Talk:PlanetPhysics/Lamellar Field

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A \htmladdnormallink{vector field}{http://planetphysics.us/encyclopedia/NeutrinoRestMass.html} \,$\vec{F} = \vec{F}(x,\,y,\,z)$,\, 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 \htmladdnormallink{curl}{http://planetphysics.us/encyclopedia/Curl.html} 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 closed contractible curve $s$ vanish. \item The vector field has a {\em scalar potential}\, $u = u(x,\,y,\,z)$\, which has continuous partial derivatives and which is up to a constant term unique in a simply connected \htmladdnormallink{domain}{http://planetphysics.us/encyclopedia/Bijective.html}; the \htmladdnormallink{scalar}{http://planetphysics.us/encyclopedia/Vectors.html} 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 force, then the potential $V$ is equal to the \htmladdnormallink{work}{http://planetphysics.us/encyclopedia/Work.html} made by the force when its point of application is displaced from $P_0$ to infinity.

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