# Which Planck numbers to use? (Ned Wright's choice?)

1. Apr 3, 2013

### marcus

Planck team published several sets of of basic cosmic parameters in their report
http://arxiv.org/pdf/1303.5076v1.pdf
see for example Table 5 on page 22.

The rightmost column of Table 5 is labeled "Planck+WP+highL+BAO".

That seems to be the set of numbers that Ned Wright chooses to report, for instance:

ΩΛ = 0.692 ± 0.010
H0 = 67.80 ± 0.77

I'll quote from Wright's "News of the universe" item where he seems to choose that column as most reliable and gives a concise summary of the numbers.
==quote==
The parameters of the 6-parameter ΛCDM model fit to Planck+WMAP polarization+SPT+ACT+BAO are ΩΛ = 0.692 ± 0.010; the baryon density = 0.416 ± 0.0045 yoctograms per cubic meter; the cold dark matter density = 2.23 ± 0.032 yoctograms per cubic meter; CDM:baryon density ratio = 5.36 ± 0.10; dark energy density = 3352 ± 125 eV/cc; H0 = 67.80 ± 0.77 km/sec/Mpc; and the age of the Universe = 13.798 ± 0.037 Gyr. The baryon density is known to 1.1% precision and the cold dark matter density is known to 1.4% precision.
==endquote==

Other people may have suggestions different from Wright's, which would be interesting to hear.

Suppose we adopt Wright's choice, what Hubbletimes are we looking at? Well the Hubble time is just 1/H. In other words H-1, the reciprocal ("one over") of the Hubble rate. So I put these blue expressions into google calculator:
1/(67.8 km/s per Mpc) and google gives me the present-day H-1 = 14.422 billion years
and
1/(67.8 km/s per Mpc)/0.692^.5 which gives the eventual H-1 = 17.337 billion years

Just for convenience, I'm inclined to round these off to 14.4 and 17.3, provisionally at least, and see how things work.
From Table 2 on page 11, I see that matter radiation equality occurs roughly around 3400.
And the discussion of curvature on page 40 indicates that it's close enough to take as flat for our purposes. The total omega had a central value of 1.001 or 1.0005 depending on which studies' results were combined.

Last edited: Apr 3, 2013
2. Apr 3, 2013

### marcus

So let't try out a history with these two Hubbletimes:
H-1now = 14.4 Gy
H-1inf = 17.3 Gy
and with Seq=3400

This table is set to begin in year 415 million (when distances are 1/12 their present size). It starts around formation time of the first galaxies and it runs to year 68 billion, at which time the cosmological constant is clearly evident in both the Hubble time and the cosmic event horizon columns. The widest point in our past (teardrop shaped) lightcone occurs at S=2.6, around year 4 billion. Emitter galaxies before that time were receding faster than c, so their light was initially swept back and only later began to make progress in our direction.

