Helmholtz decomposition

In physics and mathematics, in the area of vector calculus, Helmholtz's theorem, also known as the fundamental theorem of vector calculus, states that any sufficiently smooth, rapidly decaying vector field in three dimensions can be resolved into the sum of an irrotational (curl-free) vector field and a solenoidal (divergence-free) vector field; this is known as the Helmholtz decomposition. It is named after Hermann von Helmholtz.

This implies that any such vector field F can be considered to be generated by a pair of potentials: a scalar potential φ and a vector potential A.

Statement of the theorem

Let F be a vector field on a bounded domain V in R3, which is twice continuously differentiable, and let S be the surface that encloses the domain V. Then F can be decomposed into a curl-free component and a divergence-free component:

${\mathbf {F} }=-{\boldsymbol {\nabla }}\Phi +{\boldsymbol {\nabla }}\times {\mathbf {A} },$ where

$\Phi \left({\mathbf {r} }\right)={\frac {1}{4\pi }}\int _{V}{\frac {{\boldsymbol {\nabla }}'\cdot {\mathbf {F} }\left({\mathbf {r} }'\right)}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}{\mathrm {d} }V'-{\frac {1}{4\pi }}\oint _{S}{\mathbf {\hat {n}} }'\cdot {\frac {{\mathbf {F} }\left({\mathbf {r} }'\right)}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}{\mathrm {d} }S'$ ${\mathbf {A} }\left({\mathbf {r} }\right)={\frac {1}{4\pi }}\int _{V}{\frac {{\boldsymbol {\nabla }}'\times {\mathbf {F} }\left({\mathbf {r} }'\right)}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}{\mathrm {d} }V'-{\frac {1}{4\pi }}\oint _{S}{\mathbf {\hat {n}} }'\times {\frac {{\mathbf {F} }\left({\mathbf {r} }'\right)}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}{\mathrm {d} }S'$ If V is R3 itself (unbounded), and F vanishes faster than $1/r$ as $r\rightarrow \infty$ , then the second component of both scalar and vector potential are zero. That is,

$\Phi \left({\mathbf {r} }\right)={\frac {1}{4\pi }}\int _{\text{all space}}{\frac {{\boldsymbol {\nabla }}'\cdot {\mathbf {F} }\left({\mathbf {r} }'\right)}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}{\mathrm {d} }V'$ ${\mathbf {A} }\left({\mathbf {r} }\right)={\frac {1}{4\pi }}\int _{\text{all space}}{\frac {{\boldsymbol {\nabla }}'\times {\mathbf {F} }\left({\mathbf {r} }'\right)}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}{\mathrm {d} }V'$ Derivation

Suppose we have a vector function ${\mathbf {F} }\left({\mathbf {r} }\right)$ of which we know the curl, ${\boldsymbol {\nabla }}\times {\mathbf {F} },$ and the divergence, ${\boldsymbol {\nabla }}\cdot {\mathbf {F} }$ , in the domain and the fields on the boundary. Writing the function using delta function in the form

$\delta ^{3}\left({\mathbf {r} }-{\mathbf {r} }'\right)=-{\frac {1}{4\pi }}\nabla ^{2}{\frac {1}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}},$ ${\mathbf {F} }\left({\mathbf {r} }\right)=\int _{V}{\mathbf {F} }\left({\mathbf {r} }'\right)\delta ^{3}\left({\mathbf {r} }-{\mathbf {r} }'\right){\mathrm {d} }V'=\int _{V}{\mathbf {F} }\left({\mathbf {r} }'\right)\left(-{\frac {1}{4\pi }}\nabla ^{2}{\frac {1}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}\right){\mathrm {d} }V'=-{\frac {1}{4\pi }}\nabla ^{2}\int _{V}{\frac {{\mathbf {F} }\left({\mathbf {r} }'\right)}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}{\mathrm {d} }V'.$ Using the identity

$\nabla ^{2}{\mathbf {a} }={\boldsymbol {\nabla }}\left({\boldsymbol {\nabla }}\cdot {\mathbf {a} }\right)-{\boldsymbol {\nabla }}\times \left({\boldsymbol {\nabla }}\times {\mathbf {a} }\right).$ We get,

