Could life exist inside a black hole?

[George Jones]

If a horizon exists, so does Unruh radiation. If no horizon exists, then neither does Unruh radiation.

And there is an effect associated with cosmological horizons. This effect exists in expanding universes even when the expansion is not accelerating. I can't remember who sees what.

Regards,
George

 

[pervect]

Here's a relevant quote from Baez, again

http://math.ucr.edu/home/baez/end.html

Why does the temperature approach a particular nonzero value, and what is this value? Well, in a universe whose expansion keeps accelerating, each pair of freely falling observers will eventually no longer be able to see each other, because they get redshifted out of sight. This effect is very much like the horizon of a black hole - it's called a "cosmological horizon". And, like the horizon of a black hole, a cosmological horizon emits thermal radiation at a specific temperature. This radiation is called Hawking radiation. Its temperature depends on the value of the cosmological constant. If we make a rough guess at the cosmological constant, the temperature we get is about 10-30 Kelvin.

So there is an Unruh effect from the cosmological horizon but it is very very very small. (Far too small to explain the 3k cosmic microwave background radiation, for instance, by about thirty orders of magnitude).

Unfortunately I'm not sure exactly how this number was calculated (I'm basically just taking Baez's word for the details.)

One of many things I'm puzzled about now that I think about the problem - if the universe stops expanding at some future date, then there is no horizon - is there still an Unruh radiation "now"?

[add]I think I may have at least part of an answer to this question. $\Delta \mathrm{E} \Delta \mathrm{t}$ is constant, where E is energy and t is time. So it takes long time to observe the very low energies involved, it's not something you can observe quickly.

 

[Mike2]

Here's a relevant quote from Baez, again

http://math.ucr.edu/home/baez/end.html



So there is an Unruh effect from the cosmological horizon but it is very very very small. (Far too small to explain the 3k cosmic microwave background radiation, for instance, by about thirty orders of magnitude).

Unfortunately I'm not sure exactly how this number was calculated (I'm basically just taking Baez's word for the details.)

I followed the reference back to the book by R. M. Wald, Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics, Chapter 5, page 105.

So from your quote, may I take it that the cosmological event horizon and the cosmological constant itself are both derived from the acceleration radiation due to expansion alone? So do the existence of the horizon and the CC now serve as formal proof of the ZPE from which the acceleration radiation comes from? And has it yet been proven that the temperature associated with this CC is consistent with the energy density of Dark Energy? Have we solved the dark energy mystery now?

 

[pervect]

I followed the reference back to the book by R. M. Wald, Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics, Chapter 5, page 105.
So from your quote, may I take it that the cosmological event horizon and the cosmological constant itself are both derived from the acceleration radiation due to expansion alone? So do the existence of the horizon and the CC now serve as formal proof of the ZPE from which the acceleration radiation comes from? And has it yet been proven that the temperature associated with this CC is consistent with the energy density of Dark Energy? Have we solved the dark energy mystery now?

I think you're leaping ahead too much.

The cosmological constant indicates that there is a cosmological event horizon, at least according to standard models. (But see Space Tiger's reservations - it's possible to construct models in which there is no cosmological event horizon, for instance if the expansion deaccelerates again. We don't expect that to happen, but dark energy is not well understood, and we may not have all the details totally correct.)

If a cosmological event horizon exists. we expect that there is a very tiny amount of unruh radiation associated with it's existence.

I would not call this proof of "ZPE", because that term isn't very well defined, and there's too much nonsense written under that name. Rather I'd stick with what Baez said, that if the current expansion continues to accelerate, we'd expect the universe to reach a minimum but positive temperature due to the Unruh radiation associated with the cosmological horizon.

 

[ubavontuba]

Cool stuff guys. Keep running with it.

Keep in mind though that the cosmological event horizon (CEH) I am referring to is not real in the sense that it's in a place that can be quantified. Think of it like a rainbow. It's there and it can be observed and it has an effect on your locality, but you cannot approach it.

It looks to me like this is simply a logical deduction of relativity. As a rainbow's position is relative to the observer, so is the CEH.

 

[ubavontuba]

Pervect,

There's another way of saying something similar that is more in the spirit of true GR and perhaps simpler. The net gravitational effect of a spherically symmemtric expanding shell of matter can be shown to be zero inside the shell.

The cosmological horizon (assuming it exists, which means assuming that the universe's expansion is accelerating and will continue to accelerte) is thus not due to the effects of the matter outside - it is due to the effects of the matter (and dark energy) inside the horizon.

