The Paradoxical Nature of Black Holes

[Iggy]

Sorry it took a while for me to respond. This thread didn't show up where I check the "my content" section. I guess that proves one can't observe a black hole thread unless you're inside it

 

I wonder if any experiments have been performed that compare the atomic clock rates of two (or more) atomic clocks on satelites orbiting at different altitudes (11,000 and 22,000 km, for instance).

 

I'm not aware of any tests between purely inertial clocks with different gravitational potential.

 

A geostationary orbit has a radius of 42,164 km from the center of the Earth. (4.2164x10^7 m).

(ref. http://en.wikipedia....ionary_altitude )

 

Using these values, one second on the geostationary satellite's clock would be: t_o= t_f (1-(3/2)(8.872x10^-3 m)/(4.2164x10^7 m))^1/2 = ~0.9999999998 seconds on the distant observer's clock. Feel free to check my math, since it's very possible that my calculation isn't correct.

 

That sounds right, but I'm not sure if you were still thinking of GPS. They aren't geosynchronous.

 

By the way, the excellent reference you provided ( http://www.mathpages.../s7-03/7-03.htm ) and the previous sectin of this book ( http://www.mathpages.../s7-02/7-02.htm ) seem to argue in favor of the formation of black holes in a finite time from the distant observer's perspective.

 

This is the part I was referring to,

 

Furthermore, we should acknowledge that, even within the context of general relativity, the formal definition of a black hole may be impossible to satisfy. This is because, as discussed in the previous section, a black hole is usually defined as a region of spacetime that is not in the causal past of any point in the infinite future. Notice that this refers to the infinite future, because anything short of that could theoretically be circumvented by regions that are clearly not black holes. However, in some fairly plausible cosmological models the universe has no infinite future, because it re-collapses to a singularity in finite coordinate time. In such a universe (which could conceivably be our own), the boundary of any gravitationally collapsed region of spacetime would be contiguous with the boundary of the ultimate collapse, so it wouldn’t really be a separate black hole in the strict sense. As Wald says, "there appears to be no natural notion of a black hole in a closed Robertson-Walker universe which re-collapses to a final singularity", and further, "there seems to be no way to define a black hole in a closed universe, because it requires going to infinity, but there is no infinity in a closed universe."

 

It’s interesting that this is essentially the same objection that is often raised by people when they first hear about black holes, i.e., they reason that if it takes infinite coordinate time for any object to cross an event horizon, and if the universe is going to collapse in a finite coordinate time, then it’s clear that nothing can possess the properties of a true black hole in such a universe.

 

The link you gave mentions it as well,

 

The "surfaces" of the trousers represent future null infinity (I+) of the external region, consistent with the definition of black holes as regions of spacetime that are not in the causal past of future null infinity. (If the universe is closed, the "ceiling" from which these "stalactites" descend is at some finite height, and our future boundary is really just a single surface. In such a universe these protrusions of future infinity are not true "event horizons", making it difficult to give a precise definition of a black hole.

 

Slinkey's paradox is, in other words, sometimes formulated in a different but very similar way. That is, if the definition of a black hole invokes future infinity then it is paradoxical for the black hole to exist for a finite amount of time as with a closed universe or with Hawking radiation.

 

Hawking himself mentions what I think Slinkey is saying in his 2005 paper:

 

Black hole formation and evaporation can be thought of as a scattering process. One sends in particles and radiation from [spatial] infinity and measures what comes back out to infinity. All measurements are made at infinity, where fields are weak and one never probes the strong field region in the middle. So one can’t be sure a black hole forms, no matter how certain it might be in classical theory.

 

So, when Slinkey says,

 

The observations we have indicate something very heavy there, of that there is little doubt. However, whether the gravitating body is a black hole is another question which, if Krauss is right, is not shown by any current theory.

 

He is agreeing with Hawking and others. I think it is an open question that probably won't be solved until we get a good and confirmed quantum theory of gravity.

Edited March 29, 2011 by Iggy

[csmyth3025]

 

 

 

.........That sounds right, but I'm not sure if you were still thinking of GPS. They aren't geosynchronous.

