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Dirac Delta function

  1. Feb 25, 2005 #1
    Can someone explain me the Dirac Delta function for the function:

    [tex]\vec A = \frac{\hat r}{r^2}[/tex]
     
    Last edited: Feb 25, 2005
  2. jcsd
  3. Feb 25, 2005 #2

    James R

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    Your question doesn't seem to make sense. The Dirac Delta Function is always the same. It doesn't rely on any other function.
     
  4. Feb 25, 2005 #3
    I'm sorry, the given function is the Dirac delta function. Can someone explain it to me?
     
  5. Feb 25, 2005 #4
  6. Feb 26, 2005 #5
    Thanks for the link, Vivek. But it did not completely solve my problem. The proofs given in most texts are too mathematical. I need a more physical interpretation of the problem.
     
  7. Feb 26, 2005 #6
    what is your question?
    your question doesn't make sense at all?
     
  8. Feb 26, 2005 #7
    I believe You want to interpret its curl or div in terms of Dirac Delta Function

    [tex]\vec{\nabla} X \vec A [/tex]
     
    Last edited: Feb 26, 2005
  9. Feb 26, 2005 #8
    Yes they are mathematical because of the very definition of DDF. Strictly, it is not a function but it is considered a function. If you want good physical interpretations of its applications, get a copy of Classical Electrodynamics by Griffiths and read the first chapter (I think its called mathematical preliminaries but I'm not very sure).

    Hope that helps...

    cheers
    vivek
     
  10. Feb 26, 2005 #9

    Gokul43201

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    Perhaps you (Reshma) are asking for a proof that the charge (density) distribution that produces this field is a dirac-delta function (about the origin) ? The given field itself is not a dirac-delta.
     
  11. Feb 26, 2005 #10

    dextercioby

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    It can be proven really easily that the Green's function for an oO domain (R^{3}) for the Poisson equation:
    [tex] \Delta V(\vec{r})=f(\vec{r}) [/tex](1)

    is:[tex] G(\vec{r},\vec{r'})=\frac{1}{4\pi|\vec{r}-\vec{r}'|} [/tex](2)

    And incidentally,the field,being the -gradient of the solution of (1),can be put in connection to (2)...

    Daniel.
     
  12. Feb 26, 2005 #11
    Yes, you are right. I want an interpretation of the divergence of the given function.
     
  13. Feb 26, 2005 #12
    Yes I do have Griffith's book which has described the above function over a sphere using Green's theorem.
     
  14. Feb 26, 2005 #13

    dextercioby

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    You want the proof that the [itex] \nabla \cdot \frac{\vec{r}}{r^{3}} [/itex] is proportional (it's a "-1" the coefficient of proportionaliry,IIRC) to delta-Dirac...?

    That's a pretty delicate matter.It's not really for physicists...Any book on PDE-s should have it,when discussing Laplace & Poisson equations.

    Daniel.
     
  15. Feb 26, 2005 #14
    I am extremely sorry for stretching this thread this far :frown:
    I only want to know the proof for:

    [tex] \nabla \cdot\frac{\vec{r}}{r^{2}} [/tex]

    With a little physical interpretation...
     
  16. Feb 26, 2005 #15

    dextercioby

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    That is something else...As u can yourself check...

    Turning to the original question,i can add:except for the origin,where the fraction to whom you apply the diff.operator is not defined,the result is zero.However,as [tex] \frac{\vec{r}}{r^{3}} [/tex] is the Green function for the Poisson equation for R^{3},it can be shown that,in fact:

    [tex] \nabla\cdot\frac{-\vec{r}}{r^{3}}=-4\pi\delta(\vec{r}) [/tex]

    As for physical significance,please,check (as you probably already have) Griffiths' book.Or Jackson's...

    Daniel.
     
    Last edited: Feb 26, 2005
  17. Feb 26, 2005 #16
    [tex] \nabla \cdot \frac{\hat{r}}{r^2} =
    \frac{\partial}{\partial r} \left( r^2 \cdot \frac{1}{r^2} \right) = 0 [/tex]

    everywhere, except the origin, where we have a point of non-differentiability.

    By integrating over the volume of a sphere, and applying the divergence theorem, we see

    [tex] \int_V \nabla \cdot \frac{\hat{r}}{r^2} r^2 \sin \theta dr d\theta d\phi
    = \oint_S \frac{\hat{r}}{r^2} \cdot \hat{r} r^2 \sin \theta d\theta d\phi
    = \oint_S \sin \theta d\theta d\phi = 4\pi
    [/tex]

    independent of the radius of the sphere. Thus integration over any volume including the origin gives 4*pi, and any other volume gives zero. A function which satisfies this would be 4*pi times a delta function located at the origin. Thus, [tex] \nabla \cdot \frac{\hat{r}}{r^2} = 4 \pi \delta({\vec{r}}) [/tex]
     
  18. Feb 26, 2005 #17

    reilly

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    Physicist's are more than adequately capable of dealing with delta functions, Green's functions, and the like. About delta functions, step functions, absolute values, derivatives, all termed distributions, mathematicians told us that they only make sense when multiplied by normal functions (x, sinx,...) and integrated. (See, for example the classic by Lighthill, Fourier Analysis and Generalized Functions -- every physicist should read this book. But, the mathematician's work also provided justification for the formal algebra of step and delta functions used by engineers and physicists.

    One of the standard older attacks dealing with delta functions and Poisson's Eq. starts with Green's Thrm -- first let ((d/dx)*d/dx + (d/dy)*d/dy +(d/dz)*d/dz) == LAP, and
    dv1/dx + dv2/dy + dv3/dz == DIV v, v1 is the x component , etc.

    Green tells us A *LAP B - B * LAP A == DIV (A GRAD B - B GRAD A)

    Now integrate Green over a volume R, bounded by a closed surface S, and choose B to be the 1/r potential, and A to be a mathematically nice function, integrable, differentiable, single valued, and so forth. Rearrange so that you get

    /
    |dV *A* LAP (1/r) =
    /

    /
    |dV (1/r) LAP A + surface terms.
    /

    r is the usual radial coordinate. The rest of the work, to show that the integral on the left = -4pi *A(0), is all about limits, as r->0. Pretty standard stuff for electrodynamics.

    And, don't forget that until Dirac, there were no delta functions, so people used limits to get what we can get now in an easier fashion.

    Regards,
    Reilly Atkinson
     
    Last edited: Feb 26, 2005
  19. Feb 27, 2005 #18
    Thank you so much Kanato and Reilly!! I completely got it now.
     
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