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## Cosmo calculators with tabular output

 Quote by marcus I had another look and I think there are pros and cons about the 9th column. Multiplying by Rnow seems somewhat arbitrary. Doesn't it just scale the numbers up? I thought the notation Rnow' is a bit confusing since it gives the impression it is the derivative of Rnow,and that Rnow is changing. But Rnow is a constant. A fixed parameter of the model. Isn't da/dT what the column is really about? So couldn't you achieve the same effect by making it 100xda/dT, or 1000xda/dT? Some arbitrary multiplicative factor, in other words? Or perhaps I'm missing something.
The Hubble radius is a 'characteristic' size of the universe, so I thought multiplying by it should scale da/dT to something interesting, and it did. The problem is that the column becomes a little confusing in the context of the calculator, because it gives the recession rate (in units c) at a specific redsift (a source presently at the Hubble radius). The rest of the columns represent objects at different redshifts, detracting from the appeal of such a column.

The table below complies closely with Tamara Davids' panels (she used H0 = 70 km/s per Mpc and then 0.7 and 0.3 for the Omegas.

$${\begin{array}{|c|c|c|c|c|c|c|}\hline R_{0} (Gly) & R_{∞} (Gly) & S_{eq} & H_{0} & \Omega_\Lambda & \Omega_m\\ \hline 14&16.7&3280&69.86&0.703&0.297\\ \hline \end{array}}$$ $${\begin{array}{|r|r|r|r|r|r|r|} \hline S=z+1&a=1/S&T (Gy)&R (Gly)&D (Gly)&D_{then}(Gly)&D_{hor}(Gly)&D_{par}(Gly)&a'R_{0}\\ \hline 3.120&0.320513&3.063831&4.486962&17.510&5.612&10.723&8.992&1.000\\ \hline 2.908&0.343879&3.395474&4.944841&16.512&5.678&11.162&9.991&0.974\\ \hline 2.696&0.370920&3.789680&5.478672&15.409&5.715&11.630&11.186&0.948\\ \hline 2.484&0.402576&4.263660&6.104169&14.183&5.710&12.129&12.634&0.923\\ \hline 2.272&0.440141&4.840610&6.839559&12.813&5.639&12.658&14.416&0.901\\ \hline 2.060&0.485437&5.552535&7.704640&11.273&5.473&13.213&16.647&0.882\\ \hline 1.848&0.541126&6.443855&8.717678&9.535&5.160&13.789&19.497&0.869\\ \hline 1.636&0.611247&7.577281&9.888466&7.566&4.625&14.372&23.228&0.865\\ \hline 1.424&0.702247&9.041571&11.204956&5.332&3.745&14.943&28.254&0.877\\ \hline 1.212&0.825083&10.963724&12.613281&2.809&2.317&15.474&35.279&0.916\\ \hline 1.000&1.000000&13.528145&13.999932&0.000&0.000&15.932&45.581&1.000\\ \hline \end{array}}$$

I have changed the 9th column header to be more sensible dot{a}R_0. This corresponds with the values shown on the zoomed center panel below. The redshift of an object that is on the Hubble sphere now is actually z~1.45 or S~2.45. I got that from my old Cosmocalc_2013, with Tamara's values. The z=2.1 represents a more distant galaxy, permanently outside the Hubble sphere, but whose photons managed to reach the Hubble sphere, and hence also to reach us.

Does this make sense?

Edit: Thanks Marcus, I have corrected the z=1.45.
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 Quote by Jorrie I have changed the 9th column header to be more sensible dot{a}R_0. This corresponds with the values shown on the zoomed center panel below. The redshift of an object that is on the Hubble sphere now is actually z=1.67 or S=2.67. I got that from my old Cosmocalc_2013, with Tamara's values. The z=2.1 represents a more distant galaxy, permanently outside the Hubble sphere, but whose photons managed to reach the Hubble sphere, and hence also to reach us. ... Does this make sense?
It makes better sense with the new header!
You should probably check that the number S=2.67 is right. You might have intended, say, S=2.47, and simply misremembered. That's easy to do, memory glitch at one digit and the rest right. We should both check.

I will check using your parameters 14.0, 16.7, 3280. Let me see what I get when I put those in and look for an S that will give me the present distance D = 14.0.

