How Does the Hypersine Function Relate to the Expansion of the Universe?

[marcus]

The integrands to give comparable distance answers need to have conversion factors to get all the outputs in the same unit. That is what the
0.83coth(1.5*0.6) out front is doing---it gets all the distances in lightzeits. In the case of t=0.8 the factor .83coth(1.5* .8) is just equal to one and has no effect. Here for example is the distance integral where the model parameter is chosen to be 0.6

D_{.6}(a) = .83\coth(1.5* .6)\int_a^1 x^{-2} (\sinh^{-2}(1.5* .6)x^{-3} + 1)^{-1/2} dx

If we want to use numberempire.com or one of the other easy online integrators, then the function to be integrated would be pasted into the integrand box in this format:
.83*coth(1.5* .6)* x^(-2) (sinh(1.5* .6)^(-2)*x^(-3) +1)^(-1/2)
and then to calculate the other cases one would just change the .6 to .7, or to .9...

The limits of integration would of course be the numbers a and 1, for light that comes in today showing an initial scalefactor of a. IOW light that has been stretched by a factor of 1/a.

 

[John Bremner]

My contention is that the accelerating rate of expansion of the universe is an illusion. I don't know how to put this into mathematical or geometric terms but it seems clear to me that if we have (say) three equidistant galaxies - the first being 1 distance from Earth, the second being double that 1 distance, and the third being triple that 1 distance, and we measure the redshift at time now, and give them 10 years to account for the expansion of the universe then measure the redshift again, it would appear that the more distant galaxy were accelerating away from Earth 3 times faster (or whatever) than the closest galaxy. However, if there were merely a constant rate of expansion between us and 1, and 1 and 2, and 2 and 3, we have to sum up the constant expansion rates. because we must add the constant rate of expansion between each to the constant rate of expansion between them. Thus we get the illusion of accelerating expansion the further a galaxy (or star) is away from us. However, if the rate of expansion is constant the sum of the constants gives us the same result as assuming that some unknown force is causing the acceleration, instead of the much more plausible explanation that the force that is causing the expansion is the same everywhere. The accelerating rate of expansion is simply an illusion caused by the failure to add the expansion between objects to the same rate of expansion between more distant objects due to the constant creation of space between objects..

 

[marcus]

Hello JB, I think you are saying something very reasonable which I'll try to paraphrase as follows. The expansion RATE (properly understood as a speed-to-size ratio or a percentage growth rate) means that larger distances grow faster.

So the RATE could be constant, and if we focus on just one distance and keep track of it, its growth speed will increase over time simply because its SIZE increases over time.

The RATE could even be gradually declining, and if the decline was gradual enough we would still see the growth speed increase over time of that one distance we are tracking, because its size is increasing and the speed-to-size ratio, or percentage rate,. is nearly constant (even if slowly declining).

What we are describing is "near exponential growth" with a gradually declining growth rate.

So if I understand you, you are saying that if we think "acceleration" means increasing expansion rate then we are CONFUSED and under an ILLUSION. This is correct. You are not the only one who knows this. More or less any cosmologist or informed reader who follows the professional research literature knows this.

Actually the evidence suggests that the expansion rate has always been declining since very early times and according to standard model is expected to continue declining but more and more gradually so that it levels off at a positive rate.

At a certain point in the past (I don't recall what it is in billions of years but it was 0.36 times 17.3 billion years ago) the decline became gradual enough so that you would have seen acceleration on a distance by distance basis, as we discussed---the "near exponential growth" idea. For much of the time before that the decline in the rate was so drastic that you didn't even see the "illusory" acceleration---watching a single distance and seeing its speed change as it gets larger---its growth speed would actually decline over time even though it was getting larger because the rate was going down so drastically...

You might enjoy learning how to use Jorrie's calculator to plot curves

 

[marcus]

H is the expansion RATE which you can see has always been declining. In early times before time 0.4 very very steeply.
The blue curve (called the "scale factor" is the size of a generic distance normalized to equal 1 at the present. The present is 0.8.

