# How long can a particle survive inside of the horizon of a black hole?

By examining the causal structure of a Schwarzschild black hole, one can see that a particle in region II is unable to escape to $r=\infty$.

Such diagrams do not show, however, how long a particle can survive inside of region II without hitting the singularity. Can a particle survive indefinitely inside of the black hole? If not, is there a maximum life span of a particle inside the horizon?

• If you enjoyed this you may also enjoy No Way Back: Maximizing survival time below the Schwarzschild event horizon. – John Rennie Jul 16 '15 at 15:14
• @JohnRennie The abstract claims rockets will increase the survival time, but my analysis indicates rockets will only decrease it. Perhaps you could sketch his argument. (In his words, his result is "at odds" with mine.) – Ryan Unger Jul 16 '15 at 15:29
• I have to confess I've only skimmed the paper. It was mentioned by someone on this site in relation to an earlier question, and for some reason it stuck in my memory. That was the only reason I mentioned it. The paper is in my read carefully pile, but that pile is already large and growing daily! – John Rennie Jul 16 '15 at 16:40
• @JohnRennie I was unable to gleam much from skimming it either...I am particularly perplexed by "it appears that the error lies in the assumption that the impact onto the central singularity is the same event for the free faller and the rocketeer; if they were then the above statement would be correct and the free faller would experience the maximal proper time". It's in my "read carefully" pile as well. – Ryan Unger Jul 16 '15 at 17:00
• @annav I am asking about the classical situation. – Ryan Unger Nov 4 '16 at 16:42

The references I looked at were Misner and Lewis. Neither of them actually seems to supply a definitive answer to this question, since they both assume purely radial motion. Maybe there is some simple argument to show that purely radial motion is the best case, but I haven't seen one.

We also have to be careful because people talk about the proper time $\tau=\pi M$ (in geometrized units) as a best case, but in fact it seems to be more like an upper bound, at least for geodesic motion starting from "rest," i.e., zero velocity in Schwarzschild coordinates. If you start from rest at $r=R$, then MTW show that the proper time from there to $r=0$ (not from horizon to singularity) is $\tau=(\pi/2)R(R/2M)^{1/2}$. But a massive object can't be at rest at or inside the horizon, so $R\le 2M$ isn't possible, and we can only talk about $R=2M$ as a limiting case.

So in actual cases of interest, the observer can't be at coordinate rest as they pass through the horizon, and therefore they will always pop through with some finite speed. Once inside the horizon, Lewis shows that it's a good strategy to blast the engines as hard as possible, for a brief time, so as to match up with the trajectory corresponding to the free-falling-from-$2M$ idealized case.

So contrary to what a lot of people seem to believe:

• It is never optimal not to use your engines.

• I'm not aware of any proof that $\tau=\pi R$ is either attainable or (if nonradial motion is allowed) optimal.

References

Misner, Thorne, and Wheeler, Gravitation, p. 663

Lewis and Kwan, "No Way Back: Maximizing survival time below the Schwarzschild event horizon," https://arxiv.org/abs/0705.1029.pdf

• That's an important caveat, but I think that blurring the distinction between "maximum" and "least upper bound" is within the acceptable bounds of standard physics sloppiness. I agree that $\tau = \pi R$ is not technically obtainable with purely radial motion, but you can in principle get arbitrarily close, which is probably what most people have in mind when they make that claim. – tparker Sep 3 '18 at 1:12
• There is indeed a simple argument that purely radial motion is the best case, which ahaw95 mentions. For an arbitrary timelike trajectory (not necessarily a geodesic) in Schwarzschild spacetime, the normalization condition $U \cdot U = -1$ expands out to $$\left( \frac{dr}{d\tau} \right)^2 = \left(\frac{2GM}{r} - 1 \right) \left[ 1 + \left(\frac{2GM}{r} - 1 \right) \left( \frac{dt}{d\tau}\right)^2 + r^2 \left( \frac{d\Omega}{d\tau}\right)^2 \right].$$ – tparker Sep 3 '18 at 19:49
• So inside the horizon, where $\frac{2GM}{r} - 1$ is positive, any angular motion will necessarily cause your radius to decrease faster with proper time. – tparker Sep 3 '18 at 19:50
• $-dr / d\tau$ is like the $\gamma$ factor in special relativity. The fact that angular motion increases it is similar to how in SR, if you hold the velocity component in one given direction fixed, then any velocity in a perpendicular direction will necessarily increase the $\gamma$ factor. – tparker Sep 3 '18 at 20:20

