What is the relationship between the Hubble radius and the age of the universe?

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In summary: Qervature is.The growth rate correction is just H + Qervature, and when you do that, the strip-down Friedman equation simplifies to:H2 = Kρ + Qervature
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Cosmology is ultimately based on "the law of gravity(geometry)" i.e. the 1915 GR equation whose lefthand side is CURVATURE quantities and whose righthand is MATTER-type terms.
Cosmology uses a 1923 simplified version of GR equation called Friedman equation.

Friedman can be simpler because it assumes uniformity at any given epoch: Uniform distribution of matter (on the RHS) and uniform curvature and expansion rate (on the LHS). It is applicable at large scale because on sufficiently large scale, in any given time-slice, things do seem to be approximately even.

Let's see if we can present an intuitive version of the Friedman equation that might go part way towards satisfying non-mathy newcomers who sincerely want to understand something about modern cosmology. It might take several tries before we get the notation right.

Intuitively in POS curvature lines that start parallel will CONVERGE, like on Earth surface.
And in NEG curvature lines will DIVERGE. And that is what expansion is: negative spacetime curvature---the spreading apart of stuff that's sitting still. (if you need to, ask somebody about the Background of ancient light--and what "at CMB rest" means).

The most intuitive measure of SPATIAL curvature is the RADIUS OF CURVATURE. Call it L for the moment. A slight problem with L, as a measure, is that "flatness" or zero curvature corresponds to infinitely long radius of curvature, so I'm going to take the RECIPROCAL 1/L.

An example might be a finite highly curved universe with a RoC of a billion light years. L= 1 Gly.
So locally it looks like ordinary 3D space we are used to, but the farthest apart any pair of points can be is 3.14 Gly and if you head off straight in any direction you come back to start after 6.28 Gly.

Now I want to cancel out a factor of c,
L changes to L/c,
Gly (a billion lightyears) changes to Gy (a billion years)
so that now instead of reciprocal length, to measure curvature, I have reciprocal time
I want space curvature to be in the same terms as a fractional growth rate (such and such percentage per unit of time).

The measure of curvature is going to be the reciprocal of the TIME-IZED radius of curvature. In that example of the finite highly curved universe, instead of L=1 Gly I'm going to express the curvature by a quantity (let's call it Q for Qervature) which is 1/(L/c) = 1/Gy = 0.001/My =
0.1% per million years.
If you could travel at speed of light then in a million years you could travel 0.1% (a tenth of a percent) of the length of the radius of curvature of that universe.

For most of what we do we won't need any of this spatial curvature stuff because as far as we know the U is for all practical purposes FAPP flat. Or nearly so, anyway. So we can mostly take Q to be zero.

If spatial curvature and the "intrinsic constant" space-time curvature are small enough to be neglected then the Friedman equation becomes extremely simple.
So I'll talk about that next.
 
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Hi Marcus, since this is for laymen in cosmology (like me), I would suggest expanding or explaining all abbreviations (at least once), i.e.:

GR
LHS
RHS
POS
NEG
CMB
Gly
RoC
the U
FAPP

I know what these abbreviations mean, but probably not every layman will understand them :smile:.
All the best,
Dennis
 
  • #3
This thread could simply be a mistake. We'll see.
I have a feeling that some layish newcomers get frustrated because they are told that the results in cosmology are model dependent (that is derived using the Friedman equation with current best -fit parameters plugged in) but they never get shown the model.

I'll say what Friedman looks like WITHOUT the small curvature corrections on the LHS. Little Greek rho ρ (which looks too much like Roman p) is traditional for either matter density grams per cubic meter, or for energy density joules per cubic meter. I'll take rho to be the overall average energy density( including energy equiv of dark and ordinary matter, excluding "dark energy" which is subsumed in the intrinsic cosmo curvature constant or intrinsic minimal growth rate.)

H is the growth rate, ρ is energy density, K is a constant that converts energy densities into squared growth rates. Here's the strip-down Friedman:

H2 = Kρ

The current estimated values of H and ρ are H(now) = 1/144% per My and ρ(now) = 0.24 nanopascal = 0.24 nanojoule per cubic meter.

The constant K is Albert Einstein's fault, it is basically taken out of the original 1915 GR equation and contains the famous "ate pig by three" or 8 pi G/3
K = (8π G)/(3c2)

We have to have the c2 because of choosing to express e density in energy terms instead of mass, but otherwise it is just "ate pig over three" or words to that effect.

The interesting thing comes when you correct for the fact that there is an intrinsic longterm growth rate that the universe is SETTLING DOWN TO. Until 1998 most people thought it was zero but then it was found to be about H = 1/173% per My. At least that is the current estimate from last year's Planck mission report.
With that correction:

H2 - H2= Kρ

It is only the difference that has to balance off against the curvature caused by matter.
 
