What Happens When a Black Hole Evaporates Below its Schwarzschild Radius?

[Ascending One]

Hi,

I have a question about black holes that has to do with their evaporation. What is predicted to happen once a black hole evaporates enough matter that its Schwarzschild Radius is no longer larger than its size (i.e. the point at which it ceases to be a black hole)? Will the mass of the black hole 'snap' back into 'normal' space (from 'quantum space'), for lack of better words.

 

[cesiumfrog]

Just to be perfectly clear, how about you explain what you mean by "its size"?

 

[Ascending One]

I think what I mean is its volume, or the radius of its volume. I'm not really sure actually.

 

[mgb_phys]

As a black hole shrinks it's temperature rises, so it's final mass will be emitted as gamma rays.

 

[Ascending One]

So - are you implying that a black hole will remain a black hole until it evaporates its last bit of matter?

That really doesn't seem to make sense... wouldn't it, at some point lose enough matter that it would stop being a black before it lost all its matter?

 

[genneth]

No. Any amount of mass can be a black hole, as long as its confined to within its Schwarzschild radius.

 

[DarkMatter258]

I think the original question was what would happen once it was no longer confined to within its Schwarzschild radius... no?

 

[mgb_phys]

The Schwarzschild is proportional to mass, so as the mass decreases it's radius decreases until it 'classically' dissapears.
As it gets very small quantum effects start to matter and at some point the final mass energy escapes as a couple of gamma photons.

 

[genneth]

I've always wondered about the formation process more than the end. In particular, from the point of view of an asymptotic observer, in-falling matter will take forever before crossing the event-horizon. This includes the initial collapse. However, it will start to Hawking radiate all the time, as soon as there is curvature. In fact, it seems the black hole will evaporate before in-falling matter gets to the event horizon.

 

[mgb_phys]

The point of view of the infalling matter is irrelavant - the universe doesn't care!
Forming a black hole is easy - you just have to push matter together, closer than the Schwarzschild limit before the nuclear force can push it apart.
You only have to worry about evaporation for very small holes - this is a good thing, it stops any small holes formed by cosmic rays in the atmosphere from growing!

 

[Ascending One]

I think the original question was what would happen once it was no longer confined to within its Schwarzschild radius... no?

Yes - this was the original question.

In response to this:

The Schwarzschild is proportional to mass, so as the mass decreases it's radius decreases until it 'classically' dissapears.
As it gets very small quantum effects start to matter and at some point the final mass energy escapes as a couple of gamma photons.

Yes - I understand that the Schwarzschild radius is proportional to the mass. But I thought that the Schwarzschild radius was more of a theoretical concept - not a physical measure.

EDIT: I guess I should clarify - I know that the Schwarzschild radius can be manifested as a physical measure - namely the event horizon. But I also thought that the schwarzchild radius could be independent of the radius of the volume of the mass contained in an object.

 

[mgb_phys]

It's about the only physical measure of a black hole. It's basically the point at which sensible classical physics stops - inside this radius everything gets a bit silly.

 

[Ascending One]

But - anything has a schwarzchild radius, correct? The sun has a schwarzchild radius which is much smaller than the size of the sun. I'm saying 'size of the sun' because I'm not sure what measurement that actually is. I'm assuming that it is the 'radius of the volume of the sun.'

With a black hole, it's just that the schwarzchild radius is larger than the 'size of the black hole' before it actually turns into a black hole. At that point, what happens?

 

[genneth]

With a black hole, the mass has zero volume, since no know physical process can stop the inwards collapse. So no matter how much mass you lose to evaporation, the matter inside is still smaller than the Schwarzschild radius.

 

[mgb_phys]

Yes everything has a value for the schwarschild radius, ie the size it would have if it were to be squashed down to form a black hole.
The definition of a black hole is that it has mass 'm' inside it's schwarschild radius.
For something where this has happened the schwarschild radius = the event horizon, has a definite physical meaning.

The schwarschild radius is the boundary at which the escape volocity is greater than c, so inside this radius classical physics doesn't work, outside this radius you can still do the sums.
In practical terms not a lot actually happens as a the size goes below the radius and it becomes a black hole, the gravitiational effects of a star squeezed into a size just slightly larger than it's black hole radius are still pretty extreme.

 

[dilletante]

In practical terms not a lot actually happens as a the size goes below the radius and it becomes a black hole, the gravitiational effects of a star squeezed into a size just slightly larger than it's black hole radius are still pretty extreme.

