Inphysics andmathematics, theHelmholtz decomposition theorem or thefundamental theorem of vector calculus^{[1][2][3][4][5][6][7]} states that certain differentiablevector fields can be resolved into the sum of anirrotational (curl-free) vector field and asolenoidal (divergence-free) vector field. Inphysics, often only the decomposition of sufficientlysmooth, rapidly decayingvector fields in three dimensions is discussed. It is named afterHermann von Helmholtz.

For a vector field${\displaystyle \mathbf{F}\in C^1(V,{\mathbb{R}}^n)}$ defined on a domain${\displaystyle V\subseteq {\mathbb{R}}^n}$, a Helmholtz decomposition is a pair of vector fields${\displaystyle \mathbf{G}\in C^1(V,{\mathbb{R}}^n)}$ and${\displaystyle \mathbf{R}\in C^1(V,{\mathbb{R}}^n)}$ such that:Here,${\displaystyle \mathrm{\Phi }\in C^2(V,\mathbb{R})}$ is ascalar potential,${\displaystyle \mathrm{\nabla }\mathrm{\Phi }}$ is itsgradient, and${\displaystyle \mathrm{\nabla }\cdot \mathbf{R}}$ is thedivergence of the vector field${\displaystyle \mathbf{R}}$. The irrotational vector field${\displaystyle \mathbf{G}}$ is called agradient field and${\displaystyle \mathbf{R}}$ is called asolenoidal field orrotation field. This decomposition does not exist for all vector fields and is notunique.^[8]

The Helmholtz decomposition in three dimensions was first described in 1849^[9] byGeorge Gabriel Stokes for a theory ofdiffraction.Hermann von Helmholtz published his paper on somehydrodynamic basic equations in 1858,^[10][11] which was part of his research on theHelmholtz's theorems describing the motion of fluid in the vicinity of vortex lines.^[11] Their derivation required the vector fields to decay sufficiently fast at infinity. Later, this condition could be relaxed, and the Helmholtz decomposition could be extended to higher dimensions.^[8][12][13] ForRiemannian manifolds, the Helmholtz-Hodge decomposition usingdifferential geometry andtensor calculus was derived.^{[8][11][14][15]}

The decomposition has become an important tool for many problems intheoretical physics,^[11][14] but has also found applications inanimation,computer vision as well asrobotics.^[15]

Many physics textbooks restrict the Helmholtz decomposition to the three-dimensional space and limit its application to vector fields that decay sufficiently fast at infinity or tobump functions that are defined on abounded domain. Then, avector potential${\displaystyle A}$ can be defined, such that the rotation field is given by${\displaystyle \mathbf{R}=\mathrm{\nabla }\times \mathbf{A}}$, using thecurl of a vector field.^[16]

Let${\displaystyle \mathbf{F}}$ be a vector field on a bounded domain${\displaystyle V\subseteq {\mathbb{R}}^3}$, which is twice continuously differentiable inside${\displaystyle V}$, and let${\displaystyle S}$ be the surface that encloses the domain${\displaystyle V}$ with outward surface normal${\displaystyle {\hat{\mathbf{n}}}^{\prime }}$. Then${\displaystyle \mathbf{F}}$ can be decomposed into a curl-free component and a divergence-free component as follows:^[17]

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

and${\displaystyle {\mathrm{\nabla }}^{\prime }}$ is thenabla operator with respect to${\displaystyle {\mathbf{r}}^{\prime }}$, not${\displaystyle \mathbf{r}}$.

If${\displaystyle V={\mathbb{R}}^3}$ and is therefore unbounded, and${\displaystyle \mathbf{F}}$ vanishes faster than${\displaystyle 1/r}$ as${\displaystyle r\to \mathrm{\infty }}$, then one has^[18]

This holds in particular if${\displaystyle \mathbf{F}}$ is twice continuously differentiable in${\displaystyle {\mathbb{R}}^3}$ and of bounded support.

If${\displaystyle ({\mathrm{\Phi }}_1,{\mathbf{A}}_1)}$ is a Helmholtz decomposition of${\displaystyle \mathbf{F}}$, then${\displaystyle ({\mathrm{\Phi }}_2,{\mathbf{A}}_2)}$ is another decomposition if, and only if,

${\displaystyle {\mathrm{\Phi }}_1-{\mathrm{\Phi }}_2=\lambda }$ and${\displaystyle {\mathbf{A}}_1-{\mathbf{A}}_2={\mathbf{A}}_{\lambda }+\mathrm{\nabla }\phi ,}$

where

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

Proof: Set${\displaystyle \lambda ={\mathrm{\Phi }}_2-{\mathrm{\Phi }}_1}$ and${\displaystyle \mathbf{B}=A_2-A_1}$. According to the definitionof the Helmholtz decomposition, the condition is equivalent to

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

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

Conversely, given any harmonic function${\displaystyle \lambda }$,${\displaystyle \mathrm{\nabla }\lambda }$ is solenoidal since

${\displaystyle \mathrm{\nabla }\cdot (\mathrm{\nabla }\lambda )={\mathrm{\nabla }}^2\lambda =0.}$

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

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

The term "Helmholtz theorem" can also refer to the following. LetC be asolenoidal vector field andd a scalar field onR³ which are sufficiently smooth and which vanish faster than1/r² at infinity. Then there exists a vector fieldF such that

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

if additionally the vector fieldF vanishes asr → ∞, thenF is unique.^[18]

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 inelectrostatics, sinceMaxwell's equations for the electric and magnetic fields in the static case are of exactly this type.^[18] The proof is by a construction generalizing the one given above: we set

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

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

The Helmholtz decomposition can be generalized by reducing the regularity assumptions (the need for the existence of strong derivatives). SupposeΩ is a bounded, simply-connected,Lipschitz domain. Everysquare-integrable vector fieldu ∈ (L²(Ω))³ has anorthogonal decomposition:^[19][20][21]

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

whereφ is in theSobolev spaceH¹(Ω) of square-integrable functions onΩ whose partial derivatives defined in thedistribution sense are square integrable, andA ∈H(curl, Ω), the Sobolev space of vector fields consisting of square integrable vector fields with square integrable curl.

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

$${\displaystyle \mathbf{u}=\mathrm{\nabla }\phi +\mathbf{v}}$$

whereφ ∈H¹(Ω),v ∈ (H¹(Ω))^d.

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, theFourier transform of${\displaystyle \mathbf{F}}$, denoted as${\displaystyle \mathbf{G}}$, is guaranteed to exist. We apply the convention$${\displaystyle \mathbf{F}(\mathbf{r})=\iiint \mathbf{G}(\mathbf{k})e^{i\mathbf{k}\cdot \mathbf{r}}d{V}_k}$$

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:

Hence

A terminology often used in physics refers to the curl-free component of a vector field as thelongitudinal component and the divergence-free component as thetransverse component.^[22] This terminology comes from the following construction: Compute the three-dimensionalFourier transform${\displaystyle \hat{\mathbf{F}}}$ of the vector field${\displaystyle \mathbf{F}}$. Then decompose this field, at each pointk, into two components, one of which points longitudinally, i.e. parallel tok, the other of which points in the transverse direction, i.e. perpendicular tok. 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:

$${\displaystyle \mathbf{F}(\mathbf{r})={\mathbf{F}}_t(\mathbf{r})+{\mathbf{F}}_l(\mathbf{r})}$$$${\displaystyle \mathrm{\nabla }\cdot {\mathbf{F}}_t(\mathbf{r})=0}$$$${\displaystyle \mathrm{\nabla }\times {\mathbf{F}}_l(\mathbf{r})=\mathbf{0}}$$

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

we can get

$${\displaystyle {\mathbf{F}}_t=\mathrm{\nabla }\times \mathbf{A}=\frac{1}{4\pi }\mathrm{\nabla }\times {\int }_V\frac{{\mathrm{\nabla }}^{\prime }\times \mathbf{F}}{\left|\mathbf{r}-{\mathbf{r}}^{\prime }\right|}\mathrm{d}{V}^{\prime }}$$$${\displaystyle {\mathbf{F}}_l=-\mathrm{\nabla }\mathrm{\Phi }=-\frac{1}{4\pi }\mathrm{\nabla }{\int }_V\frac{{\mathrm{\nabla }}^{\prime }\cdot \mathbf{F}}{\left|\mathbf{r}-{\mathbf{r}}^{\prime }\right|}\mathrm{d}{V}^{\prime }}$$

so this is indeed the Helmholtz decomposition.^[23]