You can see that the blue curve slope decreases until around time 0.44 and then it gradually begins to adopt a increasing slope "near exponetual growth" shape. In this plot, time 1.0 is year 17.3 billion. that turns out to be a convenient unit of time that makes the standard cosmic model equations exceptionally simple and easy to work with.
http://www.einsteins-theory-of-relativity-4engineers.com/LightCone7z/LightConeZ.html
To get curves you just go there and tick the button that says "chart" in the "Display as" options row, and press "calculate". You will get a chart like what I copied here but with 5 curves instead of two. There is a way to deselect the other 3 curves if you want to make it easier to read, but the main thing is to do the first thing of just getting the chart. Lightcone graphic cosmology calculator is user-friendly.

 

[marcus]

The aim of this equivalent version of the standard LambdaCDM model is to have a ONE-STOP VERSION of standard cosmology (simplified by not worrying about the very early radiation-dominated era) and to have it as TRANSPARENT as possible.

You should be able to see how we arrive at the expansion age ("age of the universe" if you think there was no universe before start of expansion : ^).

That's what I mean by transparency: you should see how it is derived by fitting redshift-distance data measured using the various standard candles. Fairly directly. It should be easy.

You should be able to calculate the range of observation (the "particle horizon", the radius of the current observable region). You should be able to calculate for yourself that 46 billion ly. Or 2.7 lightzeit (in terms of our easy-to-use unit of time).
And see WHY that formula works: breaking down the light's trip into small steps each multiplied by how much the step gets expanded between then and now.

You should be able to calculate for yourself today's signal range---the "cosmic event horizon" the distance to the farthest galaxy we could reach by a signal sent today, with no limit on how long it takes to get there.
16.5 billion ly or in our terms 0.95 lightzeit. You may be able to see how the signal range is still increasing and gradually approaching a limit of exactly 1.0 natural distance unit.

It's the same integral, but with different limits: from present (sending) to infinity instead of from zero to the present(arrival).

 

[marcus]

The model really has two parameters but the first one, H_now, is a no-brainer. For nearby galaxies all the growth is effectively taking place at the present rate and a galaxy's redshift is a direct index of its distance growth speed as a fraction of c.
The second parameter is the non-trivial one obtained by fitting curve to data. The data can be redshift-luminosity, redshift-distance, scalefactor-distance. These are equivalent forms of the same information: luminosity is used to gauge distance, redshift z determines scale a = 1/(z+1) at time of emission. In this presentation we imagine the data to be (a, D) scalefactor-distance. and we want a curve D(a) that passes through those data points.

Here are some curves D_.6(a) ...D_1.0(a)
In each case the value of the curve at a is the integral from a to 1 of an integrand which depends on the parameter.
D_{.6}(a) = .83\coth(1.5* .6)\int_a^1 x^{-2} (\sinh^{-2}(1.5* .6)x^{-3} + 1)^{-1/2} dx
D_{.7}(a) = .83\coth(1.5* .7)\int_a^1 x^{-2} (\sinh^{-2}(1.5* .7)x^{-3} + 1)^{-1/2} dx
D_{.8}(a) = .83\coth(1.5* .8)\int_a^1 x^{-2} (\sinh^{-2}(1.5* .8)x^{-3} + 1)^{-1/2} dx
D_{.9}(a) = .83\coth(1.5* .9)\int_a^1 x^{-2} (\sinh^{-2}(1.5* .9)x^{-3} + 1)^{-1/2} dx
D_{1.0}(a) = .83\coth(1.5* 1.0)\int_a^1 x^{-2} (\sinh^{-2}(1.5* 1.0)x^{-3} + 1)^{-1/2} dx

 

[marcus]

I read today that the earliest known flowering plants were from 7 millizeits ago. the best fossil of one of them (Montsechia) was found just recently. It was a freshwater aquatic plant that thrived in what are now mountainous parts of Spain.
http://news.indiana.edu/releases/iu/2015/08/first-flower-angiosperms.shtml
https://www.iu.edu/~images/dams/479311_actual.jpg

Since the present, on our scale, is 797 millizeits, this puts the appearance of the first flowering plants at around 790 mz. this comes between the last two mass extinctions, that of 785 and that of 793 (which did for the non-bird dinosaurs.)
In a previous post I listed the 5 mass extinctions of the 700s.

Geologists have identified 5 major mass extinctions ...[which] occurred in the 700s. (on a millizeit scale, the present 0.797 is 797.)
They had various causes and they occurred in '71, '76, '82, '85, and '93.

