# Are we in a black hole?

by ubavontuba
Tags: black, hole
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I've been trying to find the time to put together a good summary of cosmological horizons, but I'm afraid I'm still falling short. Start by taking a look at hellfire's brief review:

Horizons

 Quote by ubavontuba Doesn't the edge of the perceivable universe lie in the future?
The edge of the perceivable universe actually lies in the past. In theory, the maximally redshifted light would reach us from the moment of the Big Bang. In practice, we can't see anything before recombination (z ~ 1100).

 Isn't it in the future for mass to move toward it?
There may exist an event horizon; that is, a comoving distance at which a light ray fired from earth at the present time would never reach. I don't know if that's what you mean. It's not the same as a black hole event horizon, but is, in some ways, analogous to it.

 Like an object entering a black hole horizon, the object should appear compressed and highly red-shifted. In other words, the edges of the universe (if we could see them) would apppear to us like a concave black hole. Again, I'm not stating it is a black hole. Eventually, mass moving away from us should appear infinitely red shifted (it disappears into the apparent event horizon). So, supposing that any of this makes sense, isn't it possible to expect black hole behavior from the boundaries of the perceivable universe?
In current theory, an infinitely redshifted "object" would sit at the particle horizon at t=0 (the Big Bang). The particle horizon is always growing with time, so nothing disappears behind it. The event horizon can shrink with time and objects can "disappear" behind it, but it doesn't represent a boundary of infinite redshift.
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 Quote by SpaceTiger The edge of the perceivable universe actually lies in the past. In theory, the maximally redshifted light would reach us from the moment of the Big Bang. In practice, we can't see anything before recombination (z ~ 1100).
So how far away is the Cosmological Event Horizon at present. I don't think this is the same as asking how far away the presently observeable horizon is because that is what it was in the past, right?

Would it be right to calculate the accelerataion of points in space from other points in space as the universe expanse as follows: Take the distance to the Cosmological Event Horizon (at present?) and say that there is a constant acceleration from zero to the speed of light over that distance over that time? Thanks.
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 Quote by SpaceTiger I've been trying to find the time to put together a good summary of cosmological horizons, but I'm afraid I'm still falling short. Start by taking a look at hellfire's brief review: Horizons
I don't think this is the link you intended :-(

 There may exist an event horizon; that is, a comoving distance at which a light ray fired from earth at the present time would never reach. I don't know if that's what you mean. It's not the same as a black hole event horizon, but is, in some ways, analogous to it.
They way I recall it is that such a horizon does in fact exist in FRW cosmologies IF the expansion of the universe is acclerating. (Current data suggests that this is the case). I suppose I should go look for some references to support my recollections.

If the expansion of the universe is deaccelerating, we will (again IIRC) see all of it eventually.
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 Quote by pervect I don't think this is the link you intended :-(
Oops, I typed the link name in place of the path (now edited)...no, I meant this link:

Horizons

 They way I recall it is that such a horizon does in fact exist in FRW cosmologies IF the expansion of the universe is acclerating. (Current data suggests that this is the case). I suppose I should go look for some references to support my recollections.
You're right, I hope I didn't give the wrong impression here...I don't have so much faith in the standard model that I would confidently extrapolate it to t->infinity.

There may be an event horizon. We are by no means confident that this is the case.

 If the expansion of the universe is deaccelerating, we will (again IIRC) see all of it eventually.
Yup. And if it accelerates for a while, then decelerates (like the post-inflation universe), then an event horizon calculated assuming $\Lambda CDM$ would be incorrect. It's for this reason (among others) that I view Mike2's entropy arguments with great suspicion.
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 Quote by Mike2 So how far away is the Cosmological Event Horizon at present. I don't think this is the same as asking how far away the presently observeable horizon is because that is what it was in the past, right?
That depends on the model, but you should be able to get the result by just integrating:

$$\int_{a_0}^{\infty}\frac{dt}{a}$$

For that integral, you would need the time dependence of the scale factor and it will give you a result in comoving coordinates (it would diverge in physical coordinates). I have to run soon, so I don't have time to do the integral.
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 Quote by pervect What you can correctly say is that the universe has a cosmological event horizon, if that's what you're getting at. Note that the the redshift and apparent disappearing of an object at the horiozn is what an outside observer sees.
I'm thinking you mean in the case of observing a regular black hole (not an extra-universe observer, observing the universe).

 An observer inside a black hole would not observe these effects, he would see light from the outside observer just fine - the horizon is a one way barrier. This barrier is "outside can't see inside" for a black hole, it's "inside can't see outside" for the universe.
Right, with an ordinary blackhole. What I'm suggesting is to take all of these effects and turn them inside-out (the infinite density is beyond the sphere, not within the sphere).

