10
$\begingroup$

Does every black hole have its mass within its Schwarzschild radius, where the mass of the black hole is the one inside its event horizon? I know that the converse is correct: every object whose mass is compressed under its Schwarzschild radius will become a black hole. But is this a necessary condition? I think that the answer to the question is positive, but I couldn't find anything online.

$\endgroup$
5
  • 7
    $\begingroup$ An accretion disk contributes to the gravitational strength of a black hole (as measured from outside), but would you call it part of the black hole? $\endgroup$
    – Qmechanic
    Apr 20 at 16:33
  • 1
    $\begingroup$ @Qmechanic no, I meant to the mass of the black hole only inside its event horizon. $\endgroup$
    – Maor Cohen
    Apr 20 at 17:23
  • 4
    $\begingroup$ If the black hole is non-charged and non-rotating... aren't the Schwarzschild radius and the event horizon radius the same thing? Then you are asking whether all the mass inside radius r must be inside radius r $\endgroup$
    – user253751
    Apr 21 at 9:55
  • $\begingroup$ I think the answers have covered this already, but I think another, clearer, way of presenting this question would be: can the event horizon of a black hole ever protrude from the Schwarszchild radius? $\endgroup$
    – geshel
    Apr 21 at 17:27
  • $\begingroup$ Related : How can anything ever fall into a black hole as seen from an outside observer? $\endgroup$
    – J...
    Apr 22 at 11:59
9
$\begingroup$

Let's consider the simplest kind of black hole, which is one with no rotation or electric charge. The quick answer to the question is "yes, the mass is entirely within the Schwarzschild radius", but I would like to elaborate a little to explain what we are saying.

In the case of a more ordinary astronomical object, such as a planet or a star, one can have both the object itself and other things in orbit around it. The same is true for a black hole: there can be other things in orbit around it. So then the question arises, how in practice do we distinguish which mass belongs to the black hole, and which mass belongs to other things in orbit? This question is easy to answer for stable orbits, where the orbiting material is well outside the Schwarzschild radius. But when material is not in a stable orbit it may move towards the event horizon (at the Schwarzschild radius) and, in a time which will depend a lot on which observer is considered, it will pass the event horizon. After that the infalling mass soon reaches the singularity. We do not know the details of what transpires in the close vicinity of the central singularity, but the large scale picture is that the mass of the black hole has then grown a little, which we can tell by measuring the orbits of anything else still in orbit, or by gravitational lensing or other methods.

The horizon of a black hole is quite a subtle place because it is not just that the force of gravity gets to be very strong, it is more a case of the direction of time itself being "steered" in the inwards direction, so that once having passed the horizon, any object falling in cannot come out again just as surely as it cannot travel back to last week while going forward in time. This special feature applies to the horizon, and it means that right at the horizon there is a sort of cut-off: nothing can possibly hold an object fixed there. The object has to be either outside, or falling in and then it must keep on falling just as it cannot help moving forward in time.

$\endgroup$
12
  • $\begingroup$ After that the infalling mass soon reaches the singularity. Can we make that statement? What does "soon" mean inside the event horizon? $\endgroup$
    – J...
    Apr 21 at 18:52
  • 1
    $\begingroup$ @J... "soon" means "after a small amount of proper time" (the number has a max value that can be calculated) $\endgroup$ Apr 21 at 19:38
  • $\begingroup$ Which would be infinitely far into the future of our outside observer's clock? $\endgroup$
    – J...
    Apr 21 at 20:11
  • $\begingroup$ @J... There are many ways to assign a time coordinate for outside observers, and among them there is a standard one (the Schwarzschild coordinates) where the infalling trajectory does not even reach the horizon until the infinite future. $\endgroup$ Apr 21 at 20:23
  • 1
    $\begingroup$ @J... I guess we should move to chat. What we observe is that all signals from the infalling matter get weaker and redder and then stop altogether, all in a finite time (any final putative emitted wavefront forever hovering at the horizon has zero measure). $\endgroup$ Apr 22 at 14:16
7
$\begingroup$

The real answer to this question is that GR tells us that the question has no definite answer. In an asymptotically flat spacetime, we do have appropriate measures of the total mass (/mass-energy) in GR. However, we do not have any well-defined way to localize where that mass is.

As an example, take the Schwarzschild spacetime with mass $m$. This is a vacuum solution, so the stress-energy is zero everywhere. How, then, do we have any mass-energy? Well, measures of total mass such as the ADM mass or whatever will give us a mass $m$, but those measures of mass are global things. They don't tell us where the mass is located.

