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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?
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.
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.
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.
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.
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.