- #1

- 2,168

- 193

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)?"

Reference: https://physics.stackexchange.com/questions/337760/a-universal-upper-limit-on-mass-within-a-radius-r

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)?"

Reference: https://physics.stackexchange.com/questions/337760/a-universal-upper-limit-on-mass-within-a-radius-r

Last edited: