Invector calculus, acomplex lamellar vector field is avector field which is orthogonal to a family of surfaces. In the broader context ofdifferential geometry, complex lamellar vector fields are more often calledhypersurface-orthogonal vector fields. They can be characterized in a number of different ways, many of which involve thecurl. Alamellar vector field is a special case given by vector fields with zero curl.

The adjective "lamellar" derives from the noun "lamella", which means a thin layer. Thelamellae to which "lamellar vector field" refers are the surfaces of constant potential, or in the complex case, the surfaces orthogonal to the vector field.^[1]

Invector calculus, acomplex lamellar vector field is avector field in three dimensions which isorthogonal to its owncurl.^[2] That is,

${\displaystyle \mathbf{F}\cdot (\mathrm{\nabla }\times \mathbf{F})=0.}$

The termlamellar vector field is sometimes used as a synonym for the special case of anirrotational vector field, meaning that^[3]

${\displaystyle \mathrm{\nabla }\times \mathbf{F}=\mathbf{0}.}$

Complex lamellar vector fields are precisely those that are normal to a family of surfaces. An irrotational vector field is locally thegradient of a function, and is therefore orthogonal to the family oflevel surfaces (theequipotential surfaces).^[4] Any vector field can be decomposed as the sum of an irrotational vector field and a complex lamellar field.^[5]

In greater generality, a vector fieldF on apseudo-Riemannian manifold is said to behypersurface-orthogonal if through an arbitrary point there is asmoothly embedded hypersurface which, at all of its points, is orthogonal to the vector field. By theFrobenius theorem this is equivalent to requiring that theLie bracket of any smooth vector fields orthogonal toF is still orthogonal toF.^[6]

The condition of hypersurface-orthogonality can be rephrased in terms of thedifferential 1-formω which is dual toF. The previously given Lie bracket condition can be reworked to require that theexterior derivativedω, when evaluated on any two tangent vectors which are orthogonal toF, is zero.^[6] This may also be phrased as the requirement that there is a smooth 1-form whosewedge product withω equalsdω.^[7]

Alternatively, this may be written as the condition that thedifferential 3-formω ∧ dω is zero. This can also be phrased, in terms of theLevi-Civita connection defined by the metric, as requiring that the totally anti-symmetric part of the 3-tensor fieldω_i∇_jω_k is zero.^[8] Using a different formulation of the Frobenius theorem, it is also equivalent to require thatω is locally expressible asλ du for some functionsλ andu.^[9]

In the special case of vector fields onthree-dimensional Euclidean space, the hypersurface-orthogonal condition is equivalent to the complex lamellar condition, as seen by rewritingω ∧ dω in terms of theHodge star operator as∗⟨ω, ∗dω⟩, with∗dω being the 1-form dual to the curl vector field.^[10]

Hypersurface-orthogonal vector fields are particularly important ingeneral relativity, where (among other reasons) the existence of aKilling vector field which is hypersurface-orthogonal is one of the requirements of astatic spacetime.^[11] In this context, hypersurface-orthogonality is sometimes calledirrotationality, although this is in conflict with the standard usage in three dimensions.^[12] Another name isrotation-freeness.^[13]

An even more general notion, in the language ofPfaffian systems, is that of acompletely integrable 1-formω, which amounts to the conditionω ∧ dω = 0 as given above.^[14] In this context, there is no metric and so there is no notion of "orthogonality".

• Beltrami vector field
• Conservative vector field

• Vector calculus

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