Equation with Dirac delta

[vermax]

Hi guys.

I play now a bit with EM fields and I have encountered some problems connected with Dirac delta. By coincidence I visited this forum and I thought I could find some help in here. Surely i have read these topics:

http://www.scienceforums.net/forum/showthread.php?t=35186&highlight=dirac+delta

http://www.scienceforums.net/forum/showthread.php?t=35132&highlight=dirac+delta

That gave me a lot but i still do not feel it. The problem is that in order to get a potential in some point from a single charge you need to just solve such thing:

[math]\square \phi = - \frac{ \rho}{\epsilon} [/math]

and there by the way you need to use such equation:

[math]\nabla \frac{ 1}{ | \vec{r} - \vec{r_0} |} = - 4 \pi \delta ( \vec{r} - \vec{r_0})[/math]

 

I tried to find where it comes from, but the only thing I've found is this:

http://www.fen.bilkent.edu.tr/~ercelebi/mp03.pdf

But I do not get it at all

I would appreciate if someone could just clear this out for me or present his own theory.

 

PS. Sorry for my mistakes or improper names for some mathematical or physical stuff but I am from Poland

[Bignose]

http://mathworld.wolfram.com/DeltaFunction.html

 

The delta function is there to mathematically represent a point-source. It is mathematical abstraction, because there is no such thing as a point-source, but very often the mathematical results computed using a point-source are exceptionally close to reality so they are used quite a bit.

 

It's biggest property is that at the value of x in the delta function that the input of the function is equal to zero, the function has a value of [math]+ \infty[/math]. Everywhere else it has a value of 0.

 

The really useful part of this come from the fact that when you integrate over the point where its input is zero, the value of that integral becomes equal to 1.

 

So. [math]\delta(x-y)=0[/math] everywhere except when x=y, then [math]\delta(0)=+\infty[/math]. And the integral property:

 

[math]\int_\Omega \delta(x) dx = 1 [/math] if x is in the domain [math]\Omega[/math]. If x isn't, then the integral evaluates to zero.

 

This integral value makes the Dirac delta function a selector of sorts because the integrand only has a value at one point.

 

e.g. [math]\int^{+\pi}_{-\pi} \sin x \delta(x-0.2) dx = \sin 0.2[/math]

 

Even though you are integrating from [math]-\pi[/math] to [math]+\pi[/math], the only point where the integrand is non-zero is at 0.2 as selected by the delta function, so the integrand less the delta function itself is the value of the integral so long as the selected value is in the domain of the integration.

 

So, in EM, if there is a point source of charge, this is represented by a delta function, as the delta function is the mathematical representation of the charge. So, when you integrate over the entire domain, there is no sources except that 1 point, and that's what comes out.

[vermax]

Thank you for your reply. I was familiar with the whole theory about Dirac delta function, but whole your text especially that example with sinus cleared things out for me perfectly now

 

But if you would be kind enough I would need some more help in proving that equation:

[math]

\nabla \frac{ 1}{ | \vec{r} - \vec{r_0} |} = - 4 \pi \delta ( \vec{r} - \vec{r_0})

[/math]

 

I have just figured out all that stuff from the link I have posted above and I think the method to achieve this:

[math] \delta( \vec{r} - \vec{r_0})= \frac{1}{4 \pi r^2 }\delta ( r - r_0) [/math]

is quite understandable now for me.

 

But according to the excersie I have:

 

[math]

\nabla \frac{ 1}{ | \vec{r} - \vec{r_0} |} = - \frac{1}{r^2} \delta ( r - r_0)

[/math]

And I have just thought that if I counted this in spherical coordinates, gradient of my function 1/r should be also presented in that form. If thats ok I would have:

[math]-\frac{1}{ | \vec{r} - \vec{r_0} |^2}= - \frac{1}{r^2} \delta ( r - r_0) [/math]

And if I integrate both sides I get:

[math]\frac{1}{ | \vec{r_0} - \vec{r} |}= \frac{1}{r_0^2} [/math]

 

So rather it is not equal? :/

Uh I am confused with this...

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Merged post follows:

Consecutive posts merged

Uh I am sorry, dunno why but I write everywhere gradient, where there should be laplasian. As far as I see it will change my last equation so that on the left side we will have 1/(r0-r)^2... Sorry one more time.