Informally speaking, in${\displaystyle {\mathbb{R}}^d}$, the Helmholtz decomposition can be expressed by${\displaystyle \mathbf{F}=-\mathrm{\nabla }\mathrm{\Phi }+\mathbf{R}}$ where${\displaystyle \mathrm{\Phi }}$ is any scalar function that solves the Poisson equation${\displaystyle -{\mathrm{\nabla }}^2\mathrm{\Phi }=f}$, where${\displaystyle f=\mathrm{\nabla }\cdot \mathbf{F}}$ is the divergence of the vector field${\displaystyle \mathbf{F}}$ in${\displaystyle {\mathbb{R}}^d}$, and${\displaystyle \mathbf{R}=\mathbf{F}+\mathrm{\nabla }\mathrm{\Phi }}$ is divergence free:${\displaystyle \mathrm{\nabla }\cdot \mathbf{R}=0}$. Thus the existence of Helmholtz decomposition a consequence of the existence of the solution of the Poisson equation${\displaystyle -{\mathrm{\nabla }}^2\mathrm{\Phi }=f}$.

The generalization to${\displaystyle d}$ dimensions cannot be done with a vector potential, since the rotation operator and thecross 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${\displaystyle |\mathbf{r}|\to \mathrm{\infty }}$ and${\displaystyle \delta >2}$.

The scalar potential is defined similar to the three dimensional case as:$${\displaystyle \mathrm{\Phi }(\mathbf{r})=-{\int }_{{\mathbb{R}}^d}\mathrm{div}(\mathbf{F}({\mathbf{r}}^{\prime }))K(\mathbf{r},{\mathbf{r}}^{\prime })\mathrm{d}{V}^{\prime }=-{\int }_{{\mathbb{R}}^d}\sum _i\frac{\mathrm{\partial }{F}_i}{\mathrm{\partial }{r}_i}({\mathbf{r}}^{\prime })K(\mathbf{r},{\mathbf{r}}^{\prime })\mathrm{d}{V}^{\prime },}$$where as the integration kernel${\displaystyle K(\mathbf{r},{\mathbf{r}}^{\prime })}$ is again thefundamental solution ofLaplace's equation, but in d-dimensional space:with${\displaystyle V_d={\pi }^{\frac{d}{2}}/\mathrm{\Gamma }(\frac{d}{2}+1)}$ the volume of the d-dimensionalunit balls and${\displaystyle \mathrm{\Gamma }(\mathbf{r})}$ thegamma 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 anantisymmetric matrix with the elements:$${\displaystyle A_{ij}(\mathbf{r})={\int }_{{\mathbb{R}}^d}\left(\frac{\mathrm{\partial }{F}_i}{\mathrm{\partial }{x}_j}({\mathbf{r}}^{\prime })-\frac{\mathrm{\partial }{F}_j}{\mathrm{\partial }{x}_i}({\mathbf{r}}^{\prime })\right)K(\mathbf{r},{\mathbf{r}}^{\prime })\mathrm{d}{V}^{\prime }.}$$Above the diagonal are${\displaystyle (\genfrac{}{}{0ex}{}{d}{2})}$ 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${\displaystyle (\genfrac{}{}{0ex}{}{d}{2})=d}$ is valid only for${\displaystyle d=3}$.

As in the three-dimensional case, the gradient field is defined as$${\displaystyle \mathbf{G}(\mathbf{r})=-\mathrm{\nabla }\mathrm{\Phi }(\mathbf{r}).}$$The rotational field, on the other hand, is defined in the general case as the row divergence of the matrix:$${\displaystyle \mathbf{R}(\mathbf{r})=\left[{\sum }_k{\mathrm{\partial }}_{r_k}{A}_{ik}(\mathbf{r});1\le i\le d\right].}$$In three-dimensional space, this is equivalent to the rotation of the vector potential.^[8][24]

