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Is the acceleration of a free-falling observer at the Schwarzschild radius of a black hole the same regardless of the numerical value of the radius (i.e. outside of a black hole, is the acceleration of the free-falling observer only a function of his distance from the Schwarzschild radius)?

Chris Laforet
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2 Answers2

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I'm going to take my life in my hands and disagree with Jerry (though actually we disagree only in terminology and not about the physics).

As with all things, the value of the acceleration for a freely falling observer depends on who you ask i.e. the figure you get depends on the coordinates you choose. There are three obvious observers you might consider:

  1. the freely falling observer
  2. the Schwarzschild observer at infinity
  3. a shell observer hovering at a fixed distance from the horizon

And of course there's the four acceleration, which is coordinate independant but isn't directly observed by anyone.

The case of the freely falling observer is easy because the acceleration they measure is zero, otherwise they wouldn't be falling freely.

The case of the Schwarzschild observer is also easy because the acceleration (and velocity) of the falling object asymptotically approaches zero as the object approaches the black hole.

The last case was analysed in twistor59's answer to What is the weight equation through general relativity?. The acceleration of the falling object measured by a shell observer is:

$$ a = \frac{GM}{r^2}\frac{1}{\sqrt{1-\frac{2GM}{c^2r}}} $$

and this goes to infinity at the event horizon.

So for all three observers the acceleration at the event horizon is independant of the mass/radius of the black hole.

However the four acceleration is given by (this is also in twistor59's answer):

$$ \mathbf{a} = (0, \frac{GM}{r^2}, 0, 0) $$

so for the freely falling observer at the event horizon:

$$ \frac{d^2r}{dt^2} = \frac{GM}{r_s^2} $$

which does depend on the black hole mass/radius. This is presumably the point in Jerry's answer.

Community
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John Rennie
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  • I'm a bit confused: isn't the four-acceleration that you give the value for a stationary observer at the event horizon? I thought the four-acceleration of a freely radially inward falling observer is identically zero. – ScroogeMcDuck Jun 02 '14 at 09:23
  • @ScroogeMcDuck: The four acceleration is $du/d\tau$, and for the freely falling observer their time is equal to the proper time. If you express $u$ using the Schwartzschild coordinates then obviously $u^1$ is not constant so $a^1 = du^1/d\tau$ is not zero. – John Rennie Jun 02 '14 at 09:26
  • Alright, I was confused because the four-acceleration is defined as $Du/d\tau$ in the Wiki article you linked to (which does vanish by the geodesic equations), and in twistor59's answer. So their definition of four acceleration is the one measured by a freely falling observer (your point 1). But the four-acceleration at the end of your answer is $Du/d\tau$ for a stationary observer at some $r$, according to twistor59's answer. Are you saying it is also $du/d\tau$ for a freely falling observer? – ScroogeMcDuck Jun 02 '14 at 12:01
  • +1. But there is one subtle point. The test particle $m$ also has a Schwarzschild radius. So it cannot reach the event horizon of $M$ and $a$ cannot approach infinity. – Hans Jan 30 '22 at 02:45
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No. The schwarzschild radius is located at $r_{S} = \frac{2GM}{c^{2}}$, and the 3-acceleration at this point (by one of physics's weird coincidences, this is identical to the newtonian expression) is given by the surface gravity $\kappa = \frac{GM}{r^{2}}$. Combining these two formulae, we find that $\kappa \propto \frac{1}{r_{S}}$, so the surface gravity depends on the schwarzschild radius.

Zo the Relativist
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