My reasoning being - lets say a rock is approaching a black hole. It would essentially stop in time for an outside observer once past the event horizon but since it would also bring along some new mass by itself, some of it should stop before reaching the event horizon, becoming the new edge of the event horizon.

Assuming that is true, if I were to feed a black hole from a single direction, it should start growing a spike, right? Going further with the same technique, you should be able to shape the black hole into a giraffe if you wanted to. Are there any flaws in this future business plan of mine?

  • $\begingroup$ Why are you saying the mass stops before reaching the event horizon? $\endgroup$ Commented May 11, 2017 at 14:24
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    $\begingroup$ You realize that the rock does actually pass the horizon in finite proper time and the horizon will never actually change shape? $\endgroup$
    – user87745
    Commented May 11, 2017 at 14:25
  • $\begingroup$ @NowIGetToLearnWhatAHeadIs because with the addition of the rocks mass, the radius of the event horizon should increase, no? Also, the center of mass should shift towards the direction the rock came from. $\endgroup$
    – user81993
    Commented May 11, 2017 at 14:30
  • $\begingroup$ For black holes in higher dimensions, see physics.stackexchange.com/q/292232/2451 $\endgroup$
    – Qmechanic
    Commented May 11, 2017 at 15:58

1 Answer 1


You are quite correct that if we drop an object into a black hole and watch it fall then we'll see it freeze at the event horizon. But this freeze occurs very close to the event horizon. In fact so close that it's barely distinguishable from the horizon. So dropping things into the black hole creates only a tiny perturbation and we couldn't use this trick to build any shape significantly different from a sphere.

If we consider the simplest case of a non-rotating black hole and drop an object from a long way away then the velocity of the infalling object is given by:

$$ v = \left(1 - \frac{r_s}{r}\right)\sqrt{\frac{r_s}{r}}c \tag{1} $$

I've discussed this before, in Will an object always fall at an infinite speed in a black hole?, and borrowing the graph from that post the velocity as a function of distance looks like:


Note that:

  1. the infall velocity peaks at about three times the event horizon radius

  2. the peak velocity is about $0.385c$ or about $115,000$ km/sec

Integrating equation (1) to get the distance as a function of time is rather messy, but we can do a quick back of the envelope calculation. If we take a Solar mass black hole then the event horizon is at about $3$ km so the peak velocity is at $9$ km. That means the infalling object is only $9$ km away and moving inwards at $115,000$ km/sec, so you'll appreciate that it's going to cross most of the $6$ km towards the event horizon pretty quickly. In fact if I do a quick and dirty numerical integration I get the following graph for time taken as a function of distance:

Distance time

The infalling object gets to within 1% of the event horizon radius in less than a millisecond.

This is the problem with your idea. Even though strictly speaking we never see the objects pass through the event horizon they very quickly get so close to it that to a distant observer they appear to have merged with it. The end result is that the horizon remains effectively spherical and we can't use your idea to build interesting shapes.

This isn't just theoretical, because we have actually observed the merger of two black holes at the LIGO gravitational wave observatory. The black holes were rotating around each other not falling directly towards each other, but even so the merger was effectively complete after about $150$ ms - that is, after $150$ ms the merged object was indistinguishable from a single spherical black hole even though the two black holes technically take an infinite time to fully merge.

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    $\begingroup$ Rather than feeding an existing black hole. It seems like it should be possible to bring a bunch of matter together all at the same time (from a particular reference frame) such that for a brief instant the collection of matter forms an event horizon in the shape of a giraffe, but then it all immediately collapses into a sphere. $\endgroup$ Commented May 11, 2017 at 18:12
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    $\begingroup$ @Shufflepants: yes, in that case the event horizon is time dependent and its shape changes with time. It will settle down to a sphere by radiating gravitational waves, and that process operates on the millisecond time scale. $\endgroup$ Commented May 11, 2017 at 18:21
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    $\begingroup$ And to add to it, as is apparent, horizons can and do indeed get deformed for very short periods of time (msecs) . These are being studied but it's not easy, most of the studies are small perturbations. When large perturbations you have to do numerical relativity. These horizons are sometimes called dynamic horizons. GR says that all the hair that can be radiated will be radiated, so only m, j and q remain. It's clear that it needs to happen fast, the sizes are not that large, maybe Kms, and the gravitational changes propagate at speeds of O(c). If we can see higher SNR we could see more of it $\endgroup$
    – Bob Bee
    Commented May 12, 2017 at 4:42
  • $\begingroup$ @JohnRennie This is the first time I have seen an actual, intuitive, well-explained answer to this. Bravo! Is the quick descent the reason that the LIGO 'chirps' do not go to lower frequencies again after seeming to approach infinity, because the waves redshift out of the detectors' sensitivity range? $\endgroup$
    – 2080
    Commented Aug 3, 2020 at 16:53
  • $\begingroup$ Small Follow up question: watching a body fall in, would we see some sort of redshifted after image that perhaps fades? Would it take ms, seconds, longer to fade? Has anyone calculated or animated this? $\endgroup$
    – RC_23
    Commented Nov 29, 2023 at 5:18

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