Helmholtz decomposition

In vector analysis, the Helmholtz decomposition of a vector field on $$ \mathbb{R}^3$$ is the decomposition of the vector field into two vector fields, one a divergence-free field and one a curl-free field. The decomposition is called after the German physiologist and physicist Hermann von Helmholtz (1821 – 1894).

Mathematical formulation
The Helmholtz decomposition may be formulated as follows. Any vector field F(r) that is sufficiently often differentiable and vanishes sufficiently fast at infinity can be written as,
 * $$ \mathbf{F} = \boldsymbol{\nabla}\times \mathbf{A} -\boldsymbol{\nabla}\Phi = \mathbf{F}_\perp(\mathbf{r})+\mathbf{F}_\parallel(\mathbf{r}) $$

with
 * $$ \begin{align} \mathbf{A}(\mathbf{r}) &= \frac{1}{4\pi} \int \frac{\boldsymbol{\nabla}'\times \mathbf{F}(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} d^3\mathbf{r}' \quad\hbox{and}\quad \mathbf{F}_\perp(\mathbf{r}) = \boldsymbol{\nabla}\times \mathbf{A}\\ \Phi(\mathbf{r}) & = \frac{1}{4\pi}  \int \frac{\boldsymbol{\nabla}'\cdot \mathbf{F}(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} d^3\mathbf{r}' \quad\hbox{and}\quad  \mathbf{F}_\parallel(\mathbf{r}) =  -\boldsymbol{\nabla}\Phi \\ \end{align} $$

The primed nabla operator &nabla;' acts on primed coordinates and the unprimed &nabla; acts on unprimed coordinates.

Note that
 * $$ \boldsymbol{\nabla}\cdot(\boldsymbol{\nabla} \times \mathbf{V}) = 0\quad\hbox{and}\quad \boldsymbol{\nabla}\times (\boldsymbol{\nabla} \Psi) = 0 $$

holds for any vector field V(r) and any scalar function &Psi;(r). Hence it follows that the first term of F is divergence-free and the second curl-free.

As a corollary follows that the specification of both the divergence and the curl of a vector field at all points of space gives the field uniquely.

A well-known example of a Helmholtz decomposition is the following form of the electric field E,
 * $$ \mathbf{E}(\mathbf{r}) = -\dot{\mathbf{A}}(\mathbf{r}) -\boldsymbol{\nabla}\Phi(\mathbf{r}), $$

where &Phi; is the electric potential and A is the (magnetic) vector potential. The dot indicates a derivative with respect to time.

Decomposition in transverse and longitudinal components
Above it was stated that a vector field F(r) with $$\mathbf{r} \in \mathbb{R}^3$$ can be decomposed in a transverse $$\scriptstyle\mathbf{F}_\perp(\mathbf{r})$$ and longitudinal component $$\scriptstyle\mathbf{F}_\parallel(\mathbf{r})$$:


 * $$ \mathbf{F}(\mathbf{r}) = \mathbf{F}_\perp(\mathbf{r})+\mathbf{F}_\parallel(\mathbf{r}), $$

where
 * $$ \boldsymbol{\nabla}\cdot \mathbf{F}_\perp(\mathbf{r}) = 0,\qquad \boldsymbol{\nabla}\times \mathbf{F}_\parallel(\mathbf{r}) = \mathbf{0}. $$

Thus, an arbitrary field F(r) can be decomposed in a part that is divergence-free, the transverse component, and a part that is curl-free, the longitudinal component. This will now be proved directly, without making the detour via the integral expressions for A(r) and &Phi;(r).

Proof of decomposition
The decomposition is formulated in r-space. By a Fourier transform the decomposition may be formulated in k-space. This is advantageous because differentiations in r-space become multiplications in k-space. We will show that divergence in r-space becomes an inner product in k-space and a curl becomes  a cross product. Thus, we define the mutually inverse Fourier transforms,
 * $$ \begin{align} \tilde{\mathbf{F}}(\mathbf{k}) & = \frac{1}{ (2\pi)^{3/2} }\int e^{-i\mathbf{k}\cdot\mathbf{r}}\, \mathbf{F}(\mathbf{r})\, d^3\mathbf{r} \\ \mathbf{F}(\mathbf{r}) & = \frac{1}{ (2\pi)^{3/2} }\int e^{i\mathbf{k}\cdot\mathbf{r}} \,\tilde{\mathbf{F}}(\mathbf{k})\, d^3\mathbf{r} \\ \end{align} $$

An arbitrary vector field in k-space can be decomposed in components parallel and perpendicular to k,
 * $$ \tilde{\mathbf{F}}_\parallel (\mathbf{k}) \equiv \hat{\mathbf{k}} \big(\hat{\mathbf{k}}\cdot \tilde{\mathbf{F}}(\mathbf{k})\big), \qquad\hbox{with}\qquad \hat{\mathbf{k}} \equiv \frac{\mathbf{k}}{|\mathbf{k}|}, $$


