How to calculate the Age of the Universe, from observations

1. May 19, 2015

marcus

By fitting redshift distance data we can estimate the present Hubble rate H0 and the longterm limit that the Hubble rate is tending towards H.

The Age (how long the universe has been expanding) can be calculated just from those two measurements, assuming the spatial-flat standard cosmic model---essentially the Friedmann equation.

Basically you take the ratio H0/H = 1.201

and solve $$x = \frac{1}{3}\ln(\frac{1.201+1}{1.201-1})$$

Then x/H is the Age in whatever units you like to use.

2. May 19, 2015

marcus

It comes out right and the reasoning is straightforward.

There is enormous supporting evidence that in broad outlines Gen.Rel. works. It is how gravity and geometry work. No mystery there.

If you believe GR then you believe the Friedmann equation that immediately derives from it with the simplifying assumption that stuff is evenly distributed. And the evidence supports spatial flat or near enough flat that for calculating the Age we can assume flat.

To an excellent approximation, we can solve the Friedmann and get that using the indicated time parameter x we have
$$H(x) = \coth(\frac{3}{2}x)$$
So all we have to do is set
$$1.201 = \coth(\frac{3}{2}x)$$
There is only one time in history when the Hubble rate (ratio to longterm) is 1.201 and that is now.
There is only one time parameter x that makes the ratio 1.201 and that is x=0.8 or more precisely 0.797.
There is only one Age which multiplied by the measured value of H gives 0.8.

It's really straightforward. Solve the Friedmann differential equation and do some simple algebra.

If you look up "hyperbolic functions" in Wikipedia it says what the inverse of the coth function is.
$$w = \coth(x)$$ for w>1 if and only if $$x = \frac{1}{2}\ln(\frac{w+1}{w-1})$$ so
$$w = \coth(\frac{3}{2}x)$$ if and only if $$x = \frac{1}{3}\ln(\frac{w+1}{w-1})$$

3. May 27, 2015

marcus

The approach to explaining/learning cosmology that I want to explore here assumes a basis in redshift-luminosity observations or in other words wave stretch (s) and distance (D) data. That was how Hubble originally determined Hnow the present-day ratio of expansion speed to distance.
One way to express Hnow is 1/14.4 ppb per year. It's a fractional growth rate. there are other equivalent expressions.
The same basis of stretch-Distance data vastly improved was used around 1998 to determine H, now estimated at around 1/17.3 ppb per year. I'll try to suggest the rough outlines of how that was done, and fill it in later.
$$H(s) =H_\infty \sqrt{((H_{now}/H_\infty)^2 - 1)s^3 + 1} = H_\infty \sqrt{((17.3/14.4)^2 - 1)s^3 + 1}$$
$$D(s) = \int_1^s \frac{ds}{H(s)}$$
So the distance, from the present observer at s=1 out to a source at s=2 is the area under the curve between 1 and 2. Likewise for source at other values of s. These curves (top to bottom) are in order of the inverse rates 16.3, 17.3, 18.3, 40, 1000. A very large number like 1000 corresponds to H so small as to be negligible.
In 1998 it was discovered that the distances corresponding to various s were LARGER than they would be for H = 0, the standard candles were DIMMER than expected. They homed in on 17.3. The top curve, green 16.3, gives areas that are too large, and those the blue 18.3 curve gives are too small. You have to pick one H that gives the right distance for each stretch s over the whole range, these being shown by the areas under the curve, out to that particular value of s.

Last edited: May 27, 2015
4. May 27, 2015

marcus

the upshot is you find Hnow and H by fitting "redshift-luminosity" (i.e. wavestretch-distance) data.

Once you have those two quantities you can tell the expansion age, and some other stuff as well.
First off, you can take their ratio:
$$\frac{H_{now}}{H_\infty} = 1.2014...$$

Solving the Friedmann equation for H(t) gives a relation between the expansion rate H and time t (measured in 17.3 billion year units)
$$\frac{H(t)}{H_\infty} = \coth(\frac{3}{2}t)$$
So we can set $$\coth(\frac{3}{2}t) = 1.2014...$$
and solve for t.
The "coth" function is easy to "un-do" or invert---it turns out that $$t = \ln(\frac{1.201 +1}{1.201 - 1})/3 = 0.797.. ≈ 0.8$$ And that, multiplied by 17.3 billion years, gives the usual figure for the age.