The most severe of these five mass extinctions was the extinction of '82, which is called the Permian-Triassic (or P-Tr) extinction. It is almost unbelievable what a large percentage of then-existing species were wiped out.

The extinction of '93 (which eliminated non-bird dinosaurs) was quite mild by comparison. Geologists are changing the name of this one: it used to be called Cretaceous-Tertiary (abbreviated K-T) but now they want to call it Cretaceous-Paleogene abbreviated K-Pg.
https://en.wikipedia.org/wiki/Extinction_event

In chronological order the five major mass extinctions are:

Extinction of '71 Ordovician-Silurian (O-S)

Extinction of '76 Late Devonian (Late-D)

Extinction of '82 Permian-Triassic (P-Tr)

Extinction of '85 Triassic-Jurasic (Tr-J)

Extinction of '93 Cretaceous-Paleogene (K-Pg)

Montsechia vidalii was contemporanious with or earlier than the other earliest known angiosperm (flowering/seed producing plant) Archaefructus sinensis---an ancient freshwater plant that lived in China. Archefructus fossils are also dated around 790 mz.

 

[Buzz Bloom]

you can probably see the place around time 0.44 in our universe's history when distance growth stopped decelerating and gradually began to accelerate.

Hi marcus:

I calculated the value of a for q = 0, and came up with a = 0.606. The equation I used is derived as follows (I use the apostrophe for d/dt):

q =(def) a'' a / a'² = - (H² + H') / H²
q = 0 → H² + H' = 0
H = H₀ ( (1-Ω_m) + Ω_m a^-3 ) ^1/2
H² = H₀² (1-Ω_m + Ω_m a^-3)
H' = H₀ (1/2) (1-Ω_m) + Ω_m a^-3 ) ^-1/2 (-3 Ω_m) a^-4 a'
H' = H₀ (-3Ω_m/2) ((1-Ω_m) + Ω_m a^-3)^-1/2 a^-3 H = (-3Ω_m/2) H₀²
H² + H' = (1-Ω_m) + Ω_m a^-3 - (3/2) Ω_m a^-3 = (1-Ω_m) -(1/2) Ω_m a^-3
q = 0 → a = [(1/2) Ω_m/(1-Ω_m)]^1/3​

The value I used for Ω_m = 0.308

Using the value 0.44 for t, I calculated a = ((1/2) (e^3t/2 - e-^3t/2))^2/3 to get a = 0.631.

Did I misunderstand how to use the hypersine?
Did you use a different value for Ω_m to calculate H_∞?
Can you suggest another reason for the discrepancy?

Regards,
Buzz

 

[marcus]

thanks for the calculation! I have to go out to an appointment and can't give a proper response now, but I'll bet your calculation is right.
My figure of 0.44 could just be approximate and give a=.631 whereas your a = .606 could be more correct (given your assumptions about the inputs, which seem reasonable---don't have time to check right now).
Delighted to see you got into this thread, Buzz.

 

[Buzz Bloom]

Hi marcus:

I did a bit more calculating and came up with the value t = 0.304 to go with the value a = 0.606. Eyeballing the chart on post #2, the inflection point could be at t = 0.3

Regards,
Buzz

 

[marcus]

Eyeballing a curve to find the inflection point can be hard esp if it is nearly linear for a considerable interval. Let's look at Lightcone calculator. It has an option where it tabulates the growth speed of a sample distance. Open the "column definition and selection" menu and look for v_gen or alternatively in the standard notation Lightcone look for a'R₀. Have to go out, back later.
0.583 1.715 0.417533 0.555714 0.287472 0.873192 1.799
0.587 1.704 0.421398 0.559710 0.285593 0.872988 1.787
0.591 1.692 0.425290 0.563708 0.283673 0.872826 1.774
0.595 1.680 0.429211 0.567708 0.281713 0.872704 1.761
0.599 1.668 0.433165 0.571709 0.279705 0.872623 1.749
0.604 1.657 0.437140 0.575712 0.277661 0.872584 1.737
0.608 1.645 0.441144 0.579714 0.275574 0.872587 1.725
0.612 1.634 0.445175 0.583716 0.273446 0.872632 1.713
Have to explain when I get back. It looks like 0.44 is right, and the a = around 0.604
and the minimum speed for this particular distance is around 0.8726

 

[marcus]

