Calculate Unit Normal Vector for Metric Tensor

In summary: 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.
  • #1
Onyx
134
4
TL;DR Summary
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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  • #3
PeterDonis said:
Unit normal vector to what?
To the hypersurface of constant ##x^0##, whatever ##x^0## may be.
 
  • #4
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.
 
  • #5
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?
 
  • #6
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?
 
  • #7
Onyx said:
How do I do what you suggested?
What is the definition of a normal vector to a surface?
 
  • #8
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.
 
  • #9
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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FAQ: Calculate Unit Normal Vector for Metric Tensor

1. What is a unit normal vector?

A unit normal vector is a vector that is perpendicular to a surface or curve at a specific point. It has a magnitude of 1 and is commonly used in mathematics and physics to represent the direction of a surface or curve.

2. What is a metric tensor?

A metric tensor is a mathematical object that describes the local geometry of a space. It is used to calculate distances, angles, and other geometric properties in a particular coordinate system.

3. Why is it important to calculate the unit normal vector for a metric tensor?

Calculating the unit normal vector for a metric tensor is important because it allows us to determine the direction of the surface or curve at a specific point. This information is crucial in many applications, such as determining the curvature of a surface or finding the normal force acting on an object.

4. How do you calculate the unit normal vector for a metric tensor?

The unit normal vector for a metric tensor can be calculated by first finding the inverse of the metric tensor, then taking the square root of the diagonal elements of the inverse matrix. The resulting vector will have a magnitude of 1 and will point in the direction of the normal to the surface or curve.

5. What are some real-world applications of calculating the unit normal vector for a metric tensor?

Calculating the unit normal vector for a metric tensor is used in various fields, such as physics, engineering, and computer graphics. Some specific applications include determining the direction of the normal force on a body in mechanics, finding the curvature of a surface in differential geometry, and creating realistic lighting effects in computer graphics.

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