Astronomy

Calculating the black hole evaporation time

Calculating the black hole evaporation time


We are searching data for your request:

Forums and discussions:
Manuals and reference books:
Data from registers:
Wait the end of the search in all databases.
Upon completion, a link will appear to access the found materials.

Black holes evaporate very, very slowly by emitting Hawking radiation, and eventually they disappear, maybe even during the lifetime of the universe ($sim 1.5 imes 10^{18}{ m s}$). That's what I recall from the lectures I attended.

I now stumbled upon a Black Hole Evaporation Time Calculator which uses this formula:

$$ au_{ m evaporation} sim left( frac{M_{ m black~hole}}{M_odot} ight)^3 imes 10^{66} { m ~years}$$

As usual, $M_odot$ is the solar mass. My questions: How is that formula deduced? In which limiting case(s) does it hold?

Edit Just as a reminder: Right now, $4.3 imes 10^{17} { m s}$ passed since the big bang.


The derivation of life time formula for black holes:

It is not difficult to approximately estimate the life time of a black hole. Since Hawking radiation is indeed a black-body radiation, the rate at which the relativistic energy/mass of black hole ($M$) radiated away by its Hawking radiation can be computed by use of the Stefan-Boltzmann law as

$$- frac{{dM}}{{dt}} = frac{{dE}}{{dt}} = Asigma {T^4},$$

where $sigma = 2.105 imes {10^{ - 33}},{ m{kg}}{ m{.}}{{ m{m}}^{{ m{ - 3}}}}{{ m{K}}^{{ m{ - 4}}}}$ is the Stefan-Boltzmann constant, $A$ is the object's surface (here black hole's horizon area, $A=4 pi r_s^2$, where $r_s=2GM$ is the horizon's radius), and $T$ is the (Hawking) temperature. Note that this is the rate at which any blackbody radiates energy, no matter what is that object. Also, in this formula, $- frac{{dM}}{{dt}}$ means that when black hole radiates, its mass decreases. Having this together with the Hawking temperature of the (Schwarzschild) black hole, i.e.,

$${T_{Hawking}} = frac{{hbar}}{{8pi G{M_ odot }{k_B}}}left( {frac{{{M_ odot }}}{M}} ight) = 6.17 imes {10^{ - 8}}left( {frac{{{M_ odot }}}{M}} ight),{{ m{K}}^circ },$$

we can evaluate the life time of the black hole ($M_odot = 2×10^{30} m{kg}$ is the solar mass). For this purpose, we shall assume that the black hole initial mass ($M$) will eventually evaporate to nothing. So, integrating the Stefan-Boltzmann law, i.e. ${ au _{{ m{life}}}} = - int_M^0 {{{(Asigma {T^4})}^{ - 1}}dM}$, yields

$${ au _{{ m{life}}}} = frac{{256{pi ^3}k_B^4}}{{3Gsigma {hbar ^4}}}{(GM)^3} = (2.095 imes {10^{67}},{ m{year}}),{left( {frac{M}{{{M_ odot }}}} ight)^3}.$$

This is much greater than the age of the universe!

In addition, I should emphasize that these computations have been performed in units in which the speed of light is set to unity ($c=1$). For example, in SI units, the Stefan-Boltzmann constant is given by $σ=5.67 imes 10^{-8},mathrm {W,m^{-2},K^{-4}}$), but in this answer we had $sigma = 2.105 imes {10^{ - 33}},{ m{kg}}{ m{.}}{{ m{m}}^{{ m{ - 3}}}}{{ m{K}}^{{ m{ - 4}}}}$.

And your final question:

In which limiting case(s) does it hold?

This estimation has been derived for nonrotating black holes. For rotating (Kerr) black holes, it is expected the order of magnitude of this estimation is approximately valid, at least in the case of slowly rotating black holes. I guess, the computation in such cases would be more difficult.

Edit: In comments, it has been discussed that there are other approximations. Of course, for example, the Hawking temperature for astrophysical black holes would be extremely small, even much smaller that the CMB radiation (see more details in my answer here, please), so black hole can absorb more matter and then they grow more and more! In order to derive that formula, you should omit this challenging part. Another approximation comes from the number of particles (scalars, photons, vectors, fermions). However, the Hawking temperature (e.g., according to surface gravity definition) is same for every type of particle, and the number of species only adds a numerical factor to the Stefan-Boltzmann law, but still the order of magnitude of the final answer is approximately valid. In addition, the particles scaping from the black hole need to pass trough the potential and this change the resulting spectrum by a greybody factor ($Gamma(Omega)<1$). In this answer, which is very common in GR books, I utilized the Hawking radiation of Schwarzschild black hole with $Gamma(Omega)=1$. For more (another) details, also see the following related SE links/papers/books:

  1. Is Hawking radiation detectable?

  2. https://www.amazon.com/Introduction-Quantum-Effects-Viatcheslav-Mukhanov/dp/0521868343

  3. https://arxiv.org/abs/2011.03486

  4. https://arxiv.org/abs/1711.01865


Can a Black Hole Become a Regular Object Through Evaporation?

“Something has always questioned me about the evaporation of black holes. The quantum effects of the vacuum and the creation of pairs on the horizon of the black hole are understandable, but I don’t understand that it can last until the black hole completely disappears. As it evaporates in its horizon, the black hole loses mass, so if its density and internal gravity decrease, does it really remain a black hole? if we push the evaporation until only a few atoms remain, we will no longer be in the presence of a dense, confined black hole ”
.
The above excerpt took a comment on this article in French, about Planck’s hypothetical stars and how they could eliminate the singularities of black holes (https://www.lefigaro.fr/sciences/2014/07/22/01008-20140722ARTFIG00221 -et-si-les-trous-noirs-finissaient-par-exploser.php? pagination = 3).

This is also my question, if this black hole evaporation by Hawking radiation is something real, wouldn’t the black hole itself lose density? because it would lose mass, so it seems to make sense that if this radiation exists, black holes over time would have a lower density, which for me contradicts the principle of black holes, where the x of the issue itself is density, and not so much the mass, since, compressing any object to infinitesimally small volumes, we would have a black hole, which could even be a meteorite, car engine or a massive star, for example.

Answer:

Since the gravitational force due to an object is proportional to its mass divided by the square of the radius of the distance at which one measures the gravitational force, you can have a black hole with any mass. It is the volume of space over which this mass exists that determines its state as a black hole (or not). Further information regarding Hawking Radiation can be found in a previous answer posted to this blog regarding the question: “Why to Black Holes Lose Mass When They Emit Hawking Radiation?”.