Hi Buzz, when I got back I used Lightcone 7z (link in my signature) to make a 20 step table between redshift z = 0.66 and 0.64
or in terms of the stretch factor 1+z between S_upper=1.66 and S_lower=1.64. Those are limits that one can set to narrow the table down to a particular time period. It looks like the minimum growth speed comes at around t = 0.4389
{\scriptsize\begin{array}{|c|c|c|c|c|c|}\hline T_{Ho} (Gy) & T_{H\infty} (Gy) & S_{eq} & H_{0} & \Omega_\Lambda & \Omega_m\\ \hline 14.4&17.3&3400&67.9&0.693&0.307\\ \hline \end{array}} {\scriptsize\begin{array}{|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|} \hline a=1/S&T (zeit)&R (lzeit)&D_{then}(lzeit)&V_{gen}/c&H(zeit^{-1}) \\ \hline 0.602&0.436076&0.574642&0.278210&0.87259076&1.740\\ \hline 0.603&0.436421&0.574992&0.278034&0.87258838&1.739\\ \hline 0.603&0.436772&0.575342&0.277851&0.87258629&1.738\\ \hline 0.604&0.437117&0.575692&0.277674&0.87258455&1.737\\ \hline 0.604&0.437469&0.576042&0.277491&0.87258310&1.736\\ \hline 0.604&0.437820&0.576392&0.277308&0.87258197&1.735\\ \hline 0.605&0.438166&0.576742&0.277130&0.87258118&1.734\\ \hline 0.605&0.438518&0.577092&0.276946&0.87258070&1.733\\ \hline 0.605&0.438864&0.577442&0.276767&0.87258054&1.732\\ \hline 0.606&0.439217&0.577792&0.276583&0.87258070&1.731\\ \hline 0.606&0.439569&0.578142&0.276398&0.87258119&1.730\\ \hline 0.606&0.439916&0.578492&0.276218&0.87258198&1.729\\ \hline 0.607&0.440269&0.578842&0.276033&0.87258311&1.728\\ \hline 0.607&0.440622&0.579192&0.275847&0.87258457&1.727\\ \hline 0.608&0.440970&0.579542&0.275666&0.87258632&1.726\\ \hline 0.608&0.441324&0.579892&0.275480&0.87258843&1.724\\ \hline 0.608&0.441672&0.580242&0.275298&0.87259081&1.723\\ \hline 0.609&0.442026&0.580592&0.275111&0.87259357&1.722\\ \hline 0.609&0.442380&0.580942&0.274924&0.87259664&1.721\\ \hline 0.609&0.442729&0.581292&0.274741&0.87259999&1.720\\ \hline 0.610&0.443083&0.581642&0.274553&0.87260372&1.719\\ \hline \end{array}}

To convert that time into years we can multiply by 17.3
Google calculator says 0.4389*17.3 = 7.59297
7.59 billion years seems to be where the inflection point comes.
when a is about 0.605.
I think you calculated it to be about that.

 

[Jorrie]

Hi marcus:

I did a bit more calculating and came up with the value t = 0.304 to go with the value a = 0.606. Eyeballing the chart on post #2, the inflection point could be at t = 0.3

Hi Buzz, further to what Marcus wrote, Lightcone7z (in my sig. as well) has a very neat graphing utility for you to visualize certain parameters. As an example, to look at the V_generic that Marcus referred to, I would open the calculator and then click 'Open Column def and Selection'. I then select 'none' at the bottom right of that block and tick 'T', 'S' and V_gen, then click 'Chart' in the yellow block above and finally Calculate.

This will produce a broad picture of V_gen, the recession rate history of a generic galaxy that is presently located on our Hubble sphere. S is only needed if we want to 'zoom in', e.g. to find the minimum point. I used the following values for a first zoom:
S_upper=10, S_lower=1 and then under 'Chart Options':
Vert min=0.8, Vert max=1, Hor min=0.2, Hor max=0.8.

It produced this graph:

Following the same method, one can zoom in further, but it may take some trial and error.

 

[Buzz Bloom]

Hi marcus and Jorre:

Thanks for the tutorial about LighCone. I will make an effort to learn how to use it.

When I woke up this morning I had an insight about the error I had made in my calculations. I had forgotten to take into account that the current value is
a = 0.8 rather than a = 1.0. I will later today recalculate and let you know what I get.

