Neumann Boundary Conditions question

[VortexLattice]

So I'm reading through Jackson's Electrodynamics book (page 39, 3rd edition), and they're covering the part about Green's theorem, where you have both [itex]\Phi[/itex] and [itex]\frac{\delta \Phi}{\delta n}[/itex] in the surface integral, so we often use either Dirichlet or Neumann BC's to eliminate one of them.

So for Dirichlet, we simply say G(x,x') = 0 and the integral simplifies a little.

But for Neumann, he says the obvious choice is to use [itex]\frac{\delta G}{\delta n} = 0[/itex] because that eliminates that part of the integral. But then he says, wait, we can't, because applying Gauss' theorem to [itex]\nabla ^2 G(x, x') = -4\pi \delta(x - x')[/itex] shows that [itex]\oint \frac{\delta G}{\delta n'} da' = -4\pi [/itex] (over the surface S), so we have to choose [itex]G(x,x') = \frac{-4\pi}{S}[/itex], where S is the total surface area of the surface S.

Why? I don't really get what he means by 'applying Gauss' theorem'... I know what the theorem is but don't see how to use it or why we can't just choose that dG/dn = 0 here.

Thanks!

 

[torquil]

So I'm reading through Jackson's Electrodynamics book (page 39, 3rd edition), and they're covering the part about Green's theorem, where you have both [itex]\Phi[/itex] and [itex]\frac{\delta \Phi}{\delta n}[/itex] in the surface integral, so we often use either Dirichlet or Neumann BC's to eliminate one of them.

So for Dirichlet, we simply say G(x,x') = 0 and the integral simplifies a little.

But for Neumann, he says the obvious choice is to use [itex]\frac{\delta G}{\delta n} = 0[/itex] because that eliminates that part of the integral. But then he says, wait, we can't, because applying Gauss' theorem to [itex]\nabla ^2 G(x, x') = -4\pi \delta(x - x')[/itex] shows that [itex]\oint \frac{\delta G}{\delta n'} da' = -4\pi [/itex] (over the surface S), so we have to choose [itex]G(x,x') = \frac{-4\pi}{S}[/itex], where S is the total surface area of the surface S.

Why? I don't really get what he means by 'applying Gauss' theorem'... I know what the theorem is but don't see how to use it or why we can't just choose that dG/dn = 0 here.

I haven't checked it out, but I have a hunch about it, so I'll take a stab at a general consideration: If you have already proven that Gauss' law is necessarily valid for the solution, but then select a boundary condition that breaks that law, it would mean that you end up with a differential equation that has no solution, of a solution that breaks with some of the assumptions you made when you determined that Gauss' law should hold. So you could say that Gauss law enforces a restriction on the allowed boundary conditions. I didn't look it up, though, since I don't have the time right now.

 

[Meir Achuz]

Gauss's law restricts the total surface integral of E.dS to equal 4 pi Q, where Q is the total charge enclosed. This puts a constraint on [tex]\partial_n\phi[/tex].
But Jackson's solution for that is wrong. The correct treatment is given on pp. 66-67 of Franklin "Classical Electromagnetism".

 

[Bill_K]

Can you explain what's wrong with Jackson's treatment, and how to correct it?

 

[Meir Achuz]

It is in that textbook, but I will try to summarize it.
The integral of [tex]\partial_n[/tex] is constrained by Gauss to equal [tex]4\pi Q[/tex].
If it does not there is no solution. An example is in heat flow. If the integral does not equal zero, then the temperature must rise and Laplace's equation will not be satisfied.
Jackson's addition will not help this. If any part of the boundary is Dirichlet, that surface will have the right E to give the right total surface integral. If the problem specifies that phi--> 0 at infiniity, then the surface integral of E at infinity will satisfy Gauss.

 

[chrisbaird]

By "applying Gauss' theorem", Jackson means you start with ∇²G(x,x′)=−4πδ(x−x′), integrate both sides over a volume V, and apply the divergence theorem to the left side to get a surface integral. This is the same thing you do in Freshman physics with the electric field, whereas here you have the Green's function.