$${\begin{array}{|c|c|c|c|c|c|c|}\hline Y_{now} (Gy) & Y_{inf} (Gy) & S_{eq} & H_{0} & \Omega_\Lambda & \Omega_m\\ \hline14.4&17.3&3400&67.92&0.693&0.307\\ \hline\end{array}}$$ $${\begin{array}{|r|r|r|r|r|r|r|} \hline S=z+1&a=1/S&T (Gy)&T_{Hub}(Gy)&D (Gly)&D_{then}(Gly)&D_{hor}(Gly)&D_{par}(Gly)\\ \hline12.000&0.083&0.415&0.624&32.112&2.676&4.050&1.179\\ \hline9.912&0.101&0.553&0.831&30.612&3.088&4.752&1.579\\ \hline8.188&0.122&0.736&1.106&28.961&3.537&5.551&2.113\\ \hline6.763&0.148&0.981&1.471&27.146&4.014&6.452&2.827\\ \hline5.586&0.179&1.306&1.954&25.153&4.503&7.454&3.779\\ \hline4.614&0.217&1.738&2.590&22.969&4.978&8.551&5.048\\ \hline3.812&0.262&2.309&3.421&20.581&5.400&9.725&6.738\\ \hline3.148&0.318&3.061&4.490&17.983&5.712&10.949&8.983\\ \hline2.601&0.385&4.043&5.832&15.181&5.837&12.177&11.953\\ \hline2.148&0.466&5.308&7.448&12.198&5.678&13.353&15.859\\ \hline1.774&0.564&6.904&9.278&9.088&5.122&14.414&20.952\\ \hline1.466&0.682&8.859&11.176&5.940&4.053&15.302&27.513\\ \hline1.211&0.826&11.168&12.943&2.867&2.368&15.986&35.847\\ \hline1.000&1.000&13.787&14.400&0.000&0.000&16.472&46.279\\ \hline0.826&1.211&16.648&15.475&-2.616&-3.167&16.793&59.176\\ \hline0.654&1.528&20.359&16.316&-5.348&-8.172&17.022&78.870\\ \hline0.518&1.929&24.220&16.789&-7.600&-14.660&17.142&103.899\\ \hline0.411&2.435&28.163&17.040&-9.423&-22.943&17.199&135.586\\ \hline0.325&3.073&32.148&17.169&-10.883&-33.447&17.223&175.631\\ \hline0.258&3.879&36.155&17.235&-12.046&-46.731&17.235&226.204\\ \hline0.204&4.897&40.173&17.267&-12.970&-63.511&17.267&290.052\\ \hline0.162&6.181&44.197&17.284&-13.703&-84.698&17.284&370.652\\ \hline0.128&7.802&48.224&17.292&-14.284&-111.444&17.292&472.392\\ \hline0.102&9.848&52.251&17.296&-14.745&-145.207&17.296&600.816\\ \hline0.080&12.431&56.280&17.298&-15.110&-187.825&17.298&762.921\\ \hline0.064&15.691&60.309&17.299&-15.399&-241.620&17.299&967.540\\ \hline0.050&19.806&64.338&17.300&-15.628&-309.523&17.300&1225.822\\ \hline0.040&25.000&68.367&17.300&-15.809&-395.235&17.300&1551.841\\ \hline\end{array}}$$
Time now (at S=1) or present age in billion years: 13.787
'T' in billion years (Gy) and 'D' in billion light years (Gly)

Last edited: Apr 3, 2013
3. Apr 3, 2013

### Jorrie

I get the impression that the WMAP9 and the Planck values fall within each others error bars, so does such a small refinement really matter? It seems more like just reciprocal confirmation.

4. Apr 3, 2013

### marcus

I agree.
Maybe there is no need for us to change from WMAP9 to Planck values.
We could keep on using the two Hubbletimes 14.0 and 16.5.
On the other hand I've noticed several other posters apparently spontaneously adopting Planck numbers. In some cases they looked similar to those I found at Ned Wright's website, where he reported the Planck results.
Not sure what's best to do.

As an example of the kind of curves the above table contains here's the scale factor (a = 1/S) plotted against cosmic time. I'm using graphs Jorrie made earlier. Hope this is OK (tell me if not).

You can see the a(t) curve starts out convex (declining slope) and has an inflection point around year 7 billion. Then it becomes very gradually concave (increasing slope).
Here's a similar graph but flipped so the scale factor a is on the horizontal axis and time is on the vertical:

Again you can see the inflection point coming right around year 7 billion.

The a(t) curve can be thought of as the growth of a generic cosmic distance (separation between two CMB stationary observers) which has been normalized to equal 1 at the present time. In concrete terms, say, it is equal to 1 billion lightyears at this moment---or some other convenient unit. The growth of other cosmic-scale distances would be proportional---similar shaped curve.

Last edited: Apr 3, 2013
5. Apr 3, 2013

### Mordred

.

probably not by very much, I think Marcus favourite toy is the calculators lol j/k Marcus. They are extremely handy tools, I'm still learning them from the numerous posts Marcus has out there, they have amazing flexibility

6. Apr 3, 2013

### marcus

Yes. It's easy to get hooked once you learn the basic routine for using it

7. Apr 3, 2013

### marcus

With the new Planck numbers (for better or for worse, just giving them a trial run) namely
14.4, 17.3, and 3400
we get the widest girth of the teardrop shaped lightcone coming almost exactly at S=2.6

So that if you set upper=12 and check "S=1" option and steps=13, you hit very close to that widest place in the lightcone. You can see that in the table back in post #2

That is where the emitter galaxy's distance is equal to the Hubble distance and the emitted photon "stands still" for a while before beginning to make progress in our direction.

