Physics puzzle

[bunny]

A semi-recent (1980s I think?) discovery in theoretical physics was the fact that black holes can emit radiation in certain circumstances and ultimately "evaporate" in a massive release of energy.

As I understand it, the prediction came from the fact that in a true vacuum, electron-positron pairs are coming into existence all the time, existing for a tiny fraction of a second, and then annihilating (repaying the energy they "borrowed" to exist in the first place). The theory goes that this may happen right next to a black hole, the antimatter crosses the event horizon and the now lone electron zooms off into the universe. The positron annihilates something or other inside the black hole and the nett result is an increase in radiation and decrease in mass of the black hole. Once this has happened enough, the black hole disappears in a flash of gamma radiation.

The puzzle is - why isnt there an equal number of positron-electron pairs going the other way? In other words - pairs where the electron falls into the hole (making it bigger) and the positron zooms off into the universe to annihilate the first electron it can find?

[Borodog]

[ QUOTE ]
The puzzle is - why isnt there an equal number of positron-electron pairs going the other way? In other words - pairs where the electron falls into the hole (making it bigger) and the positron zooms off into the universe to annihilate the first electron it can find?

[/ QUOTE ]

I think there are. This is way not my specialty, so hopefully metric can come along and back this up/correct it. A virtual particle-antiparticle pair is gravitationally boosted into a real particle-antiparticle pair near the event horizon; one of them falls in and the other flies off. It doesn't actually matter which is which. In either case there is a "hole" left near the event horizon where the energy from the p-ap came from. Energy tunnels out of the black hole, across the event horizon, to fill this space. It is that loss of energy that reduces the mass of the black hole; to an outside observer, a positron never actually makes it into the black hole to do any annihilating, so that can't possibly be it. Not only that, even if it did, the annihilation simply liberates energy that is stuck within the black hole, so the mass doesn't go down at all, even if it could get in there before infinite time had elapsed. Meanwhile, in either case for the particle that escaped, either electron or positron, there is now energy outside the event horizon that could only have, through a handshake so to speak, come from inside the hole, reducing its mass.

[Sharkey]

The gravitational field of the black hole contributes to the creation of the particles, so when one of them escapes there is a net loss of mass/energy.

Another interesting question might be: How does the matter within the horizon of the black hole interact gravitationally with anything on the outside?

[Metric]

[/ QUOTE ]The puzzle is - why isnt there an equal number of positron-electron pairs going the other way? In other words - pairs where the electron falls into the hole (making it bigger) and the positron zooms off into the universe to annihilate the first electron it can find?

[/ QUOTE ]
There is no preference of matter over antimatter in this scenario (called Hawking radiation) -- the key idea is that in order to put the liberated particle "on its mass shell" energy must come from the black hole. It works equally well with particles and anti-particles -- they emerge from the black hole with a thermal spectrum. (one must be careful not to take this explanation absolutely literally -- it's more of a thought experiment to motivate the actual calculation)

There is another, related effect which is equally interesting and does not require a black hole. In flat spacetime, if you are accelerating, there is a point (far enough behind you) beyond which not even light will be able to catch up to you -- hence an effective "event horizon" appears behind an accelerated observer. And -- guess what happens? It turns out, in analogy with Hawking radiation, that the accelerated observer sees particles (in effect, he is bathed in them) with a thermal spectrum -- even if inertial observers in the same vicinity of space cannot see or interact with them! This is called the Unruh effect.

These two effects are fascinating -- apparently there is a very deep connection between spacetime geometry, event horizons, and thermodynamics. In fact, about ten years ago, a physicist named Ted Jacobson derived the Einstein equations for general relativity from what is essentially the first law of thermodynamics generalized to situations like this! Everyone agrees that something extremely profound is going on here, but nobody quite understands what it is.

[bunny]

Cool. Thanks!

