According to general relativity, as I understand it, the space around a black hole's event horizon is distorted by gravity, such that the distance to a point approaching the event horizon from an observational point further out from the EH approaches infinity.

So if the distance to any mass inside the EH appears to be infinite, how does this mass assert gravitational effects outside the EH?

Perhaps a different question, what is the difference between the gravitational effects of a mass inside the EH of a black hole and a similar mass that is outside the cosmic event horizon of an observational point?

(edit as follows) My original intent for this question is rethink how we think about black holes. The distance to a point on the event horizon of Sag A is as far to the point on an observer's cosmic event horizon. You can travel no farther than such a point on the CEH than you could to a point on Sag A's event horizon, at least using our 4D coordinate system. BTW, I think all talk about a black hole having a singularity at its center is bunk, first of all, we can never get there to confirm or deny. Secondly, the density of Sag A if it were uniform, would mathematically would be around the density of liquid water as it exists on earth, but we would never be able to know, just as we would we would ever be able to know the density of the universe outside of our cosmic event horizons. Any answers to those two questions would be meaningless anyway, as long as there is no away around the speed of causality rule enforced by our universe.

Now I believe I have read that Hawking Radiation has been shown to exist. However, does that mean the part of whatever is outside our cosmic event horizon is evaporating as well?

  • 2
    $\begingroup$ See physics.stackexchange.com/q/937/56299 on Physics Stack Exchange. $\endgroup$ – HDE 226868 Mar 2 '17 at 18:07
  • $\begingroup$ I've added some additional comments to this question. $\endgroup$ – Robert Mashlan Apr 19 '20 at 20:05
  • $\begingroup$ That's a fairly major addition to a 3 year old question, Robert. I think it probably would be better to ask a fresh question. If you like, you can link it to this one and say it's a follow-up question. $\endgroup$ – PM 2Ring Apr 20 '20 at 4:06
  • $\begingroup$ However, I get the impression that maybe you don't fully understand (or at least don't agree with) the answers you've already been given here, and the answers in the Physics.SE question linked by HDE 226868. From Stan Liou's answer "rather than gravity having a special property that enables it to cross the horizon, in a certain sense gravity can't cross the horizon, and it is that very property that forces gravity outside of it to remain the same." $\endgroup$ – PM 2Ring Apr 20 '20 at 4:12
  • $\begingroup$ FWIW, a proper theory of quantum gravity may eliminate BH singularities. But if they do exist, they have a peculiar kind of existence: the singularity is never in the past light cone of any observer, including observers who are inside the EH. Also, we cannot give a solid proof of Hawking radiation without quantum gravity theory, and if it exists it's far too feeble to detect around stellar mass or larger BHs, even if you are very close to the BH. $\endgroup$ – PM 2Ring Apr 20 '20 at 4:26

As is the case even in Newtonian gravity, the gravitational attraction from a spherical body exterted on a far-away object is identical the that exterted by a point mass located at the the center of the body. This is the so-called shell theorem.

To an outside observer, due to the infinite time dilation at the event horizon of a black hole (BH), all mass that has ever fallen into the BH has actually never reached the event horizon, but is instead placed in a shell around the BH, still falling ever-so-slowly toward the BH.

Thus, what we feel is in fact the gravitation not from a point mass, but from a shell of matter, but there is no way for us to tell the difference.

Now you may ask, "What if we drop a massive object, e.g. a planet, into a symmetric black hole? Since it never reaches the BH, won't the effective gravitational potential be asymmetric?" What happens is that the way that space is "warped" around the black, the gravitational field is bent along with it in such a way that sufficiently far from the BH+planet, the gravitational field looks symmetric. Sort of like the following drawing shows close to (left) and far from (right) the BH.


  • $\begingroup$ Even if all the matter is collected at a shell just outside the EH, it would appear to be a very far from an observation point from further away. I don't doubt that BHs exist and exert gravitational force on nearby mostly flat space, I've seen the animations of observations of SO2 whipping around an object in the center of the galaxy that could only be a BH. The problem that I have is in understanding in how Newton's law of gravity can be applied when a BH's mass appears to be very far away, except to the infalling observer. $\endgroup$ – Robert Mashlan Mar 4 '17 at 1:52

The physical distance does not tend to infinity. What does tend to infinity is the 'coordinate distance', which is unphysical. Think about it: there are probably numerous black holes in our observable universe - how can they all be infinitely far away?

The physics of black holes is described by general relativity. It cannot be understood using any Newtonian formula, such as F = GM1M2/r^2.


Here's my understanding as a backyard astronomer...

First, gravity is more accurately defined in Einstein's STR as energy that bends space/time (because energy and mass are linked by E=mc^2)(1). So let go of Newtonian mass attracting mass. A black hole then is a huge energy source that is warping space/time initially incredibly deeply, but then tapering off according to the inverse-square law.

The event horizon (EH) is defined as the boundary where anything with mass, which includes light, has to travel more than the speed of light (c) to attain escape velocity. As c is the absolute upper limit on speed, nothing can escape and the 'hole' is 'black'. (2) Trying to calculate what is going on within the EH is fruitless, as current equations tend to return "Divide By 0"! ;)

So, with the mass of the singularity warping space/time causing what we call 'gravity', gravity itself doesn't have to escape anything - it just is.

Thus a black hole can be observed exerting a gravitational influence on objects around it, such as the observations of stars and stellar clouds in orbits around Sagittarius A* (3)
As per any gravitational system, an object may be attracted towards a black hole and then ejected, it may be in a stable orbit, or it may be spiralling inwards to eventually be torn apart by tidal forces and join the accretion disc around the EH. Here it may fall in to add to the mass of the singularity or may be radiated by the energy of the accretion disc and jets (4).

If a system consists of binary black holes, they might orbit one another until eventually spiralling into each other and merging in a cataclysmic event that the LIGO team recently detected (5)

Hope that helps a bit!

(1) http://astronomy.swin.edu.au/cosmos/M/Mass

(2) http://astronomy.swin.edu.au/cosmos/E/Event+Horizon

(3) http://astronomy.swin.edu.au/cosmos/C/Centre+Of+The+Milky+Way

(4) http://astronomy.swin.edu.au/cosmos/J/Jets

(5) http://astronomy.swin.edu.au/cosmos/G/Gravitational+Waves


Inside the event horizon, or what is thought of as the mass of the black hole is the section in which you can tell the "strength" of the pull from the singularity. The larger the event horizon (which is initially based on the star it originated from, r= 2GM/c^2, this gives the event horizon radius where G is the gravity constant and c is the speed of light) the stronger the singularity is, and the more dense the black hole is. By adding mass, or having the particle anti-particle pairs forming and then adding mass to the singularity, then the density decreases which decreases its strength therefore it decreases the radius of the event horizon. Where the event horizon is what we "see" or detect through equipment it just takes up more space, and by taking up more space, it has the greater pull on things around it. So basically by the size of the black holes event horizon, it will have a greater pull on things outside in space.

  • 1
    $\begingroup$ What do you mean by "the stronger the singularity is". The singularity is a concept, not something which can have strength. Furthermore, how does the "strength" of the singularity relate to the "density" of the black hole? And just what do you mean by the density of a black hole? $\endgroup$ – zephyr Mar 2 '17 at 15:35
  • 1
    $\begingroup$ I think that you have misunderstood my question. My conceptual problem is that the mass inside a black hole is outside the observable universe for every point outside the EH. For all except the infalling observer, this appears to be an infinite distance to the EH. F=Gm1m2/r^2 seems to imply that F approaches 0 as r approaches infinity. $\endgroup$ – Robert Mashlan Mar 2 '17 at 16:49

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.