Aparrently you may have missed this point:

My concept is that to the observer, the universe having seemingly been expanding at relativistic speed from any given point, the perceived black hole "shell" should not act inside like the interior gravitational effects of a normal gravitational sphere. This is because the mass is receding at near light speed in either perceived direction (sort of like shining two flashlights in opposite directions).

That is that from the standpoint of the observer, the energy and mass at the perceived edge of the universe on his left, let's say, is in a noncontiguos reference frame from the matter and energy on his right.

So, all the matter on his right, let's say, is going to be effected more and more by the matter at the perceived edge on his right, in a relationship to distance. The closer it is to the observer, the less it is thusly affected and conversely, the farther away it is, the faster it will seem to be accelerating outward (generally speaking).

As the universe expands, this effect should be perceived by the observer as an acceleration of the expansion of the universe. Ergo, as "dark energy." This is because the receding walls allow there to be more space for this effect to take place in, and therefore everything (including perceived motion) must seemingly get bigger (faster) too.

Do you see? We're not in a normal gravitational sphere in this case. All radii are noncontiguos (increasingly so with distance). In the normal gravitational sphere to which you refer, it's all in a single/contiguous reference frame. (Of course this is assuming that gravity truly propagates at the speed of light.)

 

[hellfire]

I am happy to see a discussion about horizons, because this will give me the oportunity to ask some questions that bother me a long time ago...

The cosmological event horizon is actually an observer dependent horizon, same as the Rindler horizon. What bothers me, specially in case of the Unruh radiation, is the following: You can get the result of an observer dependent horizon considering only the action of some field (a scalar field for example) in flat spacetime viewed by an accelerated observer. This observer will detect a thermal bath of particles. Calculations and qualitative discussions I am aware of stop at that point. However, a thermal bath of particles should create a gravitational field that should pertub the curvature making it non-vanishing. But, does this make sense? The inertial observer and the accelerated observer, both in the "same" spacetime, would measure different curvatures.

There are other things that bother me, like the fact that everywhere it is mentioned that the thermal radiation is "emitted" by the horizon. The thermal radiation from horizons is a mathematical consequence of the Bogolyubov transformations between two different vacua. In general, these lead to particle creation when evaluating the number operators in different vacua and I would expect that this is a homogeneous and isotropic and non-localized phenomenon of particle creation. These particles do not form a thermal distribution in the general case and thermal distributions appear only in case of horizons. Thus I would expect the thermal case to be a special case of the general case of particle creation between two different vacua. However, everywhere one can read that "horizons radiate", meaning that they produce the thermal distribution of particles. According to the more general case I would expect the thermal radiation in case of horizons to be also a bath of particles, a kind of "diffuse" radiation, isotropic and homogeneous, but not a radiation coming from the horizon.

 

[George Jones]

Sorry hellfire, I intended to make some comments this morning on your latest post in this thread, but I used up all my free time working on a post for the SR & GR forum, and I can procastinate no longer. To work!

I can say that many of your questions are answered in Quantum Fields in Curved Space by Birrell and Davies.

I hope come back to these issues.

Regards,
George

 

[hellfire]

I can say that many of your questions are answered in Quantum Fields in Curved Space by Birrell and Davies.
I hope come back to these issues.

Thanks George, it would be nice if you could summarize it, because I don't have any possibility to check this book.

 

[SpaceTiger]

Sorry hellfire, I intended to make some comments this morning on your latest post in this thread, but I used up all my free time working on a post for the SR & GR forum, and I can procastinate no longer.

I too should apologize for my rushed contributions to this thread. I'm in the middle of finals period (we have them after break here ), so I haven't been able to keep up with everything. When time permits, I have quite a bit to say, though I'm afraid it will contain more questions than answers, not being an expert on Hawking or Unruh radiation.

 

[George Jones]

I'm in the middle of finals period

Good luck Space Tiger!

I suspect, though, that with your intellect, you won't need much luck.

Regards,
George

 

[pervect]

Pervect,
Aparrently you may have missed this point:
Do you see? We're not in a normal gravitational sphere in this case. All radii are noncontiguos (increasingly so with distance). In the normal gravitational sphere to which you refer, it's all in a single/contiguous reference frame. (Of course this is assuming that gravity truly propagates at the speed of light.)