[Sharkey]

[ QUOTE ]
In flat spacetime, if you are accelerating, there is a point (far enough behind you) beyond which not even light will be able to catch up to you

[/ QUOTE ]

Will the light ever appear to be approaching at other than the speed of light?

[David Steele]

Interesting about the Unruh effect.

I was curious and it turns out true that it was
Dr Bill Unruh (http://en.wikipedia.org/wiki/Unruh_effect) of the UBC physics dept who discovered it.
This happens to be my alma mater.

D.

[Metric]

[ QUOTE ]
[ QUOTE ]
In flat spacetime, if you are accelerating, there is a point (far enough behind you) beyond which not even light will be able to catch up to you

[/ QUOTE ]

Will the light ever appear to be approaching at other than the speed of light?

[/ QUOTE ]
The light will never reach you if you remain accelerated, and you have no access to it (you are causally isolated from it by the horizon), so there are no measurements you can perform on it to answer this question. However, in the analysis of the problem there is no breakdown of Lorentz invariance -- special relativity is in full force throughout the problem.

[Sharkey]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
In flat spacetime, if you are accelerating, there is a point (far enough behind you) beyond which not even light will be able to catch up to you

[/ QUOTE ]

Will the light ever appear to be approaching at other than the speed of light?

[/ QUOTE ]
The light will never reach you if you remain accelerated, and you have no access to it (you are causally isolated from it by the horizon), so there are no measurements you can perform on it to answer this question. However, in the analysis of the problem there is no breakdown of Lorentz invariance -- special relativity is in full force throughout the problem.

[/ QUOTE ]

I welcome your comments on the following in that regard:

An observer begins a rest beside a laser source and then proceeds away with constant acceleration along the path of the beam.

The rate of recession of the source is found based on trip time, then the speed of the approaching light is calculated using this figure and the observed red shift.

What will be the speed of light arrived at by this method?

[Metric]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
In flat spacetime, if you are accelerating, there is a point (far enough behind you) beyond which not even light will be able to catch up to you

[/ QUOTE ]

Will the light ever appear to be approaching at other than the speed of light?

[/ QUOTE ]
The light will never reach you if you remain accelerated, and you have no access to it (you are causally isolated from it by the horizon), so there are no measurements you can perform on it to answer this question. However, in the analysis of the problem there is no breakdown of Lorentz invariance -- special relativity is in full force throughout the problem.

[/ QUOTE ]

I welcome your comments on the following in that regard:

An observer begins a rest beside a laser source and then proceeds away with constant acceleration along the path of the beam.

[/ QUOTE ]
I'm with you so far...

[ QUOTE ]
The rate of recession of the source is found based on trip time,

[/ QUOTE ]
All right, so you want the relative velocity of the emitter as a function of space-ship time?

[ QUOTE ]
then the speed of the approaching light is calculated using this figure and the observed red shift.

[/ QUOTE ]
The relativistic doppler shift also gives us the relative velocity of the emitter. What do you want to do, now?

[ QUOTE ]
What will be the speed of light arrived at by this method?

[/ QUOTE ]
At this point I'm not really following your thought experiment. Could you say a bit more about your "method?"

Are you worried about the fact that the spaceship is not in a single intertial frame of reference for the duration of the experiment, and thus that "proper" space and time intervals dx and dt as measured at the moment of light detection were not the same as the "proper" intervals over the course of the duration of the experiment? I agree that if you don't take this into account, you could arrive at strange conclusions (such as c not being constant, etc).

[Sharkey]

1. The observer starts out at rest relative to the laser to get a reference reading of its wavelength.

2. The observer then proceeds along the path of the beam, away from the source, with constant acceleration.

3. At any given instant, the speed with which the source is receding from the observer is known: i.e. acceleration * elapsed time.

4. Simultaneously, the observer can measure a red shift in the light emitted from the source.

5. The inferred velocity of the light overtaking the accelerating frame of reference is easily determined based on relative speed (3) and red shift (4).

The exact formula for step 5 escapes me at the moment, but I recall it being quite simple.