 * $$ \tilde{\mathbf{F}}_\perp (\mathbf{k}) \equiv \tilde{\mathbf{F}}(\mathbf{k}) -\tilde{\mathbf{F}}_\parallel(\mathbf{k}), $$

so that
 * $$ \tilde{\mathbf{F}} (\mathbf{k}) = \tilde{\mathbf{F}}_\perp (\mathbf{k}) + \tilde{\mathbf{F}}_\parallel (\mathbf{k}). $$

Clearly,
 * $$ \mathbf{k}\cdot \tilde{\mathbf{F}}_\perp (\mathbf{k}) = 0 \qquad\hbox{and}\qquad \mathbf{k}\times \tilde{\mathbf{F}}_\parallel (\mathbf{k}) = 0. $$

Transforming back, we get
 * $$ \mathbf{F}_\perp(\mathbf{r}) \equiv \frac{1}{ (2\pi)^{3/2} }\int  e^{i\mathbf{k}\cdot\mathbf{r}}\, \tilde{\mathbf{F}}_\perp(\mathbf{k})\, d^3\mathbf{k}, \qquad \mathbf{F}_\parallel(\mathbf{r}) \equiv  \frac{1}{ (2\pi)^{3/2} }\int  e^{i\mathbf{k}\cdot\mathbf{r}}\, \tilde{\mathbf{F}}_\parallel(\mathbf{k})\, d^3\mathbf{k}, $$

which satisfy the properties
 * $$ \begin{align} \boldsymbol{\nabla}\cdot \mathbf{F}_\perp(\mathbf{r}) &= \frac{1}{ (2\pi)^{3/2} }\int  e^{i\mathbf{k}\cdot\mathbf{r}}\, \mathbf{k}\cdot\tilde{\mathbf{F}}_\perp(\mathbf{k})\, d^3\mathbf{k}  = 0 \\ \boldsymbol{\nabla}\times \mathbf{F}_\parallel(\mathbf{r}) &=  \frac{1}{ (2\pi)^{3/2} }\int  e^{i\mathbf{k}\cdot\mathbf{r}}\, \mathbf{k}\times\tilde{\mathbf{F}}_\parallel(\mathbf{k})\, d^3\mathbf{k}  = 0. \end{align} $$

Hence we have found the required decomposition.

Integral expressions for the transverse and longitudinal components
The curl and the divergence of the vector field F(r) satisfy,
 * $$ \boldsymbol{\nabla} \times \mathbf{F}(\mathbf{r}) = \boldsymbol{\nabla} \times \mathbf{F}_\perp(\mathbf{r})\quad\hbox{and}\quad\boldsymbol{\nabla} \cdot \mathbf{F}(\mathbf{r}) = \boldsymbol{\nabla} \cdot \mathbf{F}_\parallel(\mathbf{r}). $$

Using this, we see that the following relations were stated earlier in fact:
 * $$ \begin{align} \mathbf{F}_\perp(\mathbf{r}) &= \frac{1}{4\pi}\boldsymbol{\nabla} \times \int \frac{\boldsymbol{\nabla}'\times \mathbf{F}_\perp(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} d^3\mathbf{r}' \qquad\qquad \qquad (1)\\ \mathbf{F}_\parallel(\mathbf{r}) &= -\frac{1}{4\pi}\boldsymbol{\nabla}  \int \frac{\boldsymbol{\nabla}'\cdot \mathbf{F}_\parallel(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} d^3\mathbf{r}' \qquad\qquad\qquad \quad (2)\\ \end{align} $$

They are, respectively, the perpendicular (transverse, divergence-free) and parallel (longitudinal, curl-free)  components of the field F(r). We reiterate that the operator &nabla; acts on unprimed coordinates and &nabla;' on primed coordinates. Note that the two components of F(r) are uniquely determined once the curl and the divergence of F(r) are known. The integral relations will now be proved.

Proof of integral expressions
We will confirm the integral forms, equations (1) and (2), of the components. They will be shown to lead to identities.

