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I don't know much about black holes physics and so I find the Schwarzschild equations with a few contradictions. In particular I am trying to understand this little puzzle. The Schwarzschild Newtonian gravitational field equation is expressed as follows (see http://en.wikipedia.org/wiki/Schwarzschild_radius):

$\frac { r^2 }{r_s} \frac {g}{c^2} = \frac {1}{2}$

So once a particle is close to the event horizon such that $r\to r_s$ the equation becomes:

$ r \frac {g}{c^2} = \frac {1}{2}$

But then at that point light cannot escape so I would think that $g\to c/t$ where $t=1 sec$. So the equation would now approximate to:

$r = \frac {c/t}{2}$

But this implies that $r$ is generalized for all black holes regardless the mass. If $r$ is indeed $r\equiv r_s$ then that contradicts the other Schwarzschild formula where $r$ depends on $m$:

$r_s = \frac {2 G M} {c^2}$

Also based on what we know about $r_s$ in massive black holes, a radius of 150,000 km is quite small.

What's wrong with this picture?

Qmechanic
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Luis
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1 Answers1

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In the good old Newtonian world the gravitational acceleration is just:

$$ g = \frac{GM}{r^2} $$

The equation you give is just a rewriting of this. If you substitute:

$$ r_s = \frac{2GM}{c^2} $$

into:

$$ \frac { r^2 }{r_s} \frac {g}{c^2} = \frac {1}{2} $$

you'll find it simplifies to the first equation. So there is nothing especially meaningful in this procedure.

You say:

But then at that point light cannot escape so I would think that $g \rightarrow c/t$ where $t=1$ sec.

But this is not a well motivated statement as this is not the relationship between the acceleration and the (Newtonian) escape velocity. The equation for the escape velocity is:

$$ v_e = \sqrt{\frac{2GM}{r}} $$

So if you substitute the expression for $r_s$ you end up with:

$$ v = c $$

This is the well known result that at the event horizon the escape velocity calculated from Newtonian gravity is equal to the speed of light.

However you should not take this result too seriously. If you want to understand why light can't escape from the event horizon you need to look at the general relativistic treatment. As it happens this is dicussed in the question Why is a black hole black?

John Rennie
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