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  • #4
DennisN said:
GR
LHS
RHS
POS
NEG
CMB
Gly
RoC
the U
FAPP
Dennis, good idea! Feel free to add explanations of other things that you think of. Or try alternate versions of Friedman equation to see if we can get still more intuitive.
General Relativity
Lefthandside
Righthandside
positive
negative
cosmic microwave background
gigalightyear (billion lightyears)
Radius of Curvature
the Universe
For All Practical Purposes
===

I still haven't put in the spatial curvature caused by MATTER! It is a small correction to the LHS which may in fact be ZERO. We don't know yet. But we know it is so small that it can't affect the calculations very much. If you go looking up one of the WMAP reports or Planck mission reports you will see an error bar for Ωtotal something like this:
0.995 < Ω < 1.01
the way to interpret is to say if Ω = 1 exactly then overall spatial curvature zero
but Ω could be as high as 1.01 and that means positive curvature!

In that case the current spatial RADIUS OF CURVATURE is given by the current Hubble radius divided by the square root of 0.01. It may seem a little complicated but those are the conventions. Now there's a short cut that let's us SKIP that rigamarole described earlier for "time-ifying" and taking the reciprocal, to get something in terms of a reciprocal time or growth rate that we can add to the LHS of the equation. THE SHORTCUT INVOLVES THE SCALE FACTOR. Remember that term "Q"? I'll explain in more detail in the next post, but here is the Friedman equation with both correction terms:

H2 - H2 + Q2 = Kρ

Remember that if we just consider current values as an example the current distance growth rate is H = 1/144% per million years, the longterm constant growth rate H is 1/173% per million years. K is the "ate pig over three" constant. ρ is currently about 0.24 nanojoule per cubic meter. What I didn't explain yet is Q, and if we assume a high Ω = 1.01 the current Q is going to be 1/1440 % per million years. Only a TENTH of the current H. The square root of 0.01 is a tenth. You can see it's quite small. And it gets smaller as expansion continues.
 
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  • #5
marcus said:
This thread could simply be a mistake. We'll see.
Wouldn't it be more appropriate to put something this lengthy in a blog post?
 
  • #6
Actually I think not, Bill! I'm particularly motivated by the problems raised by a newcomer currently posting quite a lot, who seems to want some contact with the standard cosmological model---to see, as a layperson, how the numbers are gotten. I'm thinking of something that might eventually be an FAQ or a sticky (like the balloon-analogy sticky). So I want this to develop in plain sight to see if we can get
contribution/crit from experts like yourself and
reactions from lay newcomers who don't know the standard presentation of the Friedman equation or find it off-putting.
I also want this, if possible, to work together with Jorrie's "Lightcone" calculator. So to speak to "look under the hood" of that version of the standard cosmic model. It uses H0 and H (or their reciprocals) as primary parameters.
We have a sticky thread on that table-making cosmic history calculator. You may have looked at it already!
 
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  • #7
This spatial curvature Q term is so small it's tempting to take it as zero, which would be consistent with the kind of Ω confidence interval we've been seeing, like:
0.995<Ω<1.01
That certainly contains the case Ω=1 which is the zero spatial curvature case.
If I take it to be zero I get to use the simpler version of the Friedman equation, namely

H2 - H2= Kρ

Let's solve that for the present-day average energy density. We use the two parameters H0 the present-day value of H, and the constant H:

If I paste this into Google,
3*c^2/(8 pi G)*((1/14400)^2 - (1/17300)^2)/ (million years)^2
I get 0.24 nanojoules per cubic meter. That then is the current value of ρ.

The tiny Q correction would not make any significant difference.

Solving the Friedman equation over time gives us the history of the SCALE-FACTOR a(t).
This is the size of a generic distance (between objects at CMB rest) normalized so that its present value a(now) = 1. Jorrie's calculator gives the scale factor as the first column in the tables that it makes.

When you solve the Friedman equation you can at each time calculate the H(t) growth rate and that tells you how fast a(t) is growing, so you advance to the next time point, updating a(t) and H(t) and the density ρ(t) as you go along. this can be done either analytically or numerically step by step using a computer.

Suppose we want to include the Q(t) term. Let's pick the 0.01 case that we looked at earlier--a high value for the spatial curvature but still consistent with current observational measurements of it. The present value Q0 = √0.01 H0 = 0.1 H0
That is why I used the value 1/1440% per million years in the earlier example.
But how does that spatial curvature evolve in time as the scale factor a(t) grows?

Think of an expanding balloon. The curvature falls off as the inverse of the scale:

Q(t) = Q0/a(t) = 0.1 H0/a(t)

At present, a(t) = 1, so it agrees with what we said. In the future when distances are TWICE what they are today and a(t) = 2, then the spatial curvature Q will be HALF its present value and its square (another measure of curvature) will of course be one QUARTER of it value now. Volumes will be 8 times present, so matter density will be 1/8.

So, to review, here is the full equation:
H2 - H2 + Q2 = Kρ

You know the present-day values of all the quantities and the equation shows how they change, allowing projection into the future and back into the past. K is a constant, H is a constant.
The matter density rho falls off as the CUBE of the scale factor a(t) as it grows, because the same amount of matter is being spread over a larger volume. The spatial curvature term Q2 falls off as the SQUARE of the scale factor, as described just now.

The point of the equation is to calculate how the scale factor a(t) itself evolves, together with its growth rate H(t).