The point of all this is that the speed of light from a source being accelerated away from is, in principle, measurable all times during the trip. For such a source to get lost behind an event horizon, the light it is emitting must, at some point, fall below c as observed from the “spaceship” in order to be outrun.

[Borodog]

[ QUOTE ]
The point of all this is that the speed of light from a source being accelerated away from is, in principle, measurable all times during the trip. For such a source to get lost behind an event horizon, the light it is emitting must, at some point, fall below c as observed from the “spaceship” in order to be escaped.

[/ QUOTE ]

I don't think this is quite correct. For example, any source falling into a black hole is "redshifted into oblivion" as it approaces the horizon; in other words it never is observed to cross the horizon, but becomes unobservable as the light from the source is redshifted to infinte wavelength in a finite time. I think this is exactly analogous to what metric was stating.

[Sharkey]

Metric said: "In flat spacetime, if you are accelerating, there is a point (far enough behind you) beyond which not even light will be able to catch up to you..."

[bunny]

[ QUOTE ]
Metric said: "In flat spacetime, if you are accelerating, there is a point (far enough behind you) beyond which not even light will be able to catch up to you..."

[/ QUOTE ]

It may be so far away though that the laser is always inside your "event horizon".

[Sharkey]

[ QUOTE ]
[ QUOTE ]
Metric said: "In flat spacetime, if you are accelerating, there is a point (far enough behind you) beyond which not even light will be able to catch up to you..."

[/ QUOTE ]

It may be so far away though that the laser is always inside your "event horizon".

[/ QUOTE ]

However far way, it will eventually catch up.

[bunny]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Metric said: "In flat spacetime, if you are accelerating, there is a point (far enough behind you) beyond which not even light will be able to catch up to you..."

[/ QUOTE ]

It may be so far away though that the laser is always inside your "event horizon".

[/ QUOTE ]

However far way, it will eventually catch up.

[/ QUOTE ]

Not necessarily - your speed will increase asymptotically towards c. I have no idea if it is true or not, but there may be a limiting point to the event horizon.

[Metric]

[ QUOTE ]
1. The observer starts out at rest relative to the laser to get a reference reading of its wavelength.

2. The observer then proceeds along the path of the beam, away from the source, with constant acceleration.

3. At any given instant, the speed with which the source is receding from the observer is known: i.e. acceleration * elapsed time.

4. Simultaneously, the observer can measure a red shift in the light emitted from the source.

5. The inferred velocity of the light overtaking the accelerating frame of reference is easily determined based on relative speed (3) and red shift (4).

[/ QUOTE ]
All I will say here is that one must be careful about using non-relativistic formuals in this analysis. E.G. v = a*t can't be taken at face value -- if you simply use the proper acceleration and "elapsed proper time" here, v would eventually exceed c -- but this doesn't physically represent something moving faster than the speed of light. You have to be very careful about exactly what you mean when you start writing down expressions.

[ QUOTE ]
The point of all this is that the speed of light from a source being accelerated away from is, in principle, measurable all times during the trip. For such a source to get lost behind an event horizon, the light it is emitting must, at some point, fall below c as observed from the “spaceship” in order to be outrun.

[/ QUOTE ]
Actually, what happens is that the redshift eventually becomes infinite and the emitter appears "frozen" on the horizon (similar to how an outside observer sees an object falling into a black hole). But from the emitter's point of view, he just keeps on emitting without any trouble -- it is only that beyond a certain time, none of the emitted light will reach the accelerating spacecraft. One can see this easily on a spacetime diagram -- the world-line of a uniformely accelerated object is hyperbolic, while light moves on 45-degree lines. Not all 45-degree lines will intercept the hyperbola.

http://en.wikipedia.org/wiki/Hyperbolic_motion_%28relativity%29

[Sharkey]

[ QUOTE ]
[ QUOTE ]
1. The observer starts out at rest relative to the laser to get a reference reading of its wavelength.