Conceptually, the idea is that relativity says mass and energy are equivalent, and gravitational fields have mass. So you would think that you could take the gravitational field $g$ and get a local measure of energy density from $g^2$. But we don't have any coordinate-independent way to define $g$. The equivalence principle tells us that for a free-falling observer, $g=0$ no matter what. Therefore we can always say that the gravitational field at some given point is zero.

$\endgroup$
7
  • $\begingroup$ This is a great answer, addressing the problem from an interesting perspective, but I'm afraid it's not what OP meant. $\endgroup$ Apr 21 at 8:15
  • 1
    $\begingroup$ It also ignores that for a stationary spacetimes we do have a quasi local measure of mass in the form of the Komar mass. The Komar mass is unambiguously constrained to be within the event horizon. $\endgroup$
    – mmeent
    Apr 21 at 8:22
  • $\begingroup$ "The stress-energy is zero everywhere" is not established because it breaks down as one approaches the singular behaviour where all such statements become questionable (along with the field eqn itself). Also "we do not have any well-defined way to localise where that mass is" is a little misleading I think, because we can make energy conservation arguments (Komar mass) for Schw. case, showing that energy falling into the horizon stays there and results in the appropriate increase in $m$. $\endgroup$ Apr 21 at 9:44
  • $\begingroup$ @mmeent Komar mass might be constrained to the event horizon, but this is not the only possible quasilocal measure of energy of black hole. There is e.g. Brown–York QL energy and the Martinez's conjecture about its properties for rotating black hole. $\endgroup$
    – A.V.S.
    Apr 21 at 14:44
  • $\begingroup$ @A.V.S. Do you agree the BH mass is inside the horizon? Nothing inside can have any effect on anything outside just like tomorrow's rain can't make you wet today. If there weren't anything inside at all, we wouldn't be able to tell a difference. It sounds to me like 8 wrong answers here. No? $\endgroup$
    – safesphere
    Apr 24 at 4:14
2
$\begingroup$

No, not all of the BH's mass needs to be inside $r_S$.
The Schwarzschild radius $r_S=\frac{2GM}{c^2}$ is the radius that a sphere of mass $M$ needs to have to be a black hole. A black hole doesn't need to have all its mass inside $r_S$, it only needs to have a fraction $M=\frac{r_Sc^2}{2G}$ of its mass inside it. The reason why the Earth isn't a black hole is because inside its Schwarzschild radius there's only a small amount of its mass. Indeed, most black holes have an accretion disc, and it's outside the Schwarzschild radius!


EDIT from comment: And if we are ignoring the acceleration disk, and defining a black hole's mass by its mass inside its event horizon, do all of a black hole's mass needs to be inside its Schwarzschild radius?
If we define the black hole's mass by the mass inside the event horizon, then the question becomes Can the event horizon be larger than the Schwarzschild radius? and the answer is No, not for any kind of black hole.

  • For a Schwarzschild BH $(Q=J=0)$ the event horizon is exactly at $r_S$.
  • For a Reissner-Nordstroem BH $(Q\neq0,J=0)$ there are two horizons, at $r_{\pm}=M\pm\sqrt{M^2-Q^2}$ in God-given-units, but both are smaller than the Schwarzschild radius.
  • For a Kerr BH $(Q=0,J\neq0)$ the two horizons are $r_{\pm}=M\pm\sqrt{M^2-a^2}$ with $a=J/M$ and again they can't be larger than the Schwarzschild radius.

So to summarize the black hole's mass, defined as the mass inside the horizon, needs to be inside the Schwarzschild radius.

$\endgroup$
3
  • 5
    $\begingroup$ The question is if mass outside a black hole is black hole mass. $\endgroup$
    – my2cts
    Apr 20 at 17:08
  • $\begingroup$ @Mauro Giliberti And if we are ignoring the acceleration disk, and defining a black hole's mass by its mass inside its event horizon, do all of a black hole's mass needs to be inside its Schwarzschild radius? $\endgroup$
    – Maor Cohen
    Apr 20 at 17:12
  • $\begingroup$ @MaorCohen please see my edited answer. $\endgroup$ Apr 20 at 19:34
1
$\begingroup$

It's a little hard to understand exactly what your question is. If you take a common definition of a black hole as 'a region of space having a gravitational field so intense that no matter or radiation can escape', then the event horizon defines the boundaries of the black hole so the question is meaningless. If it's inside the event horizon, it's part of the black hole. If it's outside the event horizon, it's not.

I assume you are thinking of some configuration of mass where for example there is an event horizon inside a neutron star and a crust of neutrons an nuclei that are actually outside the event horizon and are therefore visible to an outside observer and hiding the event horizon.

