What can you learn about cosmology from Google calculator?

[marcus]

That's a good idea. T_∞ = 17.3 billion years, is a time scale and something we use constantly.

If you walk in the door with an ordinary time T in years the first thing you do is divide by 17.3 billion years and get T/ T_∞
and if the formula gives you a distance like Hubble radius R at that time, then to get back into conventional terms you multiply by cT_∞ = 17.3 billion light years and walk out the door with the answer in billions of light years.

T_∞ and cT_∞ are how you translate between conventional scales and the world seen through hyperbolics like sinh and tanh.
So I'll try writing the formulas making that explicit.

 

[wabbit]

Sounds good to me : ) also, what's nice about large scale units is that c=1 light-year per year, so that's one constant we don't need to worry about much.

Edit : which makes me realize that c>>1 is really a very anthropocentric view. Cosomology-wise, c=1 is very natural, the universe has comparable dimensions in time and space. It's just that we are very very slow ourselves, perhaps just because we have cooled down to a very low temperature since the radiation era when everybody thought c=1 was kinda obvious. So in that sense c>>1 is a measure of how cold this universe has become, and we might be able I dare speculate to express c as some normal looking constant times some power of the ratio of the CMB temperature to the Planck temperature... Or something:)Hmm this isn't quite true our local temp is driven by the Sun not the CMB - well, something like that, just don't look too closely:)

 

[marcus]

... Another possibility is also to formulate with x but frequently add a reminder "(where $x=\frac{T}{T_\infty}$)".

That might be an even better approach! best of both ways. remarking and getting used to the change of scale when you enter and exit the hyperbolic model, but also keeping the equations light and trim.

We also should have a distinct notation for the un-normalized scale factor. It was confusing when I wrote a(x) because the scale factor is usually normalized to equal 1 at present. Let's try u(x) = sinh^2/3(1.5 x)
mnemonically, u stands for "unnormalized" so it might be easy enough to adjust to.
Keeping in mind that x_now = T_now/T_∞ = 13.787/17.3 = 0.797, we have the normalized scale factor:

a(x) = u(x)/u(x_now) = u(x)/u(.797) = u(x)/1.311

And this simplified cosmic model's most elaborate equation so far is for the present distance spanned by light emitted at time x_em:

$$D_{now}(x_{em}) = 1.311 \int_{x_{em}}^{x_{now}} \frac{cdx}{u(x)}$$

where c = 1

 

[wabbit]

About x just one more thing - i know that time=space but still, there are conventions in the back of our mind ; : ) so perhaps $\theta$ or $\tau$ could be a better name.

Regarding a(x) vs u(x) I'm mixed... I see the advantage of u(x) and I find the convention a=1 now to be confusing at times but there is also the issue of relating what's written here to what's elsewhere, and a() has that advantage of being commonly used, so is the gain in switching to u enough to balance the need to memorize one more name and how the two relate ? I just can't tell, it might even depend on who reads it and in what context.

 

[marcus]

In this thread we employ time and distance scales, provisionally dubbed T_∞ and R_∞, which are the longterm limits of the present-day Hubble time and Hubble radius scales often used in cosmology. Our scales are, in fact, the eventual these quantities are tending towards, and are based on the cosmological constant.
It turns out that the current Hubble time, is 5/6 of T_∞ , and the current Hubble radius is 5/6 of R_∞. On the other hand, the present age of universe expansion is only about 4/5 of T_∞.
T_∞ and R_∞ are estimated at 17.3 billion years and 17.3 billion light years.

On these scales of time and distance, the speed of light is one, and the present-day expansion age = x_now = 13.787/17.3 = 0.797 ≈ 4/5
Provisionally at least, we will denote times measured on this scale by x, where x = time in years divided by 17.3 billion years.
x = T/T_∞

With time scaled this way, it turns out that the size of the universe, u(x), as tracked by the size of a generic distance, follows a simple curve:

$$u(x) = \sinh^{2/3}(\frac{3}{2} x)$$

If we want to normalize this size function so that it equals one at present, we just have to compute its value at present, and divide by that.
the normalized scale factor can be called a(x)
$$a(x) = \frac{u(x)}{u(x_{now})} = \frac{u(x)}{1.311}$$

 

[marcus]

Continuing in mock tutorial form, to try out notation:
"Another nice formula in the closely approximate model we get with this scaling of time and distance is the one for H(x) the fractional growth rate at any given time x. Keep in mind that the scaled time variable x is years divided by 17.3 billion years.