2. The observer then proceeds along the path of the beam, away from the source, with constant acceleration.

3. At any given instant, the speed with which the source is receding from the observer is known: i.e. acceleration * elapsed time.

4. Simultaneously, the observer can measure a red shift in the light emitted from the source.

5. The inferred velocity of the light overtaking the accelerating frame of reference is easily determined based on relative speed (3) and red shift (4).

[/ QUOTE ]
All I will say here is that one must be careful about using non-relativistic formuals in this analysis. E.G. v = a*t can't be taken at face value -- if you simply use the proper acceleration and "elapsed proper time" here, v would eventually exceed c -- but this doesn't physically represent something moving faster than the speed of light. You have to be very careful about exactly what you mean when you start writing down expressions.

[ QUOTE ]
The point of all this is that the speed of light from a source being accelerated away from is, in principle, measurable all times during the trip. For such a source to get lost behind an event horizon, the light it is emitting must, at some point, fall below c as observed from the “spaceship” in order to be outrun.

[/ QUOTE ]
Actually, what happens is that the redshift eventually becomes infinite and the emitter appears "frozen" on the horizon (similar to how an outside observer sees an object falling into a black hole). But from the emitter's point of view, he just keeps on emitting without any trouble -- it is only that beyond a certain time, none of the emitted light will reach the accelerating spacecraft. One can see this easily on a spacetime diagram -- the world-line of a uniformely accelerated object is hyperbolic, while light moves on 45-degree lines. Not all 45-degree lines will intercept the hyperbola.

http://en.wikipedia.org/wiki/Hyperbolic_motion_%28relativity%29

[/ QUOTE ]

Fair point there; I forgot about relativize. However, that limitation really only goes to strengthen what I’m saying about the impossibility of receding from an emitter as to outrun its light.

Your second paragraph takes a peculiar tangent.

[Borodog]

It's not a tangent.

[Metric]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
1. The observer starts out at rest relative to the laser to get a reference reading of its wavelength.

2. The observer then proceeds along the path of the beam, away from the source, with constant acceleration.

3. At any given instant, the speed with which the source is receding from the observer is known: i.e. acceleration * elapsed time.

4. Simultaneously, the observer can measure a red shift in the light emitted from the source.

5. The inferred velocity of the light overtaking the accelerating frame of reference is easily determined based on relative speed (3) and red shift (4).

[/ QUOTE ]
All I will say here is that one must be careful about using non-relativistic formuals in this analysis. E.G. v = a*t can't be taken at face value -- if you simply use the proper acceleration and "elapsed proper time" here, v would eventually exceed c -- but this doesn't physically represent something moving faster than the speed of light. You have to be very careful about exactly what you mean when you start writing down expressions.

[ QUOTE ]
The point of all this is that the speed of light from a source being accelerated away from is, in principle, measurable all times during the trip. For such a source to get lost behind an event horizon, the light it is emitting must, at some point, fall below c as observed from the “spaceship” in order to be outrun.

[/ QUOTE ]
Actually, what happens is that the redshift eventually becomes infinite and the emitter appears "frozen" on the horizon (similar to how an outside observer sees an object falling into a black hole). But from the emitter's point of view, he just keeps on emitting without any trouble -- it is only that beyond a certain time, none of the emitted light will reach the accelerating spacecraft. One can see this easily on a spacetime diagram -- the world-line of a uniformely accelerated object is hyperbolic, while light moves on 45-degree lines. Not all 45-degree lines will intercept the hyperbola.

http://en.wikipedia.org/wiki/Hyperbolic_motion_%28relativity%29

[/ QUOTE ]

Fair point there; I forgot about relativize. However, that limitation really only goes to strengthen what I’m saying about the impossibility of receding from an emitter as to outrun its light.

Your second paragraph takes a peculiar tangent.

