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*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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- Thread starter dimension10
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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

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Thanks.

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- #2

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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

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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.

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You need to use two sets of coordinates: u

Then you show that x

- #5

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Bill_K said:

You need to use two sets of coordinates: u^{α}in the surface and x^{i}in the manifold. You need the relationship between the two: x^{i}_{α}≡ ∂x^{i}/∂u^{α}. You need metric tensors a^{αβ}in the surface and g^{ij}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 x^{i}_{α;β}= b_{αβ}n^{i}, where n^{i}is the unit normal vector and b_{αβ}is a symmetric surface tensor, the "second fundamental form". One can then derive the Weingarten formulas, n^{i}_{;α}= - a^{βγ}b_{βα}x^{i}_{γ}and solve for n.

Thanks.

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Bill_K said:

You need to use two sets of coordinates: u^{α}in the surface and x^{i}in the manifold. You need the relationship between the two: x^{i}_{α}≡ ∂x^{i}/∂u^{α}. You need metric tensors a^{αβ}in the surface and g^{ij}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 x^{i}_{α;β}= b_{αβ}n^{i}, where n^{i}is the unit normal vector and b_{αβ}is a symmetric surface tensor, the "second fundamental form". One can then derive the Weingarten formulas, n^{i}_{;α}= - a^{βγ}b_{βα}x^{i}_{γ}and solve for n.

How can there be a unique vector, unless the (locally isometric) product space is R^5?

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Bill_K said:

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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In the internal coordinate system we have only ε

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