# Helmholtz decomposition

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In vector analysis, the Helmholtz decomposition of a vector field on ${\displaystyle \mathbb {R} ^{3}}$ 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,

${\displaystyle \mathbf {F} ={\boldsymbol {\nabla }}\times \mathbf {A} -{\boldsymbol {\nabla }}\Phi =\mathbf {F} _{\perp }(\mathbf {r} )+\mathbf {F} _{\parallel }(\mathbf {r} )}$

with

{\displaystyle {\begin{aligned}\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{aligned}}}

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

Note that

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

holds for any vector field V(r) and any scalar function Ψ(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,

${\displaystyle \mathbf {E} (\mathbf {r} )=-{\dot {\mathbf {A} }}(\mathbf {r} )-{\boldsymbol {\nabla }}\Phi (\mathbf {r} ),}$

where Φ 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 ${\displaystyle \mathbf {r} \in \mathbb {R} ^{3}}$ can be decomposed in a transverse ${\displaystyle \scriptstyle \mathbf {F} _{\perp }(\mathbf {r} )}$ and longitudinal component ${\displaystyle \scriptstyle \mathbf {F} _{\parallel }(\mathbf {r} )}$:

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

where

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

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 Φ(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,

{\displaystyle {\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,

${\displaystyle {\tilde {\mathbf {F} }}_{\parallel }(\mathbf {k} )\equiv {\hat {\mathbf {k} }}{\big (}{\hat {\mathbf {k} }}\cdot {\tilde {\mathbf {F} }}(\mathbf {k} ){\big )},\qquad {\hbox{with}}\qquad {\hat {\mathbf {k} }}\equiv {\frac {\mathbf {k} }{|\mathbf {k} |}},}$
${\displaystyle {\tilde {\mathbf {F} }}_{\perp }(\mathbf {k} )\equiv {\tilde {\mathbf {F} }}(\mathbf {k} )-{\tilde {\mathbf {F} }}_{\parallel }(\mathbf {k} ),}$

so that

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

Clearly,

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

Transforming back, we get

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

which satisfy the properties

{\displaystyle {\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.

## Integral expressions for the transverse and longitudinal components

The curl and the divergence of the vector field F(r) satisfy,

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

Using this, we see that the following relations were stated earlier in fact:

{\displaystyle {\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}}}

They are, respectively, the perpendicular (transverse, divergence-free) and parallel (longitudinal, curl-free) components of the field F(r). We reiterate that the operator acts on unprimed coordinates and ∇' 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,

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

and insert this in

{\displaystyle {\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,

${\displaystyle \nabla ^{2}{\frac {1}{|\mathbf {r} -\mathbf {r} '|}}=-4\pi \delta (\mathbf {r} -\mathbf {r} ')}$

Hence the first term becomes (note that the unprimed nabla may be moved under the integral)

{\displaystyle {\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 (longitudinal) term we prove that the second term vanishes. To that end we introduce a shorthand notation

${\displaystyle {\boldsymbol {\nabla }}\cdot \int {\frac {{\boldsymbol {\nabla }}'\times \mathbf {F} _{\perp }(\mathbf {r} ')}{|\mathbf {r} -\mathbf {r} '|}}d^{3}\mathbf {r} '\equiv \sum _{\alpha =x,y,z}\nabla _{\alpha }\int {\frac {G_{\alpha }(\mathbf {r} ')}{|\mathbf {r} -\mathbf {r} '|}}d^{3}\mathbf {r} '.}$

Move the divergence under the integral and use

${\displaystyle \nabla _{\alpha }{\frac {1}{|\mathbf {r} -\mathbf {r} '|}}=-\nabla _{\alpha }'{\frac {1}{|\mathbf {r} -\mathbf {r} '|}}.}$

By partial integration and using that the integrand vanishes for the integral limits, we can let −∇'α act on Gα(r' ) (this trick is known as the turnover rule for the anti Hermitian operator ∇'α). Then from

${\displaystyle {\boldsymbol {\nabla }}'\cdot \mathbf {G} (\mathbf {r} ')\equiv {\boldsymbol {\nabla }}'\cdot {\big (}{\boldsymbol {\nabla }}'\times \mathbf {F} _{\perp }(\mathbf {r} '){\big )}=0,}$

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

#### Longitudinal component

From

${\displaystyle {\boldsymbol {\nabla }}\times \mathbf {F} _{\parallel }(\mathbf {r} )=0}$

follows that there is a scalar function Φ such that

${\displaystyle {\boldsymbol {\nabla }}\Phi =\mathbf {F} _{\parallel }(\mathbf {r} )\quad \Longrightarrow \quad \nabla ^{2}\Phi ={\boldsymbol {\nabla }}\cdot \mathbf {F} _{\parallel }(\mathbf {r} )={\boldsymbol {\nabla }}\cdot \mathbf {F} (\mathbf {r} )}$

We work toward an identity, using the turnover rule for the Laplace operator ∇2, which may be proved by partial integration and the assumption that the integrand vanishes at the integration limits,

{\displaystyle {\begin{aligned}\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{aligned}}}