The problem is, that the universe appears to us to be isotropic and homogeneous around any specific point, at least as long as that point is at rest relative to the CMB frame.

That's the standard assumption, and it fits well with observations. Of course we can only directly observe the universe from one particular point of view, though we can construct what we think other people should see based on our observations.

This symmetry of the universe isn't obvious if you chose some other frame. For instance, in some frame moving relative to the CMB frame, the symmetry that was present in the stationary frame becomes non-obvious, the universe appears anisotropic.

But that doesn't mean that the symmetry still isn't there, the solution in the moving frame must still appear symmetrical when viewed from the frame in which the universe is symmetrical. And we know how to switch points of view via the Lorentz transform (in SR), or generalized coordinate changes (in GR).

I don't think I can say much more without resorting to writing down equations.

 

[pervect]

I am happy to see a discussion about horizons, because this will give me the oportunity to ask some questions that bother me a long time ago...

The cosmological event horizon is actually an observer dependent horizon, same as the Rindler horizon. What bothers me, specially in case of the Unruh radiation, is the following: You can get the result of an observer dependent horizon considering only the action of some field (a scalar field for example) in flat spacetime viewed by an accelerated observer. This observer will detect a thermal bath of particles. Calculations and qualitative discussions I am aware of stop at that point. However, a thermal bath of particles should create a gravitational field that should pertub the curvature making it non-vanishing.

Here is the way I look at it. The stress-energy tensor is, and must be, the same according to both descriptions. You can actually get the stress-energy tensor when you write down the Lagrangian of your field, for instance from the Lagrangian. So we expect it to be the same.

The more "intuitive" way I look at it is this. Real particles contribute to the stress energy tensor, but so do "virtual" particles. For instance, the electric field between two charges is due entirely to virtual particles, however the electric field contributes to the stress-energy tensor as is well-known.

What happens in accelerated frames of reference as nearly as I can make out is that the booga-booga (oops, I mean Bogoliubov :-)) transforms turn some real particles into virtual particles, and some virtual particles into real particles, and that's all they do. (This comes from the fact that some postive frequences w get mapped to negative frequences w', and vica-versa, the postive frequencies corresponding to 'real' particles and the negative frequencies correspoinding to 'virtual' particles.) So the particles that you see from acceleration are not appearing from nowhere, they appear because virtual particles are being turned into real ones.

Unfortunately, it's possible I could be full of **** here, but that's the way I look at it at the moment. I haven't done any extensive calculations with quantum mechanics in curved space-time, and I find that doing detailed calculations is one of the few ways of discovering "bad" mental models.

 

[Mike2]

I think you're leaping ahead too much.
The cosmological constant indicates that there is a cosmological event horizon, at least according to standard models. (But see Space Tiger's reservations - it's possible to construct models in which there is no cosmological event horizon, for instance if the expansion deaccelerates again. We don't expect that to happen, but dark energy is not well understood, and we may not have all the details totally correct.)
If a cosmological event horizon exists. we expect that there is a very tiny amount of unruh radiation associated with it's existence.
I would not call this proof of "ZPE", because that term isn't very well defined, and there's too much nonsense written under that name. Rather I'd stick with what Baez said, that if the current expansion continues to accelerate, we'd expect the universe to reach a minimum but positive temperature due to the Unruh radiation associated with the cosmological horizon.

I wonder how they calculated that temperature. Aren't they assuming that dark energy is the same as the cosmological constant, and aren't they using that energy density to come up with an equivalent temperature? If not, can't we just do this and see if it is equal to these other calculations? If we are at least in the same order of magnitude, then isn't that encouragement to make more precise calculations?

My calculations tell me that a temperature of 10^-30K corresponds to an energy density of 7*10^-136(J/m³), which is about 128*10^-51eV/Mpc³. I'm sure this is too small to be the dark energy density, right?

 

[ubavontuba]

isotropic-anisotropic

Pervect,

The problem is, that the universe appears to us to be isotropic and homogeneous around any specific point, at least as long as that point is at rest relative to the CMB frame.

Right. I'm saying the same thing. Every reference point would view the effect the same way. There is no assymetry (to speak of) in my concept.

That's the standard assumption, and it fits well with observations. Of course we can only directly observe the universe from one particular point of view, though we can construct what we think other people should see based on our observations.

Right. Again, I'm saying the same thing as in the current model, that being that any observer anywhere will perceive themselves as being in the middle of an expanding universe. I've just built upon this a little.

