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Green's First Identity involving Electric Potential

  1. Jun 23, 2014 #1
    I am attempting to work through a paper that involves some slightly unfamiliar vector calculus, as well as many omitted steps. It begins with the potential energy due to an electric field, familiarly expressed as:
    [tex]
    U_{el} = \frac{\epsilon_r\epsilon_0}{2} \iiint_VE^2dV = \frac{\epsilon_r\epsilon_0}{2} \iiint_V(\nabla{\Phi})^2dV
    [/tex]
    I am not quite sure as to the expression for a squared gradient, but I worked from using ##\nabla(\Phi\nabla\Phi) = (\nabla\Phi)^2 + \Phi\nabla^2\Phi## so that ##(\nabla\Phi)^2 = \nabla(\Phi\nabla\Phi) - \Phi\nabla^2\Phi##. Now, I can rewrite ##U_{el}## as:
    [tex]
    U_{el} = \frac{\epsilon_r\epsilon_0}{2}\Big[\iiint_V\nabla(\Phi\nabla\Phi)dV - \iiint_V\Phi\nabla^2\Phi{dV}\Big]
    [/tex]
    I recognize the presence of Poisson's equation in the second term, but as for the first? This is where I am stuck. I am supposed to use these expressions together with the Divergence theorem so that I can invoke Green's first identity:
    [tex]
    \int_{\Omega}(\psi\nabla^2\phi + \nabla\phi\cdot\nabla\psi)dV = \oint_{\Omega}\psi(\nabla\phi\cdot{\bf{n}})dS
    [/tex]
    For two scalar functions ##\psi## and ##\phi##. Ultimately, I need to arrive at the expression:
    [tex]
    U_{el} = \frac{1}{2}\oint_S\sigma\phi_0dS + \frac{1}{2}\iiint_V\rho\Phi{dV}
    [/tex]
    Using Poisson's equation ##\nabla^2\Phi = -\frac{\rho}{\epsilon_0}## and ##\sigma = -{\epsilon_r\epsilon_0}\frac{\partial{\Phi}}{\partial{n}}## and using the fact that ##\phi_0## is the potential at the surface.

    But I do not have any dot products associated with what I did above and so I do not see how to proceed; the only way I see how this could work is understanding that there is really only one scalar function here instead of two, namely ##\Phi##, but I don't see this allowing me to make ##\nabla(\Phi\nabla\Phi) = \nabla\Phi\cdot\nabla\Phi##. Am I forgetting something simple?
     
  2. jcsd
  3. Jun 23, 2014 #2

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    You...you are right there to solving it. You already showed that ##\nabla(\Phi\nabla\Phi)=(\nabla\Phi)^2+\Phi\nabla^2\Phi##

    So I think the only point you're stuck on should be: ##(\nabla\Phi)^2=\nabla\Phi\cdot\nabla\Phi##

    Can you do it from here?

    Maybe a slightly bigger hint. I can rewrite Green's identity, if ##\phi=\psi## as such:

    $$\int_\Omega (\psi\nabla^2\phi+\nabla\phi\cdot\nabla\psi)dV=\int_\Omega (\phi\nabla^2\phi+\nabla\phi\cdot\nabla\phi)dV=\int_\Omega \nabla(\phi\nabla\phi) dV=\oint_{\partial\Omega} \phi(\nabla\phi\cdot{\bf{n}})dS$$
     
    Last edited: Jun 23, 2014
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