Have theories been formed yet on how to stop a black hole?

  • #26
15
0
No, a black hole is everything inside the Event Horizon, not just at the center. The "singularity" is just the place where our current math model breaks down. The current belief (but no evidence yet) is that it is most likely NOT a point but until there is a good theory of quantum gravity, it can't be modeled in any other way than what we currently have, which gives a most-likely-non-physical ("singularity") result.
I read in one reference that the "singularity" is actually a Planck length radius, 10-35 m, though this strikes me as a distinction without a difference.
 
  • #27
Nugatory
Mentor
12,771
5,370
Black Hole can't be huge.
What exactly do you mean by "huge" here? If you mean the size of the central singularity, that's not a meaningful concept.

If you mean the volume inside the event horizon (strictly speaking, the volume of a sphere with radius equal to the Schwarzschild radius) then it can be arbitrarily large if there's enough mass in the black hole. For example, the black hole at the center of our galaxy is probably about 250 million kilometers across. Whether that's "huge" or not depends on your perspective, but if you were in close orbit around it, you'd probably be thinking "huge".
 
  • #28
15
0
Not true. Supermassive black holes exist, having radii of millions of kilometers and densities less than water.
This doesn't sound correct as escape velocity is as much related to density as mass.
 
  • #29
Bandersnatch
Science Advisor
2,913
1,823
This doesn't sound correct as escape velocity is as much related to density as mass.
Take the escape velocity equation
##V=\sqrt{\frac{2GM}{r}}##
set ##V=c## and rearrange with ##r## on one side:
##r=\frac{2GM}{c^2}##
This is the Schwartzschild radius equation. As ##M## goes up, the radius increases linearly.
At the same time, assuming uniform density distribution of material inside, a sphere of density ##\rho## and radius ##r## will have mass:
##M =\frac{4}{3}\pi r^3\rho##
For ##M## and ##r## to change at the same rate, ##\rho## must change as ##\frac{1}{r^2}##
That is, it has to fall as the black hole grows bigger.
 
  • #30
15
0
Take the escape velocity equation
##V=\sqrt{\frac{2GM}{r}}##
set ##V=c## and rearrange with ##r## on one side:
##r=\frac{2GM}{c^2}##
This is the Schwartzschild radius equation. As ##M## goes up, the radius increases linearly.
At the same time, assuming uniform density distribution of material inside, a sphere of density ##\rho## and radius ##r## will have mass:
##M =\frac{4}{3}\pi r^3\rho##
For ##M## and ##r## to change at the same rate, ##\rho## must change as ##\frac{1}{r^2}##
That is, it has to fall as the black hole grows bigger.
If one looks at only “normal” physics escape velocity is directly related to the square root of the density and to the radius of the body. If you hold the radius constant (like for the Earth) then escape velocity is determined only by the square root of the density times a constant. But then, as always, funny things happen as one approaches the speed of light and black holes. In either case you can have densities of 1.0 though black hole dimensions get slightly grotesque even for cosmological things – Schwartzschild radii of 13 billion Km with 4 trillion solar masses (assuming my math is correct). My only point is that it sometimes is convenient to think of escape velocity in terms of the density of the base body. But this gets really screwy with black holes. For example, what is the density of a black hole with a R-schwartzschild of a million Km and any mass but all concentrated in a singularity? Density is a helpful parameter with normal things but as you point out not much help with black holes.
 

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