 Thus what's similar is the existence of a horizon in both cases, but it's not right to think of the universe as being a black hole.
Actually, I'm thinking of the extra-universe as a black hole (perceived, not real), not the universe itself.

 As far as the gravity of rapidly moving objects go, it doesn't behave like you think it does. For instance, take the following scenario spaceship1-------------->black hole spaceship2 Spaceship1 heads directly towards a black hole at a high rate of speed, passig spaceship2 which is stationary with respect to the black hole. Does the black hole appear heavier to spaceship 1 when it passes spaceship2, because of the relative motion? (Spaceship1 can consider that the black hole is moving directly towards it.) The answer as to the black hole's apparent gravity depends on the coordinate system used, but a simple measurement that's easy to do on the two spaceships is to measure the difference in gravity along the length of the ship (the tidal force). If the black hole appeared heavier, we would expect that spaceship1 would experience more tidal force than spaceship2. However, spaceship1 and spaceship2 will both measure the same tidal force in the above scenario. If spaceship1 was moving away from the black hole we would have the same result. If spaceship1 was moving up the page instead (with the black hole still to its right), it _would_ measure increased tidal forces as compared to spaceship 2.
Sure, but spaceship 1's mass will appear to change (relative to spaceship 2) as it approaches the black hole. But, I'm not talking about black holes (in general) here anyway.

I will reiterate as delicately as I can. I am not talking about black holes. I am talking about an apparent effect of relativity. There is no real black hole in question to consider.
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Mike2

 Quote by Mike2 Both Black Holes and the Observable Universe have an Event Horizon. You already know about the Event Horizon of Black Holes. The Observable Universe has a Cosmological Event Horizon. This is the distance from us where the expansion of space has caused objects at that distance to recede away from us at the speed of light so that we can no longer see them. And just like a BH, galaxies that approach the Cosmological Event Horizon appear to red shift and slow down in their motion; they freeze just before they disappear. Some have suggested that just as in the case of BH's, an entropy can be calculated for the area of the Cosmological Event Horizon with the same formula for calculating the entropy of a BH using the area of its event horizon. And they suggest that the area of the Cosmological Event Horizon constraines the entropy inside it just as in the case for a BH. The area is emmense for the Cosmological Event Horizon. But that suggests some interesting consequences for an accelerating universe. If the expansion of the universe is accelerating, then the distance to the Cosmological Event Horizon is shrinking, and the area associated with the Cosmological Event Horizon is decreasing as well. That would mean that the entropy enclosed inside the Observable Universe is decreasing. And we should expect some form of structures to emerge to compensate for the decreasing entropy of the collapsing Cosmological Event Horizon. Not only that, but we are losing material behind the Cosmological Event Horizon as galaxies recede away from us. So there seems to be a need for more complex and stable structures required of ever decreasing amounts of material. I find it interesting that life arose on earth at just about the same time that the universe started it accelerated expansion. But what can we expect of the future if this process continues? Will some people have to rise from the dead to make up for this decreasing entropy? I wonder.
It looks like you're getting it. I'm not too sure about the decreasing size of the cosmological event horizon though. I don't think the horizon is even real (in the sense that you can visit it). The size of the event horizon should always be determined by the relative speed of a lot of galaxies way out there. Where a large percentage of them approach c will determine the boundary (it needn't even be smooth).
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Mike2,

 Quote by Mike2 Just had a thought. The Unruh effect assigns a temperature to acceleration, and points in space are accelerating away from each other - the farther away points of space get, the faster they move away from us. And there must be an energy density for a given temperature. So does someone want to do the calculation: What is the energy density associated with the expansion of space? Is it anywhere close to the dark energy density? If no one else wants to venture this question, I'll try to get to it when I get time. Thanks.
I think this is a step in the right direction.
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 Quote by SpaceTiger That depends on the model, but you should be able to get the result by just integrating: $$\int_{a_0}^{\infty}\frac{dt}{a}$$ For that integral, you would need the time dependence of the scale factor and it will give you a result in comoving coordinates (it would diverge in physical coordinates). I have to run soon, so I don't have time to do the integral.
I think one can get an estimation about this considering that the universe will end in a phase of exponential expansion due to the action of the dark energy which will become dominant. In such a case (considering that the scale factor today is $a_0 = 1$):

$$\int_{a_0}^{\infty} \frac{dt}{a} \quad = \quad \int_{a_0}^{\infty} \frac{dt}{e^{Ht}} \quad = \quad \frac{1}{H}$$

This means that the event horizon will be at same distance than the Hubble sphere. I guess that an exact calculation for $\Omega_{\Lambda} = 0.73$ and $\Omega_M = 0.27$ must be done numerically.
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 Quote by ubavontuba Sure, but spaceship 1's mass will appear to change (relative to spaceship 2) as it approaches the black hole. But, I'm not talking about black holes (in general) here anyway. I will reiterate as delicately as I can. I am not talking about black holes. I am talking about an apparent effect of relativity. There is no real black hole in question to consider.
You've totally missed the point, partially because of my choice of example.

