 12
 6
 Homework Statement
 If ##\phi## is the potential due to a volumecharge density ##\rho##within a volume V and a surfacecharge density ##\sigma##on the conducting surface S bounding the volume V, while ##\phi '## is the potential due to another charge distribution ##\rho##and##\rho'##, prove Green's reciprocity theorem.
 Homework Equations

Green's reciprocity theorem:
## \int_V \rho \phi' dV + \int_S \sigma \phi' dS = \int_V \rho' \phi dV + \int_S \sigma' \phi dS ##
This is Jackson's 3rd edition 1.12 problem.
So, for both ## \phi ## and ## \phi' ##, I started from Green's second identity:
## \int_V ( \phi \nabla^2 \phi'  \phi' \nabla^2 \phi )dV = \int_S ( \phi \frac {\partial \phi'} {\partial n}  \phi' \frac {\partial \phi} {\partial n} ) dS ##
And used Poisson's equation ## \nabla^2 \phi =  \frac \rho \epsilon ## and Gauss's law and ## (E =  \nabla \phi) ## to get the relation ## \frac {\partial \phi} {\partial n} =  \frac \sigma \epsilon ## between the surface charge density and the electric potential, which resulted in:
## \int_V \rho \phi' dV  \int_S \sigma \phi' dS = \int_V \rho' \phi dV  \int_S \sigma' \phi dS ##
So this is where I am stuck. I have checked many solution's on the internet and they say that the key is to realize that the normals ##n## in Gauss's law equations are pointing out of the conductor (actually, Jackson says: "where ## \frac {\partial} {\partial n} ## is the normal derivative at the surface S directed outwards from inside the volume V"), whereas the normals in Green's theorem are pointing into the conductor, so you can flip the  signs to end with the identity you want to prove. This is how I picture this (not sure is this correct):
Why is this the case? This made me realize that I don't understand how to derive Green's identity as I don't get why we are considering the normals in that direction. And, if my drawing is correct, we are only changing the sign inside the conductor where ## E = 0 ##, so why does this even matter?
And finally, what is the physical meaning of this theorem? I see a relationship between the work ## W = \frac 1 2 \int \rho \phi dV ## which gets you the energy required to assemble charge distribution ##\rho## in an electric potential ##\phi##, so I read Green's reciprocity identity as "the energy required to assemble a system made of a charge distribution ##\rho## and ##\sigma## in a potential produced by ##\rho'## and ##\sigma'## is the same energy required to assemble a system made of a charge distribution ##\rho'## and ##\sigma'## in a potential produced by ##\rho## and ##\sigma##" . Is my understanding correct?
So, for both ## \phi ## and ## \phi' ##, I started from Green's second identity:
## \int_V ( \phi \nabla^2 \phi'  \phi' \nabla^2 \phi )dV = \int_S ( \phi \frac {\partial \phi'} {\partial n}  \phi' \frac {\partial \phi} {\partial n} ) dS ##
And used Poisson's equation ## \nabla^2 \phi =  \frac \rho \epsilon ## and Gauss's law and ## (E =  \nabla \phi) ## to get the relation ## \frac {\partial \phi} {\partial n} =  \frac \sigma \epsilon ## between the surface charge density and the electric potential, which resulted in:
## \int_V \rho \phi' dV  \int_S \sigma \phi' dS = \int_V \rho' \phi dV  \int_S \sigma' \phi dS ##
So this is where I am stuck. I have checked many solution's on the internet and they say that the key is to realize that the normals ##n## in Gauss's law equations are pointing out of the conductor (actually, Jackson says: "where ## \frac {\partial} {\partial n} ## is the normal derivative at the surface S directed outwards from inside the volume V"), whereas the normals in Green's theorem are pointing into the conductor, so you can flip the  signs to end with the identity you want to prove. This is how I picture this (not sure is this correct):
Why is this the case? This made me realize that I don't understand how to derive Green's identity as I don't get why we are considering the normals in that direction. And, if my drawing is correct, we are only changing the sign inside the conductor where ## E = 0 ##, so why does this even matter?
And finally, what is the physical meaning of this theorem? I see a relationship between the work ## W = \frac 1 2 \int \rho \phi dV ## which gets you the energy required to assemble charge distribution ##\rho## in an electric potential ##\phi##, so I read Green's reciprocity identity as "the energy required to assemble a system made of a charge distribution ##\rho## and ##\sigma## in a potential produced by ##\rho'## and ##\sigma'## is the same energy required to assemble a system made of a charge distribution ##\rho'## and ##\sigma'## in a potential produced by ##\rho## and ##\sigma##" . Is my understanding correct?