Helmholtz decomposition

Derivation

Solution space

If ${\displaystyle (\Phi _{1},{\mathbf {A} _{1}})}$ is a Helmholtz decomposition of ${\displaystyle \mathbf {F} }$, then ${\displaystyle (\Phi _{2},{\mathbf {A} _{2}})}$ is another decomposition if, and only if,

${\displaystyle \Phi _{1}-\Phi _{2}=\lambda \quad }$ and ${\displaystyle \quad \mathbf {A} _{1}-\mathbf {A} _{2}={\mathbf {A} }_{\lambda }+\nabla \varphi ,}$

where

• ${\displaystyle \lambda }$ is a harmonic scalar field,
• ${\displaystyle {\mathbf {A} }_{\lambda }}$ is a vector field which fulfills ${\displaystyle \nabla \times {\mathbf {A} }_{\lambda }=\nabla \lambda ,}$
• ${\displaystyle \varphi }$ is a scalar field.

Proof: Set ${\displaystyle \lambda =\Phi _{2}-\Phi _{1}}$ and ${\displaystyle {\mathbf {B} =A_{2}-A_{1}}}$. According to the definition of the Helmholtz decomposition, the condition is equivalent to

${\displaystyle -\nabla \lambda +\nabla \times \mathbf {B} =0}$.

Taking the divergence of each member of this equation yields ${\displaystyle \nabla ^{2}\lambda =0}$, hence ${\displaystyle \lambda }$ is harmonic.

Conversely, given any harmonic function ${\displaystyle \lambda }$, ${\displaystyle \nabla \lambda }$ is solenoidal since $${\displaystyle \nabla \cdot (\nabla \lambda )=\nabla ^{2}\lambda =0.}$$

Thus, according to the above section, there exists a vector field ${\displaystyle {\mathbf {A} }_{\lambda }}$ such that ${\displaystyle \nabla \lambda =\nabla \times {\mathbf {A} }_{\lambda }}$.

If ${\displaystyle {\mathbf {A} '}_{\lambda }}$ is another such vector field, then ${\displaystyle \mathbf {C} ={\mathbf {A} }_{\lambda }-{\mathbf {A} '}_{\lambda }}$ fulfills ${\displaystyle \nabla \times {\mathbf {C} }=0}$, hence ${\displaystyle C=\nabla \varphi }$ for some scalar field ${\displaystyle \varphi }$.

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

$${\displaystyle \nabla \cdot \mathbf {F} =d\quad {\text{ and }}\quad \nabla \times \mathbf {F} =\mathbf {C}$$ ;}

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

$${\displaystyle \mathbf {F} =\nabla ({\mathcal {G}}(d))-\nabla \times ({\mathcal {G}}(\mathbf {C} )),}$$

where ${\displaystyle {\mathcal {G}}}$ represents the Newtonian potential operator. (When acting on a vector field, such as ∇ × F, it is defined to act on each component.)

Weak formulation

The Helmholtz decomposition can 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:

$${\displaystyle \mathbf {u} =\nabla \varphi +\nabla \times \mathbf {A} }$$

where φ 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:

$${\displaystyle \mathbf {u} =\nabla \varphi +\mathbf {v} }$$

where φ ∈ H¹(Ω), v ∈ (H¹(Ω))^d.

Derivation from the Fourier transform

Note that in the theorem stated here, we have imposed the condition that if ${\displaystyle \mathbf {F} }$ is not defined on a bounded domain, then ${\displaystyle \mathbf {F} }$ shall decay faster than ${\displaystyle 1/r}$. Thus, the Fourier transform of ${\displaystyle \mathbf {F} }$, denoted as ${\displaystyle \mathbf {G} }$, is guaranteed to exist. We apply the convention $${\displaystyle \mathbf {F} (\mathbf {r} )=\iiint \mathbf {G} (\mathbf {k} )e^{i\mathbf {k} \cdot \mathbf {r} }dV_{k}}$$

