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

Mathematical formulation

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,

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

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.

Proof of Corollary

(To be continued)