What if you aren't familiar with the "coth" function? How can this be made less off-putting?
$$\coth(x) = \frac{e^x + e^{-x}}{e^x - e^{-x}} = \frac{e^{2x} + 1}{e^{2x} - 1}$$
So another way of writing the expansion rate H(t) = Hcoth(1.5t) is
$$H(t) = H_\infty \coth(\frac{3}{2}t) = \frac{e^{3t} + 1}{e^{3t} - 1}$$
Have to think how the"coth" (hyperbolic cotangent) function can be helpfully introduced.

Last edited: May 28, 2015
5. May 27, 2015

marcus

The function plotted here is one of the simplest you can make out of ex that has a certain symmetry: flipping it right to left (e.g. reversing the time direction) is the same as flipping it over top to bottom. And AFAIK it's the simplest such that, when interpreted as the rate of distance expansion and contraction, describes a symmetric universe that first goes thru steady contraction (constant negative expansion rate), then a swift crunch, then a rebound with an initially rapid expansion, that finally settles down to a steady rate. In other words settles on a constant "percentage" or fractional rate of exponential growth.

Last edited: May 28, 2015
6. May 27, 2015

marcus

The hyperbolic functions are trig function analogues which you get if you start with ex and make the easiest possible moves to get functions that are symmetric in various ways. People can look them up: "hyperbolic functions" in wikipedia etc. For example, recall that the cosine equals itself run backwards cos(x)=cos(-x) We can get the same symmetry from ex just by writing (ex + e-x)/2 and that is called cosh, for hyperbolic cosine. And (ex - e-x)/2 run backwards equals the negative of itself, analogous to sine(x) so it is called sinh.

Well coth(1.5t) is the one of these simple functions that has the symmetry you'd expect of the expansion rate in a collapse-rebound scenario like the one we just depicted and discussed.. I just happens to be how the distance expansion rate in our expanding universe behaves. (The expansion part we can see, we don't know that there ever was a contraction part, if the equations describe a prior contraction that just counts as an "unverified extra" in the symmetry department.)

So let's try to find the MINIMUM SET OF EQUATIONS to describe the universe process. One equation would certainly be H(t) = coth (1.5t), which by definition is also equal to $$H(t) = \frac{e^{3t}+1}{e^{3t}-1}$$ If you aren't familiar with hyperbolic functions you may prefer to see it spelled out like that.
And H(t) can be inverted or solved for t to give $$t(H) = \ln(\frac{H+1}{H-1})/3$$
So whenever we know H we know the time, and whenever we know the time we know the expansion rate H.
Here to simplify notation I'm assuming that H is expressed in H terms--as the ratio H/H

Maybe there is only one more essential equation and that is how the stretch factor (our distances compared to theirs) relates to H. IOW H(s) and the inverse function s(H). this function knows about the present. so it needs a constant in it that is specific to the present era. That normalizes it so that s=1 at the present.
Basically this function H(s) is a form of the Friedmann equation. These are all equivalent ways of saying the same thing.
$$H(s)^2 - 1 = 0.4433s^3$$
$$s^3 = \frac{H^2-1}{.4433}$$
$$H = \sqrt{.4433s^3 + 1}$$
Here 0.4433 = (Hnow/H)2 - 1
It's a number that gets used here a lot, which you know as soon as you have measured Hnow and H by fitting observational data.

Last edited: May 28, 2015
7. May 28, 2015

marcus

We're trying out the most basic approach to quantitative cosmo that I can think of. It boils down to two equations. I'll give some equivalent forms of each:
$$H(t) = \frac{e^{3t}+1}{e^{3t}-1} \ \ \ \ and \ its \ inverse \ \ \ t(H) = \ln(\frac{H+1}{H-1})/3$$
Knowing the time we can calculate the expansion rate H, and conversely, given H, we can calculate the time.
Here to simplify notation t is time scaled by H (so the present is denoted t = 0.8) and H is expressed in H terms--as the ratio H/H. Here's how the Friedmann equation together with some equivalent versions look in those terms:
$$H(s)^2 - 1 = 0.4433s^3$$
$$H = \sqrt{.4433s^3 + 1}$$
$$s^3 = \frac{H^2-1}{.4433}$$
$$s = (\frac{H^2-1}{.4433})^{1/3}$$
where 0.4433 = (Hnow/H)2 - 1
We can illustrate the expansion history by plotting the scale factor a(t) = 1/s, the size of a generic distance normalized to one at present t = 0.8.
$$a(t) = 1/s = (\frac{.4433}{H^2-1})^{1/3}$$

Last edited: May 29, 2015