Just a noob question: They say, when you dive into a heavy/large enough (supermassive) blackhole, there should be nothing wrong with you at the time of touching an event horizon.

I suppose it's a description of diving "right in" (90c) with no sideway speeds of a diving object

  1. What if a big, macroscopic, object, like a death star will fly by (near lightspeed) a supermassive blackhole horizon, touching it slightly. Like a pinch of it's hull touching the horizon Will it be like razor-sliced a bit and fly away?

  2. what will became with atoms/particles travelling and touching the horizon at tangential


2 Answers 2


The quick answer: it is not so much that part of your death star gets sliced away as that the death star as a whole will have an extremely hard time escaping even with maximum thrust, and the part below the horizon definitely will not escape. Either the ship tears itself apart or falls in.

Long answer: you approach the black hole along what normally would be a hyperbolic flyby trajectory. Sitting in the control room you do not experience any force, since you are in free-fall along a geodesic curve.

But as you get closer the structure starts to creak: different parts of the death star are at different distance from the black hole and subjected to different spacetime curvature. This produces tidal forces that try to pull and squeeze the spherical death star into an ellipsoid pointed at the black hole. You are however smug, since you choose a supermassive black hole and hence the tidal forces are not that severe, even at the horizon. Had you used a non-spherical spacecraft there would also had been a torque trying to rotate you, but again, death stars for the win.

Now you get very close to the event horizon, skimming it. The view outside looks decidedly weird (since you are inside the photon sphere and having relativistic aberration). At this point trouble appears: your free-fall trajectory does not continue back out to infinity, but spirals into the black hole! Unlike in Newtonian gravity close free-fall trajectories do not return out. The reason is that you are inside the innermost stable circular orbit (ISCO). The only trajectories in this region that do not end up in the black hole are sufficiently steeply outgoing ones.

So you turn on your super-thrusters and try to break free by increasing your velocity in the right direction. But now you have a problem: the part of the death star furthest from the black hole needs merely a lot of extra velocity to be on a trajectory that leaves the hole region. The innermost part needs to move faster than light to break free. So if you can achieve some maximum boost there are going to be parts that, had they not been connected to the superstructure but freely flying, would not escape. So your boost turns into stress across the death star. Something has to give, and no matter what boost you are giving the innermost parts through the engines and pulling metal they will not move outward since they are below the horizon.

It is at this point I strongly recommend running to the escape pods.

Exactly what the fate of the death star will be depends on a lot of things, but the simplest case is that the overall boost is insufficient and it all falls in. It might also break apart. Relativity shows that there are no perfectly rigid materials (they would need to have a speed of sound faster than light), so not even an imaginary death star will avoid getting pulled apart.

In the end, I leave you to explain your little joyride to the Emperor.

[It is worth noting that nothing looks strange locally when you cross an event horizon. It is a global property of spacetime, not something you can directly see. Extended objects crossing it merely have some parts whose future trajectories will meet the singularity, and the parts just outside may meet the singularity, but can escape if they break away somehow.]

  • $\begingroup$ will most of the team/structures be turned to a mass of blood/metal/whatever they use in their deathstar-designs by an abrupt collision, or it depends on a ratio of the part of the hull (or, in a best case, some gun or antenna sticiking out) the deathstar touched SMBH? $\endgroup$ Jan 25, 2020 at 15:50
  • 1
    $\begingroup$ Basically the forces will be felt like acceleration, so most of the interior will be subject to extreme acceleration forces as you try to change the velocity vector of every part of the ship at a high rate. It does not matter much if it is an antenna or an entire deck that is below the horizon. $\endgroup$ Jan 25, 2020 at 20:56
  • $\begingroup$ Your "quick answer" paragraph would ever-so-slightly better introduce the rest if it explicitly said that the upper part of the ship can only leave the doomed lower bit behind if the ship tears itself apart. The black hole won't help by "slicing" it. $\endgroup$ Jan 25, 2020 at 22:37
  • $\begingroup$ @SolomonSlow - Good point. $\endgroup$ Jan 26, 2020 at 12:52

I like this question. It made me think.