I like it that with the new numbers we get that crossing of the lightcone with c/H to happen at a nice round number like 2.6, not that it really matters
Actually it occurs at 2.5986, but 2.6 is close.

8. Apr 4, 2013

### Jorrie

Actually, the inflection point is a little later, around 7.6 Gy. It comes from solving the second Friedmann equation for $\ddot{a}=0$ (ignoring radiation):
$$a(t) = \left(\frac{\Omega_m}{2\Omega_\Lambda}\right)^\frac{1}{3}$$
One can then find the time from the tabular calculator by setting the S = 1/a = 1.65 for the Planck values that Marcus used.
I will leave it to Marcus to convert the Lambdas to Hubble times.

Last edited: Apr 4, 2013
9. Apr 4, 2013

### marcus

Just for fun, assume that the three defining parameters we are using are 14.4, 17.3, 3400.
Those are the present and future Hubbletimes, and Seq marking the epoch when matter and radiation densities are equal. Then assuming near flatness we can derive other things like
ΩΛ = (14.4/17.3)2 = 0.6928
Ωm = 1 - 0.6928 = 0.3072

So then using the formula you provide, and google calculator, we get the scale factor a(t) at the moment of inflection!

a(inflection) = (0.1536/0.6928)^.33333 = 0.60524...

and then S(inflection) = 1/a = 1.6522... just as you say!

then I whip out the online cosmic tabulator, put in 1.65222 and presto: year 7-something billion!

Year 7.59 billion, it turns out. The one-line table, setting "steps" = 0, is:
$${\begin{array}{|c|c|c|c|c|c|c|}\hline Y_{now} (Gy) & Y_{inf} (Gy) & S_{eq} & H_{0} & \Omega_\Lambda & \Omega_m\\ \hline14.4&17.3&3400&67.92&0.693&0.307\\ \hline\end{array}}$$ $${\begin{array}{|r|r|r|r|r|r|r|} \hline S=z+1&a=1/S&T (Gy)&T_{Hub}(Gy)&D (Gly)&D_{then}(Gly)&D_{hor}(Gly)&D_{par}(Gly)\\ \hline1.652&0.605&7.591&9.988&7.912&4.789&14.768&23.213\\ \hline\end{array}}$$

Last edited: Apr 4, 2013
10. Apr 4, 2013

### Jorrie

I suppose a more elegant way of doing it (less steps before using the tabular calculator), would be to go straight to S = 1/a and use only
$$\Omega_\Lambda = (Y_{now}/Y_{inf})^2$$
Then for near flatness:
$$S_{inflection} = \left(\frac{2\Omega_\Lambda}{1-\Omega_\Lambda}\right)^{1/3}$$

Edit: thanks Marcus, fixed typo now.

Last edited: Apr 4, 2013
11. Apr 4, 2013

### marcus

You are right. I'm happy with all the stuff we are unpacking from TabCosmo. It is like finding Easter eggs under the Christmas tree, or whatever

Fixed a typo.

Here's another one-liner, with the new Planck numbers, setting "steps" = 0.
$${\begin{array}{|c|c|c|c|c|c|c|}\hline Y_{now} (Gy) & Y_{inf} (Gy) & S_{eq} & H_{0} & \Omega_\Lambda & \Omega_m\\ \hline14.4&17.3&3400&67.92&0.693&0.307\\ \hline\end{array}}$$ $${\begin{array}{|r|r|r|r|r|r|r|} \hline S=z+1&a=1/S&T (Gy)&T_{Hub}(Gy)&D (Gly)&D_{then}(Gly)&D_{hor}(Gly)&D_{par}(Gly)\\ \hline2.600&0.385&4.044&5.833&15.177&5.837&12.179&11.957\\ \hline\end{array}}$$
I mentioned it earlier. At the precise S=2.6 mark which is around year 4 billion, you get the radius of the lightcone equaling the Hubble distance. So it is the widest bulge of the past lightcone and the epoch when the emitting galaxy is receding at exactly speed of light.

Last edited: Apr 4, 2013