Spaceship 1 is approaching a planet or a black hole - it doesn't matter. (I said black hole in the first post, but it doesn't really matter, black holes are just convenient to calculate with - and the example in my textbook that this is drawn from involves a black hole).

From the frame of spaceship 1, the planet is more massive. (Not its invariant mass, which is invariant, but it has a greater energy, and hence a greater "relativistic mass".)

(Technical note: to define the mass at all, you need an asymptotically flat space-time.)

However, the gravity of the planet/black hole is not spherically uniform in the moving frame. Thus when spaceship1, directly approaching (or fleeing) the planet, fires up its Forward mass detector (a real device, which however measures tidal forces, and not mass directly), it finds that the gravity (specifically, tidal gravity) of the planet has not increased _even though_ the relativistic mass (energy) of the planet has increased.

The same thing happens if the spaceship is moving away from the planet. The same thing happens if the planet is moving away from the spaceship.

Motion towards or away from a planet (or black hole) does not affect tidal gravity.

The same thing does _not_ happen if the spaceship is moving perpendicularly to the line connecting it to the planet - in that case, the tidal gravity actually does incease.

This can be understood (in an approximate sense) by thinking of the gravitational field of a moving object being compressed, much like the electric field of a moving charge is. (The anology is not exact, mainly because gravity isn't really a field in GR).

The bottom line is your idea does not work - there is no "infinite mass" due to the expanding universe, even when the velocity of recession is very large. The velocity of recession is on a direct line, and thus it does not affect the gravitational forces.

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.

Here's a reference (unfortunately it's in postscript - but with google you can view it as text, though it won't be quite as intelligible). The theorem is called Birkhoff's theorem.

www.ucolick.org/~burke/class/grclass.ps

 What this means, roughly, is that spherical symmetry is enough to ensure that the spacetime is static, without separately requiring it. Another way to appreciate the content of the theorem is to consider any spherical object. If you go inside it and do anything you wish that does not disturb the spherical symmetry, then the external gravitational field cannot change. This is consistent with our earlier result that there is no monopole gravitational radiation. Yet another meaning for the theorem is than inside of a spherically symmetric distribution of mass there are no gravitational effects, just as in Newtonian theory, despite the nonlinearities.
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 Quote by hellfire I think one can get an estimation about this considering that the universe will end in a phase of exponential expansion due to the action of the dark energy which will become dominant. In such a case (considering that the scale factor today is $a_0 = 1$): $$\int_{a_0}^{\infty} \frac{dt}{a} \quad = \quad \int_{a_0}^{\infty} \frac{dt}{e^{Ht}} \quad = \quad \frac{1}{H}$$ This means that the event horizon will be at same distance than the Hubble sphere. I guess that an exact calculation for $\Omega_{\Lambda} = 0.73$ and $\Omega_M = 0.27$ must be done numerically.
So for the case $\Omega_{\Lambda} = 1$ the event horizon is located at the Hubble sphere. The Hubble radius for H = 71 km/s Mpc is equal to 13,7 Gly.

I have calculated the integral for the case $\Omega_{\Lambda} = 0.73$ and $\Omega_M = 0.27$. The event horizon should be located about 15,5 Gly away from us. I had some problems with the numerical integration, but nevertheless I think this should be a good estimation (I can provide details if someone is interested).
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 Quote by Mike2 So how far away is the Cosmological Event Horizon at present. I don't think this is the same as asking how far away the presently observeable horizon is because that is what it was in the past, right?
I think what you're calling the presently observable horizon is the following. A "galaxy" is on the presently observable horizon if, at t = 0, the worldline of the galaxy intersects the past lightcone for us at t = now. You colud also mean at t = photon decoupling. The distance to the presently observable horizon is then the distance between our worldline and the galaxy's worldline as the measured using the spatial scale at t = now.

A galaxy is on what (I think) you're calling the Cosmological Event Horizon if, at t = now, the worldline of the galaxy intersects the past lightcone for us at t = infinity. I think you're asking for the spatial distance between our worldline and the galaxy's worldline.

