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Homework Help: Dirac Delta Function (electrodynamics)

  1. Apr 25, 2012 #1
    I'm having a hard time grasping when I should use this little "function". I'm using Griffith's Intro to Electrodynamics and either he doesn't touch on it enough or I'm missing the point. From what I think I understand I'm to use it when there would be a singularity in a result or calculation(?). Not sure about this. I've been looking around a lot online but can't find anything that explicitly states "You should use this here." Maybe I'll never find it. In any case, does anyone have any pointers?
  2. jcsd
  3. Apr 26, 2012 #2

    rude man

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    The Dirac delta function δ(x) is used

    as a sampling function: ∫f(x)δ(x-x0)dx over -∞ to +∞= f(x0)

    and as an impulse input to a system. It's very useful if for example a rectangular pulse is the input to a system of time constants τ such that duration << all τ. Then the pulse may be approximated by a delta function with coefficient = duration * amplitude.

    E.g. input pulse of 2V and 10us would be 2e-5δ(t).

    In any case you have to be very careful with the dimensions of δ(x), which has the dimension of 1/x.

    The Laplace transform easily handles a delta function since its transform is just 1. And the "impulse response" of a network is its output with the input = δ(t).
  4. Apr 26, 2012 #3


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    I'm a student of the upper undergraduate level EM course (we use Jackson's textbook). My limited experience tells me to use the Dirac delta "function" which is in fact a non regular distribution, when I want to have a useful mathematical way to describe WHERE the charge is, in space. More precisely when someone asks you to get the "charge distribution".
    Let's take an example. Say you have charge over the surface of a sphere. They ask you "give me the charge distribution". You could write "0 everywhere except on the surface of the sphere x^2+y^2+z^2=r^2". However this isn't very useful and not even accurate for a reason I'll explain.
    Something that must always be true is that [itex]\int _{\Omega } \rho (\vec x ) dV =Q_{\text{inside } \Omega }[/itex]. Here rho is the charge density and is written in terms of for example the Dirac distribution (maybe with a combination of the Heaviside "function" which is a regular distribution). So not only rho gives you information about where the charge lies but it tells you exactly how it is distributed. If you had answered "0 everywhere except on the surface of the sphere x^2+y^2+z^2=r^2" you wouldn't get a mathematically useful formula and you don't have the expression for rho such that when integrated over a region, it gives the total charge enclosed.
    So to answer your original question, I'd say "when they ask you the charge distribution".
  5. Apr 26, 2012 #4
    That's a good way of putting it, thanks a lot. Im slowly getting where I should place it, but its sort of a rote thing and I'm not comfortable doing it primarily off that principle.

    Thanks again!
  6. Apr 26, 2012 #5
    The delta function is a mathematical construct. In one sense, the construct allows us to model the electron as a point charge. In another sense, it is also simply a helpful mathematical tool. For example if I want to model a disc of charge in the xy plane I simply put a delta function of d(z)*H(R-r)*Q/piR^2 as the charge distribution (H Heaviside step).

    The trouble with giving an electron a delta function potential is that the electron has infinite self energy as we look smaller and smaller.
    Last edited: Apr 26, 2012
  7. Apr 26, 2012 #6


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    Are you sure it shouldn't be [itex]\rho (\vec x )=\frac{q }{\pi R^2r} \delta \left ( \theta - \frac{\pi }{2} \right ) \Theta (r -R)[/itex]? I had this problem to solve (see https://www.physicsforums.com/showthread.php?t=591217).
    I didn't know this was due to the delta function but that is indeed troublesome.
  8. Apr 26, 2012 #7
    Sure, if you want the disc in spherical coordinates. But that's slightly less intuitive so I went with cylindrical. :) (Oops, I actually did mess up the R though as I unwittingly made an annulus and just fixed it)

    The infinite self energy is more of a function of the particle being a point charge, but since a delta function models a point charge I think you could attribute it to the delta. Since V=e/R when R->0 then U->infty. The trick to this problem is to just leave it alone and let QM fix it. :)
  9. Apr 26, 2012 #8


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    Okay :)
    I see. If I remember well my prof made a comment that if we calculate the energy via [itex]\frac{1}{8\pi} \int _{R^3 } \vec E \cdot \vec E dV[/itex], for a point charge, this diverges.
    Intuitively I understand it as the energy required to form a charged sphere of radius r and taking the limit as r tends to 0. It becomes extremely hard to make it smaller and smaller due to repulsion. Eventually when you "reaches" the limit r=0, it has required infinitely energy to do so. Not sure this is really accurate though.
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