# Green's identity contradicts Helmholtz theorem

Let F be a vector field on a bounded domain $V \subset \mathbb{R}^3$, which is twice differentiable, let $S := \partial V$.

According to the Helmholtz Theorem, F can be decomposed, such that:

$$\mathbf{F} = -\nabla \Phi + \nabla \wedge \mathbf{A}$$ , where

$$\Phi (\mathbf{x}) = \frac{1}{4\pi} \int_V {\nabla ' \mathbf{F}(\mathbf{x} ') \over |\mathbf{x} - \mathbf{x}'|}dV' - \frac{1}{4 \pi} \oint_S {\langle \mathbf{n}', \mathbf{F}(\mathbf{x} ') \rangle \over | \mathbf{x} - \mathbf{x}' |}dS' \\$$ $$\mathbf{A} (\mathbf{x}) = \frac{1}{4\pi} \int_V {\nabla ' \wedge \mathbf{F}(\mathbf{x} ') \over |\mathbf{x} - \mathbf{x}'|}dV' - \frac{1}{4 \pi} \oint_S { \mathbf{n}' \wedge \mathbf{F}(\mathbf{x} ') \over | \mathbf{x} - \mathbf{x}' |}dS'$$

Applying this for instance to the static Maxwell's equations: $$\mathbf{E}(\mathbf{x} ) = -\nabla \Phi (\mathbf{x})$$ $$\nabla \mathbf{E}(\mathbf{x} ) = {\rho( \mathbf{x} ) \over \epsilon_0 }$$

we obtain:

$$\Phi (\mathbf{x}) = \frac{1}{4\pi\epsilon_0} \int_V { \rho(\mathbf{x} ') \over |\mathbf{x} - \mathbf{x}'|}dV' + \frac{1}{4 \pi} \oint_S \frac{\langle \mathbf{n}', \nabla ' \Phi(\mathbf{x} ') \rangle}{| \mathbf{x} - \mathbf{x}' |}dS' \\$$

Using Green's second identity, we can obtain a similiar result:

$$\Psi(\mathbf{x} ) = {1 \over 4\pi } \int_V { 1 \over | \mathbf{x} - \mathbf{x} ' |} \Delta \Psi(\mathbf{x} ') dV' + {1 \over 4\pi } \oint_{S} { \langle \mathbf{n '} (\mathbf{x} ' ), \nabla ' \Psi(\mathbf{x} ' ) \rangle \over | \mathbf{x} - \mathbf{x} ' | } dS' - {1 \over 4\pi } \oint_{S} \Psi(\mathbf{x} ') \langle \mathbf{n '} (\mathbf{x} ' ), \nabla ' { 1 \over | \mathbf{x} - \mathbf{x} ' | } \rangle dS'$$ $$\Psi(\mathbf{x} ) = {1 \over 4\pi\epsilon_0} \int_V { \rho(\mathbf{x} ') \over | \mathbf{x} - \mathbf{x} ' | } dV' + {1 \over 4\pi } \oint_{S} { \langle \mathbf{n '} (\mathbf{x} ' ), \nabla ' \Psi(\mathbf{x} ' ) \rangle \over | \mathbf{x} - \mathbf{x} ' | } dS' - {1 \over 4\pi } \oint_{S} \Psi(\mathbf{x} ') \langle \mathbf{n '} (\mathbf{x} ' ), \nabla ' { 1 \over | \mathbf{x} - \mathbf{x} ' | } \rangle dS'$$

I'm bothered by the additional term in the second variant.

At the first glance the different variants do not have to contradict since the potential of a field is not uniquely determined without additional conditions imposed. But for a given field $E: V\subset \mathbb{R}^3 \rightarrow \mathbb{R}^3$, $E = - \nabla \Phi = - \nabla \Psi$ implies that $$0 = {1 \over 4\pi } \nabla \oint_{S} \Psi(\mathbf{x} ') \langle \mathbf{n '} (\mathbf{x} ' ), \nabla ' { 1 \over | \mathbf{x} - \mathbf{x} ' | } \rangle dS'$$

However, this does not seem to be true in general (take a potential that is constant ($\neq 0$) on $\partial V$ for instance).

What am I missing here?

• Wow: very interesting formulation of the theorem: therefore does it hold for any $\mathbf{F}\in C^2(\bar{V})$ with $V$ bounded? I've been able to prove it to myself under the restrictive assumption that $\mathbf{F}$ is supported within $\mathring{V}$... Commented Mar 14, 2016 at 19:33
• ... Have you got a link to a rigourous and clear proof (or would you like to write one here)? I find the Wikipedia's terribly unclear: $\int_V\mathbf{F}(\mathbf{r}')\left(-\frac{1}{4\pi}\nabla^2\frac{1}{\left| \mathbf{r} -\mathbf{r}'\right|}\right)\mathrm{d}V'$ written as the other integrals (which I think to be Lebesgue/limits of Riemann integrals) while it's a symbolic notation for the Laplacian of a distribution, unexplained commutations between $\nabla$ and $\int$... I $\infty$-ly thank you in any case! Commented Mar 14, 2016 at 19:33

True. In the Helmholtz decomposition $\mathbf{F} = -\nabla \Phi + \nabla \wedge \mathbf{A}$ we can add any harmonic function to $\Phi$, because the gradient of a harmonic function is a solenoidal field, which can be absorbed by the term $\nabla \wedge \mathbf{A}$.
So, when you decompose $\mathbf{E}(\mathbf{x} ) = -\nabla \Phi (\mathbf{x})$, you recover $\Phi$ up to a harmonic function. And the term that bothers you, $$\oint_{S} \Psi(\mathbf{x} ') \left \langle \mathbf{n '} (\mathbf{x} ' ), \nabla ' { 1 \over | \mathbf{x} - \mathbf{x} ' | } \right \rangle dS'$$ is harmonic in $V$.
$$0 = {1 \over 4\pi } \Delta \oint_{S} \Psi(\mathbf{x} ') \left\langle \mathbf{n '} (\mathbf{x} ' ), \nabla ' { 1 \over | \mathbf{x} - \mathbf{x} ' | } \right \rangle dS'$$