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

Following the same steps as above, we can write$${\displaystyle F_{\mu }(\mathbf{r})={\int }_V{F}_{\mu }({\mathbf{r}}^{\prime })\frac{\mathrm{\partial }}{\mathrm{\partial }{r}_{\mu }}\frac{\mathrm{\partial }}{\mathrm{\partial }{r}_{\mu }}G(\mathbf{r},{\mathbf{r}}^{\prime }){\mathrm{d}}^d{\mathbf{r}}^{\prime }={\delta }_{\mu \nu }{\delta }_{\rho \sigma }{\int }_V{F}_{\nu }({\mathbf{r}}^{\prime })\frac{\mathrm{\partial }}{\mathrm{\partial }{r}_{\rho }}\frac{\mathrm{\partial }}{\mathrm{\partial }{r}_{\sigma }}G(\mathbf{r},{\mathbf{r}}^{\prime }){\mathrm{d}}^d{\mathbf{r}}^{\prime }}$$where${\displaystyle {\delta }_{\mu \nu }}$ is theKronecker 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 theLevi-Civita symbol${\displaystyle \epsilon }$,$${\displaystyle {\epsilon }_{\alpha \mu \rho }{\epsilon }_{\alpha \nu \sigma }=(d-2)!({\delta }_{\mu \nu }{\delta }_{\rho \sigma }-{\delta }_{\mu \sigma }{\delta }_{\nu \rho })}$$which is valid in${\displaystyle d\ge 2}$ dimensions, where${\displaystyle \alpha }$ is a${\displaystyle (d-2)}$-componentmulti-index. This gives$${\displaystyle F_{\mu }(\mathbf{r})={\delta }_{\mu \sigma }{\delta }_{\nu \rho }{\int }_V{F}_{\nu }({\mathbf{r}}^{\prime })\frac{\mathrm{\partial }}{\mathrm{\partial }{r}_{\rho }}\frac{\mathrm{\partial }}{\mathrm{\partial }{r}_{\sigma }}G(\mathbf{r},{\mathbf{r}}^{\prime }){\mathrm{d}}^d{\mathbf{r}}^{\prime }+\frac{1}{(d-2)!}{\epsilon }_{\alpha \mu \rho }{\epsilon }_{\alpha \nu \sigma }{\int }_V{F}_{\nu }({\mathbf{r}}^{\prime })\frac{\mathrm{\partial }}{\mathrm{\partial }{r}_{\rho }}\frac{\mathrm{\partial }}{\mathrm{\partial }{r}_{\sigma }}G(\mathbf{r},{\mathbf{r}}^{\prime }){\mathrm{d}}^d{\mathbf{r}}^{\prime }}$$

We can therefore write$${\displaystyle F_{\mu }(\mathbf{r})=-\frac{\mathrm{\partial }}{\mathrm{\partial }{r}_{\mu }}\mathrm{\Phi }(\mathbf{r})+{\epsilon }_{\mu \rho \alpha }\frac{\mathrm{\partial }}{\mathrm{\partial }{r}_{\rho }}{A}_{\alpha }(\mathbf{r})}$$whereNote that the vector potential is replaced by a rank-${\displaystyle (d-2)}$ tensor in${\displaystyle d}$ dimensions.

Because${\displaystyle G(\mathbf{r},{\mathbf{r}}^{\prime })}$ is a function of only${\displaystyle \mathbf{r}-{\mathbf{r}}^{\prime }}$, one can replace${\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }{r}_{\mu }}\to -\frac{\mathrm{\partial }}{\mathrm{\partial }{r}_{\mu }^{\prime }}}$, givingIntegration by parts can then be used to givewhere${\displaystyle S=\mathrm{\partial }V}$ is the boundary of${\displaystyle V}$. These expressions are analogous to those given above forthree-dimensional space.

For a further generalization to manifolds, see the discussion ofHodge decompositionbelow.

TheHodge decomposition is closely related to the Helmholtz decomposition,^[25] generalizing from vector fields onR³ todifferential forms on aRiemannian manifoldM. Most formulations of the Hodge decomposition requireM to becompact.^[26] Since this is not true ofR³, 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.

Most textbooks only deal with vector fields decaying faster than${\displaystyle |\mathbf{r}{|}^{-\delta }}$ with${\displaystyle \delta >1}$ at infinity.^[16][13][27] However,Otto Blumenthal showed in 1905 that an adapted integration kernel can be used to integrate fields decaying faster than${\displaystyle |\mathbf{r}{|}^{-\delta }}$ with${\displaystyle \delta >0}$, which is substantially less strict.To achieve this, the kernel${\displaystyle K(\mathbf{r},{\mathbf{r}}^{\prime })}$ in the convolution integrals has to be replaced by${\displaystyle K^{\prime }(\mathbf{r},{\mathbf{r}}^{\prime })=K(\mathbf{r},{\mathbf{r}}^{\prime })-K(0,{\mathbf{r}}^{\prime })}$.^[28]With even more complex integration kernels, solutions can be found even for divergent functions that need not grow faster than polynomial.^{[12][13][24][29]}

