The unit vector normal to the curvature of spacetime

In summary, the conversation discusses the concept of a normal vector to a curved region of spacetime and how it relates to classical general relativity and the differential geometry of a surface embedded in a manifold. The process of finding the coordinates of the normal vector is described, involving two sets of coordinates, metric tensors, and a covariant derivative. The uniqueness of the normal vector is also discussed, with a mention of the potential non-existence or non-uniqueness of an embedding of a flat 4-space in 5-space. Finally, the equation A_a = {\epsilon_a}^{bcde} \sqrt{-g} B_{bcde} is mentioned as a possible formulation for the normal vector in terms of contravariant vectors and
  • #1
dimension10
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Let us say there is a curved region of spacetime whose curvature is [tex]\kappa(s)[/tex]. How does one find the coordinates of the unit vector normal to a certain point on the region of spacetime? I tried searching Hamilton's principle and the general theory of relativity but I could not find any equation which describes the coordinates of the normal vector to the curved region of spacetime.

Thanks.
 
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  • #2
In classical general relativity, there no normal to spacetime curavture. Since spacetime is all there is, how could something be normal to it.

Perhaps you mean the normal to a "curved" hypersurface. For example, if a 3-dimensional hypersurface is picked out by the level set [itex]\Phi \left( x^\mu \right) = 0[/itex], then [itex]\partial \Phi / \partial x^\mu[/itex] is normal to the hypersurface.

Perhaps you're thinking of spacetime as a brane in a higher-dimensional bulk, in which case there is a normal to spacetime sticking out into the bulk.
 
  • #3
George Jones said:
Perhaps you're thinking of spacetime as a brane in a higher-dimensional bulk, in which case there is a normal to spacetime sticking out into the bulk.

Yes. That is what I am asking.
 
  • #4
There's a considerable body of work relating to the differential geometry of a surface embedded in a manifold. I can only summarize.

You need to use two sets of coordinates: uα in the surface and xi in the manifold. You need the relationship between the two: xiα ≡ ∂xi/∂uα. You need metric tensors aαβ in the surface and gij in the manifold. You need a covariant derivative that includes both sets of Christoffel symbols: one for the x's and one for the u's.

Then you show that xiα;β = bαβ ni, where ni is the unit normal vector and bαβ is a symmetric surface tensor, the "second fundamental form". One can then derive the Weingarten formulas, ni = - aβγbβαxiγ and solve for n.
 
  • #5
Bill_K said:
There's a considerable body of work relating to the differential geometry of a surface embedded in a manifold. I can only summarize.

You need to use two sets of coordinates: uα in the surface and xi in the manifold. You need the relationship between the two: xiα ≡ ∂xi/∂uα. You need metric tensors aαβ in the surface and gij in the manifold. You need a covariant derivative that includes both sets of Christoffel symbols: one for the x's and one for the u's.

Then you show that xiα;β = bαβ ni, where ni is the unit normal vector and bαβ is a symmetric surface tensor, the "second fundamental form". One can then derive the Weingarten formulas, ni = - aβγbβαxiγ and solve for n.

Thanks.
 
  • #6
Bill_K said:
There's a considerable body of work relating to the differential geometry of a surface embedded in a manifold. I can only summarize.

You need to use two sets of coordinates: uα in the surface and xi in the manifold. You need the relationship between the two: xiα ≡ ∂xi/∂uα. You need metric tensors aαβ in the surface and gij in the manifold. You need a covariant derivative that includes both sets of Christoffel symbols: one for the x's and one for the u's.

Then you show that xiα;β = bαβ ni, where ni is the unit normal vector and bαβ is a symmetric surface tensor, the "second fundamental form". One can then derive the Weingarten formulas, ni = - aβγbβαxiγ and solve for n.

How can there be a unique vector, unless the (locally isometric) product space is R^5?
 
  • #7
Phrak, You're right, this assumes we're dealing with an n-dimensional surface embedded in an n+1-dimensional space. In the general case one has a set of equations generalizing the Serret-Frenet formulas for a curve, namely there will be a principal normal, second normal, etc.
 
  • #8
Bill_K said:
Phrak, You're right, this assumes we're dealing with an n-dimensional surface embedded in an n+1-dimensional space. In the general case one has a set of equations generalizing the Serret-Frenet formulas for a curve, namely there will be a principal normal, second normal, etc.

Ok, assuming we have a embedded 4 surface in 5 dimenions-

I don't follow your math, but I would not start with contravariant vectors but densities.

[tex]A_a = {\epsilon_a}^{bcde} \sqrt{-g} B_{bcde}[/tex]

It seems that the dual of A is a normal vector to the surface at any point where B=B(w,x,y,z,t) is any 4-form. I couldn't say if this is equivalent to your formulation.

Edit: I should have specificed that B is a 4-form at a point lying in the 4-surface.
 
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  • #9
The problem is we don't have the embedding. We start given only the intrinsic geometry of the surface and try to construct an embedding. It may not exist, and if it does exist it may not be unique. For instance a flat 4-space immersed in 5-space could be rolled up like a piece of paper.

In the internal coordinate system we have only εabcd to deal with, and a 4-form is equivalent to a scalar. Constructing the normal vector requires 5-dimensional coordinates, and we aren't given the relationship between them, xiα ≡ ∂xi/∂uα.
 

1. What is the unit vector normal to the curvature of spacetime?

The unit vector normal to the curvature of spacetime is a mathematical concept used in the theory of relativity to describe the direction in which spacetime is curving at a given point. It is a unit vector, meaning it has a magnitude of 1, and it is perpendicular to the tangent plane of spacetime at that point.

2. How is the unit vector normal to the curvature of spacetime calculated?

The unit vector normal to the curvature of spacetime is calculated using the mathematical concept of a normal vector, which is perpendicular to a given surface. In the context of spacetime, the normal vector is calculated using the Riemann curvature tensor, which describes the curvature of spacetime at a specific point.

3. Why is the unit vector normal to the curvature of spacetime important?

The unit vector normal to the curvature of spacetime is important because it helps us understand the curvature of spacetime, which is a fundamental concept in the theory of relativity. It also allows us to make predictions about the behavior of particles and objects in the presence of gravitational fields.

4. How does the unit vector normal to the curvature of spacetime relate to gravity?

The unit vector normal to the curvature of spacetime is closely related to gravity because it describes the direction in which spacetime is curving due to the presence of massive objects. Mass and energy are sources of curvature in spacetime, and the unit vector normal helps us understand the strength and direction of this curvature.

5. Is the unit vector normal to the curvature of spacetime constant?

No, the unit vector normal to the curvature of spacetime is not constant. It varies depending on the location in spacetime and the distribution of mass and energy in that region. It is also affected by the motion of objects and the passage of time, as described by the theory of relativity.

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