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Understanding Green functions in electrostatics

  1. Jan 17, 2012 #1
    Hey guys! So I'm reading Jackson's Classical Electrodynamics, and I'm trying to get a sense of what Green Functions really are. The book says that "the potential due to a unit source and its image(s), chosen to satisfy homogeneous boundary conditions, is just the Green function appropriate for Dirichlet or Neumann boundary conditions. In G(x,x'), the variable x' refers to the location of the unit source, while the variable x is the point at which the potential is being evaluated." (Here, x and x' are vectors.)

    So I went through the example they gave, for a point charge in the presence of a grounded conducting sphere. They find the potential by using the method of images, which basically turns the problem into the potential due to two point charges. Then they do some (unexplained and mysterious) substitutions like "q->4πε" from the potential we found, and that gives us the Green function.

    So I guess I kind of understood that. However, in the discussion in the previous chapter, they have the potential in terms of an appropriate Green function G(x,x'), where like before x is the point at which we're evaluating the potential, but this time x' is an integration variable. This kinda confuses me in light of what was said above... The part I quoted seems to say that G(x,x') = [itex]\Phi(x)[/itex] with some substitutions.

    But in the previous section, they say [itex] \Phi(x) = \frac{1}{4\pi \epsilon}\int _V \rho(x') G_D (x,x') d^3x' - \frac{1}{4\pi} \oint _S \Phi(x') \frac{\delta G_D}{\delta n'} da'[/itex]...so now the Green function isn't just equal to the potential. So I'm a little confused as to what it actually is.

    I tried to do a problem from the book to understand this, problem 2.7. It says "Consider a potential problem in the half-space defined by z≥0, with Dirichlet boundary conditions on the plane z = 0 (and at infinity). Write down the appropriate Green function G(x,x')."

    I didn't really know what they meant here... I know Dirichlet boundary conditions means that the voltage of the plane, rather than the derivative of the voltage at the plane (the charge density). But they don't actually say what the boundary conditions are, just what type they are. I solved it for the plane being a grounded conductor in the presence of a point charge (like their example above), but I don't think this is what they meant.

    Can anyone help me? Thanks!!
     
  2. jcsd
  3. Jan 18, 2012 #2
    The Green's function tells you what the field is for a given source but encodes the solution to the entire problem. This way you simply integrate over the sources instead of solving the integral or integro-differential equations.

    Example: to find the electric field of a charge distribution, you integrate over the charge distribution with the free-space Green's function.

    Example two: To find the electric field of a charge distribution in the presence of a dielectric sphere over a conducting half-plane, you integrate over the charge distribution using the dielectric-sphere-over-a-half-plane Green's function.
     
  4. Jan 19, 2012 #3
    Let's take [itex]\Phi \rightarrow 0[/itex] at infinity for simplicity.

    1st paragraph:
    the Green's function is the response to a unit charge. that is - it's what
    the potential would be if you only had one charge.
    [itex]G(x,x') \propto \Phi(x)[/itex]

    2d paragraph:
    When you have many charges you add up the contributions from each.
    (Superposition). Thus the total potential is the potential from each
    extra charge so that:
    [itex] \Phi(x) \propto \int _V \rho(x') G_D (x,x') d^3x'[/itex]

    ----
    On Green's Functions (subtitled - why i shouldn't worry too
    much and was making this seem too complicated)
    ----
    What do we actually want to accomplish in an electrostatics
    problem? It boils down to you tell me the distribution
    of charge - I'll tell you the potential everywhere.
    The relation between those two guys is
    [itex] \nabla^2\Phi(x) = -\rho(x). [/itex]
    [where of course i'm dropping all constants to make
    my point].

    This isn't quite enough - I also need to specify any
    boundary conditions that might apply.

    So instead of solving that we solve
    [itex] \nabla^2G(x) = -\delta^3(x)[/itex]
    with our desired boundary condition.
    This G is the green's function (or sometimes
    called elementary solution or sometimes called
    unit response).

    We do this because -
    1. We've already incoded our boundary conditions in G
    2. If we know G we know the solution for any [itex]\rho[/itex]

    How does 2 work
    [tex]\Phi(x)=\int \rho(x')G(x-x') d^3x'.[/tex]

    ---

    Why does this work? Superposition.
    [or linearity of the operator [itex]\nabla^2[/itex]]
    This is all really a highfalutin way of
    restating what is done in undergraduate E-M.
     
  5. Jan 20, 2012 #4
    Thanks, this makes more sense now. But how are you supposed to find the Green's function for specific scenarios? I don't get how the boundary conditions work into this.

    Thank you very much, this also helps.

    Can you guys tell me if this sounds right, too?

    So when we do the problem of the method of images, let's take the simplest example: a point charge +Q in the presence of an infinite grounded plane. Jackson says that our Green functions will always have the form of [itex]G(x,x') = \frac{1}{\left| x - x' \right|} + F(x,x')[/itex] with [itex]\nabla^2 F(x,x') = 0[/itex].

