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?
To the hypersurface of constant ##x^0##, whatever ##x^0## may be.PeterDonis said:Unit normal vector to what?
How would you find the normal of a surface in Euclidean geometry?Onyx said:To the hypersurface of constant ##x^0##, whatever ##x^0## may be.
How do I do what you suggested?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.
What is the definition of a normal vector to a surface?Onyx said:How do I do what you suggested?
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.PeterDonis said:What is the definition of a normal vector to a surface?
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?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.
is the normal of the level surface of some function ##f##. How would you find that in Euclidean space?Onyx said:To the hypersurface of constant ##x^0##, whatever ##x^0## may be.
Is it by a cross-product?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?
Cross-product?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?
No. How would you test for orthogonality in Euclidean geometry?Onyx said:Is it by a cross-product?
If the dot-product of one vector with the other is zero.PeterDonis said:No. How would you test for orthogonality in Euclidean geometry?
Yes. Is there a dot product in spacetime?Onyx said:If the dot-product of one vector with the other is zero.
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.PeterDonis said:Yes. Is there a dot product in spacetime?
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?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.
No, I don't think so, because I'm partly thinking about EF coordinates, where I believe ##x^0## is not timelike.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?
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.Onyx said:I'm partly thinking about EF coordinates, where I believe ##x^0## is not timelike.
But isn't there still a lapse and shift?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.
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.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.Onyx said:But isn't there still a lapse and shift?