[/ QUOTE ]
It may be "peculiar", but it explains the appearance of a horizon. If you write down the world-line of an accelerated object on a space-time diagram, you will see that light rays emitted in a region covering 1/2 of spacetime will never intersect the world-line of the object. The boundary of this region is the horizon.

[Sharkey]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
1. The observer starts out at rest relative to the laser to get a reference reading of its wavelength.

2. The observer then proceeds along the path of the beam, away from the source, with constant acceleration.

3. At any given instant, the speed with which the source is receding from the observer is known: i.e. acceleration * elapsed time.

4. Simultaneously, the observer can measure a red shift in the light emitted from the source.

5. The inferred velocity of the light overtaking the accelerating frame of reference is easily determined based on relative speed (3) and red shift (4).

[/ QUOTE ]
All I will say here is that one must be careful about using non-relativistic formuals in this analysis. E.G. v = a*t can't be taken at face value -- if you simply use the proper acceleration and "elapsed proper time" here, v would eventually exceed c -- but this doesn't physically represent something moving faster than the speed of light. You have to be very careful about exactly what you mean when you start writing down expressions.

[ QUOTE ]
The point of all this is that the speed of light from a source being accelerated away from is, in principle, measurable all times during the trip. For such a source to get lost behind an event horizon, the light it is emitting must, at some point, fall below c as observed from the “spaceship” in order to be outrun.

[/ QUOTE ]
Actually, what happens is that the redshift eventually becomes infinite and the emitter appears "frozen" on the horizon (similar to how an outside observer sees an object falling into a black hole). But from the emitter's point of view, he just keeps on emitting without any trouble -- it is only that beyond a certain time, none of the emitted light will reach the accelerating spacecraft. One can see this easily on a spacetime diagram -- the world-line of a uniformely accelerated object is hyperbolic, while light moves on 45-degree lines. Not all 45-degree lines will intercept the hyperbola.

http://en.wikipedia.org/wiki/Hyperbolic_motion_%28relativity%29

[/ QUOTE ]

Fair point there; I forgot about relativize. However, that limitation really only goes to strengthen what I’m saying about the impossibility of receding from an emitter as to outrun its light.

Your second paragraph takes a peculiar tangent.

[/ QUOTE ]
It may be "peculiar", but it explains the appearance of a horizon. If you write down the world-line of an accelerated object on a space-time diagram, you will see that light rays emitted in a region covering 1/2 of spacetime will never intersect the world-line of the object. The boundary of this region is the horizon.

[/ QUOTE ]

Do you find fault with the experiment I outlined? Before digressing, I would like to know if you think it’s inapplicable somehow.

[Metric]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
1. The observer starts out at rest relative to the laser to get a reference reading of its wavelength.

2. The observer then proceeds along the path of the beam, away from the source, with constant acceleration.

3. At any given instant, the speed with which the source is receding from the observer is known: i.e. acceleration * elapsed time.

4. Simultaneously, the observer can measure a red shift in the light emitted from the source.

5. The inferred velocity of the light overtaking the accelerating frame of reference is easily determined based on relative speed (3) and red shift (4).

[/ QUOTE ]
All I will say here is that one must be careful about using non-relativistic formuals in this analysis. E.G. v = a*t can't be taken at face value -- if you simply use the proper acceleration and "elapsed proper time" here, v would eventually exceed c -- but this doesn't physically represent something moving faster than the speed of light. You have to be very careful about exactly what you mean when you start writing down expressions.

[ QUOTE ]
The point of all this is that the speed of light from a source being accelerated away from is, in principle, measurable all times during the trip. For such a source to get lost behind an event horizon, the light it is emitting must, at some point, fall below c as observed from the “spaceship” in order to be outrun.