I take this to mean then, that the radius of a star can be one foot (or one mile, doesn't matter) larger than the Schwarzschild radius. Does anyone theorize that it can be one foot less than the Schwarzschild radius, or do all theories assume zero volume of any mass that is within an event horizon? In other words, does current physics predict a sudden catastrophic collapse the moment mass is squeezed smaller than the BH radius, or is this a mathematical artifact of the "singularity" (infinity) problem of GR?

 

[genneth]

I take this to mean then, that the radius of a star can be one foot (or one mile, doesn't matter) larger than the Schwarzschild radius. Does anyone theorize that it can be one foot less than the Schwarzschild radius, or do all theories assume zero volume of any mass that is within an event horizon? In other words, does current physics predict a sudden catastrophic collapse the moment mass is squeezed smaller than the BH radius, or is this a mathematical artifact of the "singularity" (infinity) problem of GR?

All this stuff is in any textbook on the subject. You cannot expect PF members to recite what books take chapters to explain.

You seem to still be in a state confusion over the different times experienced by different points of view. For someone standing on the surface of a collapsing star, nothing extraordinary (as far as gravity goes, there's all sorts of stellar stuff flying about) happens. They can't even tell when they've passed the event horizon. However, they will measure a finite amount of time (as they would by looking at their watch), before they reach the singularity. At which point, our current physics doesn't say what happens.

From an observer far away, the surface of the star red-shifts out to black. As it happens, this is exponentially fast -- for a typical star, the time taken to go from 1.5x the event-horizon radius to the point where the last photon (taking into account the discreteness of energy) is likely to be emitted is about 10^-4 seconds. Furthermore, it becomes impossible to interact with the in-falling stellar surface. A photon launched at the stellar surface won't reach it before the stellar surface crosses the event-horizon.

Now, it's certainly possible for a star to have a size just above its Schwarzschild radius. It is even possible for it to be below -- though only for a finite amount of time as seen by the star. For everyone else, the star collapses to the size of the Schwarzschild radius, and then cannot be interacted with any more, so we call it a black hole. For external observers, there is no point in talking about stars smaller than their Schwarzschild radius, because there's no observational way to distinguish them -- they're all the same hairless hole.

 

[dilletante]

All this stuff is in any textbook on the subject. You cannot expect PF members to recite what books take chapters to explain.

Sorry for the inconvenience, I seem to have misplaced my textbooks 40 years ago.

 

[genneth]

Sorry for the inconvenience, I seem to have misplaced my textbooks 40 years ago.

I tend to get mine out from a library when I need them...

 

[dilletante]

I tend to get mine out from a library when I need them...

First, I wish to say that I appreciate your taking the time to answer my question. Neither you nor anyone else on the forum is obligated to answer anything, if you feel a question is stupid, or obvious, or too much trouble. So my appreciation for the time you took is genuine.

I find great value in being able to ask questions and have experts such as you and others share their knowledge. In this particular instance however, I felt berated for having asked a question, and am less likely to ask one in the future as a result, which saddens me.

If, as you put it, no one on this forum is going to take the time to answer questions that physics books devote chapters to, or that can be answered by researching textbooks in the library, then I suppose I have misunderstood the purpose of these forums. It was my understanding that discussions should be precisely about those things, and not about theories not found in textbooks.

You certainly have the option of ignoring a question if you find it offensive or beneath discussion, and I believe most laymen asking questions here would prefer that to condescension.

 

[SpaceTiger]

I take this to mean then, that the radius of a star can be one foot (or one mile, doesn't matter) larger than the Schwarzschild radius. Does anyone theorize that it can be one foot less than the Schwarzschild radius, or do all theories assume zero volume of any mass that is within an event horizon? In other words, does current physics predict a sudden catastrophic collapse the moment mass is squeezed smaller than the BH radius, or is this a mathematical artifact of the "singularity" (infinity) problem of GR?

The collapse of a star into a black hole is a highly non-trivial problem, particularly if the star is rotating. With a number of questionable simplifying assumptions, some textbooks will give models that are meant to simulate a collapsing star, but in reality the problem hasn't even been solved. For example, some theories suggest that a Type II supernova should accompany the collapse into a black hole, while others suggest that the collapse would occur with minimal radiation output.