 

 

 

........Slinkey's paradox is, in other words, sometimes formulated in a different but very similar way. That is, if the definition of a black hole invokes future infinity then it is paradoxical for the black hole to exist for a finite amount of time as with a closed universe or with Hawking radiation........

 

.........I think it is an open question that probably won't be solved until we get a good and confirmed quantum theory of gravity.

On the matter of clocks on satellites, I used a geostationary orbit as an example simply because it's a well defined orbit and easily referenced in Wikipedia. The main point is that the calculation shows the gravitational time dilation for orbiting clocks just as your previous post mentioned.

 

Concerning the "frozen star" model of black holes, I can't present any definitive arguement for or against this model. I just don't know that much about it. To be sure - as Slinkey said - "there's something very heavy there...". Until a more well established "frozen star" model is presented, though, I'll have to follow the logic of this argument from my previous link:

 

(1) An event horizon can grow only if the mass contained inside the horizon increases.

 

(2) Nothing crosses the event horizon in finite Schwarzschild coordinate time.

 

Item (1) is a consequence of the fact that, as in Newtonian gravity, the field contributed by a (static) spherical shell on its interior is zero, so an event horizon can't be expanded by accumulating mass on its exterior. Nevertheless, if mass accumulates near the exterior of a black hole's event horizon the gravitational radius of the combined system must eventually increase far enough to encompass the accumulated mass, leading unavoidably to the conclusion that matter from the outside must reach the interior, and it must do so in a way that is perceptible in finite coordinate time for a distant observer, which seems to directly conflict with Item 2 (and certainly seems inconsistent with the "frozen star" interpretation).

(ref. http://www.mathpages.../s7-02/7-02.htm )

 

Chris

 

 

[alan2here]

Sorry if this is a sufficently different question, or has already been answered but it has been bugging me for a while.

 

Even if they cannot see this due to the propagation of light, redshift or whatever and taking out dangers like tidal forces, presume this person is the super hero "super resilient man". Then from the point of view of one falling into a black hole the rest of the universe will speed up, for example if the person looks at a clock a long way from the black hole as they fall into it then the ticking of clock will appear to speed up.

 

From the point of view of an observer elsewhere in the universe the black holes will evaporate to nothing after a finite time but the person they see falling into the back hole will take an infinite time to reach the surface.

 

So how come the falling persons information will also be lost forever even though they never reach the black hole, from anyones point of view, before it evaporates?

Edited March 31, 2011 by alan2here

[csmyth3025]

Sorry if this is a sufficently different question, or has already been answered but it has been bugging me for a while.

 

....Even if they cannot see this due to the propagation of light, redshift or whatever and taking out dangers like tidal forces, presume this person is the super hero "super resilient man". Then from the point of view of one falling into a black hole the rest of the universe will speed up, for example if the person looks at a clock a long way from the black hole as they fall into it then the ticking of clock will appear to speed up....

If you look back to post #75 there is an equation that I think demonstrates this effect. If we make the simplifying assumptions that the freely falling observer is in a circular orbit around a non-rotating, non-charged black hole at the distance slightly more than the innermost stable circular orbit then his (proper) time compared to a distant observer"s time will vary as described:

 

In the Schwarzschild metric, free-falling objects can be in circular orbits if the orbital radius is larger than ........for a clock in a circular orbit, the formula is:

 

 

 

Where t_o = the orbiting observer's proper time, t_f = the distant observer's time, r_o = the Schwarzschild radius (the "event horizon"), and r = the radius of the orbit.

 

To further simplify the problem we can say that the black hole has one solar mass and the circular orbit is twice the Schwarzchild radius. As it turns out, the Schwarzschild radius is about 2.95 km and the radius of the orbit is about 5.9 km in this case.

 

Using the above formula, one second on the orbiting ("free-falling") observer's watch would equal:

[1-(³/₂ x ^2.95 /_5.9)]^{^0.5} = [1-(1.5 x 0.5)]^{^0.5} = [1-(0.75)]^{^0.5} = 0.25^{^0.5} = 0.5 seconds on the distant observers's clock.

 

I have to admit that this result puzzles me - and I believe it has to do with how the term "proper time" is defined using Schwarzschild coordinates. For every second the distant observer's clock ticks off, the orbiting observer's watch will tick off two seconds. It seems from this that the orbiting observer's clock is running faster than the distant observer's clock.

