I was recently studying Jackson Electrodynamics and faced some issues directly. I have studied Griffiths Electrodynamics and I knew about the uniqueness Theorem from it. But in Jackson, they proved it in a very rigorously. I liked it very much but it didn't provide an detailed explanation of it.
The book starts off by using Green's first theorem $$\oint{[\phi\nabla^2\psi - \nabla\phi\cdot\nabla\psi]d^3x} = \oint{\phi\frac{\partial{\psi}}{\partial{n}}da}$$ Where $n$ is the normal vector the surface.
Then the book says that, let's assume two solutions to the Poisson equations $\Phi_1$ and $\Phi_2$ and take a third function $U$ such that $$U = \Phi_1 - \Phi_2$$ Then since $\Phi_1$ and $\Psi_2$ follows the poissons equation.
$$\nabla^2 U = \nabla^2\Phi_1 - \nabla^2\Phi_2$$ This means that, $$\nabla^2 U = 0$$ By substituting $\phi = U$ and $\psi = U$.
Such that Green's first theorem gives $$\oint{|\nabla U|^2 d^3x'} = \oint{U\frac{\partial{U}}{\partial{n}}da'}$$ Here there are 2 types of Boundary conditions, namely Dirichlets and Neumann boundary conditions. Dirichlet specifies the Potential on the surface while Neumann specifies the Electric field. Since on the surface $U = 0$ $$\oint{|\nabla U|^2 d^3x'} = 0$$ And $$\nabla U = 0$$ Thus, $ U $ is constant inside and thus is 0. From above, $$\Phi_1 = \Phi_2$$ I have some major and minor questions regarding this proof.

  • In Griffiths, There is a second uniqueness theorem, which states that: ** In a volume $V$ surrounded by conductors and containing a specified charge density ρ, the electric field is uniquely determined if the total charge on each conductor is given (Fig. 3.6). (Another conductor can bound the region as a whole ductor, or else unbounded.)**

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  • Does the proof using Green's Theorem prove this case as well? Can you show me how?

I tried doing it myself, and eventually, I used the same method as above and got the result by only applying the boundary conditions of the outmost boundary. Where am I wrong? It will be really helpful. I really appreciate any help you can provide.

  • $\begingroup$ note that the holes where $Q_a, Q_b$ etc. reside are the exterior of the volume V and therefore the same principles apply, specially having in mind that if the holes are conductors all the charge is in the surface of them and therefore the boundary conditions over $\nabla U$ or $\Phi_{1,2}$ just change by the corresponding surface charge distributions. $\endgroup$ Commented May 7 at 15:27

1 Answer 1


The key equation inwproving the uniqueness theory is \begin{equation} \int_V|{\bf\cal E}|^2d\tau =-\sum_i\oint_{S_i}\psi{\bf\cal E}\cdot{\bf dA}, \end{equation} where $\psi$ and ${\bf\cal E}$ are the differences between two possible potentials or fields. If $\phi$ is given on any conductor, the integral will be zero. If $Q$ is given, then $\psi\oint{\bf\cal E}\cdot{\bf dA}=4\pi\psi(\Delta Q)=0,$ since $\psi$ is constant for a conductor.


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