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WannabeNewton

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- Thread starter WannabeNewton
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WannabeNewton

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HallsofIvy

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is a vector in that space, then the "derivative of f in the diretion [itex]\vec{v}[/itex]", the rate at which f is increasing in that direction, is given by [itex]D_\vec{v} f= \nabla f\cdot\vec{v}[/itex]. In particular, if [itex]\vec{v}[/itex]

is tangent to the surface f(X)= constant then f does not change: [itex]D_\vec{v} f= \nabla f\cdot \vec{v}= 0[/itex] which says directly that [itex]\nabla f[/itex] and [itex]\vec{v}[/itex]

are perpendicular. Since [itex]\vec{v}[/itex]

could be any vector in the tangent plane, [itex]\nabla f[/itex] is perpendicular to the tangent plane, i.e. normal to the surface.

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WannabeNewton

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Oh, the fact that I saw it in component form was confusing me. If I could just restate what you said, for my own clarification: since [itex]\Sigma [/itex] is specified by [itex]f(x) = const.[/itex] then on [itex]\Sigma [/itex], [itex]\triangledown _{\nu }f = 0 [/itex] for all [itex]\nu [/itex] so [itex]\xi ^{\mu } = g^{\mu \nu }\triangledown _{\nu }f = 0[/itex] identically as well on the hypersurface. Then, for all [itex]\mathbf{v}\in T_{p}(\Sigma )[/itex], [itex]\xi ^{\mu }v_{\mu } = 0[/itex] and since [itex]\mathbf{v}[/itex] lies on the tangent space to [itex]\Sigma [/itex] at p, [itex]\boldsymbol{\xi}[/itex] must be orthogonal to it so it is then normal to [itex]\Sigma [/itex]. Is this basically the gist of what you said?

Thank you so very much by the way. Cleared my head up; I think my problem was that I kept thinking of the function everywhere outside the hypersurface and not just what happens on it.

Thank you so very much by the way. Cleared my head up; I think my problem was that I kept thinking of the function everywhere outside the hypersurface and not just what happens on it.

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