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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?

ProfRob
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user6090
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4 Answers4

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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.

Luboš Motl
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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.

John Rennie
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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 01:08
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    Oops yes, my comment is only true for the Schwartzchild geometry. – John Rennie Jan 13 '12 at 10:56
  • This is an important point. If you consider just Newton's knowledge, there is nothing special with the speed of light - a rocket can slowly accelerate until it's even faster than the speed of light. But if you use Newton's gravity for the escape-velocity, and special (not general!) relativity for the fact that you can't accelerate past c (you'll need more and more energy for smaller increase in velocity), you get something that looks like a black hole. – Nadav Har'El Feb 15 '24 at 15:31
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The interesting properties of a black hole cannot be explained by Newtonian gravity. The behaviour of bodies with mass, and of light, is completely different near a compact, massive object if you use Newtonian physics rather than General Relativity.

A non-exhaustive list of features that GR predicts (and which in some cases have now been observationally confirmed) but which Newtonian physics cannot:

  1. Most importantly, an event horizon. In Newtonian physics there is a misleading numerical coincidence that the escape velocity reaches the speed of light at the Schwarzschild radius. But in Newtonian physics you could still escape just by applying a constant thrust. GR predicts that no escape is possible in any circumstances. In fact the "coincidence" that light cannot escape from the same radius for which Newtonian physics predicts an escape speed of $c$, is only true for light travelling radially. At other angles light cannot escape unless emitted from a larger starting radius. This angular dependence of the "escape speed" is another feature not predicted by Newtonian gravity.

  2. GR predicts an innermost stable circular orbit. A stable circular orbit would be possible at any radius in Newtonian physics. This behaviour is important for explaining the accretion phenomena observed for black holes in binary systems.

  3. In GR a particle with some angular momentum and lots of kinetic energy will end up falling into the black hole. In Newtonian physics it will scatter away to infinity.

  4. Newtonian physics predicts no precession of a two-body elliptical orbit. GR predicts orbital precession. This precession is measured in Mercury and other Solar System bodies, but has now been measured for stellar orbits around the Milky Way's central, supermassive black hole.

  5. Newtonian physics predicts that light travelling close to a massive body has a trajectory that curves by about half the amount predicted by GR. Even stranger effects are predicted close to the black hole including that light can orbit at 1.5 times the Schwarzschild radius. Evidence for the former was one of the first tests of GR applied to stars near the Sun. Evidence for the latter is now seen in "pictures" of the black hole at the centre of M87.

The GR approach to gravity is fundamentally and philosophically different to Newtonian gravity. For Newton, gravity is a universal force. In GR, gravity is not a force at all. Freefalling bodies are said to be "inertial". They accelerate, not because a force acts upon them, but because spacetime is curved by the presence of mass (and energy).

In most cases, where Newtonian gravitational fields are weak, the consequences of this difference are small (but measurable - e.g. the orbital precession of Mercury or gravitational time dilation in GPS clocks), but near large, compact masses, like black holes and neutron stars, the differences are stark and unavoidable.

ProfRob
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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 $\large {m=\frac{E}{c^2}}$:

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

or

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

so

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

and

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

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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