I find the answer of Anders Sandberg to be helpful and correct. I will add some pointers to the argument behind his conclusions.

Rather than a death star, I will first (1) treat a simple model consisting of two particles in free fall (not joined by any structure at all). Then (2) I supply rocket engines to these two particles (but still they are not linked to one another). Then (3) I introduce a structure such as a rod between them (and they still have rocket motors or similar). Finally, (4) I extend to a model of many linked powered particles (e.g. a death star).

  1. Two particles in free fall, on a trajectory just scraping the horizon. The particle which crosses the horizon moves inwards to the singularity. The particle not crossing it immediately carries on past, but eventually loops back on a spiral trajectory and falls in.

  2. Two non-linked particles, both with rocket engines, and one of them crosses the horizon. (By "particle" here I mean something whose gravitational effect can be neglected; it could be quite a big thing really, e.g. a spaceship). The behaviour is like (1) except now the inner particle can slow its approach to the singularity a little and the outer particle can escape. The inner particle cannot avoid arriving at the singularity if we have a Schwarzschild black hole. The outer particle will have to expend a lot of fuel in order to escape. But maybe death stars are equipped with some sort of warp drive? I will not get into that.

  3. Linked particles, both with motors. This is quite close now to the question. The main thing to notice is that the link structure (rod or whatever) between the particles merely adds somewhat to the forces on each of them. The situation is not changed qualitatively from case (2). Everything I wrote there still applies. However, one can notice that if the rocket motors are switched off, then we have the following predictions.

3.i For sure, the particle crossing the horizon does not emerge back out again.

3.ii Equally for sure, the tidal forces in the frame freely falling with the outer particle are small for a SMBH.

3.iii The forces exerted by the rod between the particles are whatever they are, i.e. easily strong enough to overcome the tidal forces, in that same frame.

It follows from the above that the outer particle must be pulled into and cross the horizon. So my conclusion for the case of free fall (no rockets) is:

(small black hole) $\Rightarrow$ large tidal forces, rod will snap
(large black hole) $\Rightarrow$ small tidal forces, rod will not snap or even get stretched by much $\Rightarrow$ if one particle crosses the horizon, the other soon follows

  1. Finally then, the death star. If it is in freefall on a trajectory that skims a horizon then I think that for a small black hole it will be ripped apart near the horizon, and for a large black hole it will be pulled in so that it all crosses the horizon without being ripped apart. It will then be spaghettified a little later, on its journey to the singularity.

If the death star uses its engine to try to escape, then it can only do so by ripping itself apart (once a part of it has crossed the horizon). How would the resident emperor experience this or explain it to himself? He would not notice much creaking of the superstructure at first (before switching on the engines), but then he realises with a gasp that there is a huge black hole off the port bow (it was only just deduced by the ship's computers by analysis of orbiting material). But if we use special relativity for a moment (a good approximation over small enough regions of spacetime) we can see that a trajectory that might appear 'glancing' in the Schwarzschild coordinates will, in the rest frame of the death star, be Lorentz-contracted in the forward direction and therefore be approaching the horizon more steeply. Also, the horizon is approaching at tremendous speed, so tremendous forces will be needed to modify the death star's trajectory in time to have enough effect. It is these forces---the ones provided by the engines---which then rip the death star apart if part of it manages to escape. Presumably that is the part with Darth Vader on it, or Luke or maybe Kylo Ren (woops, I muddled my Star Wars saga a bit there).

  • 1
    $\begingroup$ Re, "(large BH), small tidal force, rod will not snap." That directly contradicts Anders Sandberg's answer, which you said was "helpful and correct." Maybe what you meant to say was that whether or not the rod snaps at the edge of the super-massive BH depends on how hard the engine attached to the outer end of it pulls. If the outer engine pulls hard enough, then the rod will snap, and the outer end has a chance to break free. If the engine doesn't pull at all, then the "small tidal force" will not pull the rod apart. $\endgroup$ Jan 24, 2020 at 15:03
  • $\begingroup$ @SolomonSlow ok; my example at that moment was in the case of no rocket engine, but I see I had left that unclear. I have now modified the text a little. $\endgroup$ Jan 24, 2020 at 15:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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