This is what gives us our handle on redshift. There is a popular misconception that the redshift indicates the SPEED the emitter's distance is growing at the moment of emission. It does not. The redshift reflects the factor by which distances (including wavelengths themselves) have been stretched while the light has been traveling.
Traditionally written 1+z, this factor by which wavelengths are multiplied is simply the ratio a(time received)/a(time emitted)
 
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  • #8
In another thread a misconception about the relation between current age T0 and Hubble time 1/H0 (the reciprocal of the current expansion RATE) seems to have become problematical and created a conceptual block. I wanted to show the relation between the two GRAPHICALLY, since that may help. If the plot doesn't appear, please click on the thumbnail attached at the bottom of this post. You will see a plot of age T in blue and Hubble time 1/H in red, plotted along the scale factor a (by convention a = 1 at the present time, so you can see where the present-day is on the chart)
attachment.php?attachmentid=70762&d=1403363829.png

You will see that the curves HAVE to cross because the Hubbletime starts up faster and then levels off, whereas the age just keeps going up. They will cross when the expansion age is about 15 billion years, as you can see. that is a little over a billion years from now, in the future.

For more precision you can see the same thing in the table of numbers that the chart plots:

[tex]{\small\begin{array}{|c|c|c|c|c|c|}\hline R_{0} (Gly) & R_{\infty} (Gly) & S_{eq} & H_{0} & \Omega_\Lambda & \Omega_m\\ \hline 14.4&17.3&3400&67.9&0.693&0.307\\ \hline \end{array}}[/tex] [tex]{\small\begin{array}{|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|} \hline a=1/S&T (Gy)&R (Gly) \\ \hline 0.025&0.0674&0.1021\\ \hline 0.027&0.0758&0.1146\\ \hline 0.029&0.0851&0.1287\\ \hline 0.031&0.0956&0.1445\\ \hline 0.034&0.1074&0.1622\\ \hline 0.037&0.1206&0.1821\\ \hline 0.040&0.1354&0.2044\\ \hline 0.043&0.1520&0.2294\\ \hline 0.046&0.1707&0.2575\\ \hline 0.050&0.1917&0.2890\\ \hline 0.054&0.2153&0.3244\\ \hline 0.058&0.2417&0.3641\\ \hline 0.063&0.2713&0.4086\\ \hline 0.068&0.3046&0.4586\\ \hline 0.073&0.3420&0.5146\\ \hline 0.079&0.3839&0.5775\\ \hline 0.085&0.4309&0.6481\\ \hline 0.092&0.4837&0.7272\\ \hline 0.100&0.5430&0.8160\\ \hline 0.108&0.6094&0.9155\\ \hline 0.116&0.6840&1.0271\\ \hline 0.126&0.7676&1.1522\\ \hline 0.136&0.8615&1.2924\\ \hline 0.146&0.9667&1.4493\\ \hline 0.158&1.0847&1.6251\\ \hline 0.171&1.2170&1.8217\\ \hline 0.184&1.3653&2.0414\\ \hline 0.199&1.5315&2.2869\\ \hline 0.215&1.7176&2.5607\\ \hline 0.232&1.9259&2.8655\\ \hline 0.251&2.1589&3.2044\\ \hline 0.271&2.4193&3.5802\\ \hline 0.292&2.7102&3.9957\\ \hline 0.316&3.0346&4.4534\\ \hline 0.341&3.3958&4.9553\\ \hline 0.368&3.7974&5.5027\\ \hline 0.398&4.2427&6.0957\\ \hline 0.429&4.7354&6.7332\\ \hline 0.464&5.2788&7.4119\\ \hline 0.501&5.8756&8.1265\\ \hline 0.541&6.5285&8.8694\\ \hline 0.584&7.2393&9.6304\\ \hline 0.631&8.0089&10.3976\\ \hline 0.681&8.8373&11.1573\\ \hline 0.735&9.7233&11.8956\\ \hline 0.794&10.6648&12.5991\\ \hline 0.858&11.6586&13.2564\\ \hline 0.926&12.7008&13.8586\\ \hline 1.000&13.7872&14.3999\\ \hline 1.080&14.9126&14.8782\\ \hline 1.158&15.9670&15.2590\\ \hline 1.242&17.0463&15.5901\\ \hline 1.332&18.1471&15.8751\\ \hline 1.429&19.2663&16.1179\\ \hline 1.532&20.4012&16.3230\\ \hline 1.643&21.5492&16.4953\\ \hline 1.762&22.7082&16.6391\\ \hline 1.890&23.8766&16.7583\\ \hline 2.027&25.0524&16.8570\\ \hline 2.174&26.2345&16.9383\\ \hline 2.331&27.4219&17.0049\\ \hline 2.500&28.6133&17.0597\\ \hline \end{array}}[/tex]
 

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  • #9
In another thread my response to something Bill_K said, included this:
marcus said:
...
I don't believe it makes any different to real world computations (where there is always a limit on precision) whether one assumes Ω exactly = 1, or instead something like 1.000001.

That would correspond to a U which is spatially a 3D sphere. And the 3D spatial slice (a "hypersphere") would currently have a radius of curvature of 14400 billion light years.

