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Calcs. for continuous charged media don't explode?

  1. Sep 9, 2015 #1
    Title doesn't let me be sufficiently clear so let me do it here:

    The potential for a continuous charged object is ##V(\vec{r})=k\int\frac{dq}{|\vec{r}-\vec{r'}|}## and similarily for the electric field. This makes sense outside of the charged object but not inside!

    Namely, I say it doesn't make sense inside because when you are integrating you will be considering all points having charge including the charge at point r. This would make the denominator in my integral blow up!

    In the case of calculating energies for point charges it does blow up, and this problem has been called the "infinite self energy of the electron".

    But why doesn't it blow up here? Is it some weird property of integrals that evade a singular point?

    Thanks.
     
  2. jcsd
  3. Sep 9, 2015 #2

    Dale

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    For a continuous charge distribution, how much charge is at the single point r?
     
  4. Sep 9, 2015 #3
    Well there is zero charge for sure. But the ##dq## immediately adjacent to that point would still have a denominator that "blows up". But I guess it cancels with the extremely small value of ##dq##. Is that it? I think you've answered my question :D thanks.
     
  5. Sep 9, 2015 #4

    Dale

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    Yes, that is it!
     
  6. Sep 9, 2015 #5

    Geofleur

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    I am a little puzzled! Suppose the charge distribution is uniform inside so that we can write ## dq = \rho dV ##, where ## \rho ## is the constant charge density and ## dV ## is the volume element. Then we would have

    ## V(\textbf{r}) = k\rho\int\frac{dV}{|\textbf{r}-\textbf{r'}|} ##.

    For ## \textbf{r} ## inside the medium, it would seem that the integral does diverge. There even seems to be a paper over in arxiv discussing the issue as it relates to gravity: http://arxiv.org/pdf/1203.6822.pdf

    Am I misunderstanding something?
     
  7. Sep 9, 2015 #6

    Dale

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    Consider the simple case of a spherically symmetric charge distribution and calculate the potential at the center. How does dV behave as r goes to 0? What does that imply for convergence?
     
  8. Sep 9, 2015 #7

    Geofleur

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    The volume element is ## r'^2 \sin \theta dr' d\phi d\theta ##, and this conspires with the ## 1/r' ## part to give a convergent integral. But the center of a sphere seems like a special place, because the volume element doesn't go to zero elsewhere. Might it not be due to the choice of the coordinate system? What if we had calculated the potential at the center of a cube and used Cartesian coordinates?
     
  9. Sep 9, 2015 #8

    jtbell

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    The center of the (spherical) coordinate system is special. It doesn't need to be at the center of a spherical charge distribution. It's "merely" easier to evaluate the integral if the center of a spherical coordinate system coincides with the center of a spherically-symmetric charge distribution.

    For a uniform charge distribution, the integrand is exactly the same if the two "centers" do not coincide. Only the limits of integration change.
     
  10. Sep 10, 2015 #9

    Geofleur

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    I still don't think I understand. In spherical coordinates, and at the center of the coordinate system, the volume element goes to zero and also the denominator of the potential goes to zero, their combination going to zero as ## r'^2/r' = r' ##. But at a place off center, say at a radius ## r_0 ##, the relevant part of the volume element goes to ## r_0^2\sin\theta_0##, while ## 1/|r'-r_0| ## goes to ## 1/|r_0-r_0| = \infty ##. It still seems like the center of the coordinate system is "saving" the integral, which is why I was suggesting to consider the same problem, but with a cube in Cartesian coordinates. In that situation, the integral looks like it will diverge no matter what ## \textbf{r} ## we choose to evaluate ## V(\textbf{r})##, so long as it's inside the cube.
     
  11. Sep 10, 2015 #10

    Dale

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    Changing to Cartesian coordinates won't change the number or its convergence. It will just make the computation more difficult. Regardless of coordinates the volume within a distance |r-r'| goes as |r-r'|^3, so if the charge density goes as anything less sharp than |r-r'|^-2 then the integral converges.
     
  12. Sep 10, 2015 #11

    Geofleur

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    OK, I will go and try to prove convergence in all the details. Thanks for patiently explaining!
     
  13. Sep 10, 2015 #12

    Dale

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    Another way to see this is to note that this 1D integral diverges:
    $$\int_{-1}^{1} \frac{1}{||x||} dx$$

    but this very similar 2D integral converges:
    $$\int_{-1}^{1}\int_{-1}^{1} \frac{1}{||(x,y)||} dxdy$$
     
  14. Sep 10, 2015 #13

    Geofleur

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    That's amazing - not at all what I would expect!
     
  15. Oct 25, 2015 #14

    Geofleur

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    Update: I've been reading Pauli's lectures on physics and just came across the part in volume 1 where he says the same thing that DaleSpam and jtbell had been saying above. Pauli says (on pg. 15) that "This can easily be shown to be rigorously true by an exact consideration of the limit." And so here is what I came up with:

    Consider a volume element ## \Delta V = \Delta x \Delta y \Delta z ## containing an amount of charge ## \rho \Delta V ## with ## \rho ## a constant volume charge density. Consider the electric field at a point ## P ## located a distance ## r ## from this element, and take the limit ## \lim_{r \to 0} \frac{\Delta x \Delta y \Delta z}{r^2} ##. If the volume element is centered at the origin, we can write the position vector for point ## P ## as ## \mathbf{r} = C_1 \Delta x \mathbf{i} + C_2 \Delta y \mathbf{j} + C_3 \Delta z \mathbf{k} ##, where ## C_1 ##, ## C_2 ##, and ## C_3 ## are constants. For simplicity, let the volume element be a cube of side ## L ##. Then the limit becomes ## lim_{\Delta x, \Delta y,\Delta z \to 0} \frac{\Delta x \Delta y \Delta z}{C_1^2 \Delta x^2 + C_2^2 \Delta y^2 + C_3^2 \Delta z^2} = lim_{L \to 0} \frac{L^3}{L^2(C_1^2+C_2^2+C_3^2)} = 0 ##. And that's it!
     
  16. Oct 25, 2015 #15
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