That isn't possible. The upper limit for a neutron star's mass is called the Tolman-Oppenheimer-Volkoff limit and is about 2.16 times the mass of the sun. In a neutron star the gravity is so intense that it overcomes electron degeneracy pressure, forcing the electrons in a star to combine with their nuclei forming a neutron star. If the collapsed core of the star is large enough, the gravity will be so intense that it will overcome the neutron degeneracy pressure. At this point, even neutrons cannot 'hold up' any matter, so the star will collapse into a black hole with an event horizon.

If there were an event horizon inside a neutron star, there would be nothing to support the matter above it and it would collapse into the event horizon.

$\endgroup$
1
$\begingroup$

Strictly speaking, the Schwartzchild radius is only defined for a Schwartzchild black hole, which is a highly idealized geometry that doesn't exist in the real world. For this particular geometry, all of the mass is contained within the Schwartzchild radius. For a more general and realistic black hole, it's no longer entirely trivial to either define the parameter "$M$" or to specify exactly which hypersurface is being determined by a numerical radius, as this is a coordinate-dependent concept. So I don't think your question is well-defined in general.

$\endgroup$
0
$\begingroup$

Knowing the finite speed of light, but still looking at the Newtonian theory, for a given massive object, the Schwarzschild radius, as a function of the given mass, can be defined to be the radius of a sphere in which the mass must be confined so at the Schwarzschild radius the escape velocity matches the speed of light.

It is maybe a little intuitive that it turns out in general relativity that this is the same radius of a sphere in which the mass must be contained so inside and outside get casually disconnected, therefore we call it event horizon.

So yes, the mass has to be within the event horizon. There are objects that are only a little larger than their event horizon, for example neutron stars. But these are still far from being black holes, since nowhere the curvature of space-time becomes as large as in the case of black holes.

$\endgroup$
0
$\begingroup$

The answer is I believe yes you could say by definition, an object is a black-hole if all its mass is found below its Schwarzschild radius, thus possessing an event horizon. Think about arguing by contradiction. Assume that there is some massive body which presents a Schwarzschild radius the event horizon is then in its inside (coincides with the Schwarzschild radius). Then we know the inner region will be causally disconnected to the outside, that is whatever is happening inside the Schwarzschild radius cannot affect the outer layers. This separates the object effectively in two pieces, one inside region which is a black hole and an outer region which is just mass orbiting around and probably falling in eventually.

$\endgroup$
0
$\begingroup$

If the Schwarzschild radius is interpreted as defining the event horizon of an electrically neutral non-rotating spherically symmetric mass distribution then the answer is: no. The EH of a charged black hole is characterised by a larger radius and I am guessing that similar statements can be made about a rotating, charged black hole.

I have assumed that any mass outside the EH does not contribute to the black hole mass. Such mass belongs to the BH system but not to the BH itself. The rational: In the case of a large BH such as Sag A* besides the accretion system many stars are gravitationally bound to the BH system. It will be hard if not impossible to define the BH mass if we start counting these as BH mass.

$\endgroup$
1
  • $\begingroup$ @Safesphere I clarified my assumptions in the answer. $\endgroup$
    – my2cts
    Apr 21 at 13:37
-1
$\begingroup$

This is an interesting question with a bunch of good answers but if one considers the gravitational field to be part of the black hole (neither could exist without the other), there is reason to believe all the mass is located outside the event horizon. Completely outside.

According to Lynden-Bell and Katz, http://adsabs.harvard.edu/full/1985MNRAS.213P..21L, the total energy distributed in the gravitational field of a Schwarzschild black hole is mc^2.

The authors claim: "We show by physical arguments that static spherical systems have a coordinate-independent field energy density. ... the field energy outside a Schwarzschild black hole totals Mc^2. In this sense all the energy remains outside the hole."

Clearly, sense mc^2 is the whole ball game energy-wise it leaves zero for the inside. Its distribution outside doesn't really matter; what counts is that it's all outside.

$\endgroup$
1
  • $\begingroup$ My apologies if I've offended someone here. I was hoping to get some feedback on Lynden-Bell and Katz's claims or perhaps on whether or not the gravitational field is considered to be part of the black hole. John Archibald Wheeler (A Journey Into Gravity And Spacetime) says: 'A black hole is a disembodied mass, a mass without matter' and: 'Only gravitational attraction remains behind'. So if there is only the gravitational field left behind and that field accounts for all the mass, how could their be any mass inside the event horizon? $\endgroup$
    – dcgeorge
    Oct 3 at 17:19

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.