Dirac Delta function

**Reshma** · Feb25-05, 01:27 AM

Can someone explain me the Dirac Delta function for the function:

$LaTeX Code: \\vec A = \\frac{\\hat r}{r^2}$

**James R** · Feb25-05, 02:37 AM

Your question doesn't seem to make sense. The Dirac Delta Function is always the same. It doesn't rely on any other function.

**Reshma** · Feb25-05, 03:00 AM

I'm sorry, the given function is the Dirac delta function. Can someone explain it to me?

**maverick280857** · Feb25-05, 05:00 AM

The best thing an electrical engineer could do for physics: http://mathworld.wolfram.com/DeltaFunction.html (might help you)

**Reshma** · Feb26-05, 03:27 AM

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.

**vincentchan** · Feb26-05, 03:42 AM

what is your question?
your question doesn't make sense at all?

**himanshu121** · Feb26-05, 03:53 AM

Originally Posted by Reshma

Can someone explain me the Dirac Delta function for the function:

$LaTeX Code: \\vec A = \\frac{\\hat r}{r^2}$

I believe You want to interpret its curl or div in terms of Dirac Delta Function

$LaTeX Code: \\vec{\\nabla} X \\vec A$

**maverick280857** · Feb26-05, 04:12 AM

Originally Posted by Reshma

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.

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

**Gokul43201** · Feb26-05, 04:13 AM

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.

**dextercioby** · Feb26-05, 06:09 AM

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

is: $LaTeX Code: G(\\vec{r},\\vec{rsingle-quote})=\\frac{1}{4\\pi|\\vec{r}-\\vec{r}single-quote|}$ (2)

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

Daniel.

**Reshma** · Feb26-05, 07:28 AM

Originally Posted by himanshu121

I believe You want to interpret its curl or div in terms of Dirac Delta Function

$LaTeX Code: \\vec{\\nabla} X \\vec A$

Yes, you are right. I want an interpretation of the divergence of the given function.

**Reshma** · Feb26-05, 07:30 AM

Originally Posted by maverick280857

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

Yes I do have Griffith's book which has described the above function over a sphere using Green's theorem.

**dextercioby** · Feb26-05, 07:35 AM

You want the proof that the $LaTeX Code: \\nabla \\cdot \\frac{\\vec{r}}{r^{3}}$ 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.

**Reshma** · Feb26-05, 08:01 AM

Originally Posted by dextercioby

You want the proof that the $LaTeX Code: \\nabla \\cdot \\frac{\\vec{r}}{r^{3}}$ 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.

I am extremely sorry for stretching this thread this far

I only want to know the proof for:

$LaTeX Code: \\nabla \\cdot\\frac{\\vec{r}}{r^{2}}$

With a little physical interpretation...

**dextercioby** · Feb26-05, 11:50 AM

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 $LaTeX Code: \\frac{\\vec{r}}{r^{3}}$ is the Green function for the Poisson equation for R^{3},it can be shown that,in fact:

$LaTeX Code: \\nabla\\cdot\\frac{-\\vec{r}}{r^{3}}=-4\\pi\\delta(\\vec{r})$

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

Daniel.

**kanato** · Feb26-05, 01:57 PM

$LaTeX Code: \\nabla \\cdot \\frac{\\hat{r}}{r^2} = <BR>\\frac{\\partial}{\\partial r} \\left( r^2 \\cdot \\frac{1}{r^2} \\right) = 0$

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

$LaTeX Code: \\int_V \\nabla \\cdot \\frac{\\hat{r}}{r^2} r^2 \\sin \\theta dr d\\theta d\\phi<BR>= \\oint_S \\frac{\\hat{r}}{r^2} \\cdot \\hat{r} r^2 \\sin \\theta d\\theta d\\phi<BR>= \\oint_S \\sin \\theta d\\theta d\\phi = 4\\pi<BR>$

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, $LaTeX Code: \\nabla \\cdot \\frac{\\hat{r}}{r^2} = 4 \\pi \\delta({\\vec{r}})$