BTW, I am not sure I understand how the value a = 0.8 is derived. I have a guess about that, which I will also try out later.

Regards,
Buzz

 

[Jorrie]

I had forgotten to take into account that the current value is
a = 0.8 rather than a = 1.0. I will later today recalculate and let you know what I get.

BTW, I am not sure I understand how the value a = 0.8 is derived. I have a guess about that, which I will also try out later.

No, the current value of a=1 by definition. The 0.8 is for T_now (present age), because we are using a normalized timescale in the hypersine numeric model, where 17.3 Gy = 1 zeit.

 

[Buzz Bloom]

The 0.8 is for T_now (present age), because we are using a normalized timescale in the hypersine numeric model, where 17.3 Gy = 1 zeit.

Hi Jorrie:

Thanks for your post.

Sometimes my early morning insights are just senior moments. I guess I am still confused about the discrepancy in calculated values. I will return to the drawing boards later today.

Regards,
Buzz

 

[Jorrie]

The solution is right back in Marcus' posts 1 to 4; I recommend you reread those before trying to cope with it yourself...

 

[Buzz Bloom]

Hi marcus and Jorre:

I have fixed all my errors and misunderstandings in my calculations, and I now get t = 0.4394 zeit for a = 0.606 at the time when q = 0.

I do have one more question. I would like to be able to calculate t(a) for values of a at which dark energy becomes sufficiently insignificant, and also when radiation begins to become significant. I can get a solution model for when the significant mass densities are only for matter and radiation. I am thinking of making the transition from the hypersine model to the mass-radiation model using a value for a where Ω_r a^-4 = Ω_Λ. Does that seem reasonable, or would you recommend an alternative?

Regards,
Buzz

 

[Jorrie]

I am thinking of making the transition from the hypersine model to the mass-radiation model using a value for a where Ωr a^-4 = ΩΛ. Does that seem reasonable, or would you recommend an alternative?

This gives a ~ 0.1, which is reasonable. To find the optimal point, I would recommend that you calculate three values of of H(t) for a range of 'a' values: (i) the full Friedmann equation, (ii) Friedmann without radiation and (iii) Friedmann without Lambda. Then plot the error % of the latter two and see where they cross. I think it will be an interesting exercise if you would attempt this.

 

[nikkkom]

Same question, but this time the arriving light says it has been stretched by a factor of 1.5.

I want to add REIONIZATION (the second time the universe became transparent) to the timeline. We have to keep the timeline brief and sparse. It can't get heavy. But reionization is interesting.
Dense hydrogen gas is dazzling opaque if it is ionized. The free electrons scatter any kind of light. So space became transparent the first time when the gas cooled enough to form neutral hydrogen. ("recombination")

But there were no stars, so it was dark.

My understanding is that the sky was filled with uniform red glow for quite some time after recombination. Even 1000K blackbody spectrum has a significant high-energy tail in the visible (red).

 

[Buzz Bloom]

To find the optimal point, I would recommend that you calculate three values of of H(t) for a range of 'a' values: (i) the full Friedmann equation, (ii) Friedmann without radiation and (iii) Friedmann without Lambda.

Hi Jorrie:

Thank you very much for your excellent suggestion. I will be working on that for a few days.

Regards,
Buzz

 

[Buzz Bloom]

Hi marcus and Jorrie:

I completed the evaluation of H(a) using the Friedmann equation for the three cases Jorrie suggested:

i. H including terms for Ω_Λ, Ω_m, and Ω_r
ii. H_mΛ including terms just for Ω_Λ and Ω_m
iii H_mr including terms just for Ω_r and Ω_m​

For my calculations:

H₀ = 2.19727088648023E-15
Ω_Λ = 0.691906258479859
Ω_m = 0.308
Ω_r = 0.0000937415201411259​

I calculated the value of a I expected to be the error crossover between ii and iii.

a = (Ω_r / Ω_Λ)^1/4 = 0.107887513320811​

I then chose six other values of a by multiplying a by 0.9997, 0.9998, 0.9999, 1.0001, 1.0002, and 1.0003.

The following table shows the results.

Having completed this exercize, I then integrated the Friedmann equation for case iii, and I ran into a problem.