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: $${\displaystyle {\begin{aligned}G_{\Phi }(\mathbf {k} )&=i{\frac {\mathbf {k} \cdot \mathbf {G} (\mathbf {k} )}{\|\mathbf {k} \|^{2}}}\\\mathbf {G} _{\mathbf {A} }(\mathbf {k} )&=i{\frac {\mathbf {k} \times \mathbf {G} (\mathbf {k} )}{\|\mathbf {k} \|^{2}}}\\[8pt]\Phi (\mathbf {r} )&=\iiint G_{\Phi }(\mathbf {k} )e^{i\mathbf {k} \cdot \mathbf {r} }dV_{k}\\\mathbf {A} (\mathbf {r} )&=\iiint \mathbf {G} _{\mathbf {A} }(\mathbf {k} )e^{i\mathbf {k} \cdot \mathbf {r} }dV_{k}\end{aligned}}}$$

Hence

$${\displaystyle {\begin{aligned}\mathbf {G} (\mathbf {k} )&=-i\mathbf {k} G_{\Phi }(\mathbf {k} )+i\mathbf {k} \times \mathbf {G} _{\mathbf {A} }(\mathbf {k} )\\[6pt]\mathbf {F} (\mathbf {r} )&=-\iiint i\mathbf {k} G_{\Phi }(\mathbf {k} )e^{i\mathbf {k} \cdot \mathbf {r} }dV_{k}+\iiint i\mathbf {k} \times \mathbf {G} _{\mathbf {A} }(\mathbf {k} )e^{i\mathbf {k} \cdot \mathbf {r} }dV_{k}\\&=-\nabla \Phi (\mathbf {r} )+\nabla \times \mathbf {A} (\mathbf {r} )\end{aligned}}}$$

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 ${\displaystyle {\hat {\mathbf {F} }}}$ of the vector field ${\displaystyle \mathbf {F} }$. 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

$${\displaystyle {\hat {\mathbf {F} }}(\mathbf {k} )={\hat {\mathbf {F} }}_{t}(\mathbf {k} )+{\hat {\mathbf {F} }}_{l}(\mathbf {k} )}$$ $${\displaystyle \mathbf {k} \cdot {\hat {\mathbf {F} }}_{t}(\mathbf {k} )=0.}$$ $${\displaystyle \mathbf {k} \times {\hat {\mathbf {F} }}_{l}(\mathbf {k} )=\mathbf {0} .}$$

Now we apply an inverse Fourier transform to each of these components. Using properties of Fourier transforms, we derive:

$${\displaystyle \mathbf {F} (\mathbf {r} )=\mathbf {F} _{t}(\mathbf {r} )+\mathbf {F} _{l}(\mathbf {r} )}$$ $${\displaystyle \nabla \cdot \mathbf {F} _{t}(\mathbf {r} )=0}$$ $${\displaystyle \nabla \times \mathbf {F} _{l}(\mathbf {r} )=\mathbf {0} }$$

Since ${\displaystyle \nabla \times (\nabla \Phi )=0}$ and ${\displaystyle \nabla \cdot (\nabla \times \mathbf {A} )=0}$,

we can get

$${\displaystyle \mathbf {F} _{t}=\nabla \times \mathbf {A} ={\frac {1}{4\pi }}\nabla \times \int _{V}{\frac {\nabla '\times \mathbf {F} }{\left|\mathbf {r} -\mathbf {r} '\right|}}\mathrm {d} V'}$$ $${\displaystyle \mathbf {F} _{l}=-\nabla \Phi =-{\frac {1}{4\pi }}\nabla \int _{V}{\frac {\nabla '\cdot \mathbf {F} }{\left|\mathbf {r} -\mathbf {r} '\right|}}\mathrm {d} V'}$$

so this is indeed the Helmholtz decomposition.

Matrix approach

The generalization to ${\displaystyle d}$ dimensions cannot be done with a vector potential, since the rotation operator and the cross product are defined (as vectors) only in three dimensions.

Let ${\displaystyle \mathbf {F} }$ be a vector field in ${\displaystyle \mathbb {R} ^{d}}$ which decays faster than ${\displaystyle |\mathbf {r} |^{-\delta }}$ for ${\displaystyle |\mathbf {r} |\to \infty }$ and ${\displaystyle \delta >2}$.