 

This is not what's supposed to happen for a clock that's in a gravitational field!

 

I've clearly misused the formula or I'm misinterpreting the results.

 

HELP!

 

Chris

[Iggy]

Sorry if this is a sufficently different question...

 

So how come the falling persons information will also be lost forever even though they never reach the black hole, from anyones point of view, before it evaporates?

 

Right, I think that was Slinkey's question as well.

 

If you look back to post #75 there is an equation that I think demonstrates this effect. If we make the simplifying assumptions that the freely falling observer is in a circular orbit around a non-rotating, non-charged black hole at the distance slightly more than the innermost stable circular orbit then his (proper) time compared to a distant observer"s time will vary as described:

 

 

 

To further simplify the problem we can say that the black hole has one solar mass and the circular orbit is twice the Schwarzchild radius. As it turns out, the Schwarzschild radius is about 2.95 km and the radius of the orbit is about 5.9 km in this case.

 

Using the above formula, one second on the orbiting ("free-falling") observer's watch would equal:

[1-(³/₂ x ^2.95 /_5.9)]^{^0.5} = [1-(1.5 x 0.5)]^{^0.5} = [1-(0.75)]^{^0.5} = 0.25^{^0.5} = 0.5 seconds on the distant observers's clock.

 

I have to admit that this result puzzles me - and I believe it has to do with how the term "proper time" is defined using Schwarzschild coordinates. For every second the distant observer's clock ticks off, the orbiting observer's watch will tick off two seconds. It seems from this that the orbiting observer's clock is running faster than the distant observer's clock.

 

This is not what's supposed to happen for a clock that's in a gravitational field!

 

I've clearly misused the formula or I'm misinterpreting the results.

 

HELP!

 

Chris

 

You have [math]t_0 = t_f \cdot 0.5[/math]

 

so if [math]t_0 = 1[/math] then [math]t_f = 2[/math] because [math]1 = 2 \cdot 0.5[/math]. While 1 second ticks on the orbiting clock 2 tick on the distant clock. The distant clock ticks faster.

[csmyth3025]

Thanks Iggy. I forgot that t_o = t_ftimes sqrt....

 

 

If my (further) calculations are correct, a point object will orbit this black hole once every 2.471x10^-4 sec. (ref. http://www.wolframal...w.m1---&x=5&y=9 ). The inverse of this would be about 4047 revolutions per second.

 

A distant observer will count about 4047 orbits for this object every second according to his clock. An observer on the orbiting object will count about 4047 orbits every 0.5 seconds according to his watch.

 

For the same event (4047 orbits) the distant observer will think the orbiting observer's watch is running slow (0.5 seconds) while the orbiting observer will think the distant observer's clock is running fast (1 second).

 

For an orbital radius of 5.9 km, the circumference of the orbit would be 2*pi*5.9 km=37.1 km. 4047 orbits would make the distance traveled about 150,144 km. The distant observer would see the orbiting observer traveling at an orbital velocity of about 150,144 km/s.

 

Because the orbiting observer is going so fast, to him each meter of his orbit will be about 0.8657 m rather than the "real" meter of the distant observer (ref. http://www.wolframal...9%5E2%29+meters ) . He will think that his velocity is about 260,000 km/s according to his calculation (ref. http://www.wolframalpha.com/input/?i=%28150144000+meters%29%280.8657%29%2F%280.50s%29 )

 

Is there an additional shortening of the orbiting observer's perceived orbital length due to the strong gravitational field that will make his calculation match the distant observer's value?

 

Chris

 

Edited to correct math errors

Edited April 4, 2011 by csmyth3025

[csmyth3025]

Just to keep the record straight, in my previous post (#79) I initially indicated that we could use an obital radius slightly larger that the innermost stable circular orbit as a simplifying condition. I subsequently picked an orbital radius that is twice the Schwarzschild radius.

 

The innermost stable circular orbit for a black hole must be greater than three times the Schwarzschild radius, however. (ref. http://en.wikipedia....child_geodesics )

 

With r_s=2.95 km, I should have picked a larger orbital radius like 10 km (3.39 r_s) or even 9 km (3.05 r_s) in order to be consistent. I don't think this changes the math of the calculations we made, though.