For allanalytic vector fields that need not go to zero even at infinity, methods based onpartial integration and theCauchy formula for repeated integration^[30] can be used to compute closed-form solutions of the rotation and scalar potentials, as in the case ofmultivariate polynomial,sine,cosine, andexponential functions.^[8]

In general, the Helmholtz decomposition is not uniquely defined.Aharmonic function${\displaystyle H(\mathbf{r})}$ is a function that satisfies${\displaystyle \mathrm{\Delta }H(\mathbf{r})=0}$.By adding${\displaystyle H(\mathbf{r})}$ to the scalar potential${\displaystyle \mathrm{\Phi }(\mathbf{r})}$, a different Helmholtz decomposition can be obtained:

For vector fields${\displaystyle \mathbf{F}}$, decaying at infinity, it is a plausible choice that scalar and rotation potentials also decay at infinity. Because${\displaystyle H(\mathbf{r})=0}$ is the only harmonic function with this property, which follows fromLiouville's theorem, this guarantees the uniqueness of the gradient and rotation fields.^[31]

This uniqueness does not apply to the potentials: In the three-dimensional case, the scalar and vector potential jointly have four components, whereas the vector field has only three. The vector field is invariant to gauge transformations and the choice of appropriate potentials known asgauge fixing is the subject ofgauge theory. Important examples from physics are theLorenz gauge condition and theCoulomb gauge. An alternative is to use thepoloidal–toroidal decomposition.

The Helmholtz theorem is of particular interest inelectrodynamics, since it can be used to writeMaxwell's equations in the potential image and solve them more easily. The Helmholtz decomposition can be used to prove that, givenelectric current density andcharge density, theelectric field and themagnetic flux density can be determined. They are unique if the densities vanish at infinity and one assumes the same for the potentials.^[16]

Influid dynamics, the Helmholtz projection plays an important role, especially for the solvability theory of theNavier-Stokes equations. If the Helmholtz projection is applied to the linearized incompressible Navier-Stokes equations, theStokes 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${\displaystyle P\mathrm{\Delta }}$ is called theStokes operator.^[32]

In the theory ofdynamical systems, Helmholtz decomposition can be used to determine "quasipotentials" as well as to computeLyapunov functions in some cases.^[33][34][35]

For some dynamical systems such as theLorenz system (Edward N. Lorenz, 1963^[36]), a simplified model foratmosphericconvection, aclosed-form expression of the Helmholtz decomposition can be obtained:$${\displaystyle \dot{\mathbf{r}}=\mathbf{F}(\mathbf{r})=[a(r_2-r_1),r_1(b-r_3)-r_2,r_1{r}_2-c{r}_3].}$$The Helmholtz decomposition of${\displaystyle \mathbf{F}(\mathbf{r})}$, with the scalar potential${\displaystyle \mathrm{\Phi }(\mathbf{r})=\frac{a}{2}{r}_1^2+\frac{1}{2}{r}_2^2+\frac{c}{2}{r}_3^2}$ is given as:

$${\displaystyle \mathbf{G}(\mathbf{r})=[-a{r}_1,-r_2,-c{r}_3],}$$$${\displaystyle \mathbf{R}(\mathbf{r})=[+a{r}_2,b{r}_1-r_1{r}_3,r_1{r}_2].}$$

The quadratic scalar potential provides motion in the direction of the coordinate origin, which is responsible for the stablefix point for some parameter range. For other parameters, the rotation field ensures that astrange attractor is created, causing the model to exhibit abutterfly effect.^[8][37]

Inmagnetic 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).^[38] In this way, the complex shear modulus can be calculated without contributions from compression waves.

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.^[15][39]

• Clebsch representation for a related decomposition of vector fields
• Darwin Lagrangian for an application
• Poloidal–toroidal decomposition for a further decomposition of the divergence-free component${\displaystyle \mathrm{\nabla }\times \mathbf{A}}$.
• Scalar–vector–tensor decomposition
• Hodge theory generalizing Helmholtz decomposition
• Polar factorization theorem
• Helmholtz–Leray decomposition used for defining theLeray projection

• 1849 introductions
• 1849 in science
• Vector calculus
• Theorems in mathematical analysis
• Analytic geometry
• Hermann von Helmholtz
• Theorems in calculus

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