# Naked Singularity, Black hole mass limit

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• Arman777
Gold Member
I came across a question on PSE. I am not sure its a violation to ask the same question here, but there's no answer to the question in there so I wanted to ask it here.

Quoting his question,

"Since the universe has a positive cosmological constant, there is an upper limit on the mass of the black holes as evident from the so-called Schwarzschild-de Sitter metric:

$$ds^2 = -f(r)dt^2 + \dfrac{1}{f(r)} dr^2 + r^2 d\Omega_2^2$$

where, ##f(r) = 1 - \dfrac{2M}{r} - \dfrac{\Lambda}{3} r^2##.

It suggests that a singularity would be a black hole only if the mass is not greater than ##\dfrac{1}{3\sqrt{\Lambda}}## and if the mass exceeds this limit then the singularity would become naked. But if we consider the Cosmic Censorship seriously then we must expect that (since the naked singularities can't exist) no singularity can have a mass greater than ##\dfrac{1}{3\sqrt{\Lambda}}##. This suggests that there is an upper limit on mass itself (namely, ##\dfrac{1}{3\sqrt{\Lambda}}##) that can be put inside a radius of ##\dfrac{2}{3\sqrt{\Lambda}}##. Thus, within a radius ##R##, mass can't ever exceed ##\dfrac{1}{3\sqrt{\Lambda}}## if ##R\leq\dfrac{2}{3\sqrt{\Lambda}}##. (Perhaps, by perpetually shifting the origin of the coordinate set-up to cover the desired region, I can argue that within a radius ##R##, mass can't exceed ##\dfrac{n}{3\sqrt{\Lambda}}## if ##n## is the smallest possible integer solution for ##l## where ##R\leq\dfrac{2l}{3\sqrt{\Lambda}}##.) If we consider censorship seriously, then it doesn't suggest that if the mass exceeds this limit then it will form a naked singularity but it rather suggests that mass just can't exceed this limit.

This seems result seems quite interesting to me and I can't figure out as to what reason or mechanism would keep the mass from crossing this limit. What is the resolution to this question (provided it demands a resolution)?"

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Gold Member
This is definitely interesting. I don't have the time to go into it in detail right now, but I bet that if you considered a LCDM universe where the matter density was high enough that black holes in this mass range were possible, then you'd also end up with structures approaching Planck densities, indicating that quantum gravity has something to say which we don't understand how to interpret just yet.

Gold Member
f(r)=1−2M/r−Λ/3r2.
Hi Arman:

I confess I am puzzled by the math related to units. In particular, using SI units, it appears from the quoted equation that Λ has units kg/m3, based on the assumption that all terms in the RHS of the equation have the same units. I am guessing that there may be other aspects of the units that are hidden by assuming certain constants are equal to 1, e.g., G and c, but I am not familiar enough with the conventions to be sure of this.

The final mystery is that "the mass is not greater than 1/(3√Λ)." This should mean that the units of Λ has the units 1/kg2.

Regards,
Buzz

Gold Member
This is definitely interesting. I don't have the time to go into it in detail right now, but I bet that if you considered a LCDM universe where the matter density was high enough that black holes in this mass range were possible, then you'd also end up with structures approaching Planck densities, indicating that quantum gravity has something to say which we don't understand how to interpret just yet.
Hmm, so you are suggesting the limit is possible but we need quantum gravity to understand why. I cannot make much comment because the question is hard as you said.

Gold Member
Hi Arman:

I confess I am puzzled by the math related to units. In particular, using SI units, it appears from the quoted equation that Λ has units kg/m3, based on the assumption that all terms in the RHS of the equation have the same units. I am guessing that there may be other aspects of the units that are hidden by assuming certain constants are equal to 1, e.g., G and c, but I am not familiar enough with the conventions to be sure of this.

The final mystery is that "the mass is not greater than 1/(3√Λ)." This should mean that the units of Λ has the units 1/kg2.

Regards,
Buzz
I also did not understand how the unit system works.

Have made any calculation as to mass limit (in solar masses)?

Mentor
What is the resolution to this question (provided it demands a resolution)?"

I think the resolution is that the poster's assumption that the singularity becomes naked if the black hole mass ##M## exceeds the given threshold value is incorrect. Instead, I think what happens, heuristically, is that the black hole horizon and the cosmological horizon switch roles. See the comments here:

https://en.wikipedia.org/wiki/De_Sitter–Schwarzschild_metric#Horizon_properties

However, it's also possible that there's nothing to resolve, because the poster is using coordinates that don't cover the entire spacetime but only a part of it. The coordinates in which the metric is written in the OP are called "static" coordinates in the de Sitter case, and it is well known that they do not cover all of de Sitter spacetime. So it's possible that translating the question into coordinates that do cover all of de Sitter spacetime (or more precisely the analogue of such coordinates for the Schwarzschild-de Sitter case) will show that there is no issue in the first place.

Grinkle
Mentor
using SI units

The quoted metric is not written in SI units. It's written in "natural" units for GR, where ##G = c = 1##. That means the units of ##M## are length, and the units of ##\Lambda## are inverse length squared (which are curvature units).