 

I think the solution for resolving the orbiting observer's calculation of his orbital velocity with the distant observers calculation lies somewhere in the cited Wiki article on Schwarzschild geodesics. Unfortunately, my math skills are too poor to even understand what most of the equations in the article mean.

 

Chris

[Iggy]

Thanks Iggy. I forgot that t_o = t_ftimes sqrt....

 

 

If my (further) calculations are correct, a point object will orbit this black hole once every 2.471x10^-4 sec. (ref. http://www.wolframal...w.m1---&x=5&y=9 ). The inverse of this would be about 4047 revolutions per second.

 

A distant observer will count about 4047 orbits for this object every second according to his clock. An observer on the orbiting object will count about 4047 orbits every 0.5 seconds according to his watch.

 

For the same event (4047 orbits) the distant observer will think the orbiting observer's watch is running slow (0.5 seconds) while the orbiting observer will think the distant observer's clock is running fast (1 second).

 

For an orbital radius of 5.9 km, the circumference of the orbit would be 2*pi*5.9 km=37.1 km. 4047 orbits would make the distance traveled about 150,144 km. The distant observer would see the orbiting observer traveling at an orbital velocity of about 150,144 km/s.

 

That's right.

 

A circular Schwarzschild orbit is,

 

[math] \frac{r}{r_s} \left( \frac{v}{c} \right)^2 = \frac{1}{2} [/math]

 

if r_s/r is 0.5 then v would be 0.5c or about 150,000 km/s.

 

 

 

Because the orbiting observer is going so fast, to him each meter of his orbit will be about 0.8657 m rather than the "real" meter of the distant observer (ref. http://www.wolframal...9%5E2%29+meters ) . He will think that his velocity is about 260,000 km/s according to his calculation (ref. http://www.wolframalpha.com/input/?i=%28150144000+meters%29%280.8657%29%2F%280.50s%29 )

 

Is there an additional shortening of the orbiting observer's perceived orbital length due to the strong gravitational field that will make his calculation match the distant observer's value?

 

No, there wouldn't be any need, and you wouldn't expect, the two people to agree about the length of a meter or velocity. Schwarzschild coordinates, that we've been using, are not what the person orbiting the black hole would properly measure with his ruler especially if he were orbiting it.

 

I think velocity would be reciprocal between two observers who are the same radial distance from the black hole, one orbiting it and the other static. But, not with two observers who have different radial distances.

 

There is a useful wikipedia page here, http://en.wikipedia.org/wiki/Gullstrand%E2%80%93Painlev%C3%A9_coordinates

 

With r_s=2.95 km, I should have picked a larger orbital radius like 10 km (3.39 r_s) or even 9 km (3.05 r_s) in order to be consistent. I don't think this changes the math of the calculations we made, though.

 

I think it should matter to some degree.

 

The closer you get to event horizon, the more Newtonian approximations break down.

 

For example, you found a velocity of 150144 km/s for an orbit at r=2r_s. We figured time would be dilated by a factor of 2.

 

If we decompose time dilation into velocity and gravitational potential we'd probably get something close to two, but not exactly because it is so near to the event horizon. Let's see,

 

Velocity time dilation would be,

 

[math]\frac{t_0}{\sqrt{1 - v^2/c^2}} = \frac{1}{\sqrt{1 - .5^2/1^2}} = 1.15[/math]

 

Gravity,

 

[math]\frac{t_0}{\sqrt{1 - 2 \cdot M / r}}[/math]

 

where M = G x m(sun) / c^2

 

[math]\frac{1}{\sqrt{1 - 2 \cdot 1480 / 5900}} = 1.42[/math]

 

Combining gravity and velocity (1.15 * 1.42) gives 1.63, so not exactly 2. The closer you get to the event horizon the more Newtonian approximations break down.

Edited April 5, 2011 by Iggy

[csmyth3025]

Thanks for the math Iggy. I can follow your calculation. It's obvious that these sorts of relativistic calculations can get very complicated very quickly. I tip my hat to anyone with the math skill to wade through them and not get lost.

 

Chris

[alan2here]

Does an answer exist to my question or is it an as off yet scientifically non understood one?

Edited April 18, 2011 by alan2here

[Iggy]

Sorry if this is a sufficently different question, or has already been answered but it has been bugging me for a while.

 

Even if they cannot see this due to the propagation of light, redshift or whatever and taking out dangers like tidal forces, presume this person is the super hero "super resilient man". Then from the point of view of one falling into a black hole the rest of the universe will speed up, for example if the person looks at a clock a long way from the black hole as they fall into it then the ticking of clock will appear to speed up.

 

From the point of view of an observer elsewhere in the universe the black holes will evaporate to nothing after a finite time but the person they see falling into the back hole will take an infinite time to reach the surface.

 

So how come the falling persons information will also be lost forever even though they never reach the black hole, from anyones point of view, before it evaporates?

 

Until a confirmed theory successfully combines gravity and particle physics, both of which are necessary in any discussion of black hole evaporation, we cannot claim to know for sure the answer to the question. There has, however, been a lot of good theoretical research done into the information loss problem and one proposed solution which I linked earlier is similar to what you are describing.

[alan2here]

The gravity on the surface of a black hole is infinite.

 

So presumably one could replace the person in my above example with any matter, once the black hole was above a cirtain mass then the matter would never be able to reach the surface from the point of view of an outside observer as all of time would have passed before it could do so.

 

However, when converted to sound, the gravity waves of matter spiralling into black holes are predicted to sound like an increasing frequency until collision, so presumably gravity increasing to infinity counters this.

 

In other words it dosn't matter how slow time is for the matter very close to a black hole from the point of view of the rest of the universe it can still make increasing amounts of progress from the point of view of the rest of the universe due to gravity also being verry large.

 

So is it true to say that for a person, who would have increasing amounts of force up to infinite pulling them into the black hole countered by time dilation giving them much more time than is normal relative to the rest of the universe before they reach the black holes surface, that they might see even a large black hole evaporate before the reach it, but for something heavier relative to the black hole itself there is more likely to be a collision?

Edited April 19, 2011 by alan2here

[Spyman]

The gravity on the surface of a black hole is infinite.

No, if a 10 solar mass star would turn into a Black Hole then its gravity at the Event Horizon would be from 10 solar mass.

 

 

So is it true to say that for a person, who would have increasing amounts of force up to infinite pulling them into the black hole countered by time dilation giving them much more time than is normal relative to the rest of the universe before they reach the black holes surface, that they might see even a large black hole evaporate before the reach it, but for something heavier relative to the black hole itself there is more likely to be a collision?

No, the core of the Black Hole is deeper in the gravity well than the infalling observer and is subjected to even more time dilation and will persist longer.

Edited April 20, 2011 by Spyman

[MigL]

Take the case of two astronauts, one a distant observer to a black hole and the other unfortunate one falling into the black hole. There has been a suggestion that the distant observer never actually sees the unfortunate astronaut fall through the event horizon because time slows to zero and he will first see the evaporation of the black hole.

 

As the unfortunate astronaut falls nearer the event horizon the distant observer will see time slow, but he will also see a stretching of the visible wavelengths, ie. his image of the infalling astronaut will become redder and redder until it goes to infrared, then microwaves, long radio waves and finally stretched to infinity at the event horizon. The light has lost all its energy and since you cannot destroy mass/energy, it must have been 'absorbed' by the black hole. So, yes, things actually do fall into a black hole, even to an outside observer.

 

From the infalling astronauts point of view, the question is moot. Time does not slow down for him and he crosses the event horizon at which point the only thing in his future is the ( possible ) singularity.

 

A black hole's evaporation is a quantum mechanical phenomena, having to do with virtual particle creation. The infalling astronaut and observer is a GR phenomena. Don't mix up the two. You are trying to use the two models at boundary ponts where one result may not agree with the other.

[alan2here]

Take the case of two astronauts, one a distant observer to a black hole and the other unfortunate one falling into the black hole. There has been a suggestion that the distant observer never actually sees the unfortunate astronaut fall through the event horizon because time slows to zero and he will first see the evaporation of the black hole.

Hooray, yes, that's what I was saying.

