# Extrinsic curvature

1. Aug 9, 2010

### mersecske

In the definition of the extrinsic curvature, there is the normal vector.
It depends on the sign of the normal vector?
Because a normal vector can be directed in two ways.
For example the curvature of a circle on the plane
has different curvature from inside and outside!
But this is analogue to the extrinsic curvature?

2. Aug 9, 2010

### bcrowell

Staff Emeritus
What cases do you have in mind here? The curvature of a one-dimensional curve embedded in a plane? The mean curvature of a two-dimensional surface embedded in three dimensions? In the first case, http://en.wikipedia.org/wiki/Curvature#One_dimension_in_two_dimensions:_Curvature_of_plane_curves defines a curvature and a signed curvature. In the second case, the sign does appear to depend on the choice of normal: http://en.wikipedia.org/wiki/Mean_curvature#Surfaces_in_3D_space

Of course, extrinsic curvature isn't something we typically care about in GR -- you can't even define any measures of extrinsic curvature if all you're given is a metric expressed in some coordinates.

3. Aug 10, 2010

### mersecske

Second fundamental form of a timelike hypersurface in 4D.

4. Aug 10, 2010

### dodelson

Actually, extrinsic curvature turns out to be pretty important in the Hamiltonian/ADM formulation of GR. The general idea is that we can foliate spacetime with spacelike hypersurfaces and specify initial data on one of them in the form of a spatial metric h. Einstein's equations then determine the evolution of the extrinsic curvature of the metric h.

Extrinsic curvature is also important for GR on manifolds with boundary, i.e. AdS. It turns out that we need to supplement the Einstein-Hilbert action with a boundary term in order to get a well-defined variational principle; the boundary term, called the Gibbons-Hawking term, happens to be the trace of the extrinsic curvature! You can use this boundary term to define a notion of energy on the boundary, the Brown-York stress tensor, which is intimately related to the extrinsic curvature.

In fact, it doesn't matter which normal we choose. Suppose we have two normal vectors n and m which are orthogonal to a spacelike hypersurface Sigma with 3-metric h. Since both n and m are normal to Sigma, their derivatives along a direction tangent to Sigma must agree. But the extrinsic curvature is defined as the Lie derivative of h along the normal vector, evaluated on Sigma, so it doesn't matter whether we choose n or m.

-Matt

5. Aug 10, 2010

### mersecske

On the page:

http://en.wikipedia.org/wiki/Second_fundamental_form

There is:

"The sign of the second fundamental form
depends on the choice of direction of n
(which is called a co-orientation of the hypersurface)"

I think also that:
if we change the orientation of n ->
the extrinsic curvature change from K to -K
Am I right?

Note that I need extrinsic curvature
in the formalism of junction of thin timelike shells
The junction condition is

[K] = K_out-K_in = S (kind of surface energy tensor)

But most of the literature do not note
that K is not determined without fixing the direction of the normal vector!

6. Oct 22, 2010

### mersecske

The extrinsic curvature definition has an ambiguity in the literature.
In some papers

$$K_{ab} = \frac{1}{2}h^c_ah^d_b\mathcal{L}_nh_{cd}$$

and sometimes the signum differs:

$$K_{ab} = -\frac{1}{2}h^c_ah^d_b\mathcal{L}_nh_{cd}$$

I know that it does not metter, it is just a definition, but which one is better?
I would like to use the common one, which is used in Euclidean 3D space.
Which definition gives back the 3D results, which is used in grammar school?