This symmetry of the universe isn't obvious if you chose some other frame. For instance, in some frame moving relative to the CMB frame, the symmetry that was present in the stationary frame becomes non-obvious, the universe appears anisotropic.

Exactly. This is my point concerning noncontiguous frames of reference. To us, a point a few billion light years away seems anisotropic in respect to the rest of the universe. However, would we to be where that point is, we'd see that it is isotropic and apparently at the center of an expanding universe (just like what we see here).

But that doesn't mean that the symmetry still isn't there, the solution in the moving frame must still appear symmetrical when viewed from the frame in which the universe is symmetrical. And we know how to switch points of view via the Lorentz transform (in SR), or generalized coordinate changes (in GR).

That's what I'm saying. Everything in regards to this still applies. The difference is that I feel I have taken it a step further (not refuted it).

I don't think I can say much more without resorting to writing down equations.

Yeah, those pesky equations. I wish I knew a couple of good physicists looking to publish. I think a heck of a good paper could come out of this...

 

[Mike2]

As I said, (in a universe with flat spatial sections like ours appears to have),

r(t) = a(t)R

where: r is the physical distance between worldlines of comoving observers; R is the (constant) coordinate distance between comoving observers; and a is the scale factor.

Thus,

r' = a'R = a'r/a = Hr,

where H = a'/a. Note that both a' and a depend on time, so that the Hubble "constant" is also a function of time. Then,

r'' = a''R = H'r + r'H,

and H' = a''/a - (a'/a)^2.

In our universe, H' < 0 and a''>0.

Regards,
George

I calculated the acceleration of objects at distances where the recession rate is the speed of light,

r'=rH gives r=r'/H, and in this case r'=c, so

r=c/H

And if we assume for simplicity that H is a constant so that H'=0, then

r''=r'H = (rH)H = rH² = (c/H)*H².

When I plug this into derive the Unruh temeprature, and then plug this into get the energy density of a black body with that temperature, I get an energy density of 3.8846*10⁴¹ (eV/Mpc³).

Is this comparable to the dark energy density?

 

[Garth]

Are you sure you are calculating this correctly? If you use those units then the dimensions of density should be eV.Mpc^-3.

In which case as 1eV = 1.6022x10^-12 erg, 1 parsec = 3.0857x10¹⁸cms and c² = 8.9876x10²⁰ (cms/sec)² then
3.8846x10⁴¹ eV.Mpc^-3. = 3.8846x10⁴¹x1.6022x10^-12x8.9876^-1x10^-20x[3.0857x10²⁴]^-3gms/cc = 2.36x10^-65 gms/cc; a little less than that required for DE! (By a factor of about 10³⁶)

Garth

 

[George Jones]

What bothers me, specially in case of the Unruh radiation, is the following: You can get the result of an observer dependent horizon considering only the action of some field (a scalar field for example) in flat spacetime viewed by an accelerated observer. This observer will detect a thermal bath of particles. Calculations and qualitative discussions I am aware of stop at that point. However, a thermal bath of particles should create a gravitational field that should pertub the curvature making it non-vanishing. But, does this make sense? The inertial observer and the accelerated observer, both in the "same" spacetime, would measure different curvatures.

For now, I'm going largely to restrict my comments to this passage.

Most courses in quantum field theory and books like Peskin and Schroeder use inertial observers in a particular solution to Einstein's equation, Minkowski spacetime. In particular, the contribution from quantum fields to the stress-energy tensor in Einstein's equation is not taken into account. Stress-energy tensors of qunatum fields are considered, but they're not fed into Einstein's equation. This is a very useful approximation - quantum fields propagating in Minkowki spacetime without affecting the spacetime background as viewed from inertial frames.

This leads naturally to the idea that doing something similar usiing non-inertial frames and/or non-flat spacetimes might be interesting. Take an interesting solution to Einstein's equation, and consider quantum fields propagating through spacetime without affecting the spacetime background, i.e., don't take into account the contribution from quantum fields to the stress-energy tensor in Einstein's equation.

This results in a big payoff - the Unruh effect, Hawking radiation, and cosmological radiation. The payoff is large, but the effects themselves are usually very small.

After doing this, the "back-reaction" of the stress-energy tensor of the quantum fields on spacetime can be considered, but the methods needed to do this are often quite subtle and difficult. Generalizing renormalization of the expectation values of components of the stress-energy tensor from Minkowski spacetime to other spacetimes is not straightforward because the concept of particle, as formulated in QFT in Minkowski spacetime, often no longer applies. A field (as opposed to particle) interpretation rules! I may talk a little more about this in another post in this thread.