In effect, no one could tell it from flat :biggrin: And with that one still has isotropy.
...
Oh, I guess it makes a difference to modeling the early universe. I had forgotten about that.

And Bobie asked how you get that radius of curvature--essentially if you assume Ω = 1.000001 how do you get that the present-day radius of curvature is 14400 Gly.

Earlier in this thread we had a sample calculation showing that if Ω = 1.01 then the current radius of curvature would be 144 Gly. You take the current Hubble radius R = 14.4 Gly and DIVIDE BY THE SQUARE ROOT OF 0.01.
Rcurv = Rhubble/√0.01 = Rhubble/0.1 = 10 Rhubble

It just happens, by historical accident I guess, that observational astronomers report their measurement of spatial curvature in the form of an OMEGA confidence interval. Like with 95% confidence 0.995< Ω <1.01
So if that is the current reported result for Ω, then to get the smallest Rcurve you would take the LARGEST possible Ω, namely 1.01, and you would subtract 1, and get 0.01.
And that 0.01 would be your curvature number, your handle on today's maximum positive spatial curvature.

That is just the convention which they adhere to. They could report a minimum possible radius of curvature corresponding to the upper end of the Ω confidence interval, but they normally do not report it that way.

So in the other post I took the case that the upper end of the confidence interval was Ω = 1.000001 and the positive curvature number was 0.000001. So then the square root was 0.001 which is a thousandth. And you multiply the Hubble radius 14.4 Gly by a thousand to get 14400 Gly. This is to answer Bobie's question.
 
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  • #10
I think now it might make our version of the Friedman equation used in this thread more intuitive to rearrange it slightly. Here's what I said earlier
H2 - H2 + Q2 = Kρ

Let's see how it works to write it this way:
H2 = H2 + Kρ - Q2

This arrangement might be more natural to talk about. It is an equation to help us find how H(t) the expansion rate has changed over time (and consequently also to track the growing scale factor a(t) over time.) So since H is the main thing we are interested in, and want to calculate, we put it on the LHS.

On the RHS the first thing is the constant expansion rate H that H(t) is trending towards. In the long run that is the most important thing so it's reasonable to put it first. The other two terms are small corrections corresponding to the positive curvature effect of the average density of matter and the amount deducted from that if space is actually a finite positive curved 3D sphere. The Q term is zero if space is in fact infinite, otherwise Q is reciprocal to the hypersphere radius of curvature and shrinks towards zero as the latter increases.

Here's some earlier explanation:
marcus said:
This spatial curvature Q term is so small it's tempting to take it as zero, which would be consistent with the kind of Ω confidence interval we've been seeing, like:
0.995<Ω<1.01
That certainly contains the case Ω=1 which is the zero spatial curvature case.
If I take it to be zero I get to use the simpler version of the Friedman equation, namely

H2 - H2= Kρ

Let's solve that for the present-day average energy density. We use the two parameters H0 the present-day value of H, and the constant H:

If I paste this into Google,
3*c^2/(8 pi G)*((1/14400)^2 - (1/17300)^2)/ (million years)^2
I get 0.24 nanojoules per cubic meter. That then is the current value of ρ.

The tiny Q correction would not make any significant difference.

Solving the Friedman equation over time gives us the history of the SCALE-FACTOR a(t).
This is the size of a generic distance (between objects at CMB rest) normalized so that its present value a(now) = 1. Jorrie's calculator gives the scale factor as the first column in the tables that it makes.

When you solve the Friedman equation you can at each time calculate the H(t) growth rate and that tells you how fast a(t) is growing, so you advance to the next time point, updating a(t) and H(t) and the density ρ(t) as you go along. this can be done either analytically or numerically step by step using a computer.

Suppose we want to include the Q(t) term. Let's pick the 0.01 case that we looked at earlier--a high value for the spatial curvature but still consistent with current observational measurements of it. The present value Q0 = √0.01 H0 = 0.1 H0
That is why I used the value 1/1440% per million years in the earlier example.
But how does that spatial curvature evolve in time as the scale factor a(t) grows?

Think of an expanding balloon. The curvature falls off as the inverse of the scale:

Q(t) = Q0/a(t) = 0.1 H0/a(t)

At present, a(t) = 1, so it agrees with what we said. In the future when distances are TWICE what they are today and a(t) = 2, then the spatial curvature Q will be HALF its present value and its square (another measure of curvature) will of course be one QUARTER of it value now. Volumes will be 8 times present, so matter density will be 1/8.

So, to review, here is the full equation:
H2 - H2 + Q2 = Kρ

You know the present-day values of all the quantities and the equation shows how they change, allowing projection into the future and back into the past. K is a constant, H is a constant.
The matter density rho falls off as the CUBE of the scale factor a(t) as it grows, because the same amount of matter is being spread over a larger volume. The spatial curvature term Q2 falls off as the SQUARE of the scale factor, as described just now.

The point of the equation is to calculate how the scale factor a(t) itself evolves, together with its growth rate H(t).

