6
$\begingroup$

Since the event horizon is defined as the boundary within which the escape velocity is greater than the speed of light, and escape velocity is the speed required for that object to reach infinity away from that point, why can't light escape the event horizon even if it doesn't reach infinity away from it?

I assume this is a problem with using the Newtonian idea of escape velocity to derive the Schwarzschild radius, as although it arrives at the correct answer it describes a situation where light could leave the event horizon, even if only temporarily.

$\endgroup$
9
$\begingroup$

There is a concept in classical physics called a dark star which was tossed around as early as the 18th century. The idea is precisely what you refer to - an object whose mass is so great that the escape velocity from its surface exceeds $c$. Newton himself believed light to be particle-like, and the implication of the dark star is that light particles emitted from its surface would travel outward for some distance, reach a turning point, and then fall back in.

This is not what a black hole is. The Schwarzschild radius $r_s = \frac{2GM}{c^2}$ is equal to the critical radius of a classical dark star only by a happy coincidence of dimensional analysis$^\dagger$; though the properties of a black hole may be superficially similar to those of a dark star, they are very, very different objects.

One way to think about what happens at the Schwarzschild radius is to note that at $r=r_s$, the nature of space and time are, in a sense, flipped around. The radial coordinate (which loosely describes the distance to the singularity at the center of the black hole) becomes "time-like" while the coordinate time becomes "space-like"; as a result, an object (or indeed, a photon) within the event horizon can no more avoid the singularity than an outside observer could avoid next Tuesday.

In contrast to the classical notion of a dark star, it's not a matter of having enough energy to escape from a black hole - it's just that once you are inside the event horizon, there are no future-directed timelike curves (which are the paths massive objects follow) which don't take you in to the singularity.


$^\dagger$By this, I mean that if you want to use Newton's constant $G$, the mass of an object $M$, and the speed of light $c$ to make a distance, your only choice is $\sim GM/c^2$ - otherwise the units wouldn't work out. The fact that the additional factor of $2$ in the Schwarzschild radius matches that of the Newtonian dark star's critical radius is coincidental.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks that was a great explanation $\endgroup$ – Dylan Winkworth Jun 8 at 20:17
  • $\begingroup$ A Penrose diagram might be helpful for the OP to visualize the reversal. $\endgroup$ – user76284 Jun 11 at 20:32
  • $\begingroup$ A great description of the interpretation by Droste-Hilbert-Finkelstein. However, in the actual solution of Karl Schwarzschild, there is no horizon, because the gravitational radius is zero: $r\equiv\sqrt{x^2+y^2+z^2}=0$ - Nothing to cross or escape from: arxiv.org/abs/physics/9905030 $\endgroup$ – safesphere Jun 11 at 23:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.