & = & \frac{2 \pi}{-\mathrm{i} q} \lim_{u \rightarrow 0} \int_0^\infty \mathrm{d} r
    \left[ \exp \left( - r \left( u + s \mathrm{i} q \right) \right)\right]_{-1}^1\\
    & = & \frac{2 \pi \mathrm{i}} {q} \lim_{u \rightarrow 0} \left[
    \frac{1}{u + s \mathrm{i} q } \right]_{-1}^1 \\
    & = & \frac{2 \pi \mathrm{i}} {q} \frac{2} {\mathrm{i} q} \\
    & = & \frac{4 \pi}{q^2 }
    \end{array}

    [/tex]
     
  12. Sep 19, 2009 #11
    Just for completeness... I think it's OK to swap limits and integrals as long as you're using a Lebesgue integral. But I don't know much further, so... if someone could help on this... (:
     
  13. Sep 19, 2009 #12
    Well obviously you can't switch the limits. First integrating, then taking the limit yields the expected result. First taking the limit, then integrating - does not converge.
    That's why I still don't know why this trick is justified.
     
  14. Sep 19, 2009 #13

    Hurkyl

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    One unfortunate fact is that physics very, very often works with distributions rather than functions -- but it seems (to me) that this is rather infrequently stated explicitly.

    If you have a convergent sequence of distributions you want to convolve with a test function, then the limit of the convolutions is, by definition, the convolution of the limit.

    This runs into ambiguity when each of the individual distributions can be expressed as a function, because the limit of the functions might not converge at all. (Or it might converge to a function that does not correspond to the limiting distribution)
     
  15. Oct 21, 2010 #14
    Having the same question right now, I reanimate this thread.

    Has anyone an answer to the quoted question?
     
  16. Nov 10, 2010 #15
    I think that the justification for the Yukawa-Potential can be thought of like that:

    The Photons, having no mass at all, have a Yukawa-Potential with a vanishing exponential part. This however leads to divergence with r becoming 0. The problem in the original integral was, that it diverges because k becomes infinite. Since k behaves inverse to r, this case also corresponds to r becoming 0. Since particles with finite masses don't have this problem of diverging potentials at r=0, the photons are "given a mass" by using the Yukawa-Potential first to calculate the integral. Afterwards the mass is stripped of again and the problem is solved.

    This is my understanding of the problem.
    Kind regards
     
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