B Calculate Unit Normal Vector for Metric Tensor

[Onyx]

TL;DR

Calculating The Unit Normal Vector For Any Metric Tensor

How do I calculate the unit normal vector for any metric tensor?

 

[PeterDonis]

Unit normal vector to what?

 

[Onyx]

Unit normal vector to what?

To the hypersurface of constant $x^0$, whatever $x^0$ may be.

 

[Orodruin]

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.

 

[Ibix]

To the hypersurface of constant $x^0$, whatever $x^0$ may be.

How would you find the normal of a surface in Euclidean geometry?

 

[Onyx]

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?

 

[PeterDonis]

How do I do what you suggested?

What is the definition of a normal vector to a surface?

 

[Onyx]

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.

 

[PeterDonis]

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?

 

[Orodruin]

More concretely, what you are looking for here according to

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?

 

[Onyx]

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?

 

[Onyx]

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?

 

[PeterDonis]

Is it by a cross-product?

No. How would you test for orthogonality in Euclidean geometry?

 

[Onyx]

No. How would you test for orthogonality in Euclidean geometry?

If the dot-product of one vector with the other is zero.

 

[PeterDonis]

If the dot-product of one vector with the other is zero.

Yes. Is there a dot product in spacetime?

 

[Onyx]

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.

 

[PeterDonis]

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?

 

[Onyx]

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.

 

[PeterDonis]

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.

 

[Onyx]

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?

 

[PeterDonis]

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.

 

[Ibix]

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.