As Space Tiger says, the coodinate distance between these worldlines is

$$\int_{a_0}^{\infty}\frac{dt}{a}.$$

The physical distance between our wordline at the galaxy's worldline is the spatial scale multplied by the coordinate distance. At t = infinity, the scale is infinite, so the physical distance between the worldlines is infinite. But I think you're asking for the physical distance between the worldlines as measured along the spatial hypersurface t = now, i.e., the coordinate distance multiplied by the scale at t = now. With appropriate normalization, this physical distance is same as the coordinate distance.

Hellfire computed this physical distance to be 15.5 gigalightyears, and this also is what I calculate.

Regards,
George
 P: 1,308 Thank you both: George Jones and hellfire, It would seem as though there is a real difference in reference frames between us and distant objects. It is not objects moving in stationary space (with respect to us) that accounts for the recession and expansion of the universe. But their space (reference frame) itself is moving with respect to us. As I understand it, that is exactly what is required to apply the Unruh effect - a frame of reference that is accelerating with respect to ours will feel a temperature whereas we would not. One question might be do these frames of reference have to be at the same physical location for the Unruh effect to apply? Or can they be separated by some distance? I've looked into what might be the acceleration of space itself as the universe expands. The Hubble constant(?) is (71km/s +/- 4km/s)/Mpc. So to get the velocity that their reference frame (space) is moving wrt to us, just multipy the distance between us by the Hubble constant. OK, since this velocity is changing with distance (not time?), it would appear that this can be interpreted as an acceleration. But what is that accleration? So far we have: r' = r*H Then: r'' = r'*H + r*H' But H'=0 (????) substituting for r' with the above, then: r'' = r*H2 There is no acceleration or velocity for points of space very near are own, but there is an acceleration for points at some distance from us, right? So the question is whether there would be an Unruh temperature associated with that acceleration. Thanks.
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 Quote by Mike2 Thank you both: George Jones and hellfire,
I was just following Space Tiger's lead.

 As I understand it, that is exactly what is required to apply the Unruh effect
.

There is a similar effect associated with the expansion of the universe. As I understand it, this was first anticipated by Schrodinger. (?)

 But what is that accleration? So far we have: r' = r*H Then: r'' = r'*H + r*H' But H'=0 (????) substituting for r' with the above, then: r'' = r*H2
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
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 Quote by Mike2 As I understand it, that is exactly what is required to apply the Unruh effect - a frame of reference that is accelerating with respect to ours will feel a temperature whereas we would not.
I don't know much about the Unruh Effect, but it seems to be an effect that occurs for observers in non-inertial frames. In SR, an accelerating observer is in such a frame. In GR, however, inertial observers can have a coordinate acceleration (free-fall towards the earth, for example). In the case of the expanding universe, the galaxies are treated as following geodesics -- that is, they're treated as being an inertial frame. Were the accelerating expansion of the universe to cause an Unruh effect, I should think it would be a violation of the equivalence principle.
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According to Baez

 1) An atom sitting on the earth will not be excited by acceleration radiation. 2) An atom freely falling near the earth will (with some very small probability) be excited by acceleration radiation. 3) An atom at constant distance from a black hole will (with some small probability) be excited by acceleration radiation. 4) An atom freely falling near the horizon of a black hole will not be excited be acceleration radiation (in the approximation where it's right at the horizon). 5) An atom freely falling or at rest far from a black hole will (with some small probability) be excited by acceleration radiation.
Why is the atom freely falling near the Earth excited by Unruh radiation? Because of the presence of a very distant Rindler horizon. If you think about the presence or absence of an event horizion, the existence/nonexistence of Unruh radiation becomes simpler. The existence of Unruh radiation depends on the existence of a horizon. If a horizon exists, so does Unruh radiation. If no horizon exists, then neither does Unruh radiation.
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 Quote by pervect Why is the atom freely falling near the Earth excited by Unruh radiation? Because of the presence of a very distant Rindler horizon. If you think about the presence or absence of an event horizion, the existence/nonexistence of Unruh radiation becomes simpler. The existence of Unruh radiation depends on the existence of a horizon. If a horizon exists, so does Unruh radiation. If no horizon exists, then neither does Unruh radiation.
I'm thinking that where there is an acceleration there will always also be an horizon. Black holes exist inside gravity wells where particles in the gravitational field experience an acceleration towards the singularity, and they also have event horizons. Particles that accelerate have an instantaneous particle horizon where, if the acceleration were to continue, at some distance behind them no light could ever reach them. And then, like wise, the existence of the cosmological event horizon would seem to prove that an acceleration exist. If the Unruh effect applies the black holes and accelerating particles, does it also apply to any acceleration due to the expansion of the universe?

ST: could you tell us why you think that this would violate the equivalence principle? Thanks.
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 Quote by pervect 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

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