Potential theory
In mechanics we could calculate a vector field from a scalar potential: $$\myv{F}_g = -\myv \grad U_g.$$
Under what circumstances can this happen? This is the question that caught Herman von Helmholtz' (1821-1894) imagination.
Helmholtz theorem
Consider a vector function $\myv F(\myv r)$,
that has a divergence... $$\myv \grad \cdot \myv F(\myv r) \equiv D(\myv r),$$
and a curl... $$\myv \grad \times \myv F(\myv r) \equiv \myv C(\myv r).$$
Helmholtz found that any vector function, $\myv F(\myv r)$, which is...
- sufficiently smooth,
- and "rapidly" decaying (vanishes faster than $1/r$ as $r\to\infty$),
can be written as the sum of 2 potential pieces...
$$\Rightarrow \myv F = -\myv \grad U + \myv \grad \times \myv W,$$
where, $U(\myv r)$ is a scalar potential function, and $\myv W(\myv r)$ is a vector potential function.
$U(\myv r)$ can always be constructed as... $$U(\myv r) \equiv \frac{1}{4\pi} \int \frac{D(\myv r ')}{\rr} d \tau '.$$
And $\myv W(\myv r)$ can be, constructed as... $$\myv W (\myv r) \equiv \frac{1}{4\pi} \int \frac{\myv C (\myv r ')}{\rr} d \tau ' .$$
Curl-less fields
$D(\myv r) \neq 0$ but $\myv C(\myv r)=0$ everywhere...
We have just seen that $\myv E$ has divergence, but its curl vanishes. All of these characteristics are true of such curl-less (or irrotational) fields:
- $\myv \grad \times \myv E=0$ everywhere,
- $\int_{\myv a}^{\myv b} \myv E \cdot d \myv l$ is path independent,
- $\oint \myv E\cdot d \myv{l}=0$ for any closed loop.
The new thing that Helmholtz' theorem tells us is that any of these also imply that $\myv E$ is the gradient of a scalar function: $$\Rightarrow \myv E = - \myv \grad V.$$
Divergence-less fields
$D(\myv r) = 0$ everywhere, but $\myv C(\myv r)\neq 0$.
We shall see before long that the magnetic field $\myv B$ has a non-vanishing curl, but its divergence vanishes. All of these characteristics are true of such divergence-less (solenoidal) fields:
- $\myv \grad \cdot \myv B = 0$ everywhere,
- $\int_{\cal S}\myv B \cdot d \myv a$ is independent of surface, for any given boundary line,
- $\oint_{\cal S}\myv B \cdot d \myv a$ = 0 for any closed surface.
The new thing that Helmholtz' theorem tells is that any of these also imply that $\myv B$ is the curl of a vector function: $$\Rightarrow \myv B = \myv \grad \times \myv A.$$
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