 

From the infalling astronauts point of view, the question is moot. Time does not slow down for him and he crosses the event horizon at which point the only thing in his future is the ( possible ) singularity.

From the falling observers point of view the rest of the universe continuously speeds up and at some point the black hole will have evaporated.

 

So you would expect the falling observer to see the black hole rushing towards him, faster and faster as he is accelerated by it. There will be a finite time from the falling observers point of view before he hits the black hole, lets say 5 minutes.

 

From the falling observers point of view soon the non-falling observers movements would be speed up more and more over time, then the movement of other solar bodies would noticeably speed up and before verry long a million years will have passed from the outside observers point of view but only a few minutes from the falling observers.

 

Lets say nothing else falls into the black hole and it completly evaporates in a million years, the outside observer still hasn't seen the falling observer reach the black hole, because it would take forever from the outside observers point of view for that to happen.

 

So the falling observer sees all this speeding up occuring around him, and his perfectly normal seeming fall towards the black hole, as if falling off a building on earth but then that the black hole is shrinking, the black hole will be gone before he reaches it.

 

If the falling observer was not a scientist he might reply when asked what just happened to him that floating though space he encountered an entity that sucked him towards it shrinking and disappearing as it did so and that he was now moving much faster through space and in the future.

Edited May 9, 2011 by alan2here

[MigL]

Don't take pieces of my reply and use them to incorrectly justify your ignorance. Things do fall into black holes and the evaporation time for black holes is possibly infinite for anything other than microscopic primordial black holes. Read some Hawking for a clear explanation of BH evaporation.

[alan2here]

I'm not using your reply to as you put it justify my ignorance. It's common knowledge that black holes evaporate and that smaller ones evaporate more quickly.

 

If this means that microscopic black holes evaporate and that larger ones take forever to do so then they must evaporate at no speed at all or infinitely slowly once beyond a cirtain size, either way this is not how black holes are normally described, and we wonder why it's hard to get the general public to understand these things.

 

If this really is the case then it has been misrepresented a lot.

[csmyth3025]

I'm not using your reply to as you put it justify my ignorance. It's common knowledge that black holes evaporate and that smaller ones evaporate more quickly.

 

If this means that microscopic black holes evaporate and that larger ones take forever to do so then they must evaporate at no speed at all or infinitely slowly once beyond a cirtain size, either way this is not how black holes are normally described, and we wonder why it's hard to get the general public to understand these things.

 

If this really is the case then it has been misrepresented a lot.

You might want to take a look at the Wikipedia article on Hawking radiation/black hole evaporation here: http://en.wikipedia....ole_evaporation

 

...For a black hole of one solar mass ( = 1.98892 × 10³⁰ kg), we get an evaporation time of 2.098 × 10⁶⁷ years...

 

...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 M must be less than 0.8% of the mass of the Earth.

 

With the exception of hypothetical primordial black holes, the smallest stellar black holes are thought to be between 1.5 and 3 solar masses:

 

If the mass of the collapsing part of the star is below a certain critical value, the end product is a compact star, either a white dwarf or a neutron star. Both these stars have a maximum mass. So if the collapsing star has a mass exceeding this limit, the collapse will continue forever (catastrophic gravitational collapse) and form a black hole.The maximum mass of a neutron star is not well known, in 1939 it was estimated at 0.7 solar masses, called the TOV limit. In 1996 a different estimate put this upper mass in a range from 1.5 to 3 solar masses.

(ref. http://en.wikipedia....llar_black_hole )

 

The most massive neutron star known is PSR J1614–2230, which is almost 2 solar masses:

 

By measuring this delay, known as the Shapiro delay, astronomers determined the mass of PSR J1614–2230 and its companion. The team performing the observations found that the mass of PSR J1614–2230 is

(ref. http://en.wikipedia..../PSR_J1614-2230 )

 

This observation doesn't rule out black holes smaller than 2 solar masses, but it leads me to think that some rather exotic process would be required to form a black hole of one solar mass or less. The "neutronium" of PSR J1614–2230 can apparently support the weight of two solar masses and there are many known neutron stars with masses greater than one solar mass.

