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3-D Dirac Delta Fnc (DDF) Evaluated

  1. Mar 17, 2006 #1
    Hi,

    After reading DJGriffiths sections on the DDF I have a question about evaluating it in regards Prob. 1.46 (a), to wit:

    "Write an expression for the electric charge density [tex]\rho(\vec{r})[/tex] of a point charge q at [tex]\vec{r}[/tex]'. Make sure that the volume integral of [tex]\rho[/tex] equals q."

    This is easily done in rectangular coordinates, namely, let [tex]\rho(\vec{r}) = q\delta(\vec{r}-\vec{r}')[/tex].

    Checking that this resultsin a volume integral of q I did:

    [tex] \int_V \rho(\vec{r}) d\tau = \int_V q\delta^3(\vec{r}-\vec{r}') d\tau = q \int_V \delta^3(\vec{r}-\vec{r}') d\tau = q \cdot 1 = q.[/tex]

    Here is my question: doing the integral as is I get the correct answer. However, when I wrote it out in spherical coord's. It seems that I should rewrite [tex]\rho(\vec{r})[/tex] as [tex]\frac{q}{4\pi}[/tex]?

    Why? Because:

    [tex] \int_V \rho(\vec{r}) d\tau = \frac{q}{4\pi} \int_V \delta^3(\vec{r}-\vec{r}') dr d\Omega = \frac{q}{4\pi} \cdot \int_0^R \delta^3(\vec{r}-\vec{r}') dr \cdot \int_0^{4\pi} d\Omega = \frac{q}{4\pi} \cdot 4\pi \cdot \int_0^R \delta^3(\vec{r}-\vec{r}') dr = q \int_0^R \delta^3(\vec{r}-\vec{r}') dr = q[/tex].

    So my question is: is this second form of evaluation correct? And if it is, then why does [tex]\rho[/tex] have two different expressions? Also, what about other curvilinear coordinates? Shouldn't his expression be independent of the geometry being used? So I suspect there is something wrong with my second evaluation. Can you not integrate over [tex]\delta^3[/tex] over one dimension?

    tia!
    -LD

    EDIT: I see it now. I was missing the fact that [tex]\delta^3(\vec{r}-\vec{r}') = \delta(r-r')\delta(\theta-\theta')\delta(\phi-\phi')[/tex] and so the integral over [tex]\Omega[/tex] is still 1 and not [tex]4\pi[/tex].

    PS: I wasn't sure if I should have deleted this post having figured out the correct answer, or leave it and post what I found out. Let me know and I will do whatever is the correct procedure here.

    thx!
    -LD
     
    Last edited: Mar 17, 2006
  2. jcsd
  3. Mar 17, 2006 #2

    HallsofIvy

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    You are integrating the delta function incorrectly in your last formula. Since it is a 3 dimensional delta, its integral over all 3 dimensions, not just r, is equal to 1.
     
  4. Mar 17, 2006 #3
    :) You caught me mid-edit! :)

    but thanks for the confirmation! :!!)

    -LD
     
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