It is my understanding that observers outside the black hole wouldn't ever see some infalling object enter the black hole, as the object's clock would be running so slow in the observer's reference frame that it would essentially freeze near the event horizon.

However, it is also my understanding that the "gravitational field" (I apologise for my classical mechanical intuition) is centred around the black hole's singularity.

Of course, in the outside observer's reference frame, the object never fell into the black hole, so its gravitational field should stay near the event horizon rather than at the singularity.

Does this mean that the gravitational field itself is different for different observers, and that for outside observers most of the mass is actually concentrated throughout the black hole's circumference (event horizon) rather than its centre (singularity), or is there some other resolution?


2 Answers 2


Although I haven’t seen any calculation of this, I think that the gravitational field of a small perturbing mass outside a black hole actually becomes centered on the hole, not on the small mass, as the mass approaches the event horizon.

I think this because something similar happens with a point charge held stationary outside a black hole. (I have seen this calculation.) As the charge is held closer and closer to the horizon, its electrostatic field becomes spherically symmetric around the hole.

My way of thinking about this is that it is part of the “no hair” aspect of black holes. If the field perturbation did not become spherically symmetric as the mass or charge approaches the horizon, then by continuity it would not be spherically symmetric when the mass or charge is inside the horizon, and we would be able to tell where the charge or mass is inside the hole. Instead, by the time it crosses the horizon (which it does, according to its own clock) it has simply added to the hole’s mass and charge.

  • $\begingroup$ Oh wow, I never considered that effect of the black hole on a charge's electric field before and its relation to the no hair theorem. I probably should've though as I have come across it before. That explains a lot. Thank you. Here's a link to a paper which mentions the effect of a black hole on an electric field if OP want to look at it. link.springer.com/article/10.1007/s10701-007-9118-8 $\endgroup$
    – Laff70
    Nov 19, 2019 at 4:03
  • $\begingroup$ Here's another image of the effect: universe-review.ca/I15-41-BHelectric.png $\endgroup$
    – Laff70
    Nov 19, 2019 at 4:20
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    $\begingroup$ "a small perturbing mass outside a black hole actually becomes centered on the hole, not on the small mass, as the mass approaches the event horizon" - This sounds right +1. The horizon is not a place, e.g. two points at the horizon are not spacelike separated. The horizon is lightlike meaning that the spacetime interval between any two points on it is zero. Anything anywhere at the horizon is simultaneously everywhere at the horizon. $\endgroup$
    – safesphere
    Nov 19, 2019 at 5:43
  • 1
    $\begingroup$ @safesphere I like that way of understanding it. Thanks! $\endgroup$
    – G. Smith
    Nov 19, 2019 at 5:48

Oh, this is an interesting calculation I once did. To start off, lets imagine we have a box with a photon of wavelength $\lambda$. Now we'll have a spherical shell of mass and shrink it down from infinity to a smaller finite size. It will envelope the box and give it time dilation $t_d$. Now the photon's wavelength hasn't changed. Its energy has though. We can show this by opening the box and letting the photon leave the system and go to infinity. We'll assume the shell of mass is transparent. At infinity, where there's no time dilation, the photon's wavelength is $\lambda/t_d$. This mean's the photon's energy is less than it used to be. $$E=\frac{hc}{\lambda}$$$$\frac{E_f}{E_i}=t_d$$ As we can see, if an object is in a shrinking spherical shell where the time dilation is increasing, it loses energy. But where does that object's energy go you ask? It goes into the shrinking shell's kinetic energy.

If we had a mass and shrunk down a spherical shell on top of it as much as we could before it couldn't collapse anymore due to gravitational time dilation, all of the mass energy in said mass would flow out of it and into the shell. If we kept throwing more shells on top of it, mass energy would flow from shell to shell. At no point though would an event horizon actually form.

I've theorized that a true Schwarzschild black hole cannot form from a spherically symmetric mass distribution. What would form would be a kind of pseudo-black hole. If left to sit for a while, it'll get closer and closer to being a true black hole. However, even if you wait forever it'll never truely get there. The outer radius will always at least be infinitesimally bigger than the Schwarzschild radius. I don't know if a quantum fluctuation could change that though I suspect not due to immense time dilation slowing down all quantum processes at the pseudo-horizon. I also don't know if at any point a part inside the collapsing pseudo-black hole would be truly inescapable. I could do that calculation though. It would just depend on the rate at which time dilation increases along a radially moving photon's path from inside the pseudo-black hole. I suppose if that did happen it'd be somewhat similar to an event horizon though. $t_d$ still wouldn't quite reach 0 though. The pseudo-black hole also wouldn't have a singularity.

I don't know if black holes can't form from non-spherically symmetric mass distributions though. Unfortunately I'm not good enough at GR currently to figure that out. I also don't know what the surface energy and momentum density for a Kerr pseudo-black hole would be. I kinda suspect that gravitational time dilation would prevent the formation of those kinds of black holes as well though. I call this the No Holes Hypothesis. Essentially no true black holes or their respective singularities would actually exist. There would only be pseudo-black holes, or pseudo-holes for short, which store virtually all their energy and momentum at their surfaces.

  • $\begingroup$ Ahh, so that does have a name/term! And here I was thinking I was the first to figure out this alternative to black holes. Thanks for showing me that! I also now think that this should apply to non-spherically symmetric systems after reading G. Smith's comment. You know those metamaterial cloaking devices? They essentially work by changing both the electric permittivity and magnetic permeability of the region around them in different directions using transformation optics. As per its Wikipedia article "The mathematics underpinning transformation optics is similar to the equations that describe $\endgroup$
    – Laff70
    Nov 19, 2019 at 5:36
  • $\begingroup$ how gravity warps space and time, in general relativity". I believe that if you were to take a region of space and alter its electric permittivity and magnetic permeability in the right way, you could effectively disjoin it from the rest of the universe. Meaning that objects passing through where the inner radius should be would go around it exactly like light goes around the region a meta-material cloaking device shields. In a way it would kind of make the interior region a separate universe. Anyways, an object passing through where the inner radius should be would actually be getting smeared $\endgroup$
    – Laff70
    Nov 19, 2019 at 5:38
  • $\begingroup$ in a way around the inner radius. I believe this might also be happening in a way when a mass approaches an "event horizon". Similarly to how the electric field of a charged particle gets smeared around a pseudo-hole it approaches, so would the mass energy and momentum of an object approaching said pseudo-hole. So even in the case of a non-spherically symmetric collapse of mass, the end result is still a Schwarzschild pseudo-hole with its mass energy evenly smeared across its surface. Not sure if the smearing would be even for a Kerr pseudo-hole though. $\endgroup$
    – Laff70
    Nov 19, 2019 at 5:39
  • $\begingroup$ Well this is a lot of information to take in. It might take me a while to fully process it. Thank you very much! I might get back to you regarding all this later on. $\endgroup$
    – Laff70
    Nov 19, 2019 at 6:42
  • $\begingroup$ Thanks for the warning, I'll save this page as a PDF. $\endgroup$
    – Laff70
    Nov 19, 2019 at 6:45

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