
#19
Jan1806, 12:12 PM

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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 



#20
Jan1806, 04:02 PM

P: 1,308

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. 



#21
Jan1806, 05:11 PM

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If the expansion of the universe is deaccelerating, we will (again IIRC) see all of it eventually. 



#22
Jan1806, 07:01 PM

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



#23
Jan1806, 07:12 PM

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[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. 



#24
Jan1806, 10:22 PM

P: 167

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. 



#25
Jan1806, 10:32 PM

P: 167

Mike2




#26
Jan1906, 12:50 AM

P: 167

Mike2,




#27
Jan1906, 03:27 AM

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[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. 



#28
Jan1906, 06:46 AM

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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 spacetime.) 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 



#29
Jan1906, 08:17 AM

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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). 



#30
Jan1906, 10:34 AM

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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 



#31
Jan1906, 11:10 AM

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*H^{2} 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. 



#32
Jan1906, 11:47 AM

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



#33
Jan1906, 11:58 AM

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#34
Jan1906, 01:05 PM

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Unruh radiation is really confusing.
According to Baez http://groups.google.com/group/sci.p...e=source&hl=en 



#35
Jan1906, 01:40 PM

P: 1,308

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


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