size of observable universe in two simple cases -- this this right?

[bcrowell]

One of the most frequently asked questions in the cosmology subforum is about how the radius of the observable universe in units of light-years can be greater than the age of the universe in years. The figures are 46 billion and 14 billion, and these presumably require quite a bit of number-crunching in order to extract them from real-world cosmological observations using realistic cosmological models. I thought I would try doing the calculation in a couple of simple cases. I would appreciate it if anyone here could check me on these results.

The two cases I tried were a flat matter-dominated universe and a flat vacuum-dominated universe. The metric and coordinates I'm using are defined by
d s^2 = d t^2 - a(t)^2d \ell^2 ,
where the spatial part is
d \ell^2 = f(r)d r^2 + r^2 d \theta^2 + r^2 \sin^2\theta d \phi^2 .

In the flat case, f=1, so a photon moving radially with ds=0 has |dr/dt|=a^{-1}, so r=\pm\int dt/a. Suppressing signs, the proper distance the photon traverses starting soon after the Big Bang is L=a\int d\ell=a\int dr=ar=a\int dt/a.

In the matter-dominated case, a \propto t^{2/3}, so I get L=3t, while in the vacuum-dominated case with a\propto e^{t/T} I get L=t. Our universe has spent most of its history being matter-dominated, so it's encouraging that the matter-dominated calculation seems to do a pretty good job of reproducing the actual ratio of 46/14=3.3 between L and t. However, I would have expected the realistic result to interpolate between 1 and 3.

Does anyone see any mistakes in my analysis?

[EDIT] Fixed the sign in a\propto e^{t/T} above, which George Jones pointed I'd given incorrectly as a\propto e^{-t/T}.

[TobyC]

Sorry for changing the topic before, it got a bit carried away. It's just reading your post made me want to know the answer to that question.

Thank you to whoever moved it.

This thread looks very interesting but I can't help sorry

:smile:

[George Jones]

Suppressing signs, the proper distance the photon traverses starting soon after the Big Bang is L=a\int d\ell=a\int dr=ar=a\int dt/a.

Yes.
In the matter-dominated case, a \propto t^{2/3}, so I get L=3t,

This what I get.
while in the vacuum-dominated case with a\propto e^{-t/T} I get L=t.

Do you mean this is the 1/a in the integral (i.e., why the minus in the exp)? I don't get this L, but I might have made a mistake.
Our universe has spent most of its history being matter-dominated, so it's encouraging that the matter-dominated calculation seems to do a pretty good job of reproducing the actual ratio of 46/14=3.3 between L and t. However, I would have expected the realistic result to interpolate between 1 and 3.

Here is a crude model.

Consider a universe that is:

1) flat matter-only for 0 < t < A;

2) flat vacuum-only for A < t;

3) such that a and its time derivative from 1) and 2) match at t = A.

I haven't worked it out, but I suspect that L/t > 3 for t > A.

[bcrowell]

Hi, George -

Thanks very much for the reply!


Do you mean this is the 1/a in the integral (i.e., why the minus in the exp)?
Oops, yeah, that should have been a plus.

I don't get this L, but I might have made a mistake.
I made at least two mistakes here. One was the sign that you pointed out. Also, I completely botched that second result, for the vacuum-dominated case. It should have been

L=\left[e^{(t_2-t_1)/T}-1\right]T

where T=\sqrt{3/\Lambda}, t_1 is the time when the photon was emitted, and t_2 is the time at which we observe it to have traveled a proper distance L. This grows faster with t_2 than the matter-dominated case, so it makes sense that in our universe, L/t2 is greater than 3.

Does this make more sense?

Thanks again for your help with straightening this out!

-Ben

[George Jones]

It should have been

L=\left[e^{(t_2-t_1)/T}-1\right]T

where T=\sqrt{3/\Lambda}, t_1 is the time when the photon was emitted, and t_2 is the time at which we observe it to have traveled a proper distance L.