The previous answers are correct in that (in geometric units) it takes maximum proper time $\pi M$ to fall from the event horizon to the singularity. While it has been pointed out that the simplest argument for this via use of the geodesic equation is not appropriate, we can alternately show that $\left\lvert\frac{\text{d}r}{\text{d}\tau}\right\rvert \geq \sqrt{\frac{r_S}{r}-1}$ (in Schwarzschild coordinates) purely by requiring that the particle's trajectory be timelike. For infalling particles, we can integrate and thereby show that the maximal lifetime of any particle entering the event horizon is $\frac{r_S}{2} = \pi M$. This avoids the problem with the justification given in some previous answers by not appealing at all to the geodesic equation (this argument appears as an exercise in Carroll's textbook).

This then agrees with the naive answer via using the geodesic equation for a particle at rest to start at the event horizon, and can be viewed as a derivation of the fact that starting at rest and free-falling maximizes a particle's lifetime. This is not in disagreement with the paper cited above - towards the end, after a quote demonstrating the misconception that attempting to accelerate away after falling into the event horizon would inevitably shorten your lifespan, the paper says:

"The results of this study show that this clearly is not the case; anyone who falls through the event horizon should fire their rockets to maximize the time they have left before impacting the central singularity. In dropping from rest at the event horizon, the firing of a rocket does not extend the time left, it only diminishes it."

It turns out, weirdly enough, to be basically $\pi$ geometric time units (normalized to the mass of the hole). I think this is demonstrated in MTW. That's if you don't try to fight your way out and just fall. Attempting to accelerate away from the singularity actually shortens your time.

• This is incorrect. One can accelerate away from the singularity and get at least a little bit of time. – Ryan Unger Nov 30 '15 at 20:10
• (This is not an attack, I had an answer that gave the answer as $\pi$, but deleted it. It comes from the incorrect assumption that geodesics maximize the proper time of an observer falling in.) – Ryan Unger Nov 30 '15 at 21:14
• That's interesting. Does the infall time then approach infinity with infinite thrust? – AGML Nov 30 '15 at 22:44
• No! There is a point where acceleration will make you fall in faster. See the paper linked on the OP for an analysis. I'm just too lazy to condense it into a post...I don't know how to solve the equations analytically or numerically and I don't want to copy+paste the results of the paper. – Ryan Unger Nov 30 '15 at 23:19
• To clarify my geodesic comment: we know that a geodesic maximizes proper time in a globally hyperbolic spacetime. Schwarzschild spacetime is globally hyperbolic. The standard thing is then to say that the proper time maximizing path into the black hole must be a geodesic. The problem with this is that when you strap on your jetpack and jump in, you're not necessarily going to land at the same spacetime point. You hit the singularity, i.e. the same point in space, but the point along the "time axis" need not be the same. Thus one can't say anything about the trajectories using geodesics. – Ryan Unger Dec 1 '15 at 0:27

A geodesic is the maximal path in spacetime. If you attempt to accelerate in any direction inside a black hole the very fact you would be on a nongeodesic path means the path length in spacetime is shorter. The path length is equivalent to the proper time along the path.

The longest proper time of any observer in a black hole is as AGML says. It is $\pi$ times the time equivalent for the mass of the black hole or $\pi GM/c^3$.

• Even though I have not read the paper linked in the comments, I can immediately say that your argument does not hold. As is also explained in the comments, your argument would be valid if the collision between an accelerated observer and the singularity occured at the same point in spacetime as it does for the geodesic observer. However, you cannot assume that such is the case: the accelerated observer may follow a path that intersects the singularity worldline at a different point. In that case there is no immediate way to make the conclusion, and the above paper claims to contradicts it. – Erik Jörgenfelt Aug 29 '16 at 16:09
• @ErikJörgenfelt I've read the paper, and the argument is pretty convincing, but not suitable for a Stack Exchange answer (it requires numerical methods and a lot of background). – Ryan Unger Aug 29 '16 at 17:04
• The endpoints aren't fixed in Kruskal spacetime. If you accelerate, your path will hit a different part of the hyperbolid $r=0$, so this argument doesn't follow. – Jerry Schirmer Feb 17 '17 at 22:21