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Mexicorn
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This is actually a question pertaining to a paper I'm trying to understand (PRB 73, 115407 (2006)), but I decided to put it here just to be safe.
The paper I'm reading involves starting with an electrostatic energy contribution, and rewriting it with a green's function solution to the Poisson equation rather than the standard 1/r form to incorporate a dielectric. The step I'm missing, though, is in how the author formulates the energy contribution at first.
The electrostatic energy contribution to the total energy is described like so:
-\int \frac{\epsilon(r)}{8\pi}\left|\nabla V(r)\right|^{2}d^{3}r +\int \rho(r) V(r)d^{3}r.
After incorporating the green's function solution (which is not relevant to my holdup), the term becomes the more familiar (to me at least):
\frac{1}{2}\int \rho(r) V(r)d^{3}r. (the sign on this term is ambiguous as it is defined as positive in the paper and negative in an online formulation written by the same author)
Starting from the top equation..
-\int \frac{\epsilon(r)}{8\pi}\left|\nabla V(r)\right|^{2}d^{3}r +\int \rho(r) V(r)d^{3}r
and plugging in the identity: \textbf{E} = - \nabla V(r), I am left with
\int \frac{\epsilon(r)}{8\pi}\nabla V(r) \cdot \textbf{E} d^{3}r+\int \rho(r) V(r)d^{3}r <br />
Then using Gauss' theorem yields:
\int \frac{\epsilon(r)}{8\pi}\nabla \cdot (V(r)\textbf{E}) d^{3}r-\int \frac{\epsilon(r)}{8\pi} V(r) (\nabla \cdot \textbf{E}) d^{3}r+\int \rho(r) V(r)d^{3}r <br />
By the Divergence theorem, the first term is equivalent to \oint (V(r) \textbf{E}) \cdot d\textbf{A} which is zero under the assumption that the potential vanishes at infinity. The last step is where I am struggling. At first I thought I would merely plug in the identity \nabla \cdot \textbf{E} = 4\pi \rho, but this still leaves me with a dangling permittivity within the integral. I tried going through everything again substituting in the electric displacement instead of the field, but then my last step will have a \rho_f instead of the total charge.
Anyone know where I'm going wrong with this?
Homework Statement
The paper I'm reading involves starting with an electrostatic energy contribution, and rewriting it with a green's function solution to the Poisson equation rather than the standard 1/r form to incorporate a dielectric. The step I'm missing, though, is in how the author formulates the energy contribution at first.
Homework Equations
The electrostatic energy contribution to the total energy is described like so:
-\int \frac{\epsilon(r)}{8\pi}\left|\nabla V(r)\right|^{2}d^{3}r +\int \rho(r) V(r)d^{3}r.
After incorporating the green's function solution (which is not relevant to my holdup), the term becomes the more familiar (to me at least):
\frac{1}{2}\int \rho(r) V(r)d^{3}r. (the sign on this term is ambiguous as it is defined as positive in the paper and negative in an online formulation written by the same author)
The Attempt at a Solution
Starting from the top equation..
-\int \frac{\epsilon(r)}{8\pi}\left|\nabla V(r)\right|^{2}d^{3}r +\int \rho(r) V(r)d^{3}r
and plugging in the identity: \textbf{E} = - \nabla V(r), I am left with
\int \frac{\epsilon(r)}{8\pi}\nabla V(r) \cdot \textbf{E} d^{3}r+\int \rho(r) V(r)d^{3}r <br />
Then using Gauss' theorem yields:
\int \frac{\epsilon(r)}{8\pi}\nabla \cdot (V(r)\textbf{E}) d^{3}r-\int \frac{\epsilon(r)}{8\pi} V(r) (\nabla \cdot \textbf{E}) d^{3}r+\int \rho(r) V(r)d^{3}r <br />
By the Divergence theorem, the first term is equivalent to \oint (V(r) \textbf{E}) \cdot d\textbf{A} which is zero under the assumption that the potential vanishes at infinity. The last step is where I am struggling. At first I thought I would merely plug in the identity \nabla \cdot \textbf{E} = 4\pi \rho, but this still leaves me with a dangling permittivity within the integral. I tried going through everything again substituting in the electric displacement instead of the field, but then my last step will have a \rho_f instead of the total charge.
Anyone know where I'm going wrong with this?