This is what I got.
This grows faster with t_2 than the matter-dominated case, so it makes sense that in our universe, L/t2 is greater than 3.

Does this make more sense?

I think so.

Your interesting has post has prompted me to have a look at something that has been nagging at me for a while, the speed of light where "speed" is in the same sense as "recession speed" for galaxies, i.e., rate of change of proper distance with respect to cosmic time.

As you did above, take r to be the comoving coordinate (constant for something moving with the Hubble flow), assume that the universe is spatially flat, and that the photon moves in the direction of increasing r. Assume further that for our galaxy r = 0, and that R \left( t \right) = a\left( t \right) r\left( t \right) is proper distance with respect to us at time t. Then, as above,

\int^{r \left( t \right)}_{r_i} dr' = \int^t_{t_i} \frac{dt'}{a\left( t' \right)}

gives

r\left( t \right) = r_i + \int^t_{t_i} \frac{dt'}{a\left( t' \right)}.

First, assume that at time t_i I fire a laser pulse, so r_i = 0. Then,

R \left( t \right) = a\left( t \right) r\left( t \right) = a \left( t \right) \int^t_{t_i} \frac{dt'}{a\left( t' \right)}

gives

\frac{dR}{dt} = 1 + \frac{da}{dt}\int^t_{t_i} \frac{dt'}{a\left( t' \right)}.

The speed of the laser pulse is great than c = 1!

Next, assume that at time t_i someone in galaxy A fires a laser pulse in the direction of increasing r, so r_i = r_A. Then,

R \left( t \right) = a\left( t \right) r\left( t \right) = R_A + a \left( t \right) \int^t_{t_i} \frac{dt'}{a\left( t' \right)}

gives

\frac{dR}{dt} = \frac{dR_A}{dt} + 1 + \frac{da}{dt}\int^t_{t_i} \frac{dt'}{a\left( t' \right)},

which guarantees that the speed of the laser pulse is both great than c = 1, and greater than the recession speed dR_A/dt of galaxy A, even when dR_A/dt > c! The speed of light is strictly great than the speed of light. Here, the first "speed of light" means [itex]dR/dt[/tex] for light, while the second "speed of light" means c.

[JesseM]

which guarantees that the speed of the laser pulse is both great than c = 1, and greater than the recession speed [itex]dR_A/dt[/tex] of galaxy A, even when [itex]dR_A/dt > c[/tex]! The speed of light is strictly great than the speed of light. Here, the first "speed of light" means [itex]dR/dt[/tex] for light, while the second "speed of light" means c.
My understanding is that if we're defining "speed" as change in proper distance over cosmological time, then the speed of a light ray at any point is just the recession speed at that point (for an object at rest relative to the Hubble flow) ±c (assuming it's moving in a radial direction), as discussed in the third paragraph here (http://en.wikipedia.org/wiki/Comoving_distance#Uses_of_the_proper_distance) (with a reference to p. 19 of this paper (http://arxiv.org/abs/astro-ph/0310808)).

[bcrowell]

Cool, looks like we've reached a consensus on the question as originally posed :-)

The following is confusing me now, though:

George gives this result:

\frac{dR}{dt} = \frac{dR_A}{dt} + 1 + \frac{da}{dt}\int^t_{t_i} \frac{dt'}{a\left( t' \right)},


But JesseM says:
My understanding is that if we're defining "speed" as change in proper distance over cosmological time, then the speed of a light ray at any point is just the recession speed at that point (for an object at rest relative to the Hubble flow) ±c (assuming it's moving in a radial direction), as discussed in the third paragraph here (http://en.wikipedia.org/wiki/Comoving_distance#Uses_of_the_proper_distance) (with a reference to p. 19 of this paper (http://arxiv.org/abs/astro-ph/0310808)).