Also note that the collapse is a result of the loss of a restoring force, not compression beyond the Schwarzschild limit. Extremely compact stars, like neutron stars and the cores of massive stars, are held together by degeneracy pressure, a quantum effect that results from the limited number of states available to particles within the star. This degeneracy pressure fails when the objects exceeds the Chandrasekhar mass, resulting in rapid gravitational collapse.

 

[dilletante]

The collapse of a star into a black hole is a highly non-trivial problem, particularly if the star is rotating. With a number of questionable simplifying assumptions, some textbooks will give models that are meant to simulate a collapsing star, but in reality the problem hasn't even been solved. For example, some theories suggest that a Type II supernova should accompany the collapse into a black hole, while others suggest that the collapse would occur with minimal radiation output.

Also note that the collapse is a result of the loss of a restoring force, not compression beyond the Schwarzschild limit. Extremely compact stars, like neutron stars and the cores of massive stars, are held together by degeneracy pressure, a quantum effect that results from the limited number of states available to particles within the star. This degeneracy pressure fails when the objects exceeds the Chandrasekhar mass, resulting in rapid gravitational collapse.

That is quite helpful information about the degeneracy pressure. It seems then that collapsed stars can be in a limited number of states of collapse, depending on mass. If I understand correctly, white dwarfs are supported by electron degeneracy pressure, neutron stars by neutron degeneracy, quark stars (if they exist) by quark degeneracy, and finally black holes lack the pressure to maintain a radius larger than the Schwarzschild radius.

 

[SpaceTiger]

That is quite helpful information about the degeneracy pressure. It seems then that collapsed stars can be in a limited number of states of collapse, depending on mass. If I understand correctly, white dwarfs are supported by electron degeneracy pressure, neutron stars by neutron degeneracy, quark stars (if they exist) by quark degeneracy, and finally black holes lack the pressure to maintain a radius larger than the Schwarzschild radius.

Looks about right. In reality, these things are always more complex than the textbooks and internet resources lead one to believe, but those are the basic models we work with. At the moment, we have direct observational evidence for the existence of white dwarfs (very convincing) and neutron stars (fairly convincing). Black holes have not been directly observed, but there is a great deal of indirect evidence for their existence. This last issue has been discussed at great length on PF, but you'd have to do an archive search to find it.

 

[Chris Hillman]

Ditto ST, except:

black holes lack the pressure to maintain a radius larger than the Schwarzschild radius.

Bad way of thinking about it: the whole point is that no pressure can sustain a body which has been compacted enough that it lies within its Schwarzschild radius (or rather a somewhat larger radius, given some reasonable assumptions). Be aware that extreme pressure is associated with a substantial contribution to the stress-energy tensor which acts (in gtr) as the source for the gravitational field.

Your discarded textbooks from 40 years ago are out of date anyway, so I'd urge you to get a more recent one. There are many excellent choices, including some inexpensive books.

 

[dilletante]

Ditto ST, except:



Bad way of thinking about it: the whole point is that no pressure can sustain a body which has been compacted enough that it lies within its Schwarzschild radius (or rather a somewhat larger radius, given some reasonable assumptions). Be aware that extreme pressure is associated with a substantial contribution to the stress-energy tensor which acts (in gtr) as the source for the gravitational field.

Your discarded textbooks from 40 years ago are out of date anyway, so I'd urge you to get a more recent one. There are many excellent choices, including some inexpensive books.

Thanks, I phrased that poorly. I have Kip Thorne's "Black Holes and Time Warps" on order -- don't know if that is my best choice but I suspect I would get lost in the math of a pure textbook.

 

[randa177]

Assuming that the black hole at the center of the Milky Way Galaxy (MWG) is Schwarzschild (non-rotating) black hole, how large is its Schwarzschild radius, in AU?

 

[Chris Hillman]

Formation of a black hole?

I've always wondered about the formation process more than the end. In particular, from the point of view of an asymptotic observer, in-falling matter will take forever before crossing the event-horizon. This includes the initial collapse.

Well, as you appear to know, this is misleading. I could try to explain the mathematical details of the Oppenheimer-Snyder collapsing dust ball model or the Vaidya collapsing shell of null dust model for the formation of a black hole, but unless you have a good math/physics background it probably makes more sense to point you at the same excellent nontechnical book I just mentioned in another thread, General Relativity from A to B by Robert Geroch, which aims to explain the geometrical picture of the formation of a black hole via the OS model.

