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

[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.