If the distribution of the source charges does not go to zero at infinity (as in the case of an infinite line charge), can we still write the most general solution of Poisson's equation $$\nabla^2\phi({\vec r})=-\frac{\varrho({\vec r})}{\varepsilon_0}$$ as $$\phi({\vec r})=\frac{1}{4\pi\varepsilon_0}\int\frac{\varrho({\vec r}) d^3{\vec r}'}{|{\vec r}-{\vec r}'|}+\phi_0(\vec{r})$$ where $\nabla^2\phi_0(\vec{r})=0$? If not, what is the solution?
$\begingroup$
$\endgroup$
5
asked Feb 25, 2023 at 19:43
-
$\begingroup$ if $\phi_0(\mathbf r)$ solves $\nabla^2 \phi_0(\mathbf r)=0$ and $lim_{r \to \infty }\phi_0(\mathbf r) =K=\text{const}$ then $\phi_0(\mathbf r)=K$ everywhere. $\endgroup$– hyportnexCommented Feb 25, 2023 at 19:57
-
1$\begingroup$ What is the general solution then? $\endgroup$– SolidificationCommented Feb 25, 2023 at 20:00
-
$\begingroup$ the integral... $\endgroup$– hyportnexCommented Feb 25, 2023 at 20:00
-
1$\begingroup$ Even if $\varrho$ (and $\phi$) does not go to at infinity? $\endgroup$– SolidificationCommented Feb 25, 2023 at 20:05
-
$\begingroup$ I missed that part of your question... anyhow the integral still holds, because the equation $\nabla^2 \phi_0=0$ by definition is independent of the charge distribution. $\endgroup$– hyportnexCommented Feb 25, 2023 at 20:09
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
From linearity, you can always add a harmonic function to the solution of the Poisson equation. In fact, this is how you obtain all the solutions of the Poisson equation. This additive harmonic function can be made unique by adding appropriate boundary conditions "at infinity." The main issue is whether there exists an inhomogeneous solution and whether it is given by your integral.
You don't need the charge distribution to be compactly supported. The decay just needs to be fast enough for the integral to be well defined. However, if you want it to be a solution of the Poisson equation, you'll need a faster decay to apply derivation under the integral sign. For example, if $\rho=O(r^{-\alpha})$ with $\alpha > 4$, then you still have a solution ($\alpha>2$ is sufficient for the integral to make sense).
A classic pathological example is uniform charge density $\rho_0$ in whole space. This time, the integral clearly makes no sense, although it is easy to exhibit a solution say: $$ V = \frac{\rho_0}{2\epsilon_0}x^2 $$
Typically for examples like the line of charge/ plane of charge etc. you invoke symmetry arguments. This allows you to consider the analogous problem in fewer dimensions with a compact supported distribution of charge.
Physically, it is often natural that your actual distribution of charge is compactly supported. When you think of infinite distributions, you are essentially idealising the situation to get rid of edge effects. This means that the correct way to approach the problem is to do the integral with the large scale cutoff and then let the cutoff go to infinity. If the result you are interested in depends on the cutoff in this limit, then the edge effects are important and the infinite distribution idealisation is not appropriate.
Hope this helps.
