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How large is the non-normalized scale factor (solution of the Friedmann equation) in today‘s universe? How it is measured?

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    $\begingroup$ A non-normalized scale factor would only make sense if the Universe is not flat, in which case you can use the radius, usually denoted $R_0$. If the Universe is open, $R_0$ is imaginary, whereas if it's closed, it's a real number. A recent paper (Di Valentino et al. 2019) suggests that the Universe is in fact closed, and in this answer on astronomy.SE I show that this would imply a radius of the Universe of $R_0 \sim 67_{-21}^{+100}$ billion light-years. $\endgroup$ – pela Nov 21 '19 at 15:02
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    $\begingroup$ @pela good comment---why not turn it into an answer? (you would need to mention $a$ as well as $R$). $\endgroup$ – Andrew Steane Nov 21 '19 at 17:04
  • $\begingroup$ @pela If you use $k=-1,0,1$ then you have to scale first $k \to k/|k|$ and $R_0 \to R_0/\sqrt{|k|}.$ Therefore, your approach gives $R_0/\sqrt{|k|}$ and not $R_0.$ $\endgroup$ – user185188 Nov 21 '19 at 19:11
  • $\begingroup$ Hmm…you're right that the formula I use in the answer linked to (i.e. $\Omega_K = -kc^2\,/\,R_0^2a(t)^2H(t)^2$) defines $k$ as either $-1$, $0$, or $+1$, so $\sqrt{|k|} = 1$ for $k=\pm1$. That should then give the absolute size $R_0$. Wouldn't you agree on this, @AndrewSteane? $\endgroup$ – pela Nov 21 '19 at 20:56
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The scale factor $a$ has an arbitrary normalization. Typically people take $a=1$ for the present time.

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  • $\begingroup$ Yes, but how large is the real physical $a(t_\text{today})$? $\endgroup$ – user185188 Nov 21 '19 at 19:12
  • $\begingroup$ @user185188 What do you mean by the "real physical" $a(t_{today})$? $\endgroup$ – probably_someone Nov 21 '19 at 23:57
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    $\begingroup$ That question doesn’t make sense. It’s a relative scale factor, not an absolute distance. $\endgroup$ – G. Smith Nov 22 '19 at 1:33
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Because we don't know whether the Universe is finite or infinite, it does not make sense to talk about an absolute size, or radius, of the Universe. Instead we use an arbitrary, dimensionless scale called the scale factor, usually denoted $a$.

As the Universe expands, all (cosmological) distance scale with $a$. We define this to be equal to unity today, and say that, for instance, when the Universe was half the (linear) size, $a$ was equal to $0.5$. This number can then be multiplied on physical scales to give physical distance.

In principle you could use a non-normalized scale factor, i.e. with dimensions of length. But you would be completely free to choose whichever distance you wish, since an infinite universe doesn't have a preferred scale.

If the cosmological principle — which states that the Universe is homogeneous and isotropic on sufficiently large scales — holds true, then there are three possible geometries for the Universe: flat, open, and closed. Only the Universe is closed is it finite. In this case you could use a non-normalized scale factor equal to the radius of the Universe, which is equal the its radius of curvature. This is usually called $R_0$.

If the Universe were open, it's infinite, but still has an associated radius of curvature; however it would be an imaginary number, so it would make less sense to use as a standard for measuring distance (though you could normalize it).

As I discuss in this answer on astronomy.SE, a recent paper by Di Valentino et al. (2019) claims to have evidence (in the Planck 2018 data) for a closed Universe, with a curvature density parameter of roughly $\Omega_K\simeq-0.04$. This would correspond to a radius of the Universe of $R_0 \sim 67_{-21}^{+100}\,\mathrm{Glyr}$, not much larger than the observable Universe.

But the important thing is that, even if the Universe is finite and we can measure its radius, this would not make $R_0$ any more "correct" as the scale factor as any other distance you can come up with. So there really is no use for a non-normalized scale factor.

Moreover, $R_0$ will always have an associated error, whereas $a$, by definition, is errorless.

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  • $\begingroup$ In my opinion the equation which defines $\Omega_K$ cannot be used to determine the scale factor. There are two unknowns $k$ and $R_0$, which cannot be independently normalized and that should be determined from one equation. $\endgroup$ – user185188 Nov 21 '19 at 21:38
  • $\begingroup$ @user185188 No, with this definition, $k$ can only take the values $-1, 0, +1$. $\endgroup$ – pela Nov 22 '19 at 0:31
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    $\begingroup$ I think this answer is correct. $\Omega_K$ is itself just a way to write the curvature $K$ in units that are convenient for some purposes. There is a well-defined and unique curvature $K$ at any given cosmic time, and hence a well-defined and unique $\sqrt{|1/K|}$ which is called a radius of curvature $R$. For positive curvature one has a 3-sphere whose volume is $2 \pi^2 R^3$. $\endgroup$ – Andrew Steane Nov 22 '19 at 18:27
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How large is the non-normalized scale factor (solution of the Friedmann equation) in today‘s universe?

Scale factor by itself does not mean anything. The important thing is the ratios of the two scale factors.

If we are describing the expansion of the universe in terms of the scale factor, the important question becomes, scale factor relative to what ?

How it is measured?

You cannot measure a scale factor. Its not a measureable quantity. What matters is that the ratios of the scale factor.

There's an equation that relates $z$ and $a(t)$,

$$1+z = \frac{a(t_0)} {a(t_e)}$$

Let us say we looked at an object and observed $z=3$ this means that

$$4 = \frac{a(t_0)} {a(t_e)}$$

Here you can see that

$$\frac{a(t_0)} {a(t_e)} = \frac{4} {1} = \frac{10^{10}} {25 \times 10^8} = \frac{0.20} {0.05} = \frac{1} {0.25}$$

For $a(t_0)$ you can choose any value. Since its not an observable quantity it does not matter what you choose. But just for simplicity we choose $a(t_0) = 1$.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – tpg2114 Nov 22 '19 at 18:06

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