By definition, the Hubble time Θ is the reciprocal of the fractional growth rate. It changes inversely as the growth rate of distances changes.
I want a distinctive notation for the Hubble time, that won't be confused with the age of the universe or the time-variable itself. I'm going to try the capital letter Theta.

At any time x, the distance growth rate H(x) and the Hubble time Θ(x) are related by:
Θ(x) = 1/H(x)
for example if the Hubble time is 10 billion years, what that means is that any given distance (between stationary points) is growing by 1/10 of its length per billion years. Or 1/10,000 of its length every million years. Or in more familiar growth rate terms, growing by 1/100 of one percent per million years.
The Hubble time is a convenient way of encoding that growth rate.
Our formula for Hubble time, showing how it grows with the expansion age of the universe, is quite simple:
$$Θ(x) = \tanh(\frac{3}{2}x)$$
To do an example, remember that x_now is about 4/5, more exactly 0.797
If we calculate tanh(1.5*.797) in google we get 0.83227 (about 5/6). We can interpret that either in time or distance terms.
To get the answer in conventional distance terms, we multiply by the eventual longterm Hubble radius R_∞ = 17.3 billion light years, and get 14.398... which rounds to 14.4 billion light years.
===================================
Earlier I had misgivings about the T notation for Hubble time. Here's part of what I wrote when I was still undecided. What follows will probably be deleted, when the issue is resolved.
In an earlier post you suggested a different symbol for a time quantity, Hubble time perhaps.
Should one use Θ = 1/H?
I just now wrote
$$T(x) = \tanh(\frac{3}{2}x)$$
the formula for Hubble time. Should that instead be
$$\Theta(x) = \tanh(\frac{3}{2}x)$$
The uppercase T risks being overused.---temperature,...---and Theta starts with "T". The uppercase Theta has what looks like a small capital H in its middle, H for Hubble time, H for hyperbolic tangent...
Maybe it would be mnemonic to define Θ(x) = 1/H(x) at time x.

 

[marcus]

I think it's mathematically most convenient to use S as the independent variable, as in Lightcone tables, or its reciprocal the normalized scale factor. But time is intuitive and familiar. So newcomers may like to see a table generated by this simplified model cosmos where the x-time (time scaled by 17.3 billion years) serves as the driving variable. (Just a sketch. I might add some other columns later.)
a(x) is the normalized scale factor at time x: sinh^2/3(1.5x)/1.311
S(x) is the stretch since that time, 1/a(x) = z+1
Θ(x) is the reciprocal growth rate, the Hubble time (equiv. Hubble radius), namely tanh(1.5x)

x-time  (Gy)    a(x)    S       Theta   (Gy)     Dnow     Dnow (Gly)
.1      1.73    .216    4.632   .149    2.58    1.331       23.03
.2      3.46    .345    2.896   .291    5.04     .971       16.80
.3      5.19    .458    2.183   .422    7.30     .721       12.47
.4      6.92    .565    1.771   .537    9.29     .525        9.08
.5      8.65    .670    1.494   .635   10.99     .362        6.26
.6     10.38    .776    1.288   .716   12.39     .224        3.87
.7     12.11    .887    1.127   .782   13.53     .103        1.78
.797   13.787  1.000    1.000   .832   14.40    0            0

Everything except Dnow was calculated by google
for example a(.1) was sinh(1.5* .1)^(2/3)/1.311
Theta(.1) was tanh(1.5* .1)
and its billion year (Gy) value was 17.3*tanh(1.5* .1)

The exception Dnow required numerical integration, which is easier than one might think.
In the Mac "Grapher" utility one simply types in 1.311 (sinh(1.5 x)^-2/3 using ^ for superscript,
clicks "integration" in the "equation" menu, and enters the limits, for instance .1 and .797.