Arman777 and Buzz Bloom
Staff Emeritus
Gold Member
From the Phys. Rev. D. paper "Global Structure of Robinson-Trautman radiative space-times with cosmological constant" by Bicak and Podolsky
https://arxiv.org/abs/gr-qc/9901018

"5 The Robinson-Trautman space-times with ##9 \Lambda m^2 >1##

In this case the corresponding Schwarzschild-de Sitter space-time (16) admits no horizon in the region r > 0 (cf. [21, 36]) so that there is only a naked singularity situated at ##r = 0##."

Mentor
In this case the corresponding Schwarzschild-de Sitter space-time (16) admits no horizon in the region r > 0

Hm, yes, I see that. But as far as I can tell, that is because the function ##f(r)## is negative in the entire range ##r < 0 < \infty##, which means the entire spacetime is like the interior of a Schwarzschild black hole: no timelike KVF and no stationary observers, everything forced to fall into the singularity in finite time. The singularity, as far as I can tell, is still spacelike, so it's not a "naked singularity" in the same sense as, for example, the singularity in super-extremal Kerr spacetime, which is timelike.

In other words, there are three possible behaviors for the function ##f(r)##:

(1) For the case with two horizons, ##9 \Lambda M^2 < 1##, ##f(r)## starts out negative at small ##r##, switches to positive at the black hole horizon, is positive for the "normal" region between the two horizons, then switches to negative again at the cosmological horizon, and is negative the rest of the way out to large ##r##.

(2) For the case with one horizon, ##9 \Lambda M^2 = 1##, ##f(r)## starts out negative at small ##r##, just reaches zero at the single horizon, then is negative again for large ##r##.

(3) For the case with no horizon, ##9 \Lambda M^2 > 1##, ##f(r)## is negative everywhere: the balancing between the two terms never allows it to become positive at all.

So my proposed resolution in post #7 is wrong, but the way that it's wrong suggests a different resolution. Consider: suppose we are in the "two horizon" regime, where ##9 \Lambda M^2 < 1##. Matter in this region could fall into the black hole and increase its mass, bringing the two horizons closer together. But if this process continues to the point where ##9 \Lambda M^2## approaches ##1##, the horizons get closer and closer together. (Note that, in the Nariai solution described in the Wikipedia article I linked to, they never actually meet--but I think this solution is no longer valid if we actually let ##9 \Lambda M^2## equal ##1##, it's only valid if ##9 \Lambda M^2## is still less than ##1## but very, very close to it.)

Now, suppose we are in a spacetime in which the horizons are approaching each other in this way. What happens to observers in the region between them when they actually touch? Well, there is only one place they can go: through the horizon. So in a universe like this, eventually everyone would end up in a region like the interior of the black hole (which, as noted above, basically becomes the entire spacetime if ##9 \Lambda M^2 > 1##. And inside the black hole, the singularity is no longer hidden by a horizon, by definition.

Whether this counts as a violation of the cosmic censorship conjecture could be debated, but certainly it doesn't violate it in the way that is intuitively suggested by the OP: we don't have a "normal universe" region that now has a naked singularity in it. There is no "normal universe" at all any more.

(It's worth noting, also, that the above analysis does not take into account quantum effects. As noted in the Wikipedia article I linked to, if we include quantum effects, both horizons radiate, and it's no longer completely clear how to distinguish them.)

Arman777
Mentor
what reason or mechanism would keep the mass from crossing this limit.

In addition to my previous comments, it's also worth noting that the whole idea of "mass falling into a black hole" implicitly assumes that the hole's spacetime is asymptotically flat. But Schwarzschild-de Sitter spacetime, unlike ordinary Schwarzschild spacetime, is not asymptotically flat. So it's not clear what "mass falling into a black hole" actually means in Schwarzschild-de Sitter spacetime.

Another way to put this is to observe that the conformal structure of Schwarzschild-de Sitter spacetime is different, not only from that of Schwarzschild spacetime, but different depending on whether ##9 \Lambda M^2## is less than, equal to, or greater than ##1##. And it's not clear how any physical process could change the conformal structure of a spacetime.

Arman777
Gold Member
Thanks for your explanations. I was looking at the PSE, and I also see another perspective which says by like this. It seems that this is an answer to another question. And I would like to also share since it seems interesting.

"The rules of classical general relativity say that when you add mass to a black hole, you get a larger black hole. If you add angular momentum to a black hole at a greater rate than that at which you add mass, it would theoretically be possible to get a Kerr black hole with ##a \gt M##, which would convert the black hole to a naked singularity, but the rules of black hole thermodyanamics say that a black hole with ##a = M## has zero temeperature, so creating a naked singularity in this way is believed to be impossible."

And continues that if it has zero tempature, no material can fall into it.

From, https://physics.stackexchange.com/questions/459067/do-black-holes-have-a-limit-of-mass/459070#459070

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Mentor
the rules of black hole thermodyanamics say that a black hole with ##a = M## has zero temeperature, so creating a naked singularity in this way is believed to be impossible

This is a bit of a misstatement. The issue is not temperature (it's hard to see why it should be impossible to add energy to a system that has zero temperature, since adding energy would just increase the temperature above zero). The issue is that, when you analyze the details of how you would try to add angular momentum to a rapidly spinning black hole, i.e., one that is close to ##a = M##, it turns out that there's no way to do it without adding enough mass to keep the hole from reaching ##a = M##. Heuristically, this is because whatever you drop into the hole has to have some minimum energy, i.e., mass, in order to be able to drop it in in a way that will increase the angular momentum of a rapidly spinning hole.

Arman777