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Poisson's equation with Green's function for point charge

  1. Aug 3, 2015 #1
    Hello! I'm having a problem with the Green's function solution of the simplest case of Poisson's equation, namely a single test charge ##q## located at ##\boldsymbol r = \boldsymbol r'##. I've read the related posts on Poisson's equation via Green's function formalism, but they do not answer my specific question:

    In Gaussian units Poisson's equation for the potential reads
    $$ \nabla^2 \varPhi(\boldsymbol r) = -4 \pi \varrho(\boldsymbol r).$$
    The Green's function of the inhomogeneous PDE must satisfy
    $$ \nabla^2 G(\boldsymbol r, \boldsymbol r') = \delta(\boldsymbol r - \boldsymbol r').$$
    Some authors [Arfken, Weber: Mathematical Methods for Physicists] use ##-\delta(\boldsymbol r - \boldsymbol r')## and get a Green's function with reversed sign, but this is solely convention. The Green's function is found to be (I know how to derive this result):
    $$ G(\boldsymbol r, \boldsymbol r') = - \frac{1}{4\pi} \frac{1}{|\boldsymbol r - \boldsymbol r'|}.$$
    The solution of the Poisson equation follows from the 2nd of Green's theorems and reads
    $$ \varPhi(\boldsymbol r) = \int \mathrm{d}^3 r' \, G(\boldsymbol r, \boldsymbol r') \cdot (-) 4 \pi \varrho(\boldsymbol r'),$$
    i.e. the convolution of the Green's function with the inhomogeneity of the PDE. So far, so good - everything's straight forward.

    Now here's the problem: If I want to use the above convolution integral to solve Poisson's equation for a single test charge ##q## at ##\boldsymbol r = \boldsymbol r'## and I naively convolute ##G(\boldsymbol r,\boldsymbol r')## with ##-4 \pi q \, \delta(\boldsymbol r - \boldsymbol r')## I get a problem doing the convolution, since this would mean
    $$ \varPhi(\boldsymbol r) = \int \mathrm{d}^3 r' \, \frac{\delta(\boldsymbol r - \boldsymbol r')}{|\boldsymbol r - \boldsymbol r'|} q \stackrel{\boldsymbol r' \to \boldsymbol r}{\longrightarrow} \infty,$$
    so obviously this naive attempt to do the convolution does not work! On the other hand the textbooks say that ##G(\boldsymbol r, \boldsymbol r')## already is by definition the potential due a point charge ##q=1## at ##\boldsymbol r'##. So my questions are:
    • What's wrong with the above convolution integral? Does this have to do something with the infinitely high self energy of a point charge?
    • How to write down the solution with Green's function formalism for the point charge, simply writing ##\varPhi(\boldsymbol r) = G(\boldsymbol r, \boldsymbol r') = - \frac{1}{4 \pi |\boldsymbol r - \boldsymbol r'|}## is wrong by a factor ##-1/4\pi## and after all - how can this be generalized from ##q = 1\,\mathrm{esu}## to arbitrary charge ##q \neq 1##?
    If one considers ## \nabla^2 G(\boldsymbol r, \boldsymbol r') = - 4 \pi q \, \delta(\boldsymbol r - \boldsymbol r')## instead of the pure delta function without the preceding factor ##-4 \pi q## as was done above then from the mathematical identity
    $$ \nabla ^2 \frac{1}{|\boldsymbol r - \boldsymbol r'|} = - 4 \pi \delta(\boldsymbol r - \boldsymbol r')$$
    $$ \Rightarrow \delta(\boldsymbol r - \boldsymbol r') = \nabla^2 \frac{-1}{4 \pi |\boldsymbol r - \boldsymbol r'|}$$
    Substitution of this result for the delta function in the above ## \nabla^2 G(\boldsymbol r, \boldsymbol r') = - 4 \pi q \, \delta(\boldsymbol r - \boldsymbol r')## one obtains
    $$ \nabla^2 G(\boldsymbol r, \boldsymbol r') = \nabla^2 \frac{q}{|\boldsymbol r - \boldsymbol r'|}$$
    or equivalently
    $$ G(\boldsymbol r, \boldsymbol r') = \frac{q}{|\boldsymbol r - \boldsymbol r'|}$$
    Now this Green's function is indeed the potential of the test charge, it contains ##q## as arbitrary strength of the point charge and no wrong prefactors ##-1/4\pi##.

    While the convolution for arbitrary charge distribution ##\varrho(\boldsymbol r')## can be found in many textbooks, the simple case for the single point charge with Green's functions is discussed nowhere despite the obvious subtlety that arises when trying to convolute it with a delta function.