Superficially, it seems like JesseM (referring back to the Davis and Lineweaver paper) is saying that \frac{dR}{dt} = \frac{dR_A}{dt} + 1, which would contradict George's equation at t>t_i. But most likely this is my own mistake because I haven't carefully analyzed all the definitions, all the steps in the derivations, etc.

[George Jones]

I probably made a mistake, but I don't think that I'll have a chance to take a serious look at this until tomorrow.

[George Jones]

I don't think that there is a contradiction between my expression and the references that JesseM cites. From the Wikipedia reference,
v_{pec} is the "peculiar velocity" measured by local observers

This means that the light pulse is local to galaxy A, i.e., that the last term in my expression is zero because t = t_i = t_A. At some time t > t_A, the light pulse will have separated from galaxy A, and will be coincident with some other galaxy, B say, and the speed of light pulse will be


\frac{dR}{dt} = \frac{dR_B}{dt} + 1,


but this doesn't contradict my earlier expression.

[JesseM]

I don't think that there is a contradiction between my expression and the references that JesseM cites. From the Wikipedia reference,


This means that the light pulse is local to galaxy A, i.e., that the last term in my expression is zero because t = t_i = t_A. At some time t > t_A, the light pulse will have separated from galaxy A, and will be coincident with some other galaxy, B say, and the speed of light pulse will be


\frac{dR}{dt} = \frac{dR_B}{dt} + 1,


but this doesn't contradict my earlier expression.
But is your expression saying that there is a unique galaxy B such that this is true, so it's not true that when light was leaving A \frac{dR}{dt} = \frac{dR_A}{dt} + 1, and also not true that if long after passing B then light then passes another galaxy C, \frac{dR}{dt} = \frac{dR_C}{dt} + 1? I think the expression in the Lineweaver paper was meant to be applicable at all times.

[George Jones]

But is your expression saying that there is a unique galaxy B such that this is true, so it's not true that when light was leaving A \frac{dR}{dt} = \frac{dR_A}{dt} + 1, and also not true that if long after passing B then light then passes another galaxy C, \frac{dR}{dt} = \frac{dR_C}{dt} + 1? I think the expression in the Lineweaver paper was meant to be applicable at all times.

I am not sure if I understand what you're asking, so my answers might not make sense.

It is true that just as light is leaving A that

\frac{dR}{dt} = \frac{dR_A}{dt} + 1

and it is true that, if long after passing galaxy B, the pulse is coincident with C then

\frac{dR}{dt} = \frac{dR_C}{dt} + 1.

One way to see this is to write an equivalent form of my expression,

\frac{dR}{dt} = \frac{dR_A}{dt} + 1 + H \left( R - R_A \right) ,

but I've run out of steam, so I won't expand on this until tomorrow.

[JesseM]

One way to see this is to write an equivalent form of my expression,

\frac{dR}{dt} = \frac{dR_A}{dt} + 1 + H \left( R - R_A \right) ,

but I've run out of steam, so I won't expand on this until tomorrow.
Ah, so for example it would work out that H(R_C - R_A) = \frac{dR_C}{dt} - \frac{dR_A}{dt}?

[George Jones]

Ah, so for example it would work out that H(R_C - R_A) = \frac{dR_C}{dt} - \frac{dR_A}{dt}?

Yes.

[George Jones]

A third simple case is a universe that has vacuum and matter components, but that has no radiation component. This case is interesting because it is a fairly good approximation to our universe. In this case (with conveniently chosen scaling)
a \left(t\right) = \sinh^{\frac{2}{3}} \left(t\right).
The proper size of the observable universe blows up as t approaches infinity, but the comoving size of the observable universe (assuming that our comoving coordinate is zero),
\lim_{t \rightarrow \infty} \int_0^t \frac{dt'}{a \left(t'\right)},
approaches a finite limit. Any galaxy that has comoving coordinate beyond this limit will never be seen by us, i.e., our universe has a cosmological event horizon. For the above scale factor, this limit is approximately 4.2