This is what gives us our handle on redshift. There is a popular misconception that the redshift indicates the SPEED the emitter's distance is growing at the moment of emission. It does not. The redshift reflects the factor by which distances (including wavelengths themselves) have been stretched while the light has been traveling.
Traditionally written 1+z, this factor by which wavelengths are multiplied is simply the ratio a(time received)/a(time emitted)
 
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  • #11
marcus said:
You will see a plot of age T in blue and Hubble time 1/H in red,
You will see that the curves HAVE to cross
Hi marcus, great thread, ain't no mistake!
I'm answering this bit here, as this is rather OT in the original thread,
an interesting graph, it would be great if you could plot also the curve of visible U.
I had enquired only after the blue line: T, I'd like how you derive the current T(now) from Friedmann equation and experimental data
 
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  • #12
Hi Bobie, I'm delighted you like the thread and hope it proves useful! Keep in mind I'm not an expert, just a retired person interested in cosmology. I watch the game from the sidelines with limited ability to explain or answer your questions. Some things I can help with, though.
bobie said:
... it would be great if you could plot also the curve of visible U...
The conventional term for the radius of the observable universe is "particle horizon". The notation used in Jorrie's lightcone calculator is Dpar. Here I've asked the calculator to plot Dpar as a function of time T (since start of expansion). So in the previous graph the horizontal axis was the SCALE FACTOR "a", a generic measure of the size of distances normalized so that a=1 at present, and the blue curve was T.

Now the horizontal axis will be T, in billions of years, and the red curve will be the radius of the observable region in the model U. (The model U is actually or for all practical purposes infinite, but we see a larger and larger part of it as time goes on and light comes in from ever more distant matter. Dpar increases for two reasons---we hear from more matter AND the size of distances to that matter grows as time goes on.)
attachment.php?attachmentid=70943&stc=1&d=1403970679.png
 

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  • #13
Notice that in both plots time goes out to 28.6 billion years. This is an arbitrary cutoff used by the Lightcone plotter and it corresponds to a time when distances will be two and a half times present size (the scale factor a = 2.5). The user can easily change that if he or she wishes, by typing in a different range.

But it is actually pretty convenient---28.6 is roughly TWICE current age, so when you look at plot the present day is roughly in the middle and you have a balance of past and future curve.

The Hubble radius is red in the previous plot and blue in this one. In both cases you can see it leveling out at around 17 Gly. You understand this but I'll repeat it in case a newcomer might be reading: the Hubble radius is NOT the radius of the observable portion of the model U, or anything like that.

The Hubble radius is a reciprocal measure of the percentage growth rate of distance size. Its time version (in Gy rather than Gly, so leveling out at around 17 billion years) is the RECIPROCAL of the growth rate H. A large Hubble radius at some given time in history corresponds to a slow fractional growth rate (a small percentage growth per million years). A small R corresponds to a rapid growth rate in the size of distances.

Also one should keep in mind that the distances we are talking about growing are those between pairs of objects *each of which is stationary with respect to the background of ancient light*. IOW each marker should be at least approximately at CMB rest.
 
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  • #14
marcus said:
Keep in mind I'm not an expert, ...with limited ability to explain or answer your questions. Some things I can help with, though.
Hi marcus, you are one ot the greatest experts in this forum, you are friendly and not patronizing, (and you appreciate symmetry), a precious friend: just the ideal tutor!
I realize I am too curious and sometimes cheeky, but I ask you only one thing: when you do not know, do not ignore my questions, just say it and I'll move on. When you've had enough, just tell me.

I asked if you know:
- how they determine T, not Hubble time (which is 1/H),
besides that,
- why is the H radius (R) going to stop at 17 Gly?
- as the ratio between Dpar and R grows, will Vnow get greater than 3.15 C, or that's a limit?

Could you please answer also my questions in the Ω=1 thread?
Thanks
 
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  • #15
bobie said:
Hi marcus, you are one ot the greatest experts in this forum, you are friendly and not patronizing, (and you appreciate symmetry), a precious friend: just the ideal tutor!
I realize I am too curious and sometimes cheeky, but I ask you only one thing: when you do not know, do not ignore my questions, just say it and I'll move on. When you've had enough, just tell me.

I asked if you know:
- how they determine T, not Hubble time (which is 1/H),
besides that,
- why is the H radius (R) going to stop at 17 Gly?
- as the ratio between Dpar and R grows, will Vnow get greater than 3.15 C, or that's a limit?

Could you please answer my questions in the Ω=1 thread?
Thanks

Bobie, you have me laughing out loud, but thanks for the kind words. I am no expert! (rather a retired guy who is a fan of cosmology and quantum GR research.) I simply don't know a lot of what you ask, or it would be a lot of work to copy and write down the formulas here. At one point a couple of years ago I think it was, Jorrie and I went over the formulas of how you calculate T from the scale factor, or vice versa. But I think his calculator actually does it numerically, step by step. It includes some fine points I never bothered to learn in detail.
He can tell you. He's also totally unpatronizing (if that is an important consideration).