The scalar potential is defined similar to the three dimensional case as: $${\displaystyle \Phi (\mathbf {r} )=-\int _{\mathbb {R} ^{d}}\operatorname {div} (\mathbf {F} (\mathbf {r} '))K(\mathbf {r} ,\mathbf {r} ')\mathrm {d} V'=-\int _{\mathbb {R} ^{d}}\sum _{i}{\frac {\partial F_{i}}{\partial r_{i}}}(\mathbf {r} ')K(\mathbf {r} ,\mathbf {r} ')\mathrm {d} V',}$$ where as the integration kernel ${\displaystyle K(\mathbf {r} ,\mathbf {r} ')}$ is again the fundamental solution of Laplace's equation, but in d-dimensional space: $${\displaystyle K(\mathbf {r} ,\mathbf {r} ')={\begin{cases}{\frac {1}{2\pi }}\log {|\mathbf {r} -\mathbf {r} '|}&d=2,\\{\frac {1}{d(2-d)V_{d}}}|\mathbf {r} -\mathbf {r} '|^{2-d}&{\text{otherwise}},\end{cases}}}$$ with ${\displaystyle V_{d}=\pi ^{\frac {d}{2}}/\Gamma {\big (}{\tfrac {d}{2}}+1{\big )}}$ the volume of the d-dimensional unit balls and ${\displaystyle \Gamma (\mathbf {r} )}$ the gamma function.

For ${\displaystyle d=3}$, ${\displaystyle V_{d}}$ is just equal to ${\displaystyle {\frac {4\pi }{3}}}$, yielding the same prefactor as above. The rotational potential is an antisymmetric matrix with the elements: $${\displaystyle A_{ij}(\mathbf {r} )=\int _{\mathbb {R} ^{d}}\left({\frac {\partial F_{i}}{\partial x_{j}}}(\mathbf {r} ')-{\frac {\partial F_{j}}{\partial x_{i}}}(\mathbf {r} ')\right)K(\mathbf {r} ,\mathbf {r} ')\mathrm {d} V'.}$$ Above the diagonal are ${\displaystyle \textstyle {\binom {d}{2}}}$ entries which occur again mirrored at the diagonal, but with a negative sign. In the three-dimensional case, the matrix elements just correspond to the components of the vector potential ${\displaystyle \mathbf {A} =[A_{1},A_{2},A_{3}]=[A_{23},A_{31},A_{12}]}$. However, such a matrix potential can be written as a vector only in the three-dimensional case, because ${\displaystyle \textstyle {\binom {d}{2}}=d}$ is valid only for ${\displaystyle d=3}$.

As in the three-dimensional case, the gradient field is defined as $${\displaystyle \mathbf {G} (\mathbf {r} )=-\nabla \Phi (\mathbf {r} ).}$$ The rotational field, on the other hand, is defined in the general case as the row divergence of the matrix: $${\displaystyle \mathbf {R} (\mathbf {r} )=\left[\sum \nolimits _{k}\partial _{r_{k}}A_{ik}(\mathbf {r} );{1\leq i\leq d}\right].}$$ In three-dimensional space, this is equivalent to the rotation of the vector potential.

Tensor approach

In a ${\displaystyle d}$-dimensional vector space with ${\displaystyle d\neq 3}$, ${\textstyle -{\frac {1}{4\pi \left|\mathbf {r} -\mathbf {r} '\right|}}}$ can be replaced by the appropriate Green's function for the Laplacian, defined by $${\displaystyle \nabla ^{2}G(\mathbf {r} ,\mathbf {r} ')={\frac {\partial }{\partial r_{\mu }}}{\frac {\partial }{\partial r_{\mu }}}G(\mathbf {r} ,\mathbf {r} ')=\delta ^{d}(\mathbf {r} -\mathbf {r} ')}$$ where Einstein summation convention is used for the index ${\displaystyle \mu }$. For example, ${\textstyle G(\mathbf {r} ,\mathbf {r} ')={\frac {1}{2\pi }}\ln \left|\mathbf {r} -\mathbf {r} '\right|}$ in 2D.

