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I Green function in electrostatics

  1. Apr 2, 2016 #1
    Hello,
    I'm taking a course in electrostatics and electrodynamics.

    We learned about finding a potentional using unique Green functions that are dependent of the geometry of the problem. Specificly on a Dirichlet problem we get the solution:

    Φ(x)=∫ρ(x')G(x,x')d3x' - (1/4π)*∫Φ(x')*(∂G(x,x')/∂n')da'


    I understand the idea of the Green function, but don't understand why on the second integral (surface integral) there is a use of the partial deriviative of G. Can anyone give me some intuition about it?
     
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  3. Apr 2, 2016 #2

    vanhees71

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    The idea of the Green's function is based on Green's integral theorem, which is an application of Gauss's theorem,
    $$\int_{V} \mathrm{d}^3 \vec{x}' \vec{\nabla}' \cdot \vec{A}(\vec{x}')=\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{A}(\vec{x}'),$$
    where ##V## is some volume and ##\partial V## its boundary surface. The orientation of the surface-normal vectors is out of the volume.

    Now set
    $$\vec{A}(\vec{x}')=G(\vec{x},\vec{x}') \vec{\nabla}' \Phi(\vec{x}')-\Phi(\vec{x}')\vec{\nabla}' G(\vec{x},\vec{x}')$$
    Then
    $$\vec{\nabla}' \cdot \vec{A}(\vec{x}')=G(\vec{x},\vec{x}') \Delta' \Phi(\vec{x}') - \Phi(\vec{x}') \Delta' G(\vec{x},\vec{x}').$$
    If the Green's function now satisfies
    $$\Delta' G(\vec{x},\vec{x}')=-4 \pi \delta^{(3)}(\vec{x}-\vec{x}')$$
    and if ##\Phi## fulfills (Gauss units)
    $$\Delta' \Phi(\vec{x}')=-4 \pi \rho(\vec{x}')$$
    you have
    $$\vec{\nabla}' \cdot \vec{A}(\vec{x}')=-4 \pi G(\vec{x},\vec{x}') \rho(\vec{x}') + 4 \pi \Phi(\vec{x}') \delta^{(3)}(\vec{x},\vec{x}').$$
    Now integrate this over the volume ##V##, you get for ##\vec{x} \in V##
    $$\int_V \mathrm{d}^3 \vec{x}' \vec{\nabla}' \cdot \vec{A}(\vec{x}') = -4 \pi \int_V \mathrm{d}^3 \vec{x}' G(\vec{x},\vec{x}') \rho(\vec{x}')+4 \pi \Phi(\vec{x}).$$
    According to Gauss's Law, this equals the surface integral
    $$-4 \pi \int_V \mathrm{d}^3 \vec{x}' G(\vec{x},\vec{x}') \rho(\vec{x}')+4 \pi \Phi(\vec{x})=\int_{\partial V} \mathrm{d}^2 \vec{f}' \cdot \vec{A}(\vec{x}') = \int_{\partial V} \mathrm{d}^2 \vec{f}' \cdot [G(\vec{x},\vec{x}') \vec{\nabla}' \Phi(\vec{x}')-\Phi(\vec{x}')\vec{\nabla}' G(\vec{x},\vec{x}')].$$
    By definition the Dirichlet problem means that you know ##\Phi(\vec{x}')## for ##\vec{x}' \in \partial V## but nothing about derivatives of ##\Phi## along the surface. Thus you define the Green's function such that it has to fulfill the boundary condition
    $$G(\vec{x},\vec{x}')|_{\vec{x}' \in \partial V}=0,$$
    then you get
    $$4 \pi \Phi(\vec{x})=4 \pi \int_V \mathrm{d}^3 \vec{x}' G(\vec{x},\vec{x}') \rho(\vec{x}')-\int_{\partial V} \mathrm{d}^2 \vec{f}' \cdot \Phi(\vec{x}') \vec{\nabla}' G(\vec{x},\vec{x}').$$
    Since further ##\mathrm{d}^2 \vec{f}' =\vec{n} \mathrm{d}^2 a##, dividing by ##4 \pi## gives your formula.
     
  4. Apr 2, 2016 #3
    vanhees71
    thank you for the explenation.
    I understand the mathematics of it and how to derive it.
    What I don't understand is the physical intuition behind the derivietive of G on the surface integral.
    G on the volume integral means (as far as I understood) summing up the impulse response of 1 charge at x, for all the volume V, witch makes good sense.
    But the surface integral says: sum up the (known) potential product with the deriviative of G in direction of the normal to S. What is the physical meaning of this?
     
  5. Apr 3, 2016 #4

    vanhees71

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    The Green's function, seen as a field ##\Phi(\vec{x})=G(\vec{x},\vec{x}')## is the solution for the situation that you have a point charge of charge ##q=1## in ##\vec{x}'## and a grounded conducting surface ##\partial V##, because it's 0 along this surface. Physically the charges within the conducting surface rearrange due to the presence of the unit charge at ##\vec{x}'## such as to fulfill this boundary condition, and thus in addition to this unit charge you have a surface charge distribution along the surface, and this surface charge distribution is given by the normal component of the electric field along the surface, but the field is the gradient of ##G##, and the contribution of this surface charge thus makes the term with the surface integral in Green's formula.
     
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