lamellar fieldLaminarField2004-10-11 14:13:382008-08-27 13:54:26Definitiongradientscalar potentialpotentialrotor% this is the default PlanetMath preamble. as your knowledge
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% define commands hereA vector field \,$\vec{F} = \vec{F}(x,\,y,\,z)$,\, defined in an open set $D$ of $\mathbb{R}^3$, is\, {\em lamellar}\, if the condition
$$\nabla\!\times\!\vec{F} = \vec{0}$$
is satisfied in every point \,$(x,\,y,\,z)$\, of $D$.
Here, $\nabla\!\times\!\vec{F}$ is the curl or {\em rotor} of $\vec{F}$.\, The condition is equivalent with both of the following:
\begin{itemize}
\item The line integrals
$$\oint_s \vec{F}\cdot d\vec{s}$$
taken around any \PMlinkescapetext{closed} contractible curve $s$ vanish.
\item The vector field has a \PMlinkescapetext{{\em scalar potential}}\, $u = u(x,\,y,\,z)$\, which has continuous partial derivatives and which is up to a \PMlinkescapetext{constant term} unique in a simply connected domain; the scalar potential means that
$$\vec{F} = \nabla u.$$
\end{itemize}
The scalar potential has the expression
$$u = \int_{P_0}^P\vec{F}\cdot d\vec{s},$$
where the point $P_0$ may be chosen freely,\, $P = (x,\,y,\,z)$.
\textbf{Note.}\, In physics, $u$ is in general replaced with\, $V = -u$.\, If the $\vec{F}$ is interpreted as a \PMlinkescapetext{force}, then the potential $V$ is equal to the work made by the \PMlinkescapetext{force} when its point of application is displaced from $P_0$ to infinity.