Your questions here seem to be about the Lightcone calculator and more generally about the standard Friedman equation model. Always remember that in a mathematical science the model is not the thing. The model is a tool to understand predict explain, but it is always approximate. It always has limits to applicability. Making intelligent use of the model requires having a sense of the limits of its applicability. The Friedman equation fails around the start of expansion. The actual universe could well have had a bounce there (essentially because quantum Nature resists infinite compression and infinite density and does not understand infinite curvature). So researchers are in process of FIXING the Friedman equation model so that it will not break down there but will continue on back in time to a prior contracting phase. And other researchers have other ideas of how to modify the model to eliminate that problem.

the model is always a provisional description, one always wants more and different kinds of data and to try for an even better fit to whatever data is available.

In 1998 they had collected some new data about supernovae and they found they got an even better fit if they introduced a LOWER LIMIT ON HOW SMALL H COULD GET.

Instead of H leveling out at zero, as the model had been doing, H would continue declining but level out at around 1/170 of a percent per million years.

Currently the best estimate of this constant is 1/173 of a percent per million years.

So that is where the 17.3 billion years comes from, that you asked about. The Hubble RADIUS is just the length version of the Hubble time. A lower limit on H corresponds to an upper limit on its reciprocal. I think you understand. Did I say that clearly enough?

does that answer your second question about "why R is going to stop at" 17 Gly or 17.3 Gly or whatever?

The original 1915 GR equation had room for a constant (which I'm told Einstein denoted by Lambda sometime back around then, 1917 maybe) which for many years people thought to be zero. And when you derive Friedman from GR equation this Lambda constant turns out to be a lower limit on how small H can get. So that is an upper limit on how large R can get. It was always latent in the equation. But for a long time people imagined that the lower limit on H was zero and that there was no upper limit on R. And it seems they were wrong, the constant that was always there in the 1915 GR equation is very small but with more data and a better fit to the data we find that it is not EXACTLY zero. I think that's basically the answer to your "why" question.

You had two more questions. I will think about them and try to answer. It would be great if Jorrie or one of the others chimed in and helped.
 
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  • #16
marcus said:
..we see a larger and larger part of it as time goes on and light comes in from ever more distant matter. Dpar increases for two reasons---we hear from more matter AND the size of distances to that matter grows as time goes on.)...
... It would be great if Jorrie or one of the others chimed in and helped.
(Nice of you to reply 'first thing in the morning', marcus, :zzz: I'll comment your last post after all questions have been answered.)

That might be wrong, marcus. I already pointed out the fact the what you call Uv= Dpar is not observable, but observed U (that argument has not been yet refuted).

But you probably do not realize that you have nearly reached rock bottom, and Uv can hardly increase. With 10.3/11.9 redshift (UDFJ) you have crossed the galaxy boundary and are in the epoch of protogalaxies (0.37 Gy), you can have just a little increase in z and you are in the epoch of stars, which probably are too faint to be detected, and then , roughly at <0.2 Gy, you can hardly hear from any matter. Is this correct?
 
  • #17
Bobie, your third question is "will Vnow get greater than 3.15 C, or that's a limit?"

That's actually easy for me to answer, so why don't I do that immediately. You realize that the model parameters are only determined to within a couple of percent. Some people say 69, some say 70, some say 68 etc etc, some people give a 95% confidence range, allowing for variation of a percent or so. That does not change the qualitative conclusions.

Jorrie's calculator uses two main parameters 14.4 and 17.3 which are the reciprocals of the present H and the future lower limit H on H that I was talking about, what the model expects H to level out to as it declines more and more slowly.

with those numbers I can say that the cosmic event horizon is now 16.472 Gly. You can see that in the table.
If a galaxy (at CMB rest, not moving relative to Background) is NEARER than that then a message that we send TODAY can eventually reach them. If it's farther then a message sent today CAN NOT reach them.

[tex]{\scriptsize\begin{array}{|c|c|c|c|c|c|}\hline R_{0} (Gly) & R_{\infty} (Gly) & S_{eq} & H_{0} & \Omega_\Lambda & \Omega_m\\ \hline 14.4&17.3&3400&67.9&0.693&0.307\\ \hline \end{array}}[/tex] [tex]{\scriptsize\begin{array}{|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|} \hline a=1/S&S&T (Gy)&R (Gly)&D_{now} (Gly)&D_{then}(Gly)&D_{hor}(Gly)&V_{now} (c)&V_{then} (c) \\ \hline 0.001&1090.000&0.0004&0.0006&45.332&0.042&0.057&3.15&66.18\\ \hline 0.003&339.773&0.0025&0.0040&44.184&0.130&0.179&3.07&32.87\\ \hline 0.009&105.913&0.0153&0.0235&42.012&0.397&0.552&2.92&16.90\\ \hline 0.030&33.015&0.0902&0.1363&38.052&1.153&1.652&2.64&8.45\\ \hline 0.097&10.291&0.5223&0.7851&30.918&3.004&4.606&2.15&3.83\\ \hline 0.312&3.208&2.9777&4.3736&18.248&5.688&10.827&1.27&1.30\\ \hline 1.000&1.000&13.7872&14.3999&0.000&0.000&16.472&0.00&0.00\\ \hline 3.208&0.312&32.8849&17.1849&11.118&35.666&17.225&0.77&2.08\\ \hline 7.580&0.132&47.7251&17.2911&14.219&107.786&17.291&0.99&6.23\\ \hline 17.911&0.056&62.5981&17.2993&15.536&278.256&17.299&1.08&16.08\\ \hline 42.321&0.024&77.4737&17.2998&16.093&681.061&17.300&1.12&39.37\\ \hline 100.000&0.010&92.3494&17.2999&16.328&1632.838&17.300&1.13&94.38\\ \hline \end{array}}[/tex]
A question you could ask is this: consider a galaxy which is not moving relative to Background and is exactly at the horizon, i.e. exactly 16.472 Gly from us. How fast is the distance to that galaxy growing?

