Question about definition of a hypersurface

  • #1
tm33333
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In my notes on general relativity, hypersurfaces are defined as in the image. What confuses me is that if f=constant, surely the partial differential is going to be zero? I'm not sure if I'm missing something, but surely the function can't be equal to a constant and its partial differential be non-zero?

thanks.
 

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  • #2
Ibix
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It's requiring that ##\partial_af## be non-zero everywhere, then saying the subset of points with the same value of ##f## define a hypersurface. Analogously, you can define a function ##f(x,y)## on a two dimensional Euclidean plane and the lines of constant ##f## are the contour lines (1d analogues to 3d hypersurfaces). The gradient on a contour isn't zero, it is perpendicular to the contour line.

(Note that geographical contour lines can close, but a closed contour line encloses at least one point where the gradient is zero, so the definition of a hypersurface excludes this possibility).
 
  • #3
Orodruin
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Well, the requirement that ##\partial_a f=0## everywhere is a bit strict. It is sufficient that it is non-zero at the hypersurface being described by the particular constant. (Although you will need the full requirement if you intend to make a foliation of the manifold.)

As an example, consider the sphere in standard Euclidean space with ##f = x^2 + y^2 + z^2##. For ##R>0##, ##f = R^2## defines a sphere of radius ##R##, which is a level surface of ##f## in ##\mathbb R^3##.
 
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  • #4
tm33333
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Thank you both. That definitely clarifies things!
 
  • #5
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Moderator's note: Thread title edited to be more descriptive of the specific question.
 
  • #6
robphy
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Thank you both. That definitely clarifies things!
I guess you can say that they broke it down for you. :wink:
 

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