# Shape Operator for Schwarzschild spacetime in 2-dim

1. Sep 12, 2012

### honeytrap

Hello:

I would like to understand how to compute the shape operator (and eigenvalues etc) for a complex example like the Schwarzschild spacetime. It's easy for a submanifold in Euclidean space, but I don't know how to do it for the more advanced examples like the schwarzschild spacetime in 2-dim.

QUESTION:

Given the Schwarzschild metric in Eddington-Finkelstein coordinates in 2-
dim:
ds^2 = -(1 - 2M/r)dv^2 + 2dvdr , which is a surface.

How can I compute the Schwarzschild
a) Shape operator (Weingarten map or second fundamental tensor)
b) unit normal vectorfield
c) Eigenvalues of the Shape operator matrix?

I know all the definitons etc., but the 2-dim Schwarzschild is not the typical
surface/ submanifold embedded in Euclidean space.
Could someone show me how to compute the steps a)-c) so that I understand it for this
example (and can apply it to other examples/metrics)?

Last edited: Sep 12, 2012
2. Sep 12, 2012

### Bill_K

Seriously, you want to describe the curvature of a 2-dimensional submanifold in a 4-dimensional one? In general that's going to be quite complicated -- a blend of the Gauss-Codazzi equations with the Serret-Frenet ones for a curve. Not simply a second fundamental form.

But anyway for the case you're describing, the (u,v) manifold in Schwarzschild spacetime, isn't the exterior curvature identically zero, by reflection symmetry? The normals to the surface would lie in the 2-sphere (θ, φ), and there would be no reason for the surface to be curved preferentially in any of those directions. I say, exteriorly speaking, it's flat.

3. Sep 12, 2012

### honeytrap

I am actually interested in the 2-dimensional (u,v)- "Schwarzschild" manifold (surface) defined by the metric above.
From the shape operator (e.g. the eigenvalues = principal curvatures) you can derive the Gaussian curvature and the mean curvature. I computed the Ricci tensor and then the Gaussian curvature for the mentioned 2-dim manifold and both of them are not zero. Thus the eigenvalues (principal curvatures) of the shape operator matrix cannot be zero either. (Contradiction to flatness?)

I am interested in the eigenvalues and eigenvectors, but don't know how to compute them. I obviously need first the shape operator matrix.
Any help is appreciated!

4. Sep 12, 2012

### Ben Niehoff

If the shape operator includes the mean curvature, than there's no way you can compute it purely intrinsically! It depends upon the embedding.

Yes, having just looked up the definition of the shape operator, it certainly depends on the embedding. Intrinsically-defined surfaces don't have normal vectors at all.

5. Sep 12, 2012

### honeytrap

This means then that it doesn't make sense to calculate the shape operator at all?
Or is there a canonical embedding that could be picked (e.g. 3-dim Euclidean/ Minkowski space)?
The computation would depend on the embedding and thus the result would not deliver any general information that could be useful for a discussion?

6. Sep 12, 2012

### Bill_K

I assumed that since it was a subspace of Schwarzschild, you were embedding it in Schwarzschild! In which case my remark about the reflection symmetry applies.