 

The mass of the black hole in GRO J0422+32 falls in the range 3.66 to 4.97 solar masses.^[5] This is the smallest yet found for any black hole, and near the theoretical upper mass limit (~2.7 solar masses) for a neutron star.

(ref. http://en.wikipedia..../GRO_J0422%2B32 )

 

Discounting the absorption of Cosmic Background Radiation, a one solar mass black hole would take about 1.5x10⁵⁷ times the present age of the universe to evaporate. This is close enough to infinity to suit me.

 

Chris

[alan2here]

Thanks.

 

So the least massive black hole that would take a finite time to evaporate taking into account the cosmic background radiation that keeps falling into the back hole would be a fraction of a solar mass and 0.8 times the mass of the earth?

 

Discounting that he can't actually see it due to the redshift of light bouncing off it then how much time would an outside observer have to wait to see a falling object that started outside the black holes event horizon reach the surface of a large black hole?

Edited May 10, 2011 by alan2here

[MigL]

I'll repeat, the distant observer sees the light coming out of the gravitational well more and more red-dhifted, until it finally disappears. But the infalling astronaut falls at the acceleration due to gravity just like anywhere else. He may see the 'end' of time when looking back out of the event horizon but the ONLY thing in his future is the ( possible ) singularity.

[alan2here]

This makes sence given that all except unnaturally small black holes don't evaporate away and also answers some others questions.

 

It would also mean that cosmic background radiation for a given place is constant throughout all of time or at least that the amount the black hole is being fed is always enough to prevent it from evaporating entirely.

[csmyth3025]

This makes sence given that all except unnaturally small black holes don't evaporate away and also answers some others questions.

 

It would also mean that cosmic background radiation for a given place is constant throughout all of time or at least that the amount the black hole is being fed is always enough to prevent it from evaporating entirely.

Although the cosmic background radiation will, as far as I know, exist throughout all of time - it will be "stretched" more and more as the universe continues to expand. The "temperature" of this radiation will become less and less, eventually approaching (but not quite reaching) absolute zero.

 

There will eventually come a time when the "temerature" of the cosmic background radiation will be so low that it will be even less than a few trillionths of a degree Kelvin. At this temperature even large black holes will be "warmer" than the background. (I'm guessing at this, so if anyone can make the appropriate calculation they may want to correct me on this temperature range)

 

Once the cosmic background radiation is "cooler" than the Hawking radiation emitted by a black hole it will no longer be absorbing more radiation than it emits and will thus be able to ever so slowly evaporate.

 

A supermassive black hole with a mass of 10¹¹ (100 billion) solar masses will evaporate in around 2×10⁹⁹ years

(ref. http://en.wikipedia....wiki/Big_Freeze )

 

Chris

[alan2here]

Thanks. There appears to be some uncertainty here but if due to the speading of background radiation. Presuming the universe continues to expand for a while. Then if even large black holes evaporate, then it brings me back to my earlier point about ever reaching the surface of a black hole if you fall into it.

Edited May 13, 2011 by alan2here

[imatfaal]

Chris

 

There is a great calculator here A black hole of 4*10^6 solar masses (similar to that at centre of milky way) has a Hawking Radiation temperature of around 10^-14 Kelvin.

 

Alan

 

It would also mean that cosmic background radiation for a given place is constant throughout all of time or at least that the amount the black hole is being fed is always enough to prevent it from evaporating entirely.

I havent read through the thread but to make it clear - if the background temp is higher than the temp of the black hole the blackhole not only doesnt evaporate, it increases in mass; this increase in mass causes its temperature to DROP, which in turn causes it to gain even more mass (ie slight positive feedback). The same applies when the blackhole temperature goes above background, it loses mass which causes it to gain temperature, which in turn increases the rate of evaporation. ie it will not reach an equilibrium, it will either get bigger or smaller.

then it brings me back to my earlier point about ever reaching the surface of a black hole if you fall into it.

I haven't read what went before - but whilst a distant observer in an accelerating frame will never see the poor guy fall through an event horizon - from the frame of the victim there is no change and no noticeable affect of the event horizon. Poor old victim will sail straight through and get ripped up by the tidal gravitational forces

[alan2here]

Read what I said in previous posts. I would appreciate your feedback.