# Both positive and negative curvature?

• anorlunda
In summary, all curvature invariants are finite at the horizon, but they're not always the same size.
anorlunda
Staff Emeritus
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.

anorlunda said:
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).

anorlunda said:
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.

anorlunda said:
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.

pervect said:
All of the curvature components are finite.
Not in standard Schwarzschild coordinates. See L.22 here (http://onlinelibrary.wiley.com/doi/10.1002/9783527622061.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.

DaleSpam said:
Not in standard Schwarzschild coordinates. See L.22 here (http://onlinelibrary.wiley.com/doi/10.1002/9783527622061.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.

bcrowell said:
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.

DaleSpam said:
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.

Oops, I'd missed your #3 :-)

My bad - I shold have calculated them rather than going frommemory.

## 1. What is positive and negative curvature?

Positive and negative curvature refer to the shape of a 2-dimensional surface. Positive curvature is when the surface curves outward, like a sphere, while negative curvature is when the surface curves inward, like a saddle.

## 2. How can positive and negative curvature be measured?

Positive and negative curvature can be measured using mathematical concepts such as Gaussian curvature and sectional curvature. These measurements allow scientists to quantify the amount of curvature on a given surface.

## 3. What are some examples of objects with positive and negative curvature?

Examples of objects with positive curvature include spheres, balloons, and the Earth's surface. Examples of objects with negative curvature include saddles, Pringles chips, and some types of lenses used in optics.

## 4. How does positive and negative curvature affect the behavior of light?

Positive and negative curvature can affect the behavior of light by bending the path of light rays. In positive curvature, light rays will converge towards the center of curvature, while in negative curvature, light rays will diverge away from the center.

## 5. What are the real-world applications of studying positive and negative curvature?

Studying positive and negative curvature has many real-world applications, such as in architecture and engineering for designing structures that can withstand different types of curvature. It is also important in fields like physics and cosmology for understanding the shape of the universe and how it affects the behavior of matter and light.

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