Since for a approaching zero, I expected t ∝ a², I was quite surprised when the integral did not behave that way. I am hoping someone can help me find what's wrong in my integration.

(1) H = (da/dt)/a = H₀ (Ω_m a^-3 + Ω_r) ^a-4)1/2
(2) dt = da (1 / H₀) a / (Ω_m a + Ω_r)^1/2
(3) t(a) = (1 / H₀) ∫ a da / (Ω_m a + Ω_r)^1/2
​

Here, not trusting my integration skills, I used my 1957 edition of the CRC Standard Mathematical Tables, Integral #111 on pg 283. The following I copied from the CRC changing only the variable letters and notation:

(4) ∫ a da / (p + q a) = (-2 (2 p - q a) / (3 q²)) (p + q a)^1/2​

Since we want the value of the integral to be zero for a = zero, the constant

4 p^3/2 / 3 q²​

must be added to the integral.

Substituting Ω_r for p and Ω_m for q produces

(5) t(a) = (1 / H₀) (-2 (2 Ω_r - Ω_m a) / (3 Ω_m²)) (Ω_r + Ω_m a)^1/2 + 4 Ω_r^3/2 / 3 Ω_m²​

From this one can see that

t(a) ∝ a​

Where did I go wrong?

Regards,
Buzz

 

[Buzz Bloom]

Hi marcus and Jorrie:

Where did I go wrong?

Well, I had another early morning insight, and this time it turned out to be OK. I expanded the integral in a power series, and the coefficient of the linear term canceled out to zero.

My next task is to see if the integral will give a "correct" value for t corresponding to the value of a at recombination.

Regards,
Buzz

 

[Jorrie]

Hi Buzz, you are doing interesting work, more than I actually suggested (which was just to compare H(t) for the different scenarios).

I'm not confident that you can use the power series to integrate for t all the way to a~0, t~0, because at best it must be an approximation. Around recombination, both matter and radiation have played a significant role and AFAIK, no analytical solution exists for the integral at that epoch. It would however be interesting to see what result you get.
--
Jorrie

 

[Buzz Bloom]

I'm not confident that you can use the power series

Hi Jorrie:

Thanks for your post.

I only wanted to see the constant, linear, and quadratic terms of the power series to conform t varies as a² near a = 0. I have little confidence at my age that I can still do math without making a mistake, and the integral looked like t varied linearly with a for small a. I will use the integral to calculate values of t(a) for a < 0.1078875.

Regards,
Buzz

 

[timmdeeg]

Hi Marcus, Jorrie made me aware of the "Hypersine model", which seems very valuable to give a better understanding of how the variables play together and evolve in time.
I have a question regarding the plot in #4, where you show a, H and the reciprocal of H, which should be the Hubble length, right? It seems that a grows faster than 1/H up to roughly 0.1 time. Shouldn't 1/H grow faster than a the whole period of deceleration and then inverted? I'm a bit confused, could you please explain?

 

[marcus]

Hi Tim, a(t) is a dimensionless number (pure, unitless)
whereas R = c/H is expressed in that plot in light zeit units---one lzeit is 17.3 billion light years.
So comparing them and their slopes is a bit "apples and oranges".

I'm puzzled by your post. I don't see why 1/H should "grow faster than a the whole period of deceleration and then inverted?"

 

[timmdeeg]

I'm puzzled by your post. I don't see why 1/H should "grow faster than a the whole period of deceleration and then inverted?"

Now I'm puzzled too..
I think we do agree that during decelerated expansion the Hubble sphere grows faster than the universe. Shouldn't this be reflected by comparing the slopes of the Hubble length and the scale factor accordingly? Yes, the scale factor is dimensionless, but if it doubles, the universe doubles while the Hubble length increases by a factor > 2. That's my reasoning, but perhaps I'm mistaken.

 

[Jorrie]

Yes, the scale factor is dimensionless, but if it doubles, the universe doubles while the Hubble length increases by a factor > 2.

I think the phase that you are thinking of has to do with our past light/future cone, shown as 'D_then' in the graph below. It reaches its maximum distance where the Hubble radius (R) crosses the past light cone. This is the first time photons from the CMB started to make headway towards us (in proper distance terms). Before that time, they were moving away.

