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Author Topic: Does Hawking radiation prevent the formation of singularities?  (Read 3128 times)

Colin Earl

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Colin Earl  asked the Naked Scientists:
   
Question: Why does Hawking (black-body) radiation not prevent the formation of singularities?  The following argument for why singularities cannot form seems too obvious to be correct, but I cannot see the flaw. I would be very happy to offer $25 to the first solid explanation of why it is wrong.    Hawking radiation has a temperature T, given by the formula T=hc/(4 * 3.141… * R * k) (see http://en.wikipedia.org/wiki/Hawking_radiation (3D))  So T is proportional to 1/R where R is the radius of a black hole event horizon. However, the horizon is not a hard physical boundary, so the Hawking effect presumably continues to operate and drive ever increasing temperatures as matter falls towards the center of the hole. 
The amount of energy produced by a black body at a given temperature is proportional to T**4 and exerts the same outward pressure on incoming particles as that
 which balances gravity in our sun.
 If T is proportional to 1/R and if the amount of energy produced is proportional to T**4, incoming matter will there experience a repulsive force proportional to 1/R**4 which will eventually balance a gravitational pull that only increases at a rate of 1/R**2. So equilibrium will be reached before incoming matter reaches the hole center and no singularity will ever form. This argument also provides a cut-off-point that allows gravity to be combined with ordinary quantum mechanics without encountering singularities. 

Of course, it all hinges on the choice we made to use a formulation where T depends on 1/R, rather than depending on 1/M where M is the mass of the hole. If this is the reason why it is invalid, it must be that process which generates Hawking radiation on the event horizon is not dependent on the radius of curvature at that point but on some other characteristic of the hole that distinguishes that point from one on another hole of different mass. If so, what property is it? If not, what exactly is the error in the above argument?     Best regards,  


What do you think?
« Last Edit: 24/10/2011 02:30:04 by _system »


 

Offline yor_on

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A Hawking radiation is presumed as a result of the Black Holes event horizon already existing, as far as I know? You can't assume the radiation to come into play before the event horizon.  A event horizon is also a observer dependent phenomena, also depending on what type of Black Hole you refer too. A 'gravitationally accelerating/free falling observer' won't see a Hawking radiation.

We had a nice discussion about Hawking radiation and Unruh radiation here.  They seem as very similar effects to me.

Read it and see what you think.
 

Offline MikeS

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Colin Earl  asked the Naked Scientists:
   
Question: Why does Hawking (black-body) radiation not prevent the formation of singularities?  The following argument for why singularities cannot form seems too obvious to be correct, but I cannot see the flaw. I would be very happy to offer $25 to the first solid explanation of why it is wrong.    Hawking radiation has a temperature T, given by the formula T=hc/(4 * 3.141… * R * k) (see http://en.wikipedia.org/wiki/Hawking_radiation (3D))  So T is proportional to 1/R where R is the radius of a black hole event horizon. However, the horizon is not a hard physical boundary, so the Hawking effect presumably continues to operate and drive ever increasing temperatures as matter falls towards the center of the hole. 
The amount of energy produced by a black body at a given temperature is proportional to T**4 and exerts the same outward pressure on incoming particles as that
 which balances gravity in our sun.
 If T is proportional to 1/R and if the amount of energy produced is proportional to T**4, incoming matter will there experience a repulsive force proportional to 1/R**4 which will eventually balance a gravitational pull that only increases at a rate of 1/R**2. So equilibrium will be reached before incoming matter reaches the hole center and no singularity will ever form. This argument also provides a cut-off-point that allows gravity to be combined with ordinary quantum mechanics without encountering singularities. 

Of course, it all hinges on the choice we made to use a formulation where T depends on 1/R, rather than depending on 1/M where M is the mass of the hole. If this is the reason why it is invalid, it must be that process which generates Hawking radiation on the event horizon is not dependent on the radius of curvature at that point but on some other characteristic of the hole that distinguishes that point from one on another hole of different mass. If so, what property is it? If not, what exactly is the error in the above argument?     Best regards,  


What do you think?

Hawkin radiation only operates outside the event horizon and is a very small effect.  What happens inside the event horizon is completely unknown. 
 

Offline MikeS

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A Hawking radiation is presumed as a result of the Black Holes event horizon already existing, as far as I know? You can't assume the radiation to come into play before the event horizon.  A event horizon is also a observer dependent phenomena, also depending on what type of Black Hole you refer too. A 'gravitationally accelerating/free falling observer' won't see a Hawking radiation.

We had a nice discussion about Hawking radiation and Unruh radiation here.  They seem as very similar effects to me.

Read it and see what you think.

yor_on

Why?  As I understand it they would up to the event horizon.
 