 

[mgb_phys]

Assuming that the black hole at the center of the Milky Way Galaxy (MWG) is Schwarzschild (non-rotating) black hole, how large is its Schwarzschild radius, in AU?

R = 2GM/c^2

Assuming the black hole is around 2.5 million solar masses
G = 6 E-11 m/kg/s^2, Msun = 2E30kg C=3E8 m/s

R = 1.5E-27 * 5E36 = 7.5E9 m = 0.05 AU ( seems pretty small ? )

 

[Chris Hillman]

seems pretty small ?

That's what [post=1497780]I got too[/post] (except that I used a smaller figure for the estimated mass of Sag A*).

 

[DaveC426913]

Yes everything has a value for the schwarschild radius,

Minor poiint: Everything potentially has a S.radius - if it were shrunk. While a compactified object of the sun's mass could have a S.radius, the sun itself does not.

Maximum gravitational curvature is acquired at the sun's surface. Any deeper points have less curvature, just like any other solid spheroidal object.

 

[Chris Hillman]

Static Spherically Symmetric Perfect Fluids (ssspf), Anyone?

We are still talking about gtr, aren't we?

Maximum gravitational curvature is acquired at the sun's surface. Any deeper points have less curvature, just like any other solid spheroidal object.

If you mean path curvature of the world lines of the fluid particles, while the details depend upon what ssspf solution you adopt for the interior (in favorable cases this comes down to choice of equation of state), in general that is not quite correct: the acceleration vector of the fluid particles always points radially outward as you would expect; its magnitude is of course the "surface gravity" at the surface and typically increases to a maximum inside the surface and then decreases to zero at the center--- as must happen by symmetry. (There are some solutions in which the maximal path curvature does occur at the surface, but they are the exception.)

If you mean spacetime curvature or curvature of the spatial hyperslices (the ones orthogonal to the world lines of the fluid particles), then in general the maximal curvatures occur at the center. In particular, if you consider the three-dimensional Riemann tensor of the spatial hyperslices orthogonal to the static timelike congruence (which corresponds to the world lines of bits of fluid inside and to the world lines of static observers who use their rocket engines to "hover" outside), then with components taken wrt the natural frame field, typically 0 < r_{2323} = r_{2424} < r_{3434} with both increasing to a maximum at the center, where they agree.

The simplest model of an isolated object is probably the Schwarzschild stellar model constructed by K. Schwarzschild a few months after Einstein published his field equations. KS "matched" his vacuum solution across the world sheet of a static sphere of a certain radius to his "constant density" ssspf solution. Then if we consider the static congruence, the hyperslices orthogonal to this congruence consist of part of the Flamm paraboloid matched to a spherical cap, which you can visualize embedded in E^3 with the tangents agreeing at the sphere where the matching is carried out, which physically corresponds to the zero pressure surface. In this model, the density is indeed constant, and the pressure increases from zero at the surface to a maximum at the center of the star. Here is a Wikipedia article which was unfortunately left incomplete when I left Wikipedia, but which does include a plot of this embedding. See also [post=146274]this post[/post]. While it is not obvious from the plot, the curvature is in fact uniform and constant in the interior, but this is exceptional. For most ssspf solutions, the curvature in the interior is nonuniform, although at the center it does approximate the Schwarzschild fluid (by the remark at the end of the preceding paragraph).

I once started to write a review paper of ssspf solutions (a task for which I cheerfully confess myself utterly lacking in qualification!), so I am familiar with the behavior of pressure and density wrt "radius" (typically the Schwarzschild radial coordinate is used) for two dozen or so distinct ssspf solutions. On the order of a hundred such solutions have been published (the general ssspf is in some sense "known"), but Kayll Lake has shown in his own review that most of these were wrong or physically unacceptable for any values of their parameters. Lake's review predates important advances by Matt Visser and his coworkers. I also found simple formulas for the tidal tensor and hyperslice curvature of the relativistic polytrope, which has often been considered intractable. Unfortunately motivating my derivation requires knowledge of Lie's methods of symmetry analysis.

Among the "good" solutions, one whose virtues stand out is also one of the very earliest such solutions discovered, the Tolman IV ssspf solution. This turns out to admit an equation of state and provides a fairly good empirical match to observations of neutron stars, the most compact astrophysical objects (in the sense of "made of stuff") which we currently know about, but not to ordinary stars.

I could say a lot more about this subject, so much so that it would definitely belong in another thread (in the relativity forum).

