Reading through Jackson: Gauss Theorem

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SUMMARY

The discussion focuses on the application of Gauss's theorem as presented in Jackson's "Classical Electrodynamics." Specifically, it addresses the equation ∇'^{2}G=-4πδ(x-x') and its relation to the surface integral ∮(∂G/∂n')da' = -4π. The Green function G is defined as 1/|x-x'| + F, where F has a zero Laplacian. The conversation highlights the connection between the surface integral and Gauss's law, confirming that the author is indeed applying Gauss's theorem in this context.

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  • Knowledge of vector calculus, including divergence and surface integrals.
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thelonious
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Homework Statement



I'm reading through Jackson and ran into the following:

An application of Gauss's theorem to

∇'^{2}G=-4πδ(x-x')

shows that

\oint(\partialG/\partialn')da'= -4∏

where G is a Green function given by 1/|x-x'| + F, and F is a function whose Laplacian is zero.

(Sec. 1.10, Formal Solution of Electrostaic Boundary Value Problem)

Homework Equations



Divergence theorem?
Gauss's theorem?

The Attempt at a Solution



I don't see how to arrive at the surface integral. This looks a bit like an application of the divergence theorem because of the surface integral term. It also looks something like Gauss's law in differential form. Is this what the author means by applying Gauss's theorem?
 
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Nothing too mysterious going on. Gauss's law tells us
$$\int_V \nabla\cdot (\nabla G)\,dv = \oint_S [(\nabla G)\cdot\hat{n}]\,dS$$ The integrand of the surface integral is simply the directional derivative in the ##\hat{n}## direction, which is equal to ∂G/∂n, where n is the coordinate along the direction of ##\hat{n}##.
 
Thanks -- what was I thinking... G is a 1/r potential...
 

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