Helmholtz decomposition

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

As a corollary follows that we must specify both its divergence and its curl at all points of space in order to define a vector function on $\scriptstyle \mathbb {R} ^{3}$ uniquely,

Mathematical formulation of the Helmholtz decomposition

A vector field F(r) with $\scriptstyle \mathbf {r} \in \mathbb {R} ^{3}$ can be written as follows:

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

where

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

Thus, the arbitrary field F(r) can be decomposed in a part that is divergence-free, $\scriptstyle \mathbf {F} _{\perp }(\mathbf {r} )$, and a part that is curl-free, $\scriptstyle \mathbf {F} _{\parallel }(\mathbf {r} )$.

Proof of decomposition

The decomposition is formulated in r-space. By a Fourier transformation 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,

${\begin{aligned}{\tilde {\mathbf {F} }}(\mathbf {k} )&={\frac {1}{(2\pi )^{3/2}}}\int e^{-i\mathbf {k} \cdot \mathbf {r} }\,\mathbf {F} (\mathbf {r} )\,d^{3}\mathbf {r} \\\mathbf {F} (\mathbf {r} )&={\frac {1}{(2\pi )^{3/2}}}\int e^{i\mathbf {k} \cdot \mathbf {r} }\,{\tilde {\mathbf {F} }}(\mathbf {k} )\,d^{3}\mathbf {r} \\\end{aligned}}$

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

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

Hence we have found the required decomposition.

Mathematical formulation corollary

We assume that the curl and the divergence of an arbitrary vector field F(r)

${\boldsymbol {\nabla }}\times \mathbf {F} (\mathbf {r} )={\boldsymbol {\nabla }}\times \mathbf {F} _{\perp }(\mathbf {r} )\quad {\hbox{and}}\quad {\boldsymbol {\nabla }}\cdot \mathbf {F} (\mathbf {r} )={\boldsymbol {\nabla }}\cdot \mathbf {F} _{\parallel }(\mathbf {r} )$

are given. Then

${\begin{aligned}\mathbf {F} _{\perp }(\mathbf {r} )&={\frac {1}{4\pi }}{\boldsymbol {\nabla }}\times \int {\frac {{\boldsymbol {\nabla }}'\times \mathbf {F} _{\perp }(\mathbf {r} ')}{|\mathbf {r} -\mathbf {r} '|}}d^{3}\mathbf {r} '\qquad \qquad \qquad (1)\\\mathbf {F} _{\parallel }(\mathbf {r} )&=-{\frac {1}{4\pi }}{\boldsymbol {\nabla }}\int {\frac {{\boldsymbol {\nabla }}'\cdot \mathbf {F} _{\parallel }(\mathbf {r} ')}{|\mathbf {r} -\mathbf {r} '|}}d^{3}\mathbf {r} '\qquad \qquad \qquad \quad (2)\\\end{aligned}}$

are, respectively, the perpendicular (divergence-free) and parallel (curl-free) components. The operator ∇ acts on unprimed coordinates and ∇' acts on primed coordinates. Note that the two components of F(r) are indeed uniquely determined once the curl and the divergence of F(r) are known.

Proof of corollary

We will confirm the integral forms, equations (1) and (2), of the components. We will show that they lead to identities.

For the perpendicular component we note that for any vector A,

${\boldsymbol {\nabla }}\times {\big (}{\boldsymbol {\nabla }}\times \mathbf {A} {\big )}={\boldsymbol {\nabla }}({\boldsymbol {\nabla }}\cdot \mathbf {A} )-\nabla ^{2}\mathbf {A}$

and insert this in

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

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)

${\begin{aligned}&-{\frac {1}{4\pi }}\int {\Big (}{\boldsymbol {\nabla }}'\times \mathbf {F} _{\perp }(\mathbf {r} '){\Big )}\nabla ^{2}{\Big (}{\frac {1}{|\mathbf {r} -\mathbf {r} '|}}{\Big )}d^{3}\mathbf {r} '=\int {\Big (}{\boldsymbol {\nabla }}'\times \mathbf {F} _{\perp }(\mathbf {r} '){\Big )}\delta (\mathbf {r} -\mathbf {r} ')d^{3}\mathbf {r} '\\&={\boldsymbol {\nabla }}\times \mathbf {F} _{\perp }(\mathbf {r} ),\end{aligned}}$

so that we indeed end up with an identity.

Before turning to the parallel 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 −∇'_α act on G_α(r' ) (this trick is known as the turnover rule for the anti Hermitian operator ∇'_α). 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.

(To be continued)