 

[aman malik]

can anybodu pls. tell me what really is schwarzchild radius??

 

[Chris Hillman]

Hi, aman,

In curved spacetimes, there are many coordinate systems we may and do use to represent a given exact solution (model of some specific physical scenario involving gravitation and perhaps other physics).

The Schwarzschild radial coordinate, usually written r, is a particular coordinate which can be defined in spherically symmetric spacetimes and whose geometric meaning is comparable to the usual radial coordinate as used in the polar spherical coordinates familiar from vector calculus in euclidean three-space.

The Schwarzschild vacuum is a Lorentzian four-manifold which is an exact vacuum solution of the Einstein field equation (EFE) and which provides a simple model of a black hole. In the most commonly used coordinate chart, its line element is
<br /> ds^2 = -(1-2m/r) \, dt^2 + \frac{dr^2}{1-2m/r} <br /> + r^2 \, \left( d\theta^2 + \sin(\theta)^2 \, d\phi^2 \right),<br /><br /> -\infty &lt; t &lt; \infty, \; 2 m &lt; r &lt; \infty, \; 0 &lt; \theta &lt; \pi, \; -\pi &lt; \phi &lt; \pi<br />
where r is the Schwarzschild radial coordinate. As you can see, the case m=0 reduces to the usual polar spherical chart for Minkowski vacuum (flat spacetime; no matter or fields present)
<br /> ds^2 = -dt^2 + dr^2<br /> + r^2 \, \left( d\theta^2 + \sin(\theta)^2 \, d\phi^2 \right),<br /><br /> -\infty &lt; t &lt; \infty, \; 0 &lt; r &lt; \infty, \; 0 &lt; \theta &lt; \pi, \; -\pi &lt; \phi &lt; \pi<br />

It turns out that in gtr, any uncharged nonrotating black hole can be modeled by the Schwarzschild vacuum solution. The Schwarzschild radius of an uncharged nonrotating black hole is the value of the Schwarzschild radial coordinate which locates a sphere known as the event horizon, which forms the inner boundary of the chart above, namely r=2 m. Here, m is a parameter describing the model which turns out to correspond to mass (for example, the mass you would use in Keplers laws, far from the hole).

There is a similar model of a charged nonrotating black hole known as the Reissner-Nordstrom electrovacuum, which can also be written using a Schwarzschild radial coordinate, and which also has an event horizon. The Schwarzschild radius of such a hole is again the value of the Schwarzschild radial coordinate which locates the event horizon. It is approximately
r \approx 2 m - \frac{q^2}{2 m}
where m is the mass and q is the charge of the hole (I assumed that m \gg q, a condition which holds for astrophysical black holes). This approximation is derived from the exact value, the positive root of r^2 - 2 m r + q^2 = 0.

A bit more about coordinates: in the theory of manifolds, a coordinate is simply an increasing function u defined on some local neighborhood U of our manifold. That is, u has nonzero gradient; in symbols, we require that du \neq 0 on U. If we have another coordinate defined on a neighborhood in a two-manifold M, and if du, \; dv, \; du \wedge dv are all nonzero on U, then u,v define a coordinate chart with domain U on M. Here, the condition du \wedge dv \neq 0 ensures that the gradients of u,v are never aligned inside U, which means that the level surfaces (here one dimensional) of the coordinates, u = u_0, \, v=v_0, are nowhere tangent inside U. Similarly in higher dimensions. These conditions ensure that each point in U has a unique "address" as a tuple of values of the coordinate functions. (Indeed, a street address in Manhattan is pretty much an integer valued pair of numbers which are the values of two coordinates. One can imagine locating a tree on the sidewalk by using a tuple of real numbers.)

If there are symmetries present, it is possible to define in a "coordinate free manner" particular functions which can serve as "preferred coordinates" which respect the symmetries and which have convenient interpretations. In the case of the Schwarzschild radial coordinate on a static spherically symmetric spacetime, such a spacetime contains a family of nested static spheres labeled by different values of r, and each of these has surface area A = 4 \pi \, r^2. That is, the defining characteristic of the Schwarzschild radial coordinate is that has the same relation to surface area of these spheres as in euclidean three-space, but it does not have the usual relationship to distance measured along radii from :"the origin". Indeed, "the origin" might not exist at all in a curved but spherically symmetric space, and in fact does not in the Schwarzschild vacuum or Reissner-Nordstrom electrovacuum.

So now you know!