Well of course to find the speed (as of today) you just divide the distance today by today's value of R. So we just do the simple division 16.472/14.4 and we get about 1.144 c.

That is the future upper limit that you asked about.

What that means is that if TODAY the distance to a stationary galaxy is increasing faster than about 1.14 c then *we can't contact them* with a message we send now. It is too late. We can see them and they can see us--we know about each other's existence--but it is too late to send any new information.

and if the distance to an unmoving galaxy is increasing slower than 1.14 c then that means they are nearer to us than 16.472 Gly and *we can* send a message today which will eventually reach them although it might take a very long time.

For a concrete example, look at the last row of the table. It talks about a galaxy which if we send a message today it won't get there until year 92.35 billion! It will get there but by that time distances between unmoving points will have grown to 100 times their present size! That is WAY in the future. The table tells you the PRESENT speed the distance to that galaxy is growing. That is an example of what the Vnow quantity means. (the recession speed today of a partner we communicate with today, either as receiver or sender depending on whether it's in the partner is in past or future). So in this case, looking at the last row of the table, the present distances is growing at 1.13 c, almost at the 1.14 limit! So the message we sent today won't get there until year 92 billion, a long time from now!

As you can see, Vnow is not anything very earth-shaking. It is just the recession speed corresponding to a particular distance which varies row by row in the table depending on whatever distance the communication partner happens to be at. the 3.15 c happens in the first row of the table because that talks about the time when the ancient light of the CMB originated and distances were 1/1090 of their present size. The 3.15 c is the current recession speed of the ancient hot gas whose glow we are now finally receiving today.
 
  • #18
marcus said:
As you can see, Vnow is not anything very earth-shaking. It is just the recession speed corresponding to a particular distance which varies row by row in the table depending on whatever distance the communication partner happens to be at. the 3.15 c happens in the first row of the table because that talks about the time when the ancient light of the CMB originated and distances were 1/1090 of their present size. The 3.15 c is the current recession speed of the ancient hot gas whose glow we are now finally receiving today.
As we push towards t0, Vnow limits around 3.2, i.e. if we could observe all the way to the BB (say via gravitational waves), the area where the waves originated will presently be 46 billion years away from us, receding at proper speed of about 3.2c. The exact value is obviously sensitive to the actual parameters, which may vary from analysis to analysis.

For Bobie's question about the determination of T, a starting point may be the equations for T(S) in the "Advanced User" section of the LightCone Wiki: http://cosmocalc.wikidot.com/advanced-user. It depends pretty straight-forwardly on the parameters derived from observations. How the parameters are derived from the observations is a complicated story...

-J
 
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  • #19
marcus said:
Instead of H leveling out at zero, as the model had been doing, H would continue declining but level out at around 1/170 of a percent per million years.Currently the best estimate of this constant is 1/173 of a percent per million years. So that is where the 17.3 billion years comes from, that you asked about.
does that answer your second question ...?
I need to check some data, first, please confirm or correct what is wrong:

PlanckM found that 1Mpc espands 67.8 Km/s, that means that 1 cm increases by H = 2.2*10-18 cm/s, 1/H = TH = 4.54*1017 s = 14.42 Gy

Hubble radius RH = TH*C = 14.4 Gly is the distance from here covered by a point moving at speed (uniform or average) C for TH years.
This implies that all energy/matter within the Hubble sphere now was enclosed in ≈1 cm at BB, right?
Age of U T0 BB is 13.78Gy, this means that the average speed of the points at RH has been RH/ TBB≈1.045 C, right?

When TBB will be 28.8 Gly the points at distance R should expand, they say, at an increased and increasing rate, so presumably, the distance should be more than double, H should be less than half (≈1*10-18)

nevertheless RH will be 17.3 Gly ,TH will be 17.3 Gy , increased by 2.9 Gy/Gly in 14.4 Gy? what has really been the recession speed (0.2 )?

What H will PM find in 14.4 Gy from now?
Can the U expand if the RH stands still?
 
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  • #20
bobie said:
When TBB will be 28.8 Gly the points at distance R should expand, they say, at an increased and increasing rate, so presumably, the distance should be more than double, H should be less than half (≈1*10-18)

nevertheless RH will be 17.3 Gly ,TH will be 17.3 Gy , increased by 2.9 Gy/Gly in 14.4 Gy? what has really been the recession speed (0.2 )?

What H will PM find in 14.4 Gy from now?
Can the U expand if the RH stands still?