    Many thanks in advance - any help is appreciated!
    Pierre
     
  2. jcsd
  3. Aug 3, 2015 #2

    Orodruin

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    I do not understand why you are surprised by this. The typical potential for a point charge varies as 1/r (r being the distance from the charge). Here, you are solving for the potential of a point charge and obtain that it is infinite when the distance goes to zero. There is nothing peculiar here, just the normal behaviour of a point charge.

    But you do get the correct factor using the convolution! That it is trivial to perform due to the delta does not change the fact that you need the other factors there.

    I would claim that there is no subtlety. The convolution is trivial and you get back the Green's function multiplied by whatever factors were in front of the delta in the charge density.
     
  4. Aug 3, 2015 #3
    Thanks for the reply!

    I agree with you that this should be a trivial problem. But as far as I see the convolution cannot be performed (being singular) and is also not giving the desired result ##\varPhi(\boldsymbol r) = \frac{q}{|\boldsymbol r - \boldsymbol r'|}##. At least I do not see how to arrive at that result by doing the convolution.

    I guess my latter version is correct and Green's functions for pure delta function, i.e. ##\nabla ^2 G(\boldsymbol r, \boldsymbol r') = \delta(\boldsymbol r - \boldsymbol r')## source terms (and different differential operators) are probably tabulated so that one can easily compute the particular solution for arbitrary sources ##\varrho(\boldsymbol r')## by convoluting with the inhomogeneity like this
    $$ \varPhi(\boldsymbol r) = - \int \mathrm{d}^3 r \, G(\boldsymbol r, \boldsymbol r') 4 \pi \varrho(\boldsymbol r')$$
    $$ = \int \mathrm{d}^3 r' \frac{1}{4 \pi |\boldsymbol r - \boldsymbol r'|} 4 \pi \varrho(\boldsymbol r')$$
    $$ = \int \mathrm{d}^3 r' \frac{\varrho(\boldsymbol r')}{|\boldsymbol r - \boldsymbol r'|}$$
    which is correct being the very definition of potential.

    So the mistake in my first attempt was I guess to use this tabulated Green's function in the second line of my first post instead of
    $$\nabla^2 G(\boldsymbol r,\boldsymbol r') = -4\pi q \, \delta(\boldsymbol r - \boldsymbol r')$$
    as I did later (second half of my post). Then all factors are correct AND the Green's function is already the potential due to the point charge ##q## at ##\boldsymbol r'## - no further convolution needed (and probably also not indicated). I guess my first attempt to evaluate the Green's function for delta-source without prefactor ##-4\pi q## and then trying to convolute it with ##-4\pi q \delta(\boldsymbol r - \boldsymbol r')## is simply not the correct procedure, the Green's function being already the potential of the test charge (despite the fact that, as you say the factors cancel as they sould do).

    In the meantime I also solved the modified Helmholtz's equation for a point charge ##q## at ##\boldsymbol r'##, i.e. inhomogeneity ##-4\pi q \delta(\boldsymbol r - \boldsymbol r')## (which I was usually up to - Poisson was just intended as a warm-up)
    $$ D_\mathrm{op} G(\boldsymbol r, \boldsymbol r') = -4 \pi q \delta(\boldsymbol r - \boldsymbol r')$$
    with ##D_\mathrm{op} = \nabla^2 + k^2## and compared the Green's function I obtained by doing inverse Fourier transform and residue integrals with the one tabulated for $$D_\mathrm{op} G(\boldsymbol r, \boldsymbol r') = \delta(\boldsymbol r - \boldsymbol r')$$ being
    $$G(\boldsymbol r, \boldsymbol r') = - \frac{\exp(-k |\boldsymbol r - \boldsymbol r'|)}{4\pi|\boldsymbol r - \boldsymbol r'|}$$ and it differs from my solution again by a factor ##-4\pi q## as it should be.

    So as I understand it now, if one is simply interested in a point charge (and not an arbitrary charge distribution), instead of taking the tabulated Green's functions for ##D_\mathrm{op} G(\boldsymbol r, \boldsymbol r') = \delta(\boldsymbol r - \boldsymbol r')## and convoluting it with ##-4\pi q \,\delta(\boldsymbol r - \boldsymbol r')## one has to determine the Green's function for
    $$D_\mathrm{op} G(\boldsymbol r, \boldsymbol r') = -4 \pi q \delta(\boldsymbol r - \boldsymbol r')$$ which already is the potential of the charge by definition. Do you agree? At least I don't know how to arrive at the result by convolution. Thanks!
     
