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In the simple explanation that a black hole appears when a big star collapses under missing internal pressure and huge gravity, I can't see any need to invoke relativity. Is this correct?

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up vote 13 down vote accepted

By a coincidence, the radius of a "Newtonian black hole" is the same as the radius of the Schwarzschild black hole in general relativity. We demand the escape velocity $v$ to be the speed of light $c$, so the potential energy $GMm/R = mc^2/2$, i.e. $$ R = \frac{2GM}{c^2} $$ The agreement, especially when it comes to the numerical factor of $2$, is a coincidence. But one must appreciate that these are totally different theories. In particular, there's nothing special about the speed $c$ in the Newtonian (nonrelativistic) gravity. To be specific, objects are always allowed to move faster than $c$ which means that they may always escape the would-be black hole. There are no real black holes (object from which nothing can escape) in Newton's gravity.

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You can escape from a Newtonian black hole. The escape velocity may be c, but you could still escape at sublight speeds with a powerful enough rocket and enough fuel. By contrast, once you've crossed the event horizon of a real black hole there is nothing you can do to avoid hitting the singularity.

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You should say there is nothing you can do to move toward smaller r. It is not clear that anything massive hits the singularity for realistic rotating or charged black holes. – Ron Maimon Jan 13 '12 at 1:08
Oops yes, my comment is only true for the Schwartzchild geometry. – John Rennie Jan 13 '12 at 10:56

You might enjoy my paper giving the Newtonian force $F$ necessary to exactly produce the behavior of a test particle near a black hole through the usual $F=ma$ equation:

The force of gravity in Schwarzschild and Gullstrand-Painlevé coordinates Int.J.Mod.Phys.D18:2289-2294,2009

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Yes, black hole can be explained by newtonian gravity. But with some assumptions.

From Newton equation of energy conservation for free fall from the infinity with initial speed of object equal to zero:

$\large {mc^2=E-\frac{GMm}{R}}$, where $\large {E=\frac{mc^2}{\sqrt{1-v^2/c^2}}}$

In assumption of $\large {m=\frac{E}{c^2}}$:

$\large {mc^2=E-\frac{GM}{R}\frac{E}{c^2}}$


$\large {mc^2=E-\frac{R_{g*}}{R}{E}}$, where $\large {R_{g*}=GM/c^2}$


$\large {mc^2=E\left(1-{R_{g*}}/{R}\right)}$


$\large {E=\frac{mc^2}{1-R_{g*}/R}}$

If $R=R_{g*}$, then $E=\infty$, - event horizon of Newtonian black hole

The similar expression in General Relativity:

$\large {E=\frac{mc^2}{\sqrt{1-R_{S}/R}}}$, where $\large {R_{S}=2GM/c^2}$

See also Newtonian Gravity with no any singularity

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This answer (v1) seems to promise in words that it will only use newtonian theory, and then proceed by writing down various ad-hoc non-newtonian equations. – Qmechanic Jan 12 '12 at 21:10
@Qmechanic, you are right, it's a post-newtonian gravity, where energy $(E/c^2)$ is attracted, not mass – voix Jan 12 '12 at 21:28
It's not Newtonian if the first line uses $E=mc^2$... – David Richerby Apr 23 '14 at 8:19

protected by Qmechanic Feb 19 '14 at 19:38

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