Yes, to the last. RH is not a way of designating a volume of space or a quantity of matter. It is a way of describing a percentage RATE OF DISTANCE INCREASE. When RH has stabilized at 17.3 Gly, then distances (between non-moving objects) will be growing EXPONENTIALLY at a constant percentage growth rate of 1/173 of one percent per million years.

Can the U expand if RH (which is a rate designation) is constant? Yes! Exponentially. Like money in the bank at a fixed percentage rate of interest.

Some people make a false connection in their minds between TH and the "age of the universe" and between RH and "radius of observable region". This is like a short circuit in some electrical wiring in your house. It can burn your house down. :smile: Getting rid of that wrong connection is the first step in any understanding. It sounds like you are on the verge of realizing this! I am glad, if you are...
 
  • #21
Bobie, you ask what H will be at a time in the future, namely 14.4 Gy from now. I can't tell you precisely, of course, but I can easily give you a rough idea. I just click on the "Lightcone" link in my signature and in the Slower box, where it says 0.01 (which takes us too far into the future), I type a guess of 0.4. And I press calculate:
[tex]{\scriptsize\begin{array}{|c|c|c|c|c|c|}\hline R_{0} (Gly) & R_{\infty} (Gly) & S_{eq} & H_{0} & \Omega_\Lambda & \Omega_m\\ \hline 14.4&17.3&3400&67.9&0.693&0.307\\ \hline \end{array}}[/tex] [tex]{\scriptsize\begin{array}{|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|} \hline a=1/S&S&T (Gy)&R (Gly)&D_{now} (Gly)&D_{then}(Gly)&D_{hor}(Gly)&V_{now} (c)&V_{then} (c) \\ \hline 0.001&1090.000&0.0004&0.0006&45.332&0.042&0.057&3.15&66.18\\ \hline 0.002&501.112&0.0013&0.0022&44.654&0.089&0.122&3.10&41.20\\ \hline 0.004&230.379&0.0046&0.0072&43.602&0.189&0.261&3.03&26.32\\ \hline 0.009&105.913&0.0153&0.0235&42.012&0.397&0.552&2.92&16.90\\ \hline 0.021&48.692&0.0501&0.0759&39.639&0.814&1.153&2.75&10.72\\ \hline 0.045&22.386&0.1621&0.2445&36.120&1.614&2.350&2.51&6.60\\ \hline 0.097&10.291&0.5223&0.7851&30.918&3.004&4.606&2.15&3.83\\ \hline 0.211&4.731&1.6741&2.4969&23.266&4.918&8.402&1.62&1.97\\ \hline 0.460&2.175&5.2154&7.3341&12.398&5.700&13.279&0.86&0.78\\ \hline 1.000&1.000&13.7872&14.3999&0.000&0.000&16.472&0.00&0.00\\ \hline 2.175&0.460&26.2467&16.9390&8.590&18.685&17.177&0.60&1.10\\ \hline 2.500&0.400&28.6133&17.0597&9.606&24.014&17.203&0.67&1.41\\ \hline \end{array}}[/tex]

It says that around year 28 billion, when distances between pairs of non-moving points have grown to 2.5 times their present size, the RH will be 17.0597 Gly.

Let's call that 17.06. You understand that RH is just a way of talking about the Hubble constant (a constantly changing percentage growth rate). To say it will be 17.06 billion light years at that future time is just a CODE way of saying H will be 1/170.6 of one percent per million years.

You might want to see that expressed in conventional units "km/s per Mpc". In that case notice that 1/170.6 of one percent is the fraction 1/17060
so put this into the google window:
"(1/17060 per million years) in km/s per Mpc"
If you paste that in without the quotes you should get around 57.3 km/s per Mpc.

That 57.3 km/s per Mpc is what the percentage growth rate of 1/170.6 % per million years translates into. Google knows how to convert units, and how to convert physical quantities in one set of units into the same quantity in other units, so it can save a lot of trouble.

Anyway the answer to your question of what H will be, that far into the future, is around 57 km/s per Mpc.
 
  • #22
marcus said:
Can the U expand if RH (which is a rate designation) is constant? Yes.

If R should grow only from 14.4 to 17.28 (instead of doubling) when TBB doubles , recession speed should suddenly grow by 5, doesn't that mean the sky switches off, no more light ever reaching us?
 
  • #23
marcus said:
Some people make a false connection in their minds between TH and the "age of the universe" and between RH and "radius of observable region".

This Lightcone calculator graph gives some perspective on how different these concepts are.
attachment.php?attachmentid=71015&stc=1&d=1404219766.png


It is easy to see that the relationship between cT and R (Hubble radius) is almost linear, but not quite. R flattens out at 17.3 Gly, while T obviously increases without limit.
D_now is also known as the 'lightcone in comoving coordinates' and D_then as the 'lightcone in proper distance coordinates' in e.g. Davis (link in my sig). Both represent, in their respective units, the distance against time of a photon/graviton coming from the limit of the theoretically observable universe (emitted shortly after the BB).
 

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FAQ: What is the relationship between the Hubble radius and the age of the universe?

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