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A Help with proving Thomson's theorem

  1. Jun 10, 2017 #1
    This is out of Jackson's electrodynamics, problem 1.15.

    I am trying to prove the theorem that if a number of surfaces are fixed in position and a given total charge is placed on each surface, then the electrostatic energy in the region bounded by the surfaces is an absolute minimum when the charges are placed so that every surface is an equipotential, as happens when they are conductors.

    The electrostatic energy is given by the square of the field as

    ##W=\frac{\epsilon_0}{2}\int|E|^2d^3x.##

    In terms of the potential this is

    ##W=\frac{\epsilon_0}{2}\int\nabla\Phi\cdot\nabla\Phi d^3x.##

    We seek an extremum in the energy by varying the energy to first order

    ##W(\Phi + \delta\Phi)-W(\Phi) = \epsilon_0\int\nabla\delta\Phi\cdot\nabla\Phi.##

    Any two scalars ##\phi## and ##\psi## satisfy

    ##\int_v [\nabla \phi \cdot \nabla \psi + \phi\nabla^2\psi]d^3x = \int_s \phi \frac{\partial\psi}{\partial n} da##

    where the surface integral is taken over a the surface bounding the volume over which the volume integral is taken, and the derivative with respect to ##n## indicates a derivative in the direction normal to the surface. Thus the variation in the energy may be written

    ##W(\Phi+\delta\Phi)-W(\Phi) = \epsilon_0\bigg[ \int_s\delta\Phi\frac{\partial\Phi}{\partial n} da - \int_v \delta\Phi\nabla^2\Phi d^3x\bigg]##

    We assume that the region bounded by the surface charge is charge-free, and so the potential satisfies Laplace's equation (##\nabla^2\Phi=0##) in this region and the volume integral vanishes.

    ##\epsilon_0 \int_s\delta\Phi\frac{\partial\Phi}{\partial n} da = 0##

    This must be true for arbitrary ##\delta\Phi##, and therefore the integrand itself must be zero at all points on the surface

    ##\frac{\partial\Phi}{\partial n} = 0.##

    This is where I run into trouble. Assuming that the steps up to this point are correct, I do not understand how to interpret this condition physically. It seems to be saying that the electric field through the bounding surface must be zero at all points on the surface. But how does that translate to every surface being an equipotential? Any hints?

    Thanks!
     
  2. jcsd
  3. Jun 11, 2017 #2

    Charles Link

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    If you use Gauss law on a sufrace just inside the surface, ## \int E \cdot \, \hat{n} dA=0 ## , because ## E ## perdendicular to ## \hat{n} dA ## is 0 everywhere because ## \frac{\partial{\Phi}}{\partial{n}}=0 ## everywhere. With this feature along with ## \nabla^2 \Phi=0 ## everywhere, we must conclude that ## \Phi= ## constant, because this is a solution to the differential equation that satisfies the boundary condition. By the uniqueness theorem, it must be the only solution. ## \\ ## One point of confusion is that I don't think ## \frac{\partial{\Phi}}{\partial{n}} ## is necessarily zero on both sides of the surface (inside and out.) I'd need to study it further, but I think the integrals must be computed inside the surfaces, and ## \frac{\partial{\Phi}}{\partial{n}} ## is simply inside (such as with a conductor), because I believe (I'm not certain), that the volume integral involving ## W ## is done inside the surfaces.
     
    Last edited: Jun 11, 2017
  4. Jun 11, 2017 #3
    Gotta love the uniqueness theorem. So just to be clear, the boundary condition in combination with the fact that the potential satisfies Laplace's equation in the volume, imply not only that ##\Phi=##constant on the bounding surface, but also that ##\Phi=## constant throughout the entire volume, since this is a solution to the differential equation satisfying the boundary condition?

    I think that makes sense. If you wanted to say something about ##\frac{\partial \Phi}{\partial n}## on the outer surface then I think you would have to extend the volume integral to include the surface charge, then ##\frac{\partial\Phi}{\partial n} ##is evaluated on the surface bounding the volume containing the surface charge. In this case it is no longer correct to say that ##\nabla^2\Phi = 0 ## in the whole volume, so you get a nonzero contribution from the volume integral and things stop working.
     
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