 

[marcus]

Now I'm puzzled too..
I think we do agree that during decelerated expansion the Hubble sphere grows faster than the universe. Shouldn't this be reflected by comparing the slopes of the Hubble length and the scale factor accordingly? Yes, the scale factor is dimensionless, but if it doubles, the universe doubles while the Hubble length increases by a factor > 2. That's my reasoning, but perhaps I'm mistaken.

Tim I think I understand better now. Of course the ratio of a(t) to Hubble length tells you whether a'(t) is increasing or decreasing.
By definition H = a'/a so a'(t) = a(t)H(t) = a(t)/R(t) forgetting about factors of the speed of light and setting R = 1/H.

So if that ratio a/R is increasing then a' is increasing and if a/R is decreasing then a' is decreasing.

The trouble is with the words "F(x) grows faster than G(x)"

It is not true that the ratio F/G increasing implies the slope of F is greater than the slope of G. F/G increasing is not equivalent to F' > G'.

Example on the interval [0, 1/2) consider F(x) = x and G(x) = x²

F'(x) = 1 which is always greater than G'(x) = 2x on that interval.
However the ratio F/G = 1/x is always DECREASING.

WHEN YOU SAY "we do agree that during decelerated expansion the Hubble sphere grows faster than the universe" I have to say NO.
we do not agree because that sounds to me like the SLOPE of one curve is steeper than the other----R'(t) > a'(t) ---- and that is simply not true during decelerated expansion.

But it is true that during decelerated expansion the RATIO of a/R is decreasing. That is another way of interpreting the words "a grows slower than R".

The trouble is "a grows slower than R" is ambiguous.

 

[timmdeeg]

I think the phase that you are thinking of has to do with our past light/future cone, shown as 'D_then' in the graph below.

Not really, the past light-cone depends on how the universe expands, but doesn't show the expansion itself. Meanwhile marcus has answered and I will be busy with that.

 

[Jorrie]

WHEN YOU SAY "we do agree that during decelerated expansion the Hubble sphere grows faster than the universe" I have to say NO. we do not agree because that sounds to me like the SLOPE of one curve is steeper than the other----R'(t) > a'(t) ---- and that is simply not true during decelerated expansion.

But it is true that during decelerated expansion the RATIO of a/R is decreasing. That is another way of interpreting the words "a grows slower than R".

I guess you meant to write " the RATIO of da/dR is decreasing"?
Even that would not be generally true, because the change from decreasing to increasing in that ratio happens at the peak of the light cone curve, as I have shown above. This is well before the desired inflection point for a(t), around t=4 Gy.

 

[timmdeeg]

Marcus, thanks for answering in some detail which should make it easier to clarify things.

F'(x) = 1 which is always greater than G'(x) = 2x on that interval.

Yes, we compare the slopes. F'(x) > G'(x) means that F(x) grows faster than G(x) regarding said interval and conversely G(x) grows faster than F(x), if x > 0.5. Hopefully you do agree on that.

WHEN YOU SAY "we do agree that during decelerated expansion the Hubble sphere grows faster than the universe" I have to say NO.
we do not agree because that sounds to me like the SLOPE of one curve is steeper than the other----R'(t) > a'(t) ---- and that is simply not true during decelerated expansion.

Lets compare the slopes of a and R in the plot of #4, using the wording as above:

Time Interval [0, 0.1]: a grows faster than R, -- a'(t) > R'(t)--. -> acceleration
Time Interval [0.1, 0.4]: R grows faster than a. -> deceleration. At t = 0.4 (very roughly) deceleration turns to acceleration again.
Time Interval [0.4, infinite]: a grows faster than R. -> acceleration

I'm confused, because if I understand the plot correctly(??) the universe though being matter dominated starts to expand accelerated. Why?

https://en.wikipedia.org/wiki/Hubble_volume

For example in a decelerating Friedmann universe the Hubble sphere expands faster than the Universe and its boundary overtakes light emitted by receding galaxies so that light emitted at earlier times by objects outside the Hubble sphere still may eventually arrive inside the sphere and be seen by us.[3] Conversely, in an accelerating universe, the Hubble sphere expands more slowly than the Universe, and bodies move out of the Hubble sphere.

Mentioning "decelerating Friedmann universe" could imply Lambda = 0. Does this make the difference, as above "matter dominated" means just a large ratio of matter density to Lambda density? But still, why then should the universe start to to expand decelerated at t = 0.1 after the matter has already been diluted?
Sorry, this all seems to make not much sense.