Your point about the consistency required between the views different observers take for the Rindler spacetime Unruh effect is a good one. In this Unruh effect, suppose the quantum field in is the vacuum of the inertial observer. Then, the expectation values of the components of the stress energy tensor for this observer are all zero. Since the stress-energy tensor is a tensor, the values of the components of the stress energy tensor are all zero in every coordinate system, including the coordinate system of the accelerated observer.

What is happening here? How does the non-accelerated observer feel a temperature? To explain this, I'll quote a passage from Birrell and Davies about how an idealized accelerated particle detector reponds to the inertial vacuum.

"The explanation comes from a consideration of the agency that brings about the acceleration of the detector in the first place. As the detector accelerates, its coupling to the field causes the emission of quanta, which produces a resistance against the accelerating force. The work done by the external force to overcome this resistance supplies the missing energy that feeds into the field via the quanta emitted from the detector, and also into the detector which simulaneously makes upward transitions. But as far as the detector is concerned, the net affect is the absorption of thermally distributed quanta."

Of the three effects, I have worked through (a few years ago) the Unruh effect in some detail, looked at some of the details for Hawking radiation from eternal Schwarzschild black holes, and hardly looked at all at cosmological radiation.

As I said before, cosmological radiation is not just associated with expanding universes the have positive acceleration. I think that the point that John Baez makes is as follows.

As pervect noted, cosmological radition is presently very small - much smaller than the cosmic background radiation. As our universe expands, the scale factor (according to present models) will go to infinity, causing the CMB temperature to tend to zero. At some point in the very distant future, cosmological (Unruh) radiation will dominate the CMB radiation.

Regards,
George

 

[Mike2]

For now, I'm going largely to restrict my comments to this passage.

Thank you, I appreciate your efforts here.

I'm following my suspicion that dark energy as well as dark matter may be caused by this acceleration radiation. I've done the calculation that shows that even if all of space has a radiation calculated by assuming that it has the acceleration that the edge of the observable universe has (though places half that distance from us would not be accelerating that much) - this has turned out to be far less than the energy density of the cosmological constant. So now I'm grasping at straws, and it occurs to me that it seems these "acceleration radiation" effects assume one frame of reference accelerating with respect to one other. OK, but we know that every place in the universe is accelerating away from EVERY other place. Not only is a distant point accelerating away from us, but it is accelerating away from every other point as well. Does this mean that we might have to integrate that effect to account for the many different reference frames that each point is accelerating away from? Thanks.

 

[Garth]

Astrophysics is the science of understanding 'what goes on up there' (astro-) by understanding 'what goes on down here' (-physics) - in the laboratory. Rather than beginning to understand the Unruh effect of the expansion of the universe start with the laboratory.

The Unruh effect is observed for non-inertial observers, we are non-inertial observers. This effect predicts that empty vacuum in a suported laboratory should have a non-negative density in that frame of reference.

I make this Unruh vacuum density to be ~ 10^-113 gms/cc, as a point of interest SSC predicts a vacuum density near the Earth of ~ 10^-9 gms/cc, caused by reconciling the divergence of its two field equation solutions for a gravitational field.

Garth

 

[hellfire]

Thank you George. I was thinking about my question you tried to answer and I am not sure to understand this:

"The explanation comes from a consideration of the agency that brings about the acceleration of the detector in the first place. As the detector accelerates, its coupling to the field causes the emission of quanta, which produces a resistance against the accelerating force. The work done by the external force to overcome this resistance supplies the missing energy that feeds into the field via the quanta emitted from the detector, and also into the detector which simulaneously makes upward transitions. But as far as the detector is concerned, the net affect is the absorption of thermally distributed quanta."

However, I came to another conclusion: The question is not correctly formulated, because it is unclear that spacetime with a scalar field in it should be flat at all. I mean, the usual treatment of Unruh radiation starts with a scalar field in flat spacetime. Radiation in the accelerated frame should induce a back-reaction. However, even in the inertial frame we do not know how to compute the quantum corrections to the flat geometry. Nevertheless spacetime is assumed to be flat (as usual in QFT) per definition. Same should then apply for the accelerated frame. Does this "answer" make sense?

