1
$\begingroup$

We know that every object that has mass, also has a Schwarzschild radius $r_s$: $$r_s = \frac{2Gm}{c^2}$$ With $G$ being Newton's gravitational constant, $m$ the mass of the object and $c$ the speed of light in vacuum. When we say that the object has spherical shape, its volume $V$ is given by $$V=\frac43\pi r^3$$ This volume also equals the ratio of mass to mass density, so substituting the Schwarzschild radius for $r$ gives as a 'Schwarzschld density' $\varrho_s$: $$V=\frac43\pi r_s^3\overset{!}{=}\frac m\varrho_s\Longleftrightarrow\varrho_s = \frac{3}{4\pi}\frac{1}{m^2}\left(\frac{c^2}{2G}\right)^3 $$ Numerical evaluation of the constants gives therefore $$\varrho_s = \frac{1}{m^2}\cdot 7.29\times10^{79}\:\mathrm{\frac{kg^3}{m^3}}$$ Now let's assume in a gedankenexperiment that I built a machine capable of increasing my planet's density. I increase it to the very brink of collapse, but stop the machine so that the Schwarzschild radius is just barely smaller than the radius of my planet neutron star. Then, I call a friend of mine over for a beer and a snack.

What happens when my friend visits me with his rocket, but right before he can initiate the breaking maneuver I turn my machine back on and turn my 'home' to a black hole?

Would he hit some kind of rigid surface? Would he even see where he's going?

$\endgroup$
  • $\begingroup$ The important ratio is not $\frac{E_m(r)}{r^3}$ or mass density. The important ratio is $\frac{E_m(r)}{r}$, r being the radial Schartzschild coordinate. This ratio cannot excess $\frac{c^2}{2G}$. The limiting case corresponds to a black hole. $\endgroup$ – Trimok Jul 11 '13 at 17:42
2
$\begingroup$

The laws of physics are local so if you change something on one side of your celestial body, it won't immediately affect other parts of the celestial body. In fact, your friend won't experience anything special even after the signals about the turning on of your machine get to your friend. He will just find himself on the surface of a collapsing neutron star and very quickly afterwords, this surface will belong to the interior of the black hole whose formation has already become inevitable.

But his crossing the event horizon – the surface of the black hole from which he can't escape anymore, not even in principle – doesn't mean that he hits any wall or any irregular region of space, for that matter. The life continues as it always did, at least for a while before your friend hits the huge-curvature region near the black hole singularity at the center. The creation of the black hole singularity becomes unavoidable at the moment when you decide that the density is increased so that the black hole is formed. But the singularity is a finite distance – well, it becomes a finite time – away from the surface of the former neutron star.

Exactly one year ago, four authors – AMPS – have proposed that the conclusion we have been labeling as wrong for many decades is right. They declared that the surface of a black hole actually has a "firewall" that burns any object that dares to penetrate it by high-energy quanta. But most physicists in the field think that their argument is unwarranted and ultimately flawed but even if you imagined that this argument is right, they only say that a sufficiently old black hole – one whose 1/2 or so has already evaporated, and this is only possible for a truly old black hole – possesses such a firewall. A black hole in the state of birth – when the original neutron star just barely collapses or a similar period afterwords – surely doesn't have any firewall yet.

See also e.g.

http://motls.blogspot.com/2008/11/why-can-anything-ever-fall-into-black.html
Why can anything ever fall into a black hole

for an explanation of the black hole causal diagram.

$\endgroup$
  • $\begingroup$ "The life continues as it always did, at least for a while before your friend hits the huge-curvature region near the black hole singularity at the center." I like how understated the part where he's ripped apart by tidal forces once he does get there is ;) $\endgroup$ – Kyle Oman Jul 11 '13 at 13:35

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.