7
$\begingroup$

I am having some trouble with Green functions in electrostatics

What is the meaning of this trick:

Given $$\vec{\nabla}^2 V(\vec{r}) = \frac{-1}{\varepsilon_0}\rho(\vec{r}) = \frac{-1}{\varepsilon_0} \int \delta ^3(\vec{r}-\vec{r}')\rho(\vec{r}') d^3 \vec{r}' = \frac{1}{4 \pi \varepsilon_0} \int \vec{\nabla}^2G(\vec{r},\vec{r}')\rho(\vec{r}') d^3 \vec{r}'.$$

I see $$\vec{\nabla}^2 V(\vec{r}) = \vec{\nabla}^2\left[\frac{1}{4 \pi \varepsilon_0} \int G(\vec{r},\vec{r}')\rho(\vec{r}') d^3 \vec{r}'\right]$$ and so $$V(\vec{r}) = \frac{1}{4 \pi \varepsilon_0} \int_V G(\vec{r},\vec{r}')\rho(\vec{r}') d^3 \vec{r}'$$ where $G$ is a greens function satisfying $$\vec{\nabla}^2 G(\vec{r},\vec{r}') = - 4 \pi \delta(\vec{r} - \vec{r}').$$

Why is this not a solution? It seems like it is, and further it is general (applies to elliptic, parabolic, hyperbolic), the problem is specifying boundary conditions. If this is a solution it should turn into the following in some obvious easy way:

Using Greens identities I can show

$$V(\vec{r}) = \frac{1}{4 \pi \varepsilon_0} \int_VG(\vec{r},\vec{r}')\rho(\vec{r}') d^3 \vec{r}' + \frac{1}{4 \pi}\int_{S}\left[G(\vec{r},\vec{r}')\frac{\partial V}{\partial n'} - V(\vec{r})\frac{\partial G}{\partial n'} \ \right] \ dS'$$

Why are the extra terms missing from the first solution I gave?

It is said that this is not a solution because it over-specifies the boundary conditions, but once we specify Dirichlet or Neumann it does become one. What is the intuition for this statement though, would not over-specification only help us? Jackson does not make sense to me on this point.

$\endgroup$
  • $\begingroup$ Could you please add how exactly you used Green's identities? $\endgroup$ – Frederic Brünner May 7 '14 at 17:04
  • $\begingroup$ To save writing it out, it's just the same idea as here pdf physics.bu.edu/~pankajm/LN/LNGF2.pdf or Jackson 1.8 - 1.10 $\endgroup$ – bobby May 7 '14 at 17:39
  • 1
    $\begingroup$ @bobby overspecification means that you've actually given so many constraints on the solution that there isn't anything that solves it. If you specify $\phi$ on the boundary then $\frac{\partial\phi}{\partial n}$ is already determined by the Laplace equation, so unless you happen to guess that exact form in your Neumann boundary condition your specification of the value and normal derivative on the boundary will be incompatible. $\endgroup$ – Robert Mastragostino May 7 '14 at 19:37
  • $\begingroup$ Thanks, so the idea is: take $\frac{d^2}{dx^2}V(x) = 1$ then $V' = x + a$ and $V = \frac{x^2}{2} + ax + b$. If $V(0) = 0$ and $V(1) = 1$ then $b = 0$ and $a = \frac{1}{2}$ so $V = \frac{x^2}{2}+\frac{x}{2}$ implying $V' = x + \frac{1}{2}$. If I try to find a solution $V$ to $V'' = 0$ satisfying the Dirichlet conditions $V(0) = 0$, $V(1) = 1$ AND the Neumann boundary conditions $V'(0) = 6$, $V'(1) = 10$ I can't do it. Including, e.g., $\frac{\partial V}{\partial n'}$ in the eq means specifying Neumann boundary conditions so to avoid it we choose $G$ so that $G = 0$ on $S$ to get rid of it. $\endgroup$ – bobby May 7 '14 at 20:26
  • $\begingroup$ Any idea as to why the first method gives one apparent solution while the other method gives another? $\endgroup$ – bobby May 7 '14 at 20:27
2
$\begingroup$

In the first method (the method of the Green function) you have removed the Laplacian operator from both sides of the equation. That is not a warranted operation in general case, for the potential $V$ and the integral of $\rho$ over the domain considered may differ by some non-vanishing function $u(x,y,z)$ that obeys $$ \Delta u = 0. $$ Only in special case $u$ may actually vanish. When some condition on the solution $V$ on the boundary of the region is specified, the function $u$ may not vanish.

$\endgroup$
  • $\begingroup$ Indeed, $\nabla^2 u = \nabla^2 v$ is not equivalent to $u = v$. $u$ could be a harmonic function and $v$ identically zero. $\endgroup$ – Arthur Suvorov Jun 16 '14 at 23:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.