# Can a black hole decrease in spacetime curvature?

Since a black hole goes to a singularity, theoretically wouldn't added mass to that point decrease the spacetime curvature by increasing of the circumference, and then not have a loss in information.

pervect
Staff Emeritus
Since a black hole goes to a singularity, theoretically wouldn't added mass to that point decrease the spacetime curvature by increasing of the circumference, and then not have a loss in information.

Spacetime curvature is not represented by a single number, but a rather complex structure consisting of many numbers which are called "components" of the spacetime curvature.

There isn't any clear meaning to "decreasing" or "increasing" a set of numbers. What do you call it if one member of the set increases, while the other decreases? To take a simple example, it's not meaningful to ask whether [1,1] is "greater" or "lesser" than [2, 1/2], though one can meaningfully say they aren't equal.

There are some additional complexities. The set of numbers that completely describes spacetime curvature is called the Riemann curvature tensor. There are other smaller sets of numbers that also describe spacetime curavature that are important to the theory that are not, however, as complete as the Riemann tensor, but the term "spacetime curvature" is sufficiently vague that it's not clear which set of numbers one is talking about. For the time being, though, we can simplify things by assuming that one is talking about the most complete representation of spacetime curvature, the full set of numbers in the Riemann tensor, and not some important but less general set.

Description of the "mass" of some collection of matter is not, by itself, sufficient to compute all the numbers in the Riemann tensor. One needs additional information about the composition of the matter, for instance it's momentum, and the value of any internal pressures, to compute the Riemann tensor.

If I had a small clump of clay and gave it a spherical orientation adding more clay a.k.a mass to the sphere would increase its circumference. As we both know, curvature becomes more intense as circumference decreases. Therefore, bigger cicurmference compared to increase mass will decrease curvature, right?

PeterDonis
Mentor
As we both know, curvature becomes more intense as circumference decreases.

For a simple 2-sphere in Euclidean 3-space, yes. But for a spherically symmetric gravitating mass in 4-dimensional spacetime, it's more complicated than that. The simplistic reasoning you are trying to use will not work for this case.

First, as pervect pointed out, spacetime curvature in 4-d spacetime is not a single number. It's a rank 4 tensor, the Riemann tensor. In 4-d spacetime this tensor has twenty independent components, so it takes twenty numbers to describe spacetime curvature in the general case.

Furthermore, each of those twenty numbers applies at a single point of spacetime; the numbers can change from point to point. So you have to specify where in spacetime you want to measure the curvature.

In the case of a black hole, or more generally a spherically symmetric gravitating mass, the spacetime is highly symmetric, so, for example, the curvature of spacetime anywhere on a given 2-sphere will be the same. But it will change from one 2-sphere to another. So you have to specify which 2-sphere (or, equivalently, which radial coordinate ##r##) you want to know the spacetime curvature on. And that curvature will still not be just a single number--it is not the same as the spatial curvature of the 2-sphere (which is a single number).

PeterDonis
Mentor
Is this an "A" question?

Not really, no. I've reclassified the thread as "I".

Is it OTT to use the "report" button in theses situations?

If you mean, to report a thread that you think is not correctly classified (should not be "A", for example), yes, you can use the report button for that.