Following the same steps as above, we can write $${\displaystyle F_{\mu }(\mathbf {r} )=\int _{V}F_{\mu }(\mathbf {r} '){\frac {\partial }{\partial r_{\mu }}}{\frac {\partial }{\partial r_{\mu }}}G(\mathbf {r} ,\mathbf {r} ')\,\mathrm {d} ^{d}\mathbf {r} '=\delta _{\mu \nu }\delta _{\rho \sigma }\int _{V}F_{\nu }(\mathbf {r} '){\frac {\partial }{\partial r_{\rho }}}{\frac {\partial }{\partial r_{\sigma }}}G(\mathbf {r} ,\mathbf {r} ')\,\mathrm {d} ^{d}\mathbf {r} '}$$ where ${\displaystyle \delta _{\mu \nu }}$ is the Kronecker delta (and the summation convention is again used). In place of the definition of the vector Laplacian used above, we now make use of an identity for the Levi-Civita symbol ${\displaystyle \varepsilon }$, $${\displaystyle \varepsilon _{\alpha \mu \rho }\varepsilon _{\alpha \nu \sigma }=(d-2)!(\delta _{\mu \nu }\delta _{\rho \sigma }-\delta _{\mu \sigma }\delta _{\nu \rho })}$$ which is valid in ${\displaystyle d\geq 2}$ dimensions, where ${\displaystyle \alpha }$ is a ${\displaystyle (d-2)}$-component multi-index. This gives $${\displaystyle F_{\mu }(\mathbf {r} )=\delta _{\mu \sigma }\delta _{\nu \rho }\int _{V}F_{\nu }(\mathbf {r} '){\frac {\partial }{\partial r_{\rho }}}{\frac {\partial }{\partial r_{\sigma }}}G(\mathbf {r} ,\mathbf {r} ')\,\mathrm {d} ^{d}\mathbf {r} '+{\frac {1}{(d-2)!}}\varepsilon _{\alpha \mu \rho }\varepsilon _{\alpha \nu \sigma }\int _{V}F_{\nu }(\mathbf {r} '){\frac {\partial }{\partial r_{\rho }}}{\frac {\partial }{\partial r_{\sigma }}}G(\mathbf {r} ,\mathbf {r} ')\,\mathrm {d} ^{d}\mathbf {r} '}$$

We can therefore write $${\displaystyle F_{\mu }(\mathbf {r} )=-{\frac {\partial }{\partial r_{\mu }}}\Phi (\mathbf {r} )+\varepsilon _{\mu \rho \alpha }{\frac {\partial }{\partial r_{\rho }}}A_{\alpha }(\mathbf {r} )}$$ where $${\displaystyle {\begin{aligned}\Phi (\mathbf {r} )&=-\int _{V}F_{\nu }(\mathbf {r} '){\frac {\partial }{\partial r_{\nu }}}G(\mathbf {r} ,\mathbf {r} ')\,\mathrm {d} ^{d}\mathbf {r} '\\A_{\alpha }&={\frac {1}{(d-2)!}}\varepsilon _{\alpha \nu \sigma }\int _{V}F_{\nu }(\mathbf {r} '){\frac {\partial }{\partial r_{\sigma }}}G(\mathbf {r} ,\mathbf {r} ')\,\mathrm {d} ^{d}\mathbf {r} '\end{aligned}}}$$ Note that the vector potential is replaced by a rank-${\displaystyle (d-2)}$ tensor in ${\displaystyle d}$ dimensions.