    So in this simple problem, the omnipresent part of G(x,x'), the [itex]\frac{1}{\left| x - x' \right|}[/itex] part, is what the potential would be if there was just the point charge, nothing else. But, because there is the infinite plane there, it produces its own field in response, and that's where the F(x,x') part comes from, right?

    Still don't get it completely though... Looking again at the problem, I don't know what they want for G(x,x'). They tell us where the boundary conditions are specified but not what they actually are. Isn't that necessary to find G?

    Thanks!
     
  6. Jan 21, 2012 #5

    vanhees71

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    In electrostatics you want to solve Poisson's Equation for the potential (in Gauss's units as in the good old 2nd edition of Jackson),

    [tex]\Delta \Phi=-4 \pi \rho.[/tex]

    The idea of the Green's function is in a way to invert the Laplace operator in terms of an integral kernel, i.e.,

    [tex]\Phi(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' G(\vec{x},\vec{x}') \rho(\vec{x}').[/tex]

    In order to make this work, obviously you must have

    [tex]\Delta_{x} G(\vec{x},\vec{x}')=-4 \pi \delta^{(3)}(\vec{x}-\vec{x}').[/tex]

    This obviously is the solution of the problem of a unit point charge, sitting at [itex]\vec{x}'[/itex], and one solution is the Coulomb potential for a unit charge, i.e.,

    [tex]G_0(\vec{x},\vec{x}') =\frac{1}{|\vec{x}-\vec{x}'|}.[/tex]

    The formal proof that this fulfills the above equation can be found in Jackson.

    Now, [itex]G_0[/itex] is not a unique solution. Since the equation is linear you can always add any solution of the homogeneous equation, the Laplace or potential equation. It can be shown that the solution is made unique only if one gives appropriate boundary conditions at sufficiently smooth surfaces. For the Coulomb solution the boundary conditions are just given that the potential should go to 0 at infinity and that the only singularity of the potential can be at the place, where the point charge is sitting.

    Another example are boundary conditions given by the presence of conductors or dielectrics. For conductors, the field in the interior of the medium must vanish, i.e., the potential must be constant. Also the tangential components of the electric field along the surface must be continuous. Another constraint may be that the conductor carries a certain surface charge, etc. Thus you need a Green's function that satisfies this boundary condition.

    Of course, you can make the Ansatz

    [tex]G(\vec{x},\vec{x}')=G_0(\vec{x},\vec{x}')+F(\vec{x},\vec{x}'),[/tex]

    where

    [tex]\Delta_x F(\vec{x},\vec{x}')=0.[/tex]

    Physically that's to solve the problem of having a unit charge at [itex]\vec{x}'[/itex] and finding a solution for the Laplace equation for [itex]F[/itex] such that the boundary conditions of the given setup are fulfilled (see Jackson for details on the variety of boundary conditions possible for Poisson's equation).

    The method of image charges uses the idea to generate a potential, [itex]F[/itex], in a certain region of space by taking a charge distribution located outside of this region. The true solution of the problem outside of this region must be found otherwise. Of course these auxilliary charges are not real charges but only mathematical tricks to find a solution. There are not too many problems that can be solved in this way since you have to be able to guess the appropriate auxilliary charges, which is possible for very symmetric conditions.

    As an example take the problem of the half-space. Let the region [itex]z<0[/itex] be filled with a conducting medium, and we are looking for the Green's function. Inside the medium, the Green's function must be constant, i.e., we choose

    [tex]G(\vec{x},\vec{x}')=0 \quad \text{for} \quad z<0.[/tex]

    For [itex]z>0[/itex] we make the ansatz that [itex]F[/itex] is the field of a mirror charge [itex]q'[/itex] sitting at [itex]\vec{y}[/itex] with [itex]y_z<0[/itex] such that the boundary condition is fulfilled. One condition is that the Green's function itself is continuous at the boundary, i.e., we must have

    [tex]G(\vec{x},\vec{x}')=0 \quad \text{for} \quad z=0.[/tex]

    With our mirror-charge ansatz the solution for [itex]z>0[/itex] reads

    [tex]G(\vec{x},\vec{x}')=\frac{1}{|\vec{x}-\vec{x}'|}+\frac{q}{|\vec{x}-\vec{y}|}.[/tex]

    It's of sufficient to assume also [itex]z'>0[/itex] since there cannot be any free charge distributions inside the conductor. So obviously the boundary condition is fulfilled, if you choose [itex]\vec{y}=(x',y',-z')[/itex] and [itex]q=-1[/itex], because then obviously the boundary condition is fulfilled. Thus, the Greens' function is

    [tex]G(\vec{x},\vec{x}')=\frac{1}{\vec{x}-\vec{x}'}-\frac{1}{\vec{x}-\vec{y}} \quad \text{with} \quad \vec{y}=(x',y',-z').[/tex]

    You should check that the continuity condition for the tangential components of [itex]\vec{E}=-\vec{\nabla}_x \vec{G}(\vec{x},\vec{x}')[/itex] is fulfilled. You can also figure out that the total charge on the surface vanishies, i.e., you have a grounded conductor here.

