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Question about Dirac Delta function

  1. Aug 9, 2013 #1
    In page 555, Appendix B of Intro to electrodynamics by D Griffiths:
    [tex]\nabla\cdot \vec F=-\nabla^2U=-\frac{1}{4\pi}\int D\nabla^2\left(\frac{1}{\vec{\vartheta}}\right)d\tau'=\int D(\vec r')\delta^3(\vec r-\vec r')d\tau'=D(\vec r) [/tex]
    where ##\;\vec{\vartheta}=\vec r-\vec r'##.

    Is it supposed to be equal to ##D(\vec r')## as equation (1.98) in page 50 stated ##\int f(\vec r)\delta(\vec r-\vec a)d\tau'=f(\vec a)##?

    Thanks
     
  2. jcsd
  3. Aug 9, 2013 #2

    vanhees71

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    I don't precisely understand, what's your question. The most non-trivial statement is that
    [tex]\Delta_{\vec{x}} \frac{1}{4 \pi |\vec{x}-\vec{x}'|}=\delta^{(3)}(\vec{x}-\vec{x}').[/tex]
    To prove this, it is most simple to solve for the Green's function of the Laplace operator, i.e.,
    [tex]\Delta G(\vec{x})=-\delta(\vec{x}).[/tex]
    One way is to use the method of Fourier transformation, i.e., write
    [tex]G(\vec{x})=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{k}}{(2 \pi)^3} \tilde{G}(\vec{k})\exp(\mathrm{i} \vec{k} \cdot \vec{x}).[/tex]
    Then you find
    [tex]\Delta G(\vec{x})=-\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{k}}{(2 \pi)^3}\vec{k}^2 \tilde{G}(\vec{k})\exp(\mathrm{i} \vec{k} \cdot \vec{x}).[/tex]
    On the other hand, the Dirac [itex]\delta[/itex] distribution (it's not a function but a distribution!) has the Fourier representation
    [tex]\delta^{(3)}(\vec{x})=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{k}}{(2 \pi)^3}\exp(\mathrm{i} \vec{k} \cdot \vec{x}).[/tex]
    This implies
    [tex]\tilde{G}(\vec{k})=\frac{1}{\vec{k}^2}.[/tex]
    Now you have to do the Fourier transform. Unfortunately the integral is not absolutely convergent. This we cure by restricting the integration in momentum space to the ball of radius [itex]\Lambda[/itex],
    [tex]G(\vec{x})=\int_{B_{\Lambda}} \frac{\mathrm{d}^3 \vec{k}}{(2 \pi)^3} \frac{\exp(\mathrm{i} \vec{k} \cdot \vec{x})}{\vec{k}^2}.[/tex]
    We introduce spherical coordinates with polar axis in direction of [itex]\vec{x}[/itex]. Then we can do the angular integrals immediately (substituting [itex]u=\cos \vartheta[/itex])
    [tex]G(\vec{x})=\frac{1}{4 \pi^2} \int_0^{\Lambda} \mathrm{d} K \int_{-1}^1 \mathrm{d} u \exp(\mathrm{i} K r u)=\frac{1}{2 \pi^2 r} \int_0^{\Lambda} \frac{\sin(k r)}{k}.[/tex]
    Now we can let [itex]\Lambda \rightarrow \infty[/itex] and use the standard integral
    [tex]\int_0^{\infty} \mathrm{d} k \frac{\sin(k r)}{k}=\frac{\pi}{2}[/tex]
    to finally obtain
    [tex]G(\vec{x})=\frac{1}{4\pi |\vec{x}|}.[/tex]
     
  4. Aug 9, 2013 #3

    HallsofIvy

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    No. In that last equation, the integration is with respect to the "primed" values and the result is f of the unprimed a. In the equation you are asking about, the integration is again with respect to the "primed" variables and the result is again a function of the unprimed variable.
     
  5. Aug 9, 2013 #4
    Thanks for your response. That's exactly my question. I don't understand why using ##D(\vec r)## instead of ##D(\vec r')##?

    I don't understand why ##\int D(\vec r')\delta^3(\vec r-\vec r')d\tau'=D(\vec r) ## where integration with respect to the "primed" variables gives a function of the unprimed variable? Is it like this:
    [tex]\int D(\vec r')\delta^3(\vec r-\vec r')d\tau'=-\int D(\vec r')\delta^3(\vec r'-\vec r)d\tau'=D(\vec r)[/tex]
    as ##\delta(-x)=-\delta(x)\Rightarrow\;\delta^3(\vec r-\vec r')=-\delta^3(\vec r'-\vec r)##

    Thanks for your help.
     
    Last edited: Aug 9, 2013
  6. Aug 9, 2013 #5
    Its because the primed r inside of the D is just a tool for you to note that you are integrating that function over all space, while weighting it with the 3-D point function.

    Once you do the integral, it takes on the value of the function D evaluated at the point r (unprimed). The function at all other points outside of that the integral is zero because its weight with the Dirac function
     
  7. Aug 10, 2013 #6
    how did you got that?Delta function is an even function.What you have written holds for δ'(-x).
     
  8. Aug 10, 2013 #7
    I double check my notes, I was wrong.
    [tex]\delta(-x)=\delta(x)\; and \; \delta(\vec r-\vec r')=\delta(\vec r'-\vec r)[/tex]

    Does this work better?
     
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