  5. Aug 3, 2015 #4
    sorry I meant ##D_\mathrm{op} = \nabla^2 - k^2##, the plus is for the case of the original (not the modified) Helmholtz equation.
     
  6. Aug 4, 2015 #5

    Orodruin

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    When ##\vec r \neq \vec r'##, the convolution is trivial. When ##\vec r = \vec r'##, the convolution is singular and it should be singular. There is no problem here.

    This is the definition of a Green's function, so of course this is what you find tabulated! The Green's function is the solution to the differential equation with a delta function (without factors in front) as the inhomogeneity.

    No, this was not your mistake. The Green's function is what it is. Your problem is not accepting that the convolution behaves exactly as you would expect it to - being well defined with the correct value for ##\vec r \neq \vec r'## and singular for ##\vec r = \vec r'##. (If you want to get technical about it, the convolution is a convolution between distributions and results in a distribution, because they are both singular, this requires a bit more maths than regular convolutions between functions.)

    What you should be doing is to simply take ##\rho = q \delta(\vec r - \vec r')##, which will give you the ##4\pi## from Poisson's equation in your chosen units. You will then have to do the convolution of ##4\pi q\delta(\vec r - \vec r')##, which will give you the correct result as explained above.

    The entire point of a Green's function of any linear differential operator ##D## is that it solves
    $$
    DG = - \delta(\vec r - \vec r')
    $$
    and that you can then construct any inhomogeneity from the right-hand side and obtain the general solution
    $$
    u = - \int G(\vec r, \vec r') f(\vec r') dV' \quad \Longrightarrow \quad Du = \int f(\vec r') DG\, dV' = f(\vec r).
    $$
    Of course, to get the correct result, you have to insert the full inhomogeneity into the convolution. In particular, if you have an inhomogeneity ##f = 4\pi q \delta(\vec r - \vec r_0)##, you will obtain the solution
    $$
    u(\vec r) = -4\pi q G(\vec r, \vec r_0).
    $$
    You should never normalise the Green's function to a particular charge. If you do, you will have to divide by that charge if you want to compute the solution for any other charge distribution.
     
  7. Aug 4, 2015 #6
    First let me thank you a lot for taking your time and posting such an extensive and valuable answer! You are right, using the standard Green's function for a pure delta source without factors in front is desirable, so that one does not have to adapt it for different inhomogeneities. I found a way to work with normalised Green's functions and accepting them being the potential of the point charge already, which they are by definition. But I admit not being fully satisfied with this approach why I started this thread originally. Your advice not to use normalised Green's functions for specific problems and to do the convolution with the standard Green's functions for delta-source instead seems good to me and if it works it will be my method of choice.

    Unfortunately, I still don't understand a single point: you wrote
    I agree with the latter, the singularity at ##\boldsymbol r = \boldsymbol r'## is evident. As for ##\boldsymbol r \neq \boldsymbol r'## I have to say that at least for me it's not trivial. I have
    $$ \Delta \varPhi(\boldsymbol r) = - 4 \pi q \delta(\boldsymbol r - \boldsymbol r'), \qquad \Delta G(\boldsymbol r, \boldsymbol r') = \delta(\boldsymbol r - \boldsymbol r').$$
    This gives the Green's function
    $$G(\boldsymbol r, \boldsymbol r') = - \frac{1}{4 \pi |\boldsymbol r - \boldsymbol r'|}$$
    Now evaluating the convolution with the inhomogeneity ##f(\boldsymbol r') = - 4 \pi q \delta(\boldsymbol r - \boldsymbol r')## according to you gives
    $$\varPhi(\boldsymbol r) = \int \mathrm{d}^3 r' \, G(\boldsymbol r, \boldsymbol r') f(\boldsymbol r') = \int \mathrm{d}^3 r' \, \frac{q \delta(\boldsymbol r - \boldsymbol r')}{|\boldsymbol r - \boldsymbol r'|} = \begin{cases} \text{undefined} & \boldsymbol r = \boldsymbol r'\\ \frac{q}{|\boldsymbol r - \boldsymbol r'|} & \boldsymbol r \neq \boldsymbol r' \end{cases}$$
    For the latter case ##\boldsymbol r \neq \boldsymbol r'## this would require an integration over all space except for an ##\varepsilon##-sphere around ##\boldsymbol r## but there, if I'm not completely wrong, the delta-function is 0, or to be more precise the integral over the delta function with ##\boldsymbol r## not being in the integration range has to be zero or to put it in a 1-D picture to simplify matters:
    $$ \int_{\mathbb{R}} \mathrm{d} x' \, q \frac{\delta(x - x')}{|x - x'|} \stackrel{x \neq x'}{=} \int_{-\infty}^{x-\epsilon} \mathrm{d} x \, q \frac{\delta(x - x')}{|x - x'|} + \int_{x+\epsilon}^{\infty} \mathrm{d} x \, q \frac{\delta(x - x')}{|x - x'|} \stackrel{?}{=} 0 + 0$$
    However you say it is trivially to be seen that the above integral gives ##q/|x-x'|##. I don't know a single valid operation to get the desired ##q/|\boldsymbol r - \boldsymbol r'|## instead of 0, which does not mean I don't believe you - it is just that I don't know how you arrive there: do you evaluate a principal value integral or deform the ##x##-axis to a closed contour taking a residue? Sorry if my question is stupid. If it is trivial probably I am just missing some simple point or I'm lacking mathematical rigor - after all I'm not dealing with convolutions of distributions every day - I have to admit that.
     
