Helmholtz's Theorem

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $R$ be a region of ordinary space.

Let $\mathbf V$ be a vector field over $R$.

Let $\mathbf V$ be both non-conservative and non-solenoidal.

Then $\mathbf V$ can be decomposed into the sum of $2$ vector fields:

one being conservative, with scalar potential $S$, but not solenoidal
one being solenoidal, with vector potential $\mathbf A$, but not conservative.

Thus $\mathbf V$ satifies the partial differential equations:

\(\text {(1)}: \quad\) \(\ds \operatorname {div} \mathbf V = \operatorname {div} \grad S\) \(=\) \(\ds \nabla^2 S \ne 0\)
\(\text {(2)}: \quad\) \(\ds \curl \mathbf V = \curl \curl \mathbf A\) \(=\) \(\ds \nabla^2 \mathbf A \ne \bszero\)

where:

$\operatorname {div}$ denotes the divergence operator
$\grad$ denotes the gradient operator
$\curl$ denotes the curl operator
$\nabla^2$ denotes the Laplacian.


Proof

Let us write:

$\mathbf V = \grad S + \curl \mathbf A$

where:

$S$ is a scalar field
$\mathbf A$ is a vector field chosen to be solenoidal



Then:

\(\ds \operatorname {div} \mathbf V\) \(=\) \(\ds \operatorname {div} \grad S + \operatorname {div} \curl \mathbf A\)
\(\ds \) \(=\) \(\ds \operatorname {div} \grad S\) Divergence of Curl is Zero
\(\ds \) \(=\) \(\ds \nabla^2 S\) Laplacian on Scalar Field is Divergence of Gradient
\(\ds \) \(\ne\) \(\ds 0\) as $\mathbf V$ is not solenoidal

and:

\(\ds \curl \mathbf V\) \(=\) \(\ds \curl \grad S + \curl \curl \mathbf A\)
\(\ds \) \(=\) \(\ds \curl \curl \mathbf A\) Curl of Gradient is Zero
\(\ds \) \(=\) \(\ds \grad \operatorname {div} \mathbf A - \nabla^2 \mathbf A\) Curl of Curl is Gradient of Divergence minus Laplacian
\(\ds \) \(=\) \(\ds -\nabla^2 \mathbf A\) as $\mathbf A$ is solenoidal: $\operatorname {div} \mathbf A = 0$
\(\ds \) \(\ne\) \(\ds \bszero\) as $\mathbf A$ is not conservative




$\blacksquare$


Source of Name

This entry was named for Hermann Ludwig Ferdinand von Helmholtz.


Sources