Thread: Collapse and evaporation of black holes in proper time and coordinate time

This is my first post in ATM so I hope I'm following all the rules correctly. Remind me if necessary.

In this thread, I will first handle the collapse to a black hole in proper time and coordinate time. After I established that, I will progress to the evaporation of black holes by Hawking radiation, again in proper time and coordinate time. Mass would always be distributed as Schwarzschild objects: spherically symmetrical, non-rotating and uncharged. Schwarzschild objects might be static or collapsing under their own gravity (another case would be expanding, but I'm leaving that out right now). Particle movement will generally be considered to be radial, unless I specify something else for a specific example.

The usual perspective in which the collapse to a black hole is watched is by using the comoving coordinate system, which uses an infalling observer's proper time as its time coordinate. In this perspective, the infalling observer reaches the singularity after finite proper time. This is mathematically correct, but it is only one perspective on the process, the one of the infalling observer.

Another perspective is using Schwarzschild (or standard) coordinates. This is often used for calculating objects' movements, e.g. those of planets on relativistic Kepler orbits, or for photons of a signal bouncing off Venus when the path of the signal passes close by the Sun, so the photons' Shapiro delay due to the gravitational field of the Sun becomes larger than otherwise.

The relationship between proper time and coordinate time is affected by time dilation, both by the strength of the gravitational field and the relative movement compared to the origin of the coordinate system.
(Latex for equations doesn't seem to work at the moment)

Watched by a resting observer in flat spacetime (where proper time passes with the same speed as coordinate time), an object's movement takes longer than this object's proper time clock would indicate. (For an observer on Earth, Earth's time dilation would have to be considered to calculate the time measured by it, but this will usually be much less time dilation than the moving object's time dilation)

So considering all this, when using the Schwarzschild coordinate system, an infalling observer will not reach the singularity in any finite coordinate time - this is not a contradiction to reaching it after finite proper time, as due to the growing time dilation, the observer doesn't reach that specific finite proper time either.

For a specific exact calculation of infalling particle movement within a collapsing object, I prefer this work by Professor Shuang-Nan Zhang and his student Yuan Liu: http://www.sciencedirect.com/science. 7026930900851X

While this is for specific starting assumptions, other assumptions also lead to particles, including those of the collapsing object itself, not reaching the singularity in finite coordinate time, so at any finite coordinate time, the singularity wouldn't exist (yet). Note that the infalling particles still reach the event horizon after finite coordinate time the shell of particles still remaining outside the event horizon will converge towards zero over time.

Before continuing to evaporation, I first have to establish that there is dr/dt symmetry, or that in Schwarzschild coordinates, the "coordinate light speed" (Schwarzschild distance / Schwarzschild time) will be exactly the same inwards as outwards at any point in spacetime the actual value of dr/dt would depend on this point in spacetime. (Light speed as proper distance / proper time would still be c, of course)

For this, there are four sub-cases: a) in the vacuum outside of a static mass distribution, b) inside a static mass distribution, c) outside a collapsing mass distribution, d) inside a collapsing mass distribution.

Static mass distributions are their own temporal mirror images as General Relativity in time reversal invariant, and as the temporal mirror image of a photon moving outwards at a specific spacetime point at a specific dr/dt is a photon moving inwards at the same point with the same dr/dt, dr/dt-symmetry is given for the cases a and b.

Then, Birkhoff's theorem shows that the gravitational field outside of a collapsing Schwarzschild object is the same as that of a static Schwarzschild object of the same mass, so since dr/dt-symmetry is given in case a, it's also given in case c.

For case d, inside a collapsing mass distribution, I cannot directly conclude there is dr/dt-symmetry, only by analogy: with a rotating Kerr black hole, there clearly is an asymmetry regarding light movement with or against the direction of rotation. But this asymmetry is still larger than zero in the vacuum outside the Kerr black hole, so if the analogy between these symmetries holds, then there could only be symmetry outside the object if there already was symmetry within it, so since dr/dt-symmetry is given in case c, it would also be given in case d - and with this, in the entire Schwarzschild spacetime. It would be best if case d could be directly derived from General Relativity, but unfortunately that's beyond my mathematical abilities.

(It's late, or rather, early. More in a later posting)

Thanks for everyone participating,
Frank

correct me if I'm wrong, but I think you're referring to a black hole with the event horizon at its Schwarzschild radius. And according to the comoving coordinate system, something like that would form after finite proper time. However, if you're starting with the Schwarzschild coordinate system and follow particles during the star's collapse, the Schwarzschild coordinate system dutifully shows their movement during all finite Schwarzschild coordinate time. (Including the time after they pass inside the event horizon, but I'll get to that in part 2, which I had started, then had to stop, then continued later today, and will post in probably an hour or so)

So the point where the coordinate singularity would invalidate the usage of the Schwarzschild coordinate system doesn't happen within finite coordinate time, and so the rules of normal General Relativity still apply, where the Schwarzschild coordinate system is a normal acceptable perspective.

Hi FrankWSchmidt and welcome to the forum.
As Shaula pointed out Schwarzschild co-ordinates are not valid for analysing matter falling into a black hole.

There is a more basic reason why Schwarzschild coordinates cannot be used. You are looking at matter collapsing to create a black hole. The Schwarzschild solution is for the gravitational field outside of a massive body. There is a restriction on the coordinate r to be greater than the radius of the body. So the coordinates do not exist inside a body collapsing to make a black hole.

A problem is that a collapsing ball of matter is not the time inverse of a black hole evaporating via Hawking radiation. The OP may be confusing evaporation with white holes which are the time inverse of eternal black holes in GR. Black holes that form through gravitational collapse have no time inverse. In the OP language there is no "dr/dt symmetry" because particles of matter enter the hole and never come out.

(Continuing with Schwarzschild coordinate perspective. I'll eventually return to the observer's proper time perspective as it still continues to exist, but not right now)

dr/dt-symmetry is essentially the rejection of the intuitive assumption that "gravity sucks". Instead, it curves spacetime, and while to a Schwarzschild coordinate perspective all non-radial paths get curved more and more towards the mass with increasing curvature, there is still a way for a photon to go outwards.

This doesn't mean there is no event horizon in the sense of a border to a region of space from which signals and information cannot escape to infinity. It just looks different when seen in Schwarzschild coordinate perspective. As the collapse of a dying star continues, the infalling matter of the star's interior is subject to continuously decreasing dr/dt values, so that eventually radially outgoing signals from a particle will continue to remain outgoing, but will slow down and never reach the star's Schwarzschild radius - and will so fail to escape to infinity. Outside the Schwarzschild radius, dr/dt will not converge towards zero but to a non-zero amount, so if a signal can reach that point it would be able to escape. With dr/dt-symmetry, the same effect applies inwards in the same way, and as matter is always slower than light, it also will have a limited range of inwards movement within finite coordinate time. This would be the reason for the effect mathematically described by Liu and Zhang, that infalling particles don't reach the singularity within finite coordinate time.

Continuing to black hole evaporation: It's meanwhile widely accepted that black holes will eventually evaporate within extremely long, but still finite timescales. This has been generally combined with a collapse description in the comoving coordinate system, as that is pretty much the only coordinate system used when dealing with black holes. However, the comoving coordinate system is excellent when answering questions about what the observer experiences and measures, but it is still its inside perspective. I think it is this decision to combine the inside perspective of collapse with the external observation of escaping Hawking radiation that causes paradoxes later on.

The perspective of the Schwarzschild coordinate system meanwhile allows to calculate the movement of particles, including photons. A photon of Hawking radiation would have to be emitted at some point of coordinate time and show up in the Schwarzschild coordinate system, and the black hole would lose mass at that point. I think the most sensible hypothesis is that since Hawking radiation escapes on the outside, this mass loss would also occur at the surface.

At this point in coordinate time the collapse is still ongoing, and this collapse will continue just as before on the inside where mass isn't lost yet (so both infalling matter and outgoing radiation and information would move infinitesimally slow). However, at the surface things change, as mass loss means the Schwarzschild radius decreases as well. Extremely slowly of course, but still faster than the particle movement on the inside. So the inwards-moving Schwarzschild-radius would reach infalling matter and outgoing radiation (outgoing radiation is possible due to dr/dt-symmetry, but couldn't escape to infinity before). At that point, the "coordinate speed" of outgoing radiation would no longer remain infinitesimally slow, and it could escape. This would resolve the information loss paradox, because there would be a way in which information could escape. However, it would be very unlikely that much energy would escape due to this, because of the extreme redshift and because a very small deviation from a radially outgoing path would put a photon on a path that would lead back into the black hole due to the extreme curvature.

dr/dt-symmetry combined with the possibility to escape would also remove the firewall paradox as the quantum fields inside and outside the black hole wouldn't be completely ripped apart by the collapse, just extremely stretched to the point that information could only move for a limited Schwarzschild distance within finite Schwarzschild time. During the mass loss in the evaporation, this stretching eventually decreases allowing all quantum fields again being reachable by each other when the black hole has completely evaporated.

Under normal conditions, such an escape would be impossible because the event horizon would be inescapable. But the math that guarantees this inescapability relies on the black hole not losing math, which is usually a given. Once a black hole loses mass, the border stops being an inescapable event horizon and shifts to the above behavior. If in some way the black hole would stop losing mass, then it would become an inescapable event horizon again.

Now, the entire situation has a proper time perspective as well. But as the evaporation changed the effects of the collapse due to their interaction, the proper time perspective also changes from what it would have been in a pure collapse scenario where the black hole wouldn't lose mass.

For the often-used infalling astronaut these processes would be most likely deadly, if he hadn't already been killed by tidal forces or other normal collapse phenomena. It would start falling in as it usually does, pass inside the event horizon after finite coordinate time, but soon after that (in proper time) it would reach the point where it would only move infinitesimally slowly in Schwarzschild time, so the following proper time would pass in an instant, after which it would have reached the point where the astronaut would be reached by the inwards-moving Schwarzschild radius of the evaporating black hole. At this point, the particles further out would start moving faster inwards due to them being just outside the Schwarzschild radius, while the ones further in would move as slowly as before. So if the astronaut had somehow escaped being spaghettified by tidal forces, it would now be compressed. And as part of the black hole surface, he would probably also lose energy (which the black hole loses due to the Hawking radiation). But if he somehow survives he would notice the black hole would be losing mass compared to what he would expect to measure in a "normal" collapse without evaporation.

The question remains to how the black hole evaporation would eventually end. My personal guess is that the extreme conditions the matter experiences in this altered scenario would eventually reach the point where proton-decay-like reactions might happen, allowing matter-antimatter-annihilation to energy happening afterwards, so there would be no remnant left after the evaporation finishes, just photons and neutrinos.

I hope that now I covered all the points I wanted to cover.


  • R < 3 R S , there are no stable orbits - all matter gets sucked in.
  • At R = 1.5 R S , photons would orbit in a circle!

Jill is orbiting the black hole in a starship at a safe distance away in a stable circular orbit. She watches Jack fall in by monitoring the incoming flashes from his laser beacon.

  • He sees the ship getting further away.
  • He flashes his blue laser at Jill once a second by his watch.
  • Each laser flash take longer to arrive than the last
  • Each laser flash become redder and fainter than the one before it.
  • His blue laser flash every second by his watch
  • The outside world looks oddly distorted (positions of stars have changed since he started).
  • Jack's laser flashing about once every hour.
  • The laser flashes are now shifted to radio wavelengths, and
  • the flashes are getting fainter with each flash.
  • One last flash from Jack's laser after a long delay (months?)
  • The last flash is very faint and at very long radio wavelengths.
  • She never sees another flash from Jack.
  • The universe appear to vanish as he crosses the event horizon
  • He gets shredded by strong tides near the singularity and crushed to infinite density.
  • Time appears to stand still at the event horizon as seen by a distant observer.
  • Time flows as it always does as seen by an infalling astronaut.
  • Light emerging from near the black hole is Gravitationally Redshifted to longer (red) wavelengths.

Take a Virtual Trip to a Black Hole or Neutron Star. Pictures & movies by relativist Robert Nemiroff at the Michigan Technical University.


Hawking Radiation - Black Hole Evaporation

When particles escape, the black hole loses a small amount of its energy and therefore of its mass (mass and energy are related by Einstein's equation E = mc²).

The power emitted by a black hole in the form of Hawking radiation can easily be estimated for the simplest case of a nonrotating, non-charged Schwarzschild black hole of mass . Combining the formulas for the Schwarzschild radius of the black hole, the Stefan–Boltzmann law of black-body radiation, the above formula for the temperature of the radiation, and the formula for the surface area of a sphere (the black hole's event horizon), equation derivation:

Black hole surface gravity at the horizon:

Hawking radiation has a black-body (Planck) spectrum with a temperature T given by:

Hawking radiation temperature:

Schwarzschild sphere surface area of Schwarzschild radius :

Stefan–Boltzmann power law:

A black hole is a perfect black-body:

Stefan–Boltzmann–Schwarzschild–Hawking black hole radiation power law derivation:

Stefan–Boltzmann-Schwarzschild-Hawking power law:

Where is the energy outflow, is the reduced Planck constant, is the speed of light, and is the gravitational constant. It is worth mentioning that the above formula has not yet been derived in the framework of semiclassical gravity.

The power in the Hawking radiation from a solar mass black hole turns out to be a minuscule 9 × 10−29 watts. It is indeed an extremely good approximation to call such an object 'black'.

Under the assumption of an otherwise empty universe, so that no matter or cosmic microwave background radiation falls into the black hole, it is possible to calculate how long it would take for the black hole to dissipate:

Given that the power of the Hawking radiation is the rate of evaporation energy loss of the black hole:

Since the total energy E of the black hole is related to its mass M by Einstein's mass-energy formula:

We can then equate this to our above expression for the power:

This differential equation is separable, and we can write:

The black hole's mass is now a function M(t) of time t. Integrating over M from (the initial mass of the black hole) to zero (complete evaporation), and over t from zero to :

The evaporation time of a black hole is proportional to the cube of its mass:

The time that the black hole takes to dissipate is:

Where is the mass of the black hole.

The lower classical quantum limit for mass for this equation is equivalent to the Planck mass, .

Planck mass quantum black hole Hawking radiation evaporation time:

For a black hole of one solar mass ( = 1.98892 × 1030 kg), we get an evaporation time of 2.098 × 1067 years—much longer than the current age of the universe at 13.73 ± 0.12 x 109 years.

But for a black hole of 1011 kg, the evaporation time is 2.667 billion years. This is why some astronomers are searching for signs of exploding primordial black holes.

However, since the universe contains the cosmic microwave background radiation, in order for the black hole to dissipate, it must have a temperature greater than that of the present-day black-body radiation of the universe of 2.7 K = 2.3 × 10−4 eV. This implies that must be less than 0.8% of the mass of the Earth.

Cosmic microwave background radiation universe temperature:

Hawking total black hole mass:

Where, is the total Earth mass.

So, for instance, a 1-second-lived black hole has a mass of 2.28 × 105 kg, equivalent to an energy of 2.05 × 1022 J that could be released by 5 × 106 megatons of TNT. The initial power is 6.84 × 1021 W.

Black hole evaporation has several significant consequences:

  • Black hole evaporation produces a more consistent view of black hole thermodynamics, by showing how black holes interact thermally with the rest of the universe.
  • Unlike most objects, a black hole's temperature increases as it radiates away mass. The rate of temperature increase is exponential, with the most likely endpoint being the dissolution of the black hole in a violent burst of gamma rays. A complete description of this dissolution requires a model of quantum gravity, however, as it occurs when the black hole approaches Planck mass and Planck radius.
  • The simplest models of black hole evaporation lead to the black hole information paradox. The information content of a black hole appears to be lost when it dissipates, as under these models the Hawking radiation is random (it has no relation to the original information). A number of solutions to this problem have been proposed, including suggestions that Hawking radiation is perturbed to contain the missing information, that the Hawking evaporation leaves some form of remnant particle containing the missing information, and that information is allowed to be lost under these conditions.

Read more about this topic: Hawking Radiation

Famous quotes containing the words black and/or hole :

&ldquo In it he proves that all things are true and states how the truths of all contradictions may be reconciled physically, such as for example that white is black and black is white that one can be and not be at the same time that there can be hills without valleys that nothingness is something and that everything, which is, is not. But take note that he proves all these unheard-of paradoxes without any fallacious or sophistical reasoning. &rdquo
&mdashSavinien Cyrano De Bergerac (1619�)

&ldquo But the surface of the Earth was meant for man. He wasn’t meant to live in a hole in the ground. &rdquo
&mdashEdward L. Bernds (b. 1911)


Emission process

A black hole emits thermal radiation at a temperature ,

in natural units with G, c, and k equal to 1, and where &kappa is the surface gravity of the horizon.

In particular, the radiation from a Schwarzschild black hole is black-body radiation with temperature:

where is the reduced Planck constant, c is the speed of light, k is the Boltzmann constant, G is the gravitational constant, and M is the mass of the black hole.


Black hole evaporation

1. I'm aware that black hole evaporation is due to virtual particle/antiparticle pairs popping into existence (if that's even the term for it) and having one of the pair pass beyond the event horizon. Somehow this particle "gains negative energy" as it falls into the black hole and as a result contributes a net negative energy to the black hole, decreasing the hole's energy, and therefore it's mass and size. Or so that's what I've been led to believe.

I don't understand how the particle gains negative energy. Also, why don't other particles falling into black holes contribute a net negative energy in the same fashion? I suspect the answers are related.

2. Smaller black holes evaporate faster. This seems counter-intuitive to me, because the evaporation relies on virtual particle pairs coming into existence very close to the event horizon. If the black hole is larger its event horizon has a larger area and therefore more virtual particle pairs are created in a proper way to aid the black holes evaporation. For that reason it seems larger black holes should evaporate faster - where's the fault in my logic?

#2 gazerjim

Rather than wax ignorant, I will refer the reader to

#3 jupiterzkool

On Question #1, the convention for gravitational potential energy is to assign a particle an infinite distance away to be zero energy. As a particle falls toward the gravitational source, the energy increases in the negative sense (i.e. decreases). Another way of thinking about it is that it takes energy to pull the particle away from the gravitational source.

#4 Dane B

Yes jupterzkool, but that is by convention. In other words, it's arbitrary and only represents a way of thinking out it and not necessarily the way it is.

If something is to decrease the total energy/mass of a black hole, its negative energy must be greater than the positive energy of its mass. This negative energy can't exist solely due to the arbitrary assignment made by human scientists.

Maybe this is a better way of thinking about it - according to Newton, when a ball falls toward the Earth it loses potential energy. But the Earth also falls toward the ball by an infinitesimal amount and thus loses an infinitesimal amount of potential energy. But that still doesn't seem to add up, because evaporation requires the absorption of half a virtual particle pair, whereas the above loss of potential energy happens with any object pulled into a black hole.

#5 jupiterzkool

It does not matter where one puts the zero energy level. The point is that one state have more or less energy than another. The current convention is one of mathematical convenience.

I think the entire point of black hole evaporation is that a portion of the black hole's energy has been lost, i.e. one particle less (because it has escaped).

#6 Dane B

It does not matter where one puts the zero energy level. The point is that one state have more or less energy than another.

But less energy isn't the same as negative energy.

You can (hypothetically) continue to reduce the energy of a mass until it is at absolute zero. You could even "place" this mass into an imaginary universe with no other objects and no force fields. This object now has zero thermal, kinetic, and potential energy. But it has no negative energy, and in fact it's mass represents positive energy. I don't see how you could do anything to this mass to give it negative energy, much less enough negative energy to negate the positive energy of its mass.

I think the entire point of black hole evaporation is that a portion of the black hole's energy has been lost, i.e. one particle less (because it has escaped).

But the "escaped" particle was never within the black hole's event horizon to begin with, so I don't understand how it carries away energy from the black hole when it has always been completely cut off from the other size of the event horizon.

I must be missing something fundamental, because it sounds like magic to me.

#7 gazerjim

You can (hypothetically) continue to reduce the energy of a mass until it is at absolute zero

This is all strange magic to me.

It's my understanding that a state of absolute zero energy cannot exist even in the most perfect of vacuums. Laws of physics allow energy and mass to pop into and out of existence. Perhaps (I'm waxing again ) this is another way of saying there is no zero point from which to start.

#8 jupiterzkool

It does not matter where one puts the zero energy level. The point is that one state have more or less energy than another.

But less energy isn't the same as negative energy.

You can (hypothetically) continue to reduce the energy of a mass until it is at absolute zero. You could even "place" this mass into an imaginary universe with no other objects and no force fields. This object now has zero thermal, kinetic, and potential energy. But it has no negative energy, and in fact it's mass represents positive energy. I don't see how you could do anything to this mass to give it negative energy, much less enough negative energy to negate the positive energy of its mass.

I think the entire point of black hole evaporation is that a portion of the black hole's energy has been lost, i.e. one particle less (because it has escaped).

But the "escaped" particle was never within the black hole's event horizon to begin with, so I don't understand how it carries away energy from the black hole when it has always been completely cut off from the other size of the event horizon.

I must be missing something fundamental, because it sounds like magic to me.


You're seem to be missing the point about "negative" energy. The energy of the particle falling in is mathematically negative because the convention is to place zero at infinity. If you choose to place zero at the center of the black hole, then all of the potential energy is positive.

The virtual particles are still within the gravitational pull of the black hole. Eventually, one particle will cross he the event horizon while the other flies off in the opposite direction with enough kinetic energy to overcome the influence of the black hole.

#9 Dane B

Sorry Scott, I appreciate your effort to help explain this me. I hope I'm not trying your patience, but it just doesn't add up for me.

"mathematically negative" and "by convention" has nothing to do with the physical reality, except that it is a human contrived method of representing physical reality. No matter where we choose our zero point to be, regardless of the conventions and the resulting mathematical signs, an object falling into a black hole will carry the same energy as it would if we had chosen any other zero point. I'm not arguing the usefulness or validity of the convention - but the convention doesn't explain how the measurable physcal quantities of the object change in a way that gives it, in effect, a net negative mass.

gazerjim - you're right, I overlooked the fact that the uncertainty principle forbids true absolute zero. But I believe the thought experiment still works - the mass carries as little energy as possible but it's mass is still positive and thus if it were hurled into a black hole should contribute a net positive mass.

#10 Dane B

Just for clarification - I understand that as an object becomes farther away from a gravitational source it gains positive energy, and conversely if it moves closer to a gravitational source it loses positive energy or "gains negative energy". I just don't see how an object can lose so much positive energy simply due to motion within a gravitational field that its net mass/energy is negative.

By the law of conservation of energy, is something loses energy something else must gain energy. So where is this energy going and how does it get there? One answer is the other particle in the virtual particle pair is the one that gains energy and thus the net energy within the event horizon decreases while the net energy outside increases. But how does the energy balance inside the black hole affect the energy balance outside the black hole? I didn't think any information (such as information about a particle's change in energy) could be retrieved once on the other side of the event horizon.

#11 jupiterzkool

Mathematically, we can assign any value to a particular energy state (quantum mechanics places some additional constraints). It is only the differences in two energy states that count. It is the difference that is the physically meaningful quantity, not the absolute value (whether positive or negative).

Modern physical theories are just now reaching the point where we can speculate about the physics inside a black hole and even before the big bang. However, we are a long way from validating any of the predictions.

#12 Qkslvr

Dane, This is how I think of it, Not 100% sure it's correct, but I think it is.

At the quantum soup level of space where virtual particles boil into existence, Some are the equivalent Positron and Electron, which when they collide they become energy again.

If some of those fall into the event horizon, and some are kicked out into space, Any Negative matter (like positrons) would evaporate real matter trapped in the blackhole converting mass into energy.

This BTW is quite similar to how solar cells collect electricity from light in a silicon diode.

#13 llanitedave

The concept seems simple, if subtle -- and unless I'm wrong. The virtual pairs are created from the vacuum energy of space surrounding the black hole. Under normal circumstances, the pair would either recombine, releasing that same amount of energy back into space, or they would separate from each other permanently. The odds of either event happening are fixed and constant. If the space were "empty", then the number of escaped virtual particles of each sign would be equal, and the overall energy content of space would be preserved. Those particles would be available for collision and annihilation with any other complementary particle that had escaped from its own pair-creation. So the particles that are being created and the particles that are annihilating each other, are not necessarily the same, as I understand it. However, the rate is constant and balanced, so that really makes no difference under normal conditions. Whenever two complimentary particles collide, the energy released is still equal to the energy that went into their creation.

The event horizon of a black hole eliminates that balance. Virtual pairs are created, but now, if one escapes, it can be removed from space. It's no longer available to give back the energy required for its creation. The net charge of space remains the same, because the escaped particle can be either positive or negative, and eventually it should find a partner to annihilate with. But half the energy of the particle-pair creation has been lost. Energy has to be conserved, however, so the only way to balance the books is to take that energy from the black hole that absorbed the particle. And the only source of energy the black hole has is mass. (I'm not sure how rotational energy fits into this. Does the black hole give up rotational energy before giving up mass, or are they both dissipated together?) By "paying back" the energy contained in the sucked-in particle, the black hole loses mass back into space: it decreases the local warpage of space-time.

The reason the evaporation accelerates as it progresses had me stumped for a long time, too. Now, I'm guessing that it's a matter of simple geometry: The mass of a black hole is analogous to its volume, and the surface area of the event horizon decreases more slowly than does the volume. Since the rate of particle-pair creation (and loss) over a given area of event horizon should be constant, the proportion of energy lost per volume is greater for a small black hole than a large one. (Part of me thinks that's probably an inadequate explanation, but it's a bit closer than the understanding I had before).

Hope this helps, and doesn't create even more confusion.

#14 Dane B

It is the difference that is the physically meaningful quantity, not the absolute value (whether positive or negative).

So you're saying "delta"E="delta"m times c^2

Any negative change in energy results in a negative change in mass. If the energy changes by a great enough amount, "delta"m will exceed m and effective mass will be negative. That sounds reasonable, but doesn't explain why other things that fall into a black hole don't contribute a net negative mass. Why is it only half of a virtual particle pair that has this ability?

The proper virtual particles are those created very close to the event horizon. But a particle that has been travelling across the entire universe toward the black hole should have an even greater "delta"E and therefore more of a negative mass effect.

Any Negative matter (like positrons) would evaporate real matter trapped in the blackhole converting mass into energy.

Anti-matter has positive mass. And if it annihilates other matter inside the black hole turning it into energy, that energy is still trapped in the black hole and the overall energy/mass of the blackhole remains unchanged.

The net charge of space remains the same, because the escaped particle can be either positive or negative, and eventually it should find a partner to annihilate with.

That doesn't add up. If the universe contains +2 and -2 charge the overall is neutral. But if -1 charge gets sucked into a black hole and then its anti-particle +1 charge annihilates with the remaining -1 charge, the universe has gone from nuetral to +1 charge.

But half the energy of the particle-pair creation has been lost. Energy has to be conserved, however, so the only way to balance the books is to take that energy from the black hole that absorbed the particle. And the only source of energy the black hole has is mass.

Yes, but how does the energy of a black holes mass manifest itself outside the event horizon? That's the part that sounds like magic. Hawking radiation wouldn't be taken seriously unless a reasonable mechanism for this change in energy balance had been proposed.

I feel like a jerk asking you guys for help and then arguing with you about it, but the more I explain my reasoning the more likely the fault in it will be exposed.

EDIT: Didn't realize "another word for donkey" was going to get bleeped.

#15 llanitedave

The net charge of space remains the same, because the escaped particle can be either positive or negative, and eventually it should find a partner to annihilate with.

The thing to remember is that particle/anti-particle pairs are being created all the time, and at the event horizon, a +1 is just as likely to escape as is a -1. Therefore, there should be equal numbers of "orphaned" particles of each sign traveling about, which keeps the overall charge neutral.

But half the energy of the particle-pair creation has been lost. Energy has to be conserved, however, so the only way to balance the books is to take that energy from the black hole that absorbed the particle. And the only source of energy the black hole has is mass.

I'm guessing here, but I would say the energy is manifested in the curvature of local spacetime. As the energy decreases, so does the amount of space-time curvature. So, as the black hole loses energy, the local spacetime "relaxes".

I feel like a jerk asking you guys for help and then arguing with you about it, but the more I explain my reasoning the more likely the fault in it will be exposed.

Don't apologize. It's a stimulating topic, and there's no better way to get to the bottom of it than to critically and skeptically ask questions. How else will you design your experiment?

I've learned far more from arguing while being wrong than I ever learned from being right!

#16 llanitedave

The proper virtual particles are those created very close to the event horizon. But a particle that has been travelling across the entire universe toward the black hole should have an even greater "delta"E and therefore more of a negative mass effect.

OK, try forgetting about "negative energy" for a moment. The terminology is just as confusing for me.

Look at it this way: An object falling into a black hole adds its own mass to that of the black hole.

Energy must be expended to create matter. The creation of a particle/antiparticle pair at the event horizon requires a certain amount of virtual energy. If the two particles re-annihilate immediately, the energy and particles are both considered "virtual". However, if for any reason the two particles escape, then, "real" energy has been lost and "real" matter gained. The energy has to come from somewhere. At the event horizon, it comes from the black hole. If one particle escapes from the event horizon and the other doesn't, the black hole expended the energy to make two particles, but only got one of them back. The escaping particle is the equivalent of mass generated by the black hole -- it's taking that mass away. The particle that falls back in is no longer available to interact, so it can't be said to be anything gained.

Scenario 1
1. Black hole energy creates 2 particles.
2. Particles immediately recombine, give back energy of creation.

Scenario 2
1. Black hole energy creates 2 particles.
2. Particle A falls back into event horizon, returning 1/2 of creation energy as mass.
3. Particle B flies away from event horizon, "stealing" 1/2 of creation energy as mass.

Net effect on black hole, has lost the energy of 1/2 of a particle/antiparticle creation event.

Scenario 3.
I really don't have a clue what I'm talking about. I'm a geologist, not an astrophysicist!

#17 Pess

Scenario 1
1. Black hole energy creates 2 particles.
2. Particles immediately recombine, give back energy of creation.

Hunh? Unless I missed something somewhere, virtual particle pairs have nothing to do with Black holes or EH's.

Indeed, the fabric of the Universe is permeated with virtual pairs popping in and out of existence.

It seems everyone is bending over backwards to 'explain' how this 'energy', negative or otherwise, is getting out of the Black Hole.

I further think there is some confusion as to the difference between 'energy' and 'mass'.

Black holes can and do give off copious amounts of energy in a variety of forms. For example, they probably radiate gravitational waves. There is also recent research that suggests that Black Holes may violate the 'Black Holes have no Hair' theorem and may indeed radiate like any 'ol standard black body radiation.

In any event the Pinocchio particle becomes a real boy at the expense of energy drained from the black hole. This energy may be a straightforward siphon of gravitational energy or it could be something as esoteric as 'leakage' through a folded up dimension (where most of the gravitational force is thought to hide anyways.)

#18 Dane B

The thing to remember is that particle/anti-particle pairs are being created all the time, and at the event horizon, a +1 is just as likely to escape as is a -1. Therefore, there should be equal numbers of "orphaned" particles of each sign traveling about, which keeps the overall charge neutral.

That doesn't sound kosher. If I have a chemical reaction on one side a beaker that violates the law of convservation of charge by creating a net +1 charge, and another violating reaction on the other side of the beaker that creates a -1 charge, then the overall charge in the universe remains the same. but the law of conservation of charge was broken twice. Two wrongs don't make a right.

Unless one reaction is linked to the other in a way that forbids us to "decouple" them and look at a single reaction as an independent occurence, then the conservation of charge is broken within the scope of that independent occurence. The only way I see to reconcile this is if the reactions are linked in a way that makes it inappropriate to limit the scope to a single reaction. As far as I know, the creation of charge via virtual particles near black holes are not coordinated in this manner.

EDIT: This can't be the right way to think about it, that if half of a charged virtual particle pair gets pulled into a black hole the net charge of the universe has changed and somehow needs to be balanced. If a charge falling into a black hole can be considered removed from the universe, than NO charged particle, virtual or not, could fall into a black hole without violating conservation of charge. Even if it is cut off from the rest of the universe, that charge must still be considered to exist within the universe. Not even black holes can violate the law of conservation of charge.

Pess - you dismissed a lot while explaining little. If we're really going down the wrong line of thinking here then please give a more detailed reply.

I'm talking about black hole evaporation via Hawking radiation which relies on virtual particles - I'm surprised you didn't recognize the connection. I suppose I should have been more explicit about the evaporation being via Hawking radiation, but I didn't realize there were other proposed mechanisms that I needed to distinguish from.

Energy/mass within a black hole decreases while simultaneously the energy/mass outside of the black hole increases. I'm wondering how these two processes coordinate in a way that prevents violating conservation of energy when they are apparently cut off by the event horizon.

I'm sure there are many other theories and advances in black hole theory we're not taking into account - but because we're talking about Hawking radiation that's beside the point. Hawking radiation was accepted as a plausible mechanism without the aid of these new revelations, and attempting to tie them into an explanation of Hawking radiation before I even understand the fundamentals is only going to lead to more confusion.

I'm curious what you meant about confusion with the difference between energy and mass. It seems to me that in this situation they should be treated as the same thing because this process involves the disappearance of mass being balanced by the appearance of energy. Therefore the difference between mass and energy isn't important, because we aren't considering them as seperate concepts in the context of conservation of energy in the overall process.


The Curve Becomes the Key

In 1992, Don Page and his family spent their Christmas vacation house-sitting in Pasadena, enjoying the swimming pool and watching the Rose Parade. Page, a physicist at the University of Alberta in Canada, also used the break to think about how paradoxical black holes really are. His first studies of black holes, when he was a graduate student in the ’70s, were key to his adviser Stephen Hawking’s realization that black holes emit radiation — the result of random quantum processes at the edge of the hole. Put simply, a black hole rots from the outside in.

The particles it sheds appear to carry no information about the interior contents. If a 100-kilogram astronaut falls in, the hole grows in mass by 100 kilograms. Yet when the hole emits the equivalent of 100 kilograms in radiation, that radiation is completely unstructured. Nothing about the radiation reveals whether it came from an astronaut or a lump of lead.

That’s a problem because, at some point, the black hole emits its last ounce and ceases to be. All that’s left is a big amorphous cloud of particles zipping here and there at random. It would be impossible to recover whatever fell in. That makes black hole formation and evaporation an irreversible process, which appears to defy the laws of quantum mechanics.

Hawking and most other theorists at the time accepted that conclusion — if irreversibility flouted the laws of physics as they were then understood, so much the worse for those laws. But Page was perturbed, because irreversibility would violate the fundamental symmetry of time. In 1980 he broke with his former adviser and argued that black holes must release or at least preserve information. That caused a schism among physicists. “Most general relativists I talked to agreed with Hawking,” said Page. “But particle physicists tended to agree with me.”

On his Pasadena vacation, Page realized that both groups had missed an important point. The puzzle wasn’t just what happens at the end of the black hole’s life, but also what leads up to it.

He considered an aspect of the process that had been relatively neglected: quantum entanglement. The emitted radiation maintains a quantum mechanical link to its place of origin. If you measure either the radiation or the black hole on its own, it looks random, but if you consider them jointly, they exhibit a pattern. It’s like encrypting your data with a password. The data without the password is gibberish. The password, if you have chosen a good one, is meaningless too. But together they unlock the information. Maybe, thought Page, information can come out of the black hole in a similarly encrypted form.

Page calculated what that would mean for the total amount of entanglement between the black hole and the radiation, a quantity known as the entanglement entropy. At the start of the whole process, the entanglement entropy is zero, since the black hole has not yet emitted any radiation to be entangled with. At the end of the process, if information is preserved, the entanglement entropy should be zero again, since there is no longer a black hole. “I got curious how the radiation entropy would change in between,” Page said.

Initially, as radiation trickles out, the entanglement entropy grows. Page reasoned that this trend has to reverse. The entropy has to stop rising and start dropping if it is to hit zero by the endpoint. Over time, the entanglement entropy should follow a curve shaped like an inverted V.

Page calculated that this reversal would have to occur roughly halfway through the process, at a moment now known as the Page time. This is much earlier than physicists assumed. The black hole is still enormous at that point — certainly nowhere near the subatomic size at which any putative exotic effects would show up. The known laws of physics should still apply. And there is nothing in those laws to bend the curve down.

With that, the problem got much more acute. Physicists had always figured that a quantum theory of gravity came into play only in situations so extreme that they sound silly, such as a star collapsing to the radius of a proton. Now Page was telling them that quantum gravity mattered under conditions that, in some cases, are comparable to those in your kitchen.

Page’s analysis justified calling the black hole information problem a paradox as opposed to merely a puzzle. It exposed a conflict within the semiclassical approximation. “The Page-time paradox seems to point to a breakdown of low-energy physics in a place where it has no business breaking down, because the energies are still low,” said David Wallace, a philosopher of physics at the University of Pittsburgh.

On the bright side, Page’s clarification of the problem paved the way to a solution. He established that, if entanglement entropy follows the Page curve, then information gets out of the black hole. In doing so, he transformed a debate into a calculation. “Physicists are not always so good at words,” said Andrew Strominger of Harvard University. “We do best with sharp equations.”

Now physicists just had to calculate the entanglement entropy. If they could pull it off, they’d get a straight answer. Does the entanglement entropy follow an inverted V or not? If it does, the black hole preserves information, which means particle physicists were right. If it doesn’t, the black hole destroys or bottles up information, and general relativists can help themselves to the first doughnut at faculty meetings.

Yet even though Page spelled out what physicists had to do, it took theorists nearly three decades to figure out how.


How Do Black Holes Evaporate?

The actor Stephen Hawking is best known for his cameo appearances in Futurama and Star Trek, you might surprised to learn that he’s also a theoretical astrophysicist. Is there anything that guy can’t do?

One of the most fascinating theories he came up with is that black holes, the Universe’s swiffer, can actually evaporate over vast periods of time.

Quantum theory suggests there are virtual particles popping in and out of existence all the time. When this happens, a particle and its antiparticle appear, and then they recombine and disappear again.

When this takes place near an event horizon, strange things can happen. Instead of the two particles existing for a moment and then annihilating each other, one particle can fall into the black hole, and the other particle can fly off into space. Over vast periods of time, the theory says that this trickle of escaping particles causes the black hole to evaporate.

Wait, if these virtual particles are falling into the black hole, shouldn’t that make it grow more massive? How does that cause it to evaporate? If I add pebbles to a rock pile, doesn’t my rock pile just get bigger?

It comes down to perspective. From an outside observer watching the black hole’s event horizon, it appears as if there’s a glow of radiation coming from the black hole. If that was all that was happening, it would violate the law of thermodynamics, as energy can neither be created nor destroyed. Since the black hole is now emitting energy, it needs to have given up a little bit of its mass to provide it.

Let’s try another way to think about this. A black hole has a temperature. The more massive it is, the lower its temperature, although it’s still not zero.

From now and until far off into the future, the temperature of the largest black holes will be colder than the background temperature of the Universe itself. Light from the cosmic microwave background radiation will fall in, increasing its mass.

Viewed in visible light, Markarian 739 resembles a smiling face. Inside are two supermassive black holes, separated by about 11,000 light-years. The galaxy is 425 million light-years away from Earth. Credit: Sloan Digital Sky Survey

Now, fast forward to when the background temperature of the Universe drops below even the coolest black holes. Then they’ll slowly radiate heat away, which must come from the black hole converting its mass into energy.

The rate that this happens depends on the mass. For stellar mass black holes, it might take 10^67 years to evaporate completely.

For the big daddy supermassive ones at the cores of galaxies, you’re looking at 10^100. That’s a one, followed by 100 zero years. That’s huge number, but just like any gigantic and finite number, it’s still less than infinity. So over an incomprehensible amount of time, even the longest living objects in the Universe – our mighty black holes – will fade away into energy.

One last thing, the Large Hadron Collider might be capable of generating microscopic black holes, which would last for a fraction of a second and disappear in a burst of Hawking radiation. If they find them, then Hawking might want to the acting on hold and focus on physics.

The LHC. Image Credit: CERN

Nothing is eternal, not even black holes. Over the longest time frames we’re pretty sure they’ll evaporate away into nothing. The only way to find out is to sit back and watch, well maybe it’s not the only way.

Does the idea of these celestial nightmares evaporating fill you with existential sadness? Feel free to share your thoughts with others in the comments below.

Thanks for watching! Never miss an episode by clicking subscribe.

Our Patreon community is the reason these shows happen. We’d like to thank Dana Nourie and the rest of the members who support us in making great space and astronomy content. Members get advance access to episodes, extras, contests, and other shenanigans with Jay, myself and the rest of the team. Want to get in on the action? Click here.


Hawking radiation and BH evaporation time

I don't know a lot about this topic so corrections are solicited.

My understanding is that black holes evaporate by Hawking radiation. Hawking radiation, by my reading, occurs when a pair of virtual particles emerges very close to but just outside the event horizon of the black hole. When they emerge (from quantum foam, not from the BH) one of the pair may occur closer to the event horizon, the other further from it. In some cases, this difference is just right to cause one of the virtual particles to enter the event horizon, while the other escapes into our universe.

Normally virtual particles immediately anihiliate each other, but in this case, as they are seperated, they do not immediately anihiliate. In fact, the particle which is free to enter our universe is really no different than any other particle in this regard, and may expect to have the same half-life other particles of its kind enjoy. This would be a cause for concern, since it appears to violate the conservation of mass, since a "new" particle is created and enters our universe.

To explain this apparent violation of conservation of mass, one only has to realize that the particle which enters our universe is matched by an anti-particle which goes into the closed and very small region of the BH. Because this region is closed and very small, particles entering it soon encounter the other particles that are trapped in there. On average, any particle resulting from the stripping of a virtual pair will soon encounter its anti-particle which has been stripped from another pair, and these two will anihiliate. When they do, conservation of mass is restored, since the infall anihiliation accounts for two particles radiated.

Last night while boiling out the deep fryer I was thinking about this, from the point of view of infall. Now it happens that the information ie mass, contained by a BH is proportional to its surface area, not its volume as one might assume. This is because we cannot know what goes on inside a black hole, but we can have an idea anyway of conditions on its surface. The seeming contradiction here is due to the distortion of time near an event horizon.

If we hover outside the event horizon and lob rocks into it, we could watch the rocks as they fall in. But we don't see them enter the event horizon. Instead, they seem to go slower and slower as approach the horizon, and at the same time they grow dimmer and dimmer, and their escaping photons become weaker and weaker, the energy waves longer and longer, until after a while we do not have any quanta from the rocks at all. But during the time that we can watch them, they do not enter the horizon at all, but seem to us to slow down and stop right at the horizon.

So, from our perspective outside the horizon, everything that goes into the event horizon seems to just hang there until we can't see it any more. This is why we can surmize that all the information that goes into a black hole is right there on the surface, and, as far as we are concerned, does not proceed any deeper.

Another way to look at this is to consider that time, viewed from the outside, seems to stop for the infalling object. The infalling object wouldn't see it that way, but from the outside, looking in, that is what we see. In a sense, as far as we are concerned, the infalling object becomes eternal and no longer changes in time as we do. It no longer shows any indication of ageing, moving or reacting.

So that's the puzzle. If, as far as we are concerned, the object as it infalls attains an eternal state, then it cannot, as far as we are concerned, react with its antiparticle on the inside to produce the required mass loss to counter the mass gain resulting from the radiated partner.

Our universe, then, must experience a net gain of mass and energy, upsetting the mass conservation applecart. We now have to look for a sink somewhere else where mass is lost from the universe to restore our precious conservation.

But this is not what concerns me immediately. My immediate concern has to do with the supposed loss of mass of the Black Hole to evaporation, which seems to depend upon the anihiliation of particles enclosed within the horizon. This loss, when calculated, yields an evaporation time for small black holes which is pretty fast, say about 10^-23 seconds. A small black hole doesn't have enough time, at that rate, to interact with much else in the universe. It is not likely, for example, that a small black hole would suck up the earth's atmosphere, or fall to the center of the earth and give another revival to the hollow earth theories. It just can't last long enough to get the matter it needs to stay alive.

However, the analysis above may change that scenario back again to threat status. If my reading is correct, the balanceing act of particle extinction within the black hole does not happen until sometime in the extremely distant future. If that is so, the black hole will not be seen to evaporate, but will continue to have opportunities to encounter some mighty tastey bananas at our expense for a long, long time. The first black hole we create on the planet will indeed have a chance to swallow us all up.

I should very much like to hear an argument that counters this unfortunate scenario.