B Calculate Unit Normal Vector for Metric Tensor

Onyx
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Calculating The Unit Normal Vector For Any Metric Tensor
How do I calculate the unit normal vector for any metric tensor?
 
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PeterDonis said:
Unit normal vector to what?
To the hypersurface of constant ##x^0##, whatever ##x^0## may be.
 
That’s not always possible. Since this is the relativity forum I assume your metric is indefinite which means the surface may be light-like.

If you assume a non-light-like surface: Just compute the surface normal and normalize it.
 
Onyx said:
To the hypersurface of constant ##x^0##, whatever ##x^0## may be.
How would you find the normal of a surface in Euclidean geometry?
 
Orodruin said:
That’s not always possible. Since this is the relativity forum I assume your metric is indefinite which means the surface may be light-like.

If you assume a non-light-like surface: Just compute the surface normal and normalize it.
How do I do what you suggested?
 
Onyx said:
How do I do what you suggested?
What is the definition of a normal vector to a surface?
 
PeterDonis said:
What is the definition of a normal vector to a surface?
Well, a normal vector to a surface is a vector that is orthogonal to said surface, and a unit normal vector is a normal vector of unit length.
 
Onyx said:
a normal vector to a surface is a vector that is orthogonal to said surface, and a unit normal vector is a normal vector of unit length.
Yes. So how would you find out if a given vector is orthogonal to a given surface, given that you know the metric ##g_{ab}##, the vector ##v^a##, and the surface? And given a vector, how would you find a unit vector in the same direction?
 
  • #10
More concretely, what you are looking for here according to
Onyx said:
To the hypersurface of constant ##x^0##, whatever ##x^0## may be.
is the normal of the level surface of some function ##f##. How would you find that in Euclidean space?
 
  • #11
PeterDonis said:
Yes. So how would you find out if a given vector is orthogonal to a given surface, given that you know the metric ##g_{ab}##, the vector ##v^a##, and the surface? And given a vector, how would you find a unit vector in the same direction?
Is it by a cross-product?
 
  • #12
Orodruin said:
More concretely, what you are looking for here according to

is the normal of the level surface of some function ##f##. How would you find that in Euclidean space?
Cross-product?
 
  • #13
Onyx said:
Is it by a cross-product?
No. How would you test for orthogonality in Euclidean geometry?
 
  • #14
PeterDonis said:
No. How would you test for orthogonality in Euclidean geometry?
If the dot-product of one vector with the other is zero.
 
  • #15
Onyx said:
If the dot-product of one vector with the other is zero.
Yes. Is there a dot product in spacetime?
 
  • #16
PeterDonis said:
Yes. Is there a dot product in spacetime?
Yes. I would assume, then, that finding the normal vector to the spacelike hypersurface would involve dot-producting a surface vector with a purely timelike one.
 
  • #17
Onyx said:
I would assume, then, that finding the normal vector to the spacelike hypersurface would involve dot-producting a surface vector with a purely timelike one.
Yes. But in your OP, you said "any metric tensor". That would include cases in which ##x^0## is not timelike and/or surfaces of constant ##x^0## are not spacelike. Did you intend to exclude those cases?
 
  • #18
PeterDonis said:
Yes. But in your OP, you said "any metric tensor". That would include cases in which ##x^0## is not timelike and/or surfaces of constant ##x^0## are not spacelike. Did you intend to exclude those cases?
No, I don't think so, because I'm partly thinking about EF coordinates, where I believe ##x^0## is not timelike.
 
  • #19
Onyx said:
I'm partly thinking about EF coordinates, where I believe ##x^0## is not timelike.
Not only that, but a surface of constant ##x^0## is not spacelike, it's null. That means there is no such thing as a unit normal vector to it. See post #4.
 
  • #20
PeterDonis said:
Not only that, but a surface of constant ##x^0## is not spacelike, it's null. That means there is no such thing as a unit normal vector to it. See post #4.
But isn't there still a lapse and shift?
 
  • #21
Onyx said:
But isn't there still a lapse and shift?
Lapse and shift don't even make sense for EF coordinates on Schwarzschild spacetime. They only make sense when you have one timelike and three spacelike coordinates.
 
  • #22
Onyx said:
But isn't there still a lapse and shift?
Lapse and shift rely on the concept of a vector pointing out of a spacelike plane that is normal to all vectors in the plane. You'd think you could just change the word "spacelike" to "null", but there's a problem - null vectors are normal to themselves (since ##n^an^bg_{ab}=0## is the definition of a null vector). So the "normal to the plane" concept falls apart. So there isn't a uniquely defined vector pointing out of a null plane so there's no well-defined concept of lapse and (hence) shift.
 
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