 

[George Jones]

Radiation in the accelerated frame should induce a back-reaction. However, even in the inertial frame we do not know how to compute the quantum corrections to the flat geometry. Nevertheless spacetime is assumed to be flat (as usual in QFT) per definition.

Sometimes, for simplicity, back-reaction is negelcted, but in the example I gave, there is no back-reaction caused by quantum field because the renormalized stress energy tensor of the field is zero in all frames.

Other examples could start with an excited field for the inertial observer, and then there would be back-reaction.

Regards,
George

 

[ubavontuba]

The CEH and gravity

Forum,

As I have had a recent post in this thread removed for being too speculative, I wish to resubmit the ideas presented in the removed post as questions. I will include a couple of citations in reference to these questions that I hope will allow it to get by the censors as being in the ballpark of opinions that are "currently held by the scientific community." Here goes:

In this http://physicsweb.org/articles/world/17/5/7" we read:

The aim of all this activity is, of course, to answer the question, what is the dark energy? If w is about -1, then a cosmological constant might be the solution. If w is more than -1, the right answer might be quintessence. And we cannot rule out a new twist to gravity that even Einstein did not foresee: while most theories that link gravitational and quantum physics predict novel behaviour on microscopic length scales or at very early times in the universe, few, if any, anticipate new effects on the largest length scales in the present day. And what if w is less than -1? Whatever the answer, something mysterious is at work in the cosmos.

Please note particularly where it states: "And we cannot rule out a new twist to gravity that even Einstein did not foresee:"

And in this http://en.wikipedia.org/wiki/Dark_energy" :

Other ideas
Some theorists think that dark energy and cosmic acceleration are a failure of general relativity on very large scales, larger than superclusters. It is a tremendous extrapolation to think that our law of gravity, which works so well in the solar system, should work without correction on the scale of the universe. However, most attempts at modifying general relativity have turned out either to be equivalent to theories of quintessence, or are inconsistent with observations.

Other ideas for dark energy have come from string theory, brane cosmology and the holographic principle, but have not yet proved as compelling as quintessence and the cosmological constant.

Note particularly where it states: "Some theorists think that dark energy and cosmic acceleration are a failure of general relativity on very large scales, larger than superclusters. It is a tremendous extrapolation to think that our law of gravity, which works so well in the solar system, should work without correction on the scale of the universe."

So, with these references in mind, I will reitterate the main concepts in my deleted post:

It seems to me that the current thinking in dark energy is to look for an energy density between the galaxies to explain it. My questions are:

What if the energy density between galaxies is irrelevent? What if the galaxies are apparently simply falling outward toward the CEH, rather than being forced outward from an internal pressure?

What if due to a quirk of relativity, the universe appears to be accelerating from all reference frames, but may or may not actually be accelerating? What if the apparent expansion acceleration is simply caused by relativistic effects of very distant mass in motion?

Could these proposed relativistic effects cause an apparent infinite density to the CEH that essentially causes it to behave much like a black hole event horizon in that it emits Unruh and other radiation (that we perceive as the CMBR) and gravity/acceleration (that we perceive as the effects of dark energy)?

That is, (in deference to my citations) could gravity itself be a relativistic effect in the extrenum of the cosmological event horizon?

Note: This post has been edited by ubavontuba

 

[Garth]

From my calculations above is it not clear that any Unruh coming from the CEH is many orders of magnitude smaller than that required for the CMB?

Garth

 

[ubavontuba]

From my calculations above is it not clear that any Unruh coming from the CEH is many orders of magnitude smaller than that required for the CMB?
Garth

No, but I've always had trouble paying attention in math.

Anyway, let's think about what's happening at the CEH. Isn't mass heading (falling?) into it?

Now let's think about the CEH like how we might think about black holes. In this case, let's think about black holes with halos of matter around and falling into them. They emit a lot more energy in IR than simply the Unruh and Hawking radiation, don't they?

P.S. It looks to me like the expansion model must still hold, but the universe needn't be quite as young or old as we might measure. Also, it isn't the concept of expansion that is in question, but rather the acceleration effect known as dark energy. In other words, is it possible that even a linear expansion might be perceived as an acceleration in the extrenum of relativity and the CEH?

 

[Mike2]

...
"The explanation comes from a consideration of the agency that brings about the acceleration of the detector in the first place. As the detector accelerates, its coupling to the field causes the emission of quanta, which produces a resistance against the accelerating force. The work done by the external force to overcome this resistance supplies the missing energy that feeds into the field via the quanta emitted from the detector, and also into the detector which simulaneously makes upward transitions. But as far as the detector is concerned, the net affect is the absorption of thermally distributed quanta."