  8. Aug 4, 2015 #7

    Orodruin

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    The intuitive response is to simply state that the delta is zero everywhere except one point and the value away from that point, no matter how singular, is irrelevant.

    The more stringent way is to consider the solutions as distributions rather than normal functions, but to do that you will need to study distribution theory.
     
  9. Aug 4, 2015 #8
    PROBLEM FINALLY SOLVED!

    Thanks Orodruin! Your advice was very valuable! The "subtlety" that was puzzling me the whole time was actually a mistake I made. I just found what we were discussing about in my favourite textbook on Electrodynamics. There it says to get the potential due to a point charge ##q## at ##\boldsymbol r_0## the charge density ##\varrho(\boldsymbol r) = q \, \delta(\boldsymbol r - \boldsymbol r_0)## is substituted to the integral expression for the potential
    $$\varPhi(\boldsymbol r) = \int \mathrm{d}^3 r' \, \frac{\varrho(\boldsymbol r')}{|\boldsymbol r - \boldsymbol r'|},$$
    actually the same expression as the convolution of Green's function with ##f(\boldsymbol r')##, giving
    $$\varPhi(\boldsymbol r) = \int \mathrm{d}^3 r' \, \frac{q \, \delta(\boldsymbol r' - \boldsymbol r_0)}{|\boldsymbol r - \boldsymbol r'|} = \frac{q}{|\boldsymbol r - \boldsymbol r_0|}$$
    That's it! I made the stupid mistake to plug in ##\varrho(\boldsymbol r') = q \delta(\boldsymbol r - \boldsymbol r')##, which is a point charge at the FIELD POINT ##\boldsymbol r## at which we want to evaluate the potential, ##\varPhi(\boldsymbol r)## which of course does not make any sense at all! Of course that integral looked dreadful: ##\int \mathrm{d}^3 r' \, \frac{q \, \delta(\boldsymbol r - \boldsymbol r')}{|\boldsymbol r - \boldsymbol r'|}##.

    So my conclusions are as follows:
    • One actually can solve the problem by just identifiying the Green's function with the potential of the point charge. Jackson says that clearly in his chapter on Electrostatics, although he is referring to a unit point charge ##q = 1##. But of course one can find a Green's function also for ##q \neq 1## and some other prefactors, which apart from a factor like e.g. ##-4 \pi q## does not distinguish from the standard Green's function for the pure delta source. Probably this method is not elegant since one has to fiddle around with different Green's functions for one and the same linear differential operator like ##D_\mathrm{op} = \nabla^2## or ##D_\mathrm{op} = \nabla^2 - k^2##, etc. but it definitely leads to correct results. Orodruin, if you hadn't posted a reply to my post, I would have used this method to the end of my days and I am happy you suggested doing it the elegent way.
    • Using the standard Green's function and convoluting it with the inhomogeneity ##f(\boldsymbol r') = -4\pi q \delta(\boldsymbol r - \boldsymbol r')## gives immediately the correct result if one clearly distinguishes the 3 occurring position vectors, namely (a) the field point ##\boldsymbol r##, (b) the integration domain for the sources ##\boldsymbol r'## and (c) the location of the point charge ##\boldsymbol r_0##. Setting the latter two equal carelessly was the mistake that lead to this thread, since if that convolution had worked right from the beginning I wouldn't have started to try to normalise Green's functions.
    I am very happy this thread is now completely solved. Many thanks, Orodruin!
     
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