[/ QUOTE ]
Actually, what happens is that the redshift eventually becomes infinite and the emitter appears "frozen" on the horizon (similar to how an outside observer sees an object falling into a black hole). But from the emitter's point of view, he just keeps on emitting without any trouble -- it is only that beyond a certain time, none of the emitted light will reach the accelerating spacecraft. One can see this easily on a spacetime diagram -- the world-line of a uniformely accelerated object is hyperbolic, while light moves on 45-degree lines. Not all 45-degree lines will intercept the hyperbola.

http://en.wikipedia.org/wiki/Hyperbolic_motion_%28relativity%29

[/ QUOTE ]

Fair point there; I forgot about relativize. However, that limitation really only goes to strengthen what I’m saying about the impossibility of receding from an emitter as to outrun its light.

Your second paragraph takes a peculiar tangent.

[/ QUOTE ]
It may be "peculiar", but it explains the appearance of a horizon. If you write down the world-line of an accelerated object on a space-time diagram, you will see that light rays emitted in a region covering 1/2 of spacetime will never intersect the world-line of the object. The boundary of this region is the horizon.

[/ QUOTE ]

Do you find fault with the experiment I outlined? Before digressing, I would like to know if you think it’s inapplicable somehow.

[/ QUOTE ]
You can in principle perform any experiment you like, collecting any data you like. Where you need to be careful and slip-ups can occur is where you start performing an analysis and drawing conclusions.

[Sharkey]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
1. The observer starts out at rest relative to the laser to get a reference reading of its wavelength.

2. The observer then proceeds along the path of the beam, away from the source, with constant acceleration.

3. At any given instant, the speed with which the source is receding from the observer is known: i.e. acceleration * elapsed time.

4. Simultaneously, the observer can measure a red shift in the light emitted from the source.

5. The inferred velocity of the light overtaking the accelerating frame of reference is easily determined based on relative speed (3) and red shift (4).

[/ QUOTE ]
All I will say here is that one must be careful about using non-relativistic formuals in this analysis. E.G. v = a*t can't be taken at face value -- if you simply use the proper acceleration and "elapsed proper time" here, v would eventually exceed c -- but this doesn't physically represent something moving faster than the speed of light. You have to be very careful about exactly what you mean when you start writing down expressions.

[ QUOTE ]
The point of all this is that the speed of light from a source being accelerated away from is, in principle, measurable all times during the trip. For such a source to get lost behind an event horizon, the light it is emitting must, at some point, fall below c as observed from the “spaceship” in order to be outrun.

[/ QUOTE ]
Actually, what happens is that the redshift eventually becomes infinite and the emitter appears "frozen" on the horizon (similar to how an outside observer sees an object falling into a black hole). But from the emitter's point of view, he just keeps on emitting without any trouble -- it is only that beyond a certain time, none of the emitted light will reach the accelerating spacecraft. One can see this easily on a spacetime diagram -- the world-line of a uniformely accelerated object is hyperbolic, while light moves on 45-degree lines. Not all 45-degree lines will intercept the hyperbola.

http://en.wikipedia.org/wiki/Hyperbolic_motion_%28relativity%29

[/ QUOTE ]

Fair point there; I forgot about relativize. However, that limitation really only goes to strengthen what I’m saying about the impossibility of receding from an emitter as to outrun its light.

Your second paragraph takes a peculiar tangent.

[/ QUOTE ]
It may be "peculiar", but it explains the appearance of a horizon. If you write down the world-line of an accelerated object on a space-time diagram, you will see that light rays emitted in a region covering 1/2 of spacetime will never intersect the world-line of the object. The boundary of this region is the horizon.

[/ QUOTE ]

Do you find fault with the experiment I outlined? Before digressing, I would like to know if you think it’s inapplicable somehow.

[/ QUOTE ]
You can in principle perform any experiment you like, collecting any data you like. Where you need to be careful and slip-ups can occur is where you start performing an analysis and drawing conclusions.

[/ QUOTE ]

That’s why it’s called a thought experiment: because experimental slip-ups are thoughtless.

[bunny]

I dont know the maths at all - I expect that every point you've ever visited though (including the laser) is within your horizon. That would mean you would never be able to observe something in the way you have described.