Is the cosmological constant or dark energy considered to be the due to the acceleration of spacetime? Or is it the other way around? Is the cosmological constant considered to be due to the acceleration of the expansion rate or the acceleration of a constant expansion rate as more distant objects recede more quickly as they recede?

I wonder if acceleration radiation is the same as vacuum energy of the cosmological constant? If so, then it would seem that since zero acceleration gives zero energy density, then inertial frames traveling arbitrarily close to any point in space (even with zero velocity) would feel no temperature and would prove that there is no zero point energy/vacuum energy/cosmological constant. I could use some clarification on this. Thank you.

 

[Garth]

Is the cosmological constant or dark energy considered to be the due to the acceleration of spacetime? Or is it the other way around?

The latter

Is the cosmological constant considered to be due to the acceleration of the expansion rate or the acceleration of a constant expansion rate as more distant objects recede more quickly as they recede?

The cosmological constant is one possible cause of cosmic acceleration.

I wonder if acceleration radiation is the same as vacuum energy of the cosmological constant?

As in a previous post above any acceleration radiation (if it exists in the first place) is many many orders of magnitude smaller than the CMB or the cosmological constant energy density

If so, then it would seem that since zero acceleration gives zero energy density, then inertial frames traveling arbitrarily close to any point in space (even with zero velocity) would feel no temperature and would prove that there is no zero point energy/vacuum energy/cosmological constant. I could use some clarification on this. Thank you.

The temperature we "feel" is that of the CMB radiation, 2.7⁰K, a real temperature of radiation emitted by real hot gas (at z >1000). It might indeed be the case that there is no cosmological constant but that possibility would not be proven by the non-observance of cosmological Unruh radiation.

Garth

 

[Mike2]

As in a previous post above any acceleration radiation (if it exists in the first place) is many many orders of magnitude smaller than the CMB or the cosmological constant energy

I wonder if particle creation (virtual or real) might be types of acceleration radiation - even the zero point energy or the cosmological constant. What I mean is this: force itself is described in terms of mass(energy) and the acceleration. Even with virtual particles, they are produced in pairs (and only pairs?) that move away from each other and then come back together. So even there they are accelerating with respect to each other. So there seems to be some sort of accelerating reference frames even at the local scale that produces particles (or is at least associated with particle production). Any thoughts on this? Thanks.

 

[Garth]

I wonder if particle creation (virtual or real) might be types of acceleration radiation - even the zero point energy or the cosmological constant. What I mean is this: force itself is described in terms of mass(energy) and the acceleration. Even with virtual particles, they are produced in pairs (and only pairs?) that move away from each other and then come back together. So even there they are accelerating with respect to each other. So there seems to be some sort of accelerating reference frames even at the local scale that produces particles (or is at least associated with particle production). Any thoughts on this? Thanks.

You are mixing up Quantum effects and GR gravitational effects. As we do not yet have a quantum gravity theory your speculation cannot be assessed. In virtual particle pair production there is no sum total change of momentum and no overall acceleration so there is no Unruh radiation, or are you claiming to have detected it?

Garth

 

[Mike2]

You are mixing up Quantum effects and GR gravitational effects. As we do not yet have a quantum gravity theory your speculation cannot be assessed. In virtual particle pair production there is no sum total change of momentum and no overall acceleration so there is no Unruh radiation, or are you claiming to have detected it?
Garth

I'm just marvelling at the fact that acceleration produces radiation, thus particles. OK... how many ways are there to produce "particles". It would seem to me that there would have to be only one way of producing particles. Some fundamental transformation is involved. If it is acceleration in one instance, could it be acceleration in all instances? So I remembered that virtual particles do appear in pairs, they first separate, and then come back together in a brief enough time so as to not violate the uncertainty principle. Separating and coming back together involves a change in velocity, thus it involves acceleration as well. The only alternative is to suppose particles can be produced in many different inequivalent ways, when they all end up having the same properties.

This would give us a connection between the properties of spacetime and the properties of matter, so that QFT might lead to QG.

 

[ubavontuba]

Forum,

Hey, take a look at this article in the current issue of NewScientist!

Apparently life inside a black hole, isn't quite such an absurd notion afterall.