    Of course you can have in addition an arbitrary homogeneously distributed charge on the surface without violating any of the conditions. This gives the corresponding additional field from these surface charges.

    Another example where the image-charge method works is the conducting or dielectric sphere or spherical shell (see Jackson, where this is discussed nicely).
     
  7. Jan 25, 2012 #6
    It didn't say that in the problem though, why are we supposed to assume that? It seems like it could just have an arbitrary V(x,y) on the boundary.

    Ok, assuming what you said before, why must it be constant?



    So I've been a little confused as to why we were just considering point charges with these boundary conditions. Can someone tell me if this is right? We consider the response of the boundary conditions with respect to a single point charge at x', then basically integrate over all the volume. Each [itex]ρ(x')d^3x'[/itex] is a little piece of charge, so each [itex]G(x,x')ρ(x')d^3x[/itex] is the potential due to that little piece, and they're all added because of the superposition principle.

    Is that right?

    There are a couple things I still don't get though. For example, I know for superposition, if you have two point charges in space, to find the potential at a point, you just add the potential that each point charge would generate by itself. But why is it the same when you have the point charges in the presence of a grounded conductor? Why is the charge in the conductor attracted by the two point charges just the sum of the charge that would be attracted by one point charge? I assume this is just an application of superposition, but I can't figure out how.


    Also, I'm trying to interpret the Green function for a point charge in the presence of the grounded sphere, like I mentioned above (pages 64-65 of Jackson). So, he gets from the method of images that the potential outside the sphere (from the charge and its image) is:

    [itex]\Phi(\vec{x}) = \frac{q}{4\pi\epsilon}(\frac{1}{\left|\vec{x} - \vec{x'}\right|} - \frac{a}{x' \left|\vec{x} - \frac{a^2}{x'^2}\vec{x'}\right|})[/itex]

    (for a charge q and radius a.)

    From this, he says we do that magic substitution (no idea how/why), and get this for the Green function:

    [itex]G(\vec{x},\vec{x'}) = \frac{1}{\left|\vec{x} - \vec{x'}\right|} - \frac{a}{x' \left|\vec{x} - \frac{a^2}{x'^2}\vec{x'}\right|}[/itex]

    Then he does a few things: converts the Green function to spherical coordinates; He then says that for the solution to the Poisson equation for Dirichlet conditions (the one above, in my OP), we need [itex]\frac{∂G}{∂n'}[/itex] also, and calculates that (basically, n' is x', evaluated at x' = a). Then he plugs this into the "general solution" and gets:

    [itex]\Phi(\vec{x}) = \frac{1}{4\pi}\int \Phi(a,\theta ',\phi ') \frac{a(x^2 - a^2)}{(x^2 + a^2 - 2ax cos(\gamma))^{3/2}} d\Omega '[/itex]

    Where the fraction in the integrand is from the spherical coordinates of [itex]\frac{∂G}{∂n'}[/itex] and dΩ' is an element of solid angle.

    Basically, I have two questions. First, it seems to me that this should simplify to the result for [itex]\Phi[/itex] we got before, simply from the method of images. However, because the sphere is grounded, [itex]\Phi = 0[/itex] everywhere on it, so it seems like [itex]\Phi(a, \theta',\phi') = 0[/itex] for the whole integral, which would give [itex]\Phi(\vec{x}) = 0[/itex] at places even that aren't on the sphere.

    Secondly, this is the surface integral term. But what happens to the volume integral term in the solution in my OP? It seems like [itex]\rho[/itex] is zero in most places of the volume (except at the point charge), but it seems like G(x,x') isn't, so I don't see why it completely disappears.

    I'd love it if someone could help me, I think I almost understand it!
     
  8. Jan 25, 2012 #7

    vanhees71

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    The potential must stay constant inside a conductor as well as along its surface in order to have no currents according to Ohm's Law [itex]\vec{j}=\sigma \vec{E}[/itex], i.e., inside the conductor and along its surface one must have [itex]\vec{E}=0[/itex] and [itex]\vec{E}_{\text{tang}}=0[/itex], respetively. I've just chosen the half-space as the most simple example for a problem with a conductor.

    Despite that now all of a sudden you changed to SI units your general formula for the potential is correct with the Green's function fulfilling all boundary conditions.

    The Green's function for the sphere follows from the mirror-charge method. You just make an ansatz with the mirror charge and calculate its magnitude and position from the boundary conditions.
     
  9. Jan 25, 2012 #8
    Yeah, but why, given the problem I mentioned in my OP with boundary conditions on the z = 0 plane, did you take that to mean it was constant? That's what confuses me...it seems like it could be anything.

    I see how he gets the potential for the sphere from the mirror charge method. But I don't get two things after that: first, how he just gets the Green function from the potential by that "magic substitution". Secondly, like I pointed out in my previous post, it seems like the general formula for the potential gives [itex]\Phi = 0[/itex] everywhere for a grounded sphere.

    Eh?
     
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