- #1

- 371

- 0

*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.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter dimension10
- Start date

- #1

- 371

- 0

Thanks.

- #2

George Jones

Staff Emeritus

Science Advisor

Gold Member

- 7,422

- 1,063

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

- 371

- 0

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

Bill_K

Science Advisor

- 4,155

- 201

You need to use two sets of coordinates: u

Then you show that x

- #5

- 371

- 0

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.

- #6

- 4,239

- 1

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?

- #7

Bill_K

Science Advisor

- 4,155

- 201

- #8

- 4,239

- 1

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.

Last edited:

- #9

Bill_K

Science Advisor

- 4,155

- 201

In the internal coordinate system we have only ε

Share: