Are we in a black hole?

Mike2

If you could give me the expansion rate in units of meters and seconds and the dark energy density in SI units of Joules per meter cubed, I'll see what I can do. Thanks.

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


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).


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.


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.

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?

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.

pervect

I don't think this is the link you intended :-(

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.

SpaceTiger

Oops, I typed the link name in place of the path (now edited)...no, I meant this link:

Horizons


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.


Yup. And if it accelerates for a while, then decelerates (like the post-inflation universe), then an event horizon calculated assuming [itex]\Lambda CDM[/itex] would be incorrect. It's for this reason (among others) that I view Mike2's entropy arguments with great suspicion.

SpaceTiger

That depends on the model, but you should be able to get the result by just integrating:

[tex]\int_{a_0}^{\infty}\frac{dt}{a}[/tex]

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.

ubavontuba

I'm thinking you mean in the case of observing a regular black hole (not an extra-universe observer, observing 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).

Actually, I'm thinking of the extra-universe as a black hole (perceived, not real), not the universe itself.

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.

ubavontuba

Mike2

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).

ubavontuba

Mike2,

I think this is a step in the right direction.

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 [itex]a_0 = 1[/itex]):

[tex]\int_{a_0}^{\infty} \frac{dt}{a} \quad = \quad \int_{a_0}^{\infty} \frac{dt}{e^{Ht}} \quad = \quad \frac{1}{H}[/tex]

This means that the event horizon will be at same distance than the Hubble sphere. I guess that an exact calculation for [itex]\Omega_{\Lambda} = 0.73[/itex] and [itex]\Omega_M = 0.27[/itex] must be done numerically.

pervect

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.

[add]
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

hellfire

So for the case [itex]\Omega_{\Lambda} = 1[/itex] 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 [itex]\Omega_{\Lambda} = 0.73[/itex] and [itex]\Omega_M = 0.27[/itex]. 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).

George Jones

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

[tex]\int_{a_0}^{\infty}\frac{dt}{a}.[/tex]

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

Mike2

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.

George Jones

I was just following Space Tiger's lead.

.

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

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

SpaceTiger

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.

pervect

Unruh radiation is really confusing.

According to Baez

http://groups.google.com/group/sci.p...e=source&hl=en

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.