Transverse component
For the perpendicular (transverse) component we note that for any vector V,
 * $$ \boldsymbol{\nabla} \times \big( \boldsymbol{\nabla} \times \mathbf{V} \big)= \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{V}) - \nabla^2 \mathbf{V} $$

and insert this in
 * $$ \begin{align} \boldsymbol{\nabla} \times\mathbf{F}_\perp(\mathbf{r}) &= \frac{1}{4\pi}\boldsymbol{\nabla} \times\Big(\boldsymbol{\nabla} \times \int \frac{\boldsymbol{\nabla}'\times \mathbf{F}_\perp(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} d^3\mathbf{r}' \Big) \\ &= -\frac{1}{4\pi} \nabla^2\int \frac{\boldsymbol{\nabla}'\times \mathbf{F}_\perp(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} d^3\mathbf{r}' + \frac{1}{4\pi}\boldsymbol{\nabla} \Big(\boldsymbol{\nabla} \cdot \int \frac{\boldsymbol{\nabla}'\times \mathbf{F}_\perp(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} d^3\mathbf{r}' \Big). \end{align} $$

Below we will show that second term vanishes. Use for the first term the following equation for the Dirac delta function,
 * $$ \nabla^2 \frac{1}{|\mathbf{r}-\mathbf{r}'|} = -4\pi \delta(\mathbf{r}-\mathbf{r}') $$

Hence the first term becomes (note that the unprimed nabla may be moved under the integral)
 * $$ \begin{align} &-\frac{1}{4\pi}\int \Big(\boldsymbol{\nabla}'\times \mathbf{F}_\perp(\mathbf{r}')\Big) \nabla^2 \Big( \frac{1}{|\mathbf{r}-\mathbf{r}'|} \Big) d^3\mathbf{r}' = \int \Big(\boldsymbol{\nabla}'\times \mathbf{F}_\perp(\mathbf{r}') \Big) \delta(\mathbf{r}-\mathbf{r}')  d^3\mathbf{r}' \\ &= \boldsymbol{\nabla}\times \mathbf{F}_\perp(\mathbf{r}), \end{align} $$

so that we indeed end up with an identity.

Before turning to the parallel (longitudinal) term we prove that the second term vanishes. To that end we introduce a shorthand notation
 * $$ \boldsymbol{\nabla}\cdot \int \frac{\boldsymbol{\nabla}'\times \mathbf{F}_\perp(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} d^3\mathbf{r}' \equiv \sum_{\alpha=x,y,z} \nabla_\alpha \int \frac{G_\alpha(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} d^3\mathbf{r}' . $$

Move the divergence under the integral and use
 * $$ \nabla_\alpha \frac{1}{|\mathbf{r}-\mathbf{r}'|} = - \nabla_\alpha' \frac{1}{|\mathbf{r}-\mathbf{r}'|} . $$

By partial integration and using that the integrand vanishes for the integral limits, we can let &minus;&nabla;'&alpha; act on G&alpha;(r' ) (this trick is known as the turnover rule for the anti Hermitian operator &nabla;'&alpha;). Then from
 * $$ \boldsymbol{\nabla}' \cdot \mathbf{G}(\mathbf{r}') \equiv \boldsymbol{\nabla}' \cdot  \big(\boldsymbol{\nabla}' \times \mathbf{F}_\perp(\mathbf{r}')\big) = 0, $$

(because the divergence of the curl of any vector is zero) follows the vanishing of the second term.

Longitudinal component
From
 * $$ \boldsymbol{\nabla} \times \mathbf{F}_\parallel(\mathbf{r}) = 0 $$

follows that there is a scalar function &Phi; such that
 * $$ \boldsymbol{\nabla}\Phi = \mathbf{F}_\parallel(\mathbf{r}) \quad\Longrightarrow\quad \nabla^2 \Phi = \boldsymbol{\nabla}\cdot \mathbf{F}_\parallel(\mathbf{r}) = \boldsymbol{\nabla}\cdot \mathbf{F}(\mathbf{r}) $$

We work toward an identity, using the turnover rule for the Laplace operator &nabla;2, which may be proved by partial integration and the assumption that the integrand vanishes at the integration limits,
 * $$ \begin{align} \mathbf{F}_\parallel(\mathbf{r}) &= -\frac{1}{4\pi}\boldsymbol{\nabla}\int \frac{\boldsymbol{\nabla}'\cdot \mathbf{F}(\mathbf{r}') }{|\mathbf{r}-\mathbf{r}'|} d^3\mathbf{r} = -\frac{1}{4\pi}\boldsymbol{\nabla}\int \frac{(\nabla')^2 \Phi(\mathbf{r'})}{|\mathbf{r}-\mathbf{r}'|} d^3\mathbf{r} \\ &= -\frac{1}{4\pi}\boldsymbol{\nabla}\int \Phi(\mathbf{r'})(\nabla')^2 \Big( \frac{1} {|\mathbf{r}-\mathbf{r}'|}\Big) d^3\mathbf{r} = \boldsymbol{\nabla}\int \Phi(\mathbf{r'}) \delta(\mathbf{r}-\mathbf{r}')  d^3\mathbf{r} \\ &= \boldsymbol{\nabla} \Phi(\mathbf{r}) = \mathbf{F}_\parallel(\mathbf{r}). \end{align} $$