${\mathbf {F} }\left({\mathbf {r} }\right)=-{\frac {1}{4\pi }}\left[{\boldsymbol {\nabla }}\left({\boldsymbol {\nabla }}\cdot \int _{V}{\frac {{\mathbf {F} }\left({\mathbf {r} }'\right)}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}{\mathrm {d} }V'\right)-{\boldsymbol {\nabla }}\times \left({\boldsymbol {\nabla }}\times \int _{V}{\frac {{\mathbf {F} }\left({\mathbf {r} }'\right)}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}{\mathrm {d} }V'\right)\right]$ $=-{\frac {1}{4\pi }}\left[{\boldsymbol {\nabla }}\left(\int _{V}{\mathbf {F} }\left({\mathbf {r} }'\right)\cdot {\boldsymbol {\nabla }}{\frac {1}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}{\mathrm {d} }V'\right)+{\boldsymbol {\nabla }}\times \left(\int _{V}{\mathbf {F} }\left({\mathbf {r} }'\right)\times {\boldsymbol {\nabla }}{\frac {1}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}{\mathrm {d} }V'\right)\right].$ ${\mathbf {F} }\left({\mathbf {r} }\right)=-{\frac {1}{4\pi }}\left[-{\boldsymbol {\nabla }}\left(\int _{V}{\mathbf {F} }\left({\mathbf {r} }'\right)\cdot {\boldsymbol {\nabla }}'{\frac {1}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}{\mathrm {d} }V'\right)-{\boldsymbol {\nabla }}\times \left(\int _{V}{\mathbf {F} }\left({\mathbf {r} }'\right)\times {\boldsymbol {\nabla }}'{\frac {1}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}{\mathrm {d} }V'\right)\right].$ Then using the vectorial identities

${\mathbf {a} }\cdot {\boldsymbol {\nabla }}\psi =-\psi \left({\boldsymbol {\nabla }}\cdot {\mathbf {a} }\right)+{\boldsymbol {\nabla }}\cdot \left(\psi {\mathbf {a} }\right)$ and

${\mathbf {a} }\times {\boldsymbol {\nabla }}\psi =\psi \left({\boldsymbol {\nabla }}\times {\mathbf {a} }\right)-{\boldsymbol {\nabla }}\times \left(\psi {\mathbf {a} }\right),$ we get

${\mathbf {F} }\left({\mathbf {r} }\right)=-{\frac {1}{4\pi }}\left[-{\boldsymbol {\nabla }}\left(-\int _{V}{\frac {{\boldsymbol {\nabla }}'\cdot {\mathbf {F} }\left({\mathbf {r} }'\right)}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}{\mathrm {d} }V'+\int _{V}{\boldsymbol {\nabla }}'\cdot {\frac {{\mathbf {F} }\left({\mathbf {r} }'\right)}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}{\mathrm {d} }V'\right)-{\boldsymbol {\nabla }}\times \left(\int _{V}{\frac {{\boldsymbol {\nabla }}'\times {\mathbf {F} }\left({\mathbf {r} }'\right)}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}{\mathrm {d} }V'-\int _{V}{\boldsymbol {\nabla }}'\times {\frac {{\mathbf {F} }\left({\mathbf {r} }'\right)}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}{\mathrm {d} }V'\right)\right].$ Take advantage of the divergence theorem, the equation can be rewritten as

${\mathbf {F} }\left({\mathbf {r} }\right)=-{\frac {1}{4\pi }}\left[-{\boldsymbol {\nabla }}\left(-\int _{V}{\frac {{\boldsymbol {\nabla }}'\cdot {\mathbf {F} }\left({\mathbf {r} }'\right)}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}{\mathrm {d} }V'+\oint _{S}{\mathbf {\hat {n}} }'\cdot {\frac {{\mathbf {F} }\left({\mathbf {r} }'\right)}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}{\mathrm {d} }S'\right)-{\boldsymbol {\nabla }}\times \left(\int _{V}{\frac {{\boldsymbol {\nabla }}'\times {\mathbf {F} }\left({\mathbf {r} }'\right)}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}{\mathrm {d} }V'-\oint _{S}{\mathbf {\hat {n}} }'\times {\frac {{\mathbf {F} }\left({\mathbf {r} }'\right)}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}{\mathrm {d} }S'\right)\right]$ $=-{\boldsymbol {\nabla }}\left[{\frac {1}{4\pi }}\int _{V}{\frac {{\boldsymbol {\nabla }}'\cdot {\mathbf {F} }\left({\mathbf {r} }'\right)}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}{\mathrm {d} }V'-{\frac {1}{4\pi }}\oint _{S}{\mathbf {\hat {n}} }'\cdot {\frac {{\mathbf {F} }\left({\mathbf {r} }'\right)}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}{\mathrm {d} }S'\right]+{\boldsymbol {\nabla }}\times \left[{\frac {1}{4\pi }}\int _{V}{\frac {{\boldsymbol {\nabla }}'\times {\mathbf {F} }\left({\mathbf {r} }'\right)}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}{\mathrm {d} }V'-{\frac {1}{4\pi }}\oint _{S}{\mathbf {\hat {n}} }'\times {\frac {{\mathbf {F} }\left({\mathbf {r} }'\right)}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}{\mathrm {d} }S'\right].$ Define