Because ${\displaystyle G(\mathbf {r} ,\mathbf {r} ')}$ is a function of only ${\displaystyle \mathbf {r} -\mathbf {r} '}$, one can replace ${\displaystyle {\frac {\partial }{\partial r_{\mu }}}\rightarrow -{\frac {\partial }{\partial r'_{\mu }}}}$, giving $${\displaystyle {\begin{aligned}\Phi (\mathbf {r} )&=\int _{V}F_{\nu }(\mathbf {r} '){\frac {\partial }{\partial r'_{\nu }}}G(\mathbf {r} ,\mathbf {r} ')\,\mathrm {d} ^{d}\mathbf {r} '\\A_{\alpha }&=-{\frac {1}{(d-2)!}}\varepsilon _{\alpha \nu \sigma }\int _{V}F_{\nu }(\mathbf {r} '){\frac {\partial }{\partial r_{\sigma }'}}G(\mathbf {r} ,\mathbf {r} ')\,\mathrm {d} ^{d}\mathbf {r} '\end{aligned}}}$$ Integration by parts can then be used to give $${\displaystyle {\begin{aligned}\Phi (\mathbf {r} )&=-\int _{V}G(\mathbf {r} ,\mathbf {r} '){\frac {\partial }{\partial r'_{\nu }}}F_{\nu }(\mathbf {r} ')\,\mathrm {d} ^{d}\mathbf {r} '+\oint _{S}G(\mathbf {r} ,\mathbf {r} ')F_{\nu }(\mathbf {r} '){\hat {n}}'_{\nu }\,\mathrm {d} ^{d-1}\mathbf {r} '\\A_{\alpha }&={\frac {1}{(d-2)!}}\varepsilon _{\alpha \nu \sigma }\int _{V}G(\mathbf {r} ,\mathbf {r} '){\frac {\partial }{\partial r_{\sigma }'}}F_{\nu }(\mathbf {r} ')\,\mathrm {d} ^{d}\mathbf {r} '-{\frac {1}{(d-2)!}}\varepsilon _{\alpha \nu \sigma }\oint _{S}G(\mathbf {r} ,\mathbf {r} ')F_{\nu }(\mathbf {r} '){\hat {n}}'_{\sigma }\,\mathrm {d} ^{d-1}\mathbf {r} '\end{aligned}}}$$ where ${\displaystyle S=\partial V}$ is the boundary of ${\displaystyle V}$. These expressions are analogous to those given above for three-dimensional space.

For a further generalization to manifolds, see the discussion of Hodge decomposition below.

Electrodynamics

The Helmholtz theorem is of particular interest in electrodynamics, since it can be used to write Maxwell's equations in the potential image and solve them more easily. The Helmholtz decomposition can be used to prove that, given electric current density and charge density, the electric field and the magnetic flux density can be determined. They are unique if the densities vanish at infinity and one assumes the same for the potentials.

Fluid dynamics

In fluid dynamics, the Helmholtz projection plays an important role, especially for the solvability theory of the Navier-Stokes equations. If the Helmholtz projection is applied to the linearized incompressible Navier-Stokes equations, the Stokes equation is obtained. This depends only on the velocity of the particles in the flow, but no longer on the static pressure, allowing the equation to be reduced to one unknown. However, both equations, the Stokes and linearized equations, are equivalent. The operator ${\displaystyle P\Delta }$ is called the Stokes operator.

Dynamical systems theory

In the theory of dynamical systems, Helmholtz decomposition can be used to determine "quasipotentials" as well as to compute Lyapunov functions in some cases.

For some dynamical systems such as the Lorenz system (Edward N. Lorenz, 1963), a simplified model for atmospheric convection, a closed-form expression of the Helmholtz decomposition can be obtained: $${\displaystyle {\dot {\mathbf {r} }}=\mathbf {F} (\mathbf {r} )={\big [}a(r_{2}-r_{1}),r_{1}(b-r_{3})-r_{2},r_{1}r_{2}-cr_{3}{\big ]}.}$$ The Helmholtz decomposition of ${\displaystyle \mathbf {F} (\mathbf {r} )}$, with the scalar potential ${\displaystyle \Phi (\mathbf {r} )={\tfrac {a}{2}}r_{1}^{2}+{\tfrac {1}{2}}r_{2}^{2}+{\tfrac {c}{2}}r_{3}^{2}}$ is given as:

$${\displaystyle \mathbf {G} (\mathbf {r} )={\big [}-ar_{1},-r_{2},-cr_{3}{\big ]},}$$ $${\displaystyle \mathbf {R} (\mathbf {r} )={\big [}+ar_{2},br_{1}-r_{1}r_{3},r_{1}r_{2}{\big ]}.}$$

The quadratic scalar potential provides motion in the direction of the coordinate origin, which is responsible for the stable fix point for some parameter range. For other parameters, the rotation field ensures that a strange attractor is created, causing the model to exhibit a butterfly effect.

Medical Imaging

In magnetic resonance elastography, a variant of MR imaging where mechanical waves are used to probe the viscoelasticity of organs, the Helmholtz decomposition is sometimes used to separate the measured displacement fields into its shear component (divergence-free) and its compression component (curl-free). In this way, the complex shear modulus can be calculated without contributions from compression waves.

Computer animation and robotics

The Helmholtz decomposition is also used in the field of computer engineering. This includes robotics, image reconstruction but also computer animation, where the decomposition is used for realistic visualization of fluids or vector fields.