# Both positive and negative curvature?

by anorlunda
Tags: curvature
 P: 171 What happens to the Reimann tensor at the event horizon of a black hole? Do some of the 24 components become zero or infinite? What happens to parallel transport of a vector on the surface of an event horizon that is different than on a surface outside the event horizon? I'm newly educated on general relativity and hopeful that these new tools may enable me to see things differently.
Emeritus
P: 7,599
 Quote by anorlunda What happens to the Reimann tensor at the event horizon of a black hole? Do some of the 24 components become zero or infinite? What happens to parallel transport of a vector on the surface of an event horizon that is different than on a surface outside the event horizon? I'm newly educated on general relativity and hopeful that these new tools may enable me to see things differently.
All of the curvature components are finite. For the most part they'll be m/r^3 in geometric units, i.e. they'll be equal to the tidal forces. They'll be independent of the radial infal velocity, too. (Tangential velocity will change them).
Mentor
P: 16,989
 Quote by anorlunda What happens to the Reimann tensor at the event horizon of a black hole? Do some of the 24 components become zero or infinite?
The components of a tensor depend on the coordinate system used to express them. So the answer to this question depends on which coordinate system you are using.

In standard Schwarzschild coordinates there are many zero components throughout the spacetime and there are some infinite components in the limit as you approach the horizon.

In Kruskal Szekeres coordinates there are also many zero components throughout the spacetime but there are no infinite components in the limit as you approach the horizon.

All curvature invariants are finite at the horizon. Furthermore, the larger the black hole the smaller the curvature invariants at the horizon. For a supermassive black hole there could be less tidal gravity than at the surface of the earth.

 Quote by anorlunda What happens to parallel transport of a vector on the surface of an event horizon that is different than on a surface outside the event horizon?
Nothing. The event horizon is merely an outgoing null surface, so you cannot have any outgoing timelike paths which cross it.

Mentor
P: 16,989
Both positive and negative curvature?

 Quote by pervect All of the curvature components are finite.
Not in standard Schwarzschild coordinates. See L.22 here (http://onlinelibrary.wiley.com/doi/1...2061.app12/pdf) for a list of all of the non-zero components. For example:
$${R^{\theta}}_{r \theta r}={R^{\phi}}_{r \phi r}=\frac{M}{(2M-r)r^2}$$
goes to -∞ as r goes to 2 M.
Emeritus
PF Gold
P: 5,585
 Quote by DaleSpam Not in standard Schwarzschild coordinates. See L.22 here (http://onlinelibrary.wiley.com/doi/1...2061.app12/pdf) for a list of all of the non-zero components. For example: $${R^{\theta}}_{r \theta r}={R^{\phi}}_{r \phi r}=\frac{M}{(2M-r)r^2}$$ goes to -∞ as r goes to 2 M.
But let's make sure nobody is led to believe that this should be interpreted as a curvature singularity at r=2M. This is a coordinate singularity. No physical measurement gives a result that diverges to infinity as r approaches 2M. If we chose different coordinates, we could make all of R's components finite. Even in these coordinates, actual observable quantities like the Kretschmann scalar $R_{abcd}^{abcd}$ are finite at r=2M, because the infinities "mysteriously" cancel.
Mentor
P: 16,989
 Quote by bcrowell But let's make sure nobody is led to believe that this should be interpreted as a curvature singularity at r=2M. This is a coordinate singularity. No physical measurement gives a result that diverges to infinity as r approaches 2M. If we chose different coordinates, we could make all of R's components finite. Even in these coordinates, actual observable quantities like the Kretschmann scalar $R_{abcd}^{abcd}$ are finite at r=2M, because the infinities "mysteriously" cancel.
Agreed. I specifically mentioned in post 3 that all curvature invariants are finite at the horizon, but I certainly could have been more emphatic about it.
Emeritus