$\Phi \left({\mathbf {r} }\right)\equiv {\frac {1}{4\pi }}\int _{V}{\frac {{\boldsymbol {\nabla }}'\cdot {\mathbf {F} }\left({\mathbf {r} }'\right)}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}{\mathrm {d} }V'-{\frac {1}{4\pi }}\oint _{S}{\mathbf {\hat {n}} }'\cdot {\frac {{\mathbf {F} }\left({\mathbf {r} }'\right)}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}{\mathrm {d} }S'$ ${\mathbf {A} }\left({\mathbf {r} }\right)\equiv {\frac {1}{4\pi }}\int _{V}{\frac {{\boldsymbol {\nabla }}'\times {\mathbf {F} }\left({\mathbf {r} }'\right)}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}{\mathrm {d} }V'-{\frac {1}{4\pi }}\oint _{S}{\mathbf {\hat {n}} }'\times {\frac {{\mathbf {F} }\left({\mathbf {r} }'\right)}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}{\mathrm {d} }S'$ Hence

${\mathbf {F} }=-{\boldsymbol {\nabla }}\Phi +{\boldsymbol {\nabla }}\times {\mathbf {A} }$ Another derivation from the Fourier transform

Write F as a Fourier transform :

${\vec {\mathbf {F} }}({\vec {r}})=\iiint {\vec {\mathbf {G} }}({\vec {\omega }})e^i\,{\vec {\omega }}\cdot {\vec {r}}}d{\vec {\omega }}$ The Fourier transform of a scalar field is a scalar field, and the Fourier transform of a vector field is a vector field of same dimension.

Now consider the following scalar and vector fields :

$\displaystyle \begin{array}{lll} G_\Phi(\vec{\omega}) = i\, \frac\vec{\mathbf{G}}(\vec{\omega}) \cdot \vec{\omega}}{||\vec{\omega}||^2} & \quad\quad & \vec{\mathbf{G}}_\mathbf{A}(\vec{\omega}) = i\, \frac\vec{\mathbf{G}}(\vec{\omega}) \times \vec{\omega}}{||\vec{\omega}||^2} \\ && \\ \Phi(\vec{r}) = \displaystyle \iiint G_\Phi(\vec{\omega}) e^i \, \vec{\omega} \cdot \vec{r}} d\vec{\omega} & & \vec{\mathbf{A}}(\vec{r}) = \displaystyle \iiint \vec{\mathbf{G}}_\mathbf{A}(\vec{\omega}) e^i \, \vec{\omega} \cdot \vec{r}} d\vec{\omega} \end{array}$

Hence

${\vec {\mathbf {G} }}({\vec {\omega }})=-i\,{\vec {\omega }}\,G_{\Phi }({\vec {\omega }})+i\,{\vec {\omega }}\times {\vec {\mathbf {G} }}_{\mathbf {A} }({\vec {\omega }})$ ${\begin{array}{lll}{\vec {\mathbf {F} }}({\vec {r}})&=&\displaystyle -\iiint i\,{\vec {\omega }}\,G_{\Phi }({\vec {\omega }})\,e^i\,{\vec {\omega }}\cdot {\vec {r}}}d{\vec {\omega }}+\iiint i\,{\vec {\omega }}\times {\vec {\mathbf {G} }}_{\mathbf {A} }({\vec {\omega }})e^i\,{\vec {\omega }}\cdot {\vec {r}}}d{\vec {\omega }}\\&=&-{\boldsymbol {\nabla }}\Phi ({\vec {r}})+{\boldsymbol {\nabla }}\times {\vec {\mathbf {A} }}({\vec {r}})\end{array}}$ Fields with prescribed divergence and curl

The term "Helmholtz Theorem" can also refer to the following. Let C be a solenoidal vector field and d a scalar field on R3 which are sufficiently smooth and which vanish faster than 1/r2 at infinity. Then there exists a vector field F such that

$\nabla \cdot {\mathbf {F} }=d$ and $\nabla \times {\mathbf {F} }={\mathbf {C} };$ if additionally the vector field F vanishes as r → ∞, then F is unique.