Offline Soul Surfer

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Probably it does.  Because if you consider the simple collapse of a schwartschild black hole towards the "singularity" there must be a second "event horizon"  because as the collapse continues, the stronger gravitational field prevents light getting a shorter distance from the centre than the event horizon in our universe.  This will shrink and get hotter with hawking radiation as it contracts and there must eventually be a balance with radiation outflow and radiation inflow.  The interesting point is as the black hole gets heavier the final horizon gets smaller because of the greater energy flow needed.  This is a simpler and much more physically sensible model of a "singularity"  It should be possible to calculate how small this is from the total energy in the collapsed black hole and the rate of energy flow across the horizon.  mind you it may be smaller than a planck unit!
« Last Edit: 24/10/2011 18:38:37 by Soul Surfer »
 

Offline yor_on

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What are you asking about Mike? If there is an 'absolute horizon'? Or why a free falling observer won't see a Hawking radiation? In the first case there must always be a region from where you find no light able to get out and that will be the event horizon, but it seems also a function of what kind of Black Hole you describe, as well as of where you are relative it and if you're accelerating/hovering or 'free falling'. The kind of event horizons a Rindler observer might see will behave the exact same as a 'real event horizon' and seems also to be called a 'killing horizon', but as a different observer will find it somewhere else it's a relativistic effect. But there must be a 'real Event Horizon', although it seems very tricky to define it. You just need to think of the way we ordinary find different positional definitions (time dilations and Lorentz contractions) relative 'relative motions', and mass, to see it.

The Hawking radiation is a function similar to the radiation a Rindler observer experience as he accelerates, as I understands it, that's also why a free falling observer won't find it. What is similar in both cases is 'accelerations' (equivalence principle). That just means that in both cases you're not in a geodesic, as when free falling towards that Black Holes 'center'. In the case of a Rindler observer you're in a free fall anytime that acceleration stops.
« Last Edit: 26/10/2011 00:08:01 by yor_on »
 

Offline yor_on

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What a 'real event horizon' should mean is quite hard to understand for me though. I would like to define it from the Black Hole itself but from the inside there won't be any noticeable event horizon as I understands it, and if defined from the outside it must involve a observer. Using something 'at rest' outside it, relative the gravitational potential, maybe? But that is our geodesic observer, 'free falling'. And how can I define that as the only 'true event horizon'? I can have as many identical Rindler observers I like, with all swearing to seeing the Event horizon, or would they? I'm not sure, I would assume them to see a same Unruh radiation though if being 'at rest' relative each other, and as any patch of SpaceTime should be as the next?

"An absolute horizon is only defined in an asymptotically flat spacetime a spacetime which approaches flat space as one moves far away from any massive bodies. Examples of asymptotically flat spacetimes include Schwarzschild and Kerr black holes. The FRW universe which is believed to be a good model for our universe is generally not asymptotically flat. Nonetheless, we can think of an isolated object in an FRW universe as being nearly an isolated object in an asymptotically flat universe.   

The definition of an absolute horizon is sometimes referred to as teleological, meaning that it cannot be known where the absolute horizon is without knowing the entire evolution of the universe, including the future. This is both an advantage and a disadvantage. The advantage is that this notion of a horizon is very geometrical, and does not depend on the observer, unlike apparent horizons, for example. The disadvantage is that it requires the full history (all the way into the future) of the spacetime to be known. In the case of numerical relativity, where a spacetime is simply being evolved into the future, only a finite portion of the spacetime can be known."

Maybe all event horizons, ignoring conceptual definitions, will be 'apparent Horizons'?


 

Offline MikeS

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yor_on
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 "A 'gravitationally accelerating/free falling observer' won't see a Hawking radiation."

The Hawkin radiation is all generated outside of the event horizon and is leaving the black hole.  The free falling observer is approaching the event horizon.  Why would the observer not 'see' the radiation?
 

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Check this one and see what you think Hawking radiation and Unruh radiation.

But it's a strange subject :)
 

Offline MikeS

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If the radiation and observer were both travelling at the same velocity and direction then I could understand why the radiation would remain invisible to the observer but they are not, they are going in opposite directions?
 

Offline yor_on

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Does Hawking radiation prevent the formation of singularities?
« Reply #10 on: 26/10/2011 11:36:21 »
The effect is a relativistic one, not about vectors per se, and there is no way you can be at rest with light btw, but the idea was nice :) And I'm not even sure it apply to all types of black holes? I know it apply to a Schwartzchild configuration, but that is a non rotating Black Hole, and physics seem to expect that kind to be very rare. A Kerr configuration (spinning BH) might be different? I'm not sure about what to think myself, but read Black Hole Thermodynamics and Statistical Mechanics, by Steve Carlip, and see what you make of it.
==

It is very good btw, and discuss a lot of different angles and history too.
« Last Edit: 26/10/2011 11:41:38 by yor_on »
 

Offline MikeS

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Does Hawking radiation prevent the formation of singularities?
« Reply #11 on: 27/10/2011 09:55:48 »
yor_on
I don't see this scenario as being particular relativistic.  To me it seems very simple.  If you are approaching something that is coming toward you, you will see it.  It's as simple as that.  If the radiation is emitted very close to the event horizon then it may be approaching you very slowly due to time dilation near the black hole but approach it will.  Am I missing something here?
 

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Does Hawking radiation prevent the formation of singularities?
« Reply #11 on: 27/10/2011 09:55:48 »

 

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