Green function and the boundary conditions

[lost_boy]

Hello there,

I am glad that I found this forum. Because I have a little bit trouble with theoretical physics. :yuck:

The problem is the Green function in theoretical electrodynamic.
I try to understand the difference between the Dirichlet Condition and the Neumann Condition.
I understand the mathematic behind this theory, but If possible I would like to picture this stuff for better understanding.

Dirichlet Condition shows a picture with a mirror charge, a real charge and the zero potential between it. So it means G(r,r')=0 and r' is on the surface.

Neumann Condition means dG(r,r')/ dn'= 4pi/F with r' on the surface and F is the whole surface.

Now I can't understand what's the meaning of Neumann condition - not mathematicly but as a picture .
Maybe I misunderstood this theory completely.

I hope somebody knows what I mean and can help me.

thanks a lot

the lost boy


"Mathematics? What is it ? Can I eat it ?"

 

[ZapperZ]

Hello there,

I am glad that I found this forum. Because I have a little bit trouble with theoretical physics. :yuck:

The problem is the Green function in theoretical electrodynamic.
I try to understand the difference between the Dirichlet Condition and the Neumann Condition.
I understand the mathematic behind this theory, but If possible I would like to picture this stuff for better understanding.

Dirichlet Condition shows a picture with a mirror charge, a real charge and the zero potential between it. So it means G(r,r')=0 and r' is on the surface.

Neumann Condition means dG(r,r')/ dn'= 4pi/F with r' on the surface and F is the whole surface.

Now I can't understand what's the meaning of Neumann condition - not mathematicly but as a picture .
Maybe I misunderstood this theory completely.

I hope somebody knows what I mean and can help me.

OK,let's see if we can tackle this.

The Green's function method in electrodynamics allows you to write the "field" solution that depends only on the geometry of the problem, and not on the charge distribution. So if you have solved the Green's function, then if you introduce any kind of charge distribution, the problem can be easily solved.

What this means is that each charge distribution will give you a unique solution (up to an additive constant, of course). To be able to do that, you have to be told of what the "condition" is at all the boundaries. This condition can be either the exact values of the field at the boundary (the Dirichlet BC's), or the normal gradient of the field at the boundary (the Neumann BC's).

Notice that once the normal gradient is given at the boundary, then the field can then be written as an integral equation around the boundary. This may be easier to solve than Poisson's equation.

The moral of the story is that the field can be uniquely determined given either boundary conditions (and sometime, a combination of those two). It is always good to have more than 1 way to skin a cat. :)

Zz.

 

[lost_boy]

Good. If I understand it right the essence is:

The concept of the mirror charge is usefull for both conditions. But it depends on the problem which condition helps me to solve it easily.

So I can use one of it or both to eliminate an integral term and than I can solve the other integrals with the help of the mirror charge concept.

Hopefully the way I undestood is correctly?!
If not please try it once more

Thank you
lost boy

Ps: It is no homework, but I have my final exam in about a week


------------------------------------------
"Mathematics? What is it ? Can I eat it ?"

 

[ZapperZ]

Good. If I understand it right the essence is:

The concept of the mirror charge is usefull for both conditions. But it depends on the problem which condition helps me to solve it easily.

So I can use one of it or both to eliminate an integral term and than I can solve the other integrals with the help of the mirror charge concept.

Hopefully the way I undestood is correctly?!
If not please try it once more

Thank you
lost boy

Ps: It is no homework, but I have my final exam in about a week

If we're talking about the standard method of images problem of 1 charge (and an image charge), there you have two different ways of solving it. Let's say you have a charge q at (0,0,z1). Then your image charge -q is at (0,0,-z1). You can either use the fact that the electrostatic potential is zero at the x-y plane (Dirichlet) or that the normal gradient of the potential at the x-y plane (Neumann) if you're given this. Note that this is the E-field normal to this plane. Thus, you can solve the electrostatic potential either by knowing its value over the boundary, or the E-field values over that boundary.

Good luck with the final exam and kick some ass.

Zz.

 

[lost_boy]

I understand now, obviously it has been to easy for me. :tongue2:

Good luck with the final exam and kick some ass.

I will.

Thanks.

cu LB