In other words, a vector field can be constructed with both a specified divergence and a specified curl, and if it also vanishes at infinity, it is uniquely specified by its divergence and curl. This theorem is of great importance in electrostatics, since Maxwell's equations for the electric and magnetic fields in the static case are of exactly this type. The proof is by a construction generalizing the one given above: we set

${\mathbf {F} }=-\nabla \,({\mathcal {G}}(d))+\nabla \times ({\mathcal {G}}({\mathbf {C} })),$ where ${\mathcal {G}}$ represents the Newtonian potential operator. (When acting on a vector field, such as ∇ × F, it is defined to act on each component.)

Differential forms

The Hodge decomposition is closely related to the Helmholtz decomposition, generalizing from vector fields on R3 to differential forms on a Riemannian manifold M. Most formulations of the Hodge decomposition require M to be compact. Since this is not true of R3, the Hodge decomposition theorem is not strictly a generalization of the Helmholtz theorem. However, the compactness restriction in the usual formulation of the Hodge decomposition can be replaced by suitable decay assumptions at infinity on the differential forms involved, giving a proper generalization of the Helmholtz theorem.

Weak formulation

The Helmholtz decomposition can also be generalized by reducing the regularity assumptions (the need for the existence of strong derivatives). Suppose Ω is a bounded, simply-connected, Lipschitz domain. Every square-integrable vector field u ∈ (L2(Ω))3 has an orthogonal decomposition:

${\mathbf {u} }=\nabla \varphi +\nabla \times {\mathbf {A} }$ where $\varphi$ is in the Sobolev space H1(Ω) of square-integrable functions on Ω whose partial derivatives defined in the distribution sense are square integrable, and A ∈ H(curl,Ω), the Sobolev space of vector fields consisting of square integrable vector fields with square integrable curl.

For a slightly smoother vector field u ∈ H(curl,Ω), a similar decomposition holds:

${\mathbf {u} }=\nabla \varphi +{\mathbf {v} }$ Longitudinal and transverse fields

A terminology often used in physics refers to the curl-free component of a vector field as the longitudinal component and the divergence-free component as the transverse component. This terminology comes from the following construction: Compute the three-dimensional Fourier transform of the vector field F. Then decompose this field, at each point k, into two components, one of which points longitudinally, i.e. parallel to k, the other of which points in the transverse direction, i.e. perpendicular to k. So far, we have

${\mathbf {F} }\left({\mathbf {k} }\right)={\mathbf {F} }_{t}\left({\mathbf {k} }\right)+{\mathbf {F} }_{l}\left({\mathbf {k} }\right)$ ${\mathbf {k} }\cdot {\mathbf {F} }_{t}\left({\mathbf {k} }\right)=0.$ ${\mathbf {k} }\times {\mathbf {F} }_{l}\left({\mathbf {k} }\right)={\mathbf {0} }.$ Now we apply an inverse Fourier transform to each of these components. Using properties of Fourier transforms, we derive:

${\mathbf {F} }\left({\mathbf {r} }\right)={\mathbf {F} }_{t}\left({\mathbf {r} }\right)+{\mathbf {F} }_{l}\left({\mathbf {r} }\right)$ ${\boldsymbol {\nabla }}\cdot {\mathbf {F} }_{t}\left({\mathbf {r} }\right)=0$ ${\boldsymbol {\nabla }}\times {\mathbf {F} }_{l}\left({\mathbf {r} }\right)={\mathbf {0} }$ we can get

${\mathbf {F} }_{t}={\boldsymbol {\nabla }}\times {\mathbf {A} }={\frac {1}{4\pi }}{\boldsymbol {\nabla }}\times \int _{V}{\frac {{\boldsymbol {\nabla }}'\times {\mathbf {F} }}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}{\mathrm {d} }V'$ ${\mathbf {F} }_{l}=-{\boldsymbol {\nabla }}\Phi =-{\frac {1}{4\pi }}{\boldsymbol {\nabla }}\int _{V}{\frac {{\boldsymbol {\nabla }}'\cdot {\mathbf {F} }}{\left|{\mathbf {r} }-{\mathbf {r} }'\right|}}{\mathrm {d} }V'$ so this is indeed the Helmholtz decomposition.