 

[marcus]

Hi Tim, I posted this before I saw your post #83, so it doesn't respond to what you just said but to something earlier.

Now I'm puzzled too..
I think we do agree that during decelerated expansion the Hubble sphere grows faster than the universe. Shouldn't this be reflected by comparing the slopes of the Hubble length and the scale factor accordingly? Yes, the scale factor is dimensionless, but if it doubles, the universe doubles while the Hubble length increases by a factor > 2. That's my reasoning, but perhaps I'm mistaken.

Tim, did my comment make sense to you. I think you are giving a non-standard interpretation to the words "hubble sphere grows faster than a".
You are not comparing slopes, which would be a usual interp.
You are talking as if you mean the RATIO, namely a/R, is decreasing.
You say for example that the scale factor DOUBLES and R MORE THAN DOUBLES. Mathematically that means a/R decreases.

But that is not equivalent to saying the slopes are in the relation a' < R' which I think is how most people would tend to hear words like R grows faster than a.

That is the verbal ambiguity I was talking about in my comment.

 

[marcus]

Marcus, thanks for answering in some detail which should make it easier to clarify things.

Yes, we compare the slopes. F'(x) > G'(x) means that F(x) grows faster than G(x) regarding said interval and conversely G(x) grows faster than F(x), if x > 0.5. Hopefully you do agree on that.

Lets compare the slopes of a and R in the plot of #4, using the wording as above:

Time Interval [0, 0.1]: a grows faster than R, -- a'(t) > R'(t)--. -> acceleration
Time Interval [0.1, 0.4]: R grows faster than a. -> deceleration. At t = 0.4 (very roughly) deceleration turns to acceleration again.
Time Interval [0.4, infinite]: a grows faster than R. -> acceleration
...

This shows that you must never verbally interpret deceleration by saying "a grows slower than R"
That statement is first of all meaningless because they don't have the same units. They are incommensurable.
But more importantly, it will give people the impression that you mean something about the SLOPES (a' < R') which is simply not true.
So you screw people up if you say things like "deceleration means the universe grows slower than the Hubble radius", if they believe your words.

Look at that plot #4 for example. You can see that a(t) starts out decelerating because it is convex upward. But it is obviously not true that a'<R'

I think if you want a mathematical condition that would correspond truthfully to deceleration you could say for example that "The ratio a/R is decreasing"

That is true during deceleration because (up to factors of c) R = 1/H and so a/R = aH
and since H is a'/a what we have here is a/R = a'
So if the ratio a/R is decreasing then a' is decreasing, which is what people normally associate with deceleration.

 

[Jorrie]

I guess you meant to write " the RATIO of da/dR is decreasing"?
Even that would not be generally true, because the change from decreasing to increasing in that ratio happens at the peak of the light cone curve, as I have shown above. This is well before the desired inflection point for a(t), around t=4 Gy.

Sorry Marcus, I read you wrong; you are right about a/R that decreases during acceleration and increases during accelerating expansion. I hope I did not compound Tim's problem with this!

 

[timmdeeg]

I think if you want a mathematical condition that would correspond truthfully to deceleration you could say for example that "The ratio a/R is decreasing"

That is true during deceleration because (up to factors of c) R = 1/H and so a/R = aH
and since H is a'/a what we have here is a/R = a'
So if the ratio a/R is decreasing then a' is decreasing, which is what people normally associate with deceleration.

Very true and it confirms, what I was thinking previously.
https://www.physicsforums.com/threa...e-ratio-hubble-length-to-scale-factor.842617/ #4

The ratio I'm asking for is $1/Ha$. Replacing $H$ by $(da/dt)/a$ yields $1/(da/dt)$. Therefore $1/Ha$ should increase as long as the universe expands decelerated and decrease during accelerated expansion then.

I think, in the meantime I was misled by the wording "in a decelerating Friedmann universe the Hubble sphere expands faster than the Universe", causing me to think in terms of slopes, which however yields a wrong result, as is obvious from my post 83.

Marcus, I'm very thankful that you brought me back on the right track. Thanks for your efforts! Sorry, it took a while thought you mentioned the ambiguity of the wording "grows faster" and the like a few times.