I get S=2.454 using your numbers.

$${\begin{array}{|c|c|c|c|c|c|c|}\hline R_{now} (Gly) & R_{∞} (Gly) & S_{eq} & H_{0} & \Omega_\Lambda & \Omega_m\\ \hline14&16.7&3280&69.86&0.703&0.297\\ \hline\end{array}}$$ $${\begin{array}{|r|r|r|r|r|r|r|} \hline S=z+1&a=1/S&T (Gy)&R (Gly)&D (Gly)&D_{then}(Gly)&D_{hor}(Gly)&D_{par}(Gly)\\ \hline2.454&0.407498&4.338413&6.201108&13.998&5.704&12.202&12.864\\ \hline\end{array}}$$

Using numbers that we were using earlier 14.0, 16.5, 3280 it's more like 2.43 (but about the same.)
$${\begin{array}{|c|c|c|c|c|c|c|}\hline R_{now} (Gly) & R_{∞} (Gly) & S_{eq} & H_{0} & \Omega_\Lambda & \Omega_m\\ \hline14&16.5&3280&69.86&0.72&0.28\\ \hline\end{array}}$$ $${\begin{array}{|r|r|r|r|r|r|r|} \hline S=z+1&a=1/S&T (Gy)&R (Gly)&D (Gly)&D_{then}(Gly)&D_{hor}(Gly)&D_{par}(Gly)\\ \hline2.430&0.411523&4.522759&6.430132&14.028&5.773&12.278&13.434\\ \hline\end{array}}$$
 Recognitions: Gold Member Science Advisor I think I know now what the vertical dashed line labeled z=1.67 is supposed to be. With your numbers 14.0, 16.7, 3280, we get S=2.61 for the intersection of lightcone with Hubble radius. That is, a galaxy we are observing today which was receding at c in the past when it emitted the light. THAT is a galaxy which was subsequently inside the Hubble sphere, and then later was again outside. $${\begin{array}{|c|c|c|c|c|c|c|}\hline R_{now} (Gly) & R_{∞} (Gly) & S_{eq} & H_{0} & \Omega_\Lambda & \Omega_m\\ \hline14&16.7&3280&69.86&0.703&0.297\\ \hline\end{array}}$$ $${\begin{array}{|r|r|r|r|r|r|r|} \hline S=z+1&a=1/S&T (Gy)&R (Gly)&D (Gly)&D_{then}(Gly)&D_{hor}(Gly)&D_{par}(Gly)\\ \hline2.6104&0.383083&3.970&5.7192&14.929&5.7192&11.828&11.737\\ \hline\end{array}}$$ So the vertical line for that galaxy does slice off a bit of the side bulge of the Hubble radius curve, just the way it appears in the figure. First it is outside Hubble sphere, then the sphere expands more rapidly than the galaxy is receding, and takes it in (for a while). Then its recession begins to dominate and it exits. But that galaxy is not NOW at the Hubble radius. Your calculator says that its current distance is 14.929 Gly, not 14.0 Gly. So instead of being labeled "z=1.67" the vertical dashed line probably wants to be labeled "z=1.61" or S=2.61, and to be moved slightly over to the right so that it passes exactly thru the intersection of lightcone with Hubble radius. It will still slice off some of the bulge, on its way up, though slightly less of it. OOPS! EDIT EDIT EDIT! I see you relabeled that to say z=1.45. Now it makes sense, talking about a galaxy which is at comoving distance (now distance) Rnow = 14.0 Gly. So multiplying that by the scale factor a(t) we get the past distance history of that galaxy D(t) = Rnow a(t) OK so that is a sample proper distance history. And you are going to take the slope of that. And the slope should decline at first and then start increasing---the distance growth curve should have an inflection point where the slope is at a minimum. Which, as I recall, it does. Yes! I checked on your table. S=1.636 is where the table minimum of the slope comes. Which is around year 7.6 billion. So that looks quite good. So I can see a real pedagogical benefit. This is making a lot of sense now. I still don't have a definite opinion whether the 9th column pedagogical benefits outweigh the cost of having a more elaborate table. Probably it depends on who one expects to be the user.

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 Quote by marcus I think I know now what the vertical dashed line labeled z=1.67 is supposed to be. With your numbers 14.0, 16.7, 3280, we get S=2.61 for the intersection of lightcone with Hubble radius. That is, a galaxy we are observing today which was receding at c in the past when it emitted the light. THAT is a galaxy which was subsequently inside the Hubble sphere, and then later was again outside. So the vertical line for that galaxy does slice off a bit of the side bulge of the Hubble radius curve. First it is outside Hubble sphere, then the sphere expands more rapidly than the galaxy is receding, and takes it in (for a while). Then its recession begins to dominate and it exits. But that galaxy is not NOW at the Hubble radius. Your calculator says that its current distance is 14.929 Gly, not 14.0 Gly.
I'll have to think about this a little more. A dotted vertical line represents a constant co-moving distance and, I think, a constant redshift over time. Galaxies below z ~ 1.67 must have entered the Hubble radius of the time and later exited it again. Now if the recession speed "then" must have been c when the galaxy entered the Hubble distance and again when it leaves it, there must be a single redshift that satisfies this condition for such galaxies. I could not find such a solution through the calculator, so now I'm a little confused.

What am I missing?