Scale factor and age of the universe

Question

Why is the cosmological scale factor (expansion rate of the universe) not simply the time $t$, i.e. the age of the universe?

Answer 1

The expansion of the universe is described by the Friedmann equations, which is a particular set of solutions of the General-Relativistic Einstein field equations. As you may know, these equations describe how spacetime is curved by matter and energy.

As a result, the scale factor $a(t)$ depends on the density of radiation, matter and energy. This also means that the expansion rate $\dot{a}(t)$ depends on how these densities change as the universe expands.

And this change is different for each of these densities: the matter density decreases a $\rho_\text{M}\sim a^{-3}$ (this is simply conservation of matter in a co-moving volume). The radiation density scales as $\rho_\text{R}~a^{-4}$ (the extra factor is because photons lose energy due to redshift). And the dark energy density $\rho_\Lambda$ remains constant (at least in the Standard Model).

If we put all this together, it turns out that $$ \frac{\dot{a}}{a} = H_0\sqrt{\Omega_{R,0}a^{-4} + \Omega_{M,0}a^{-3} + \Omega_{K,0}a^{-2} + \Omega_{\Lambda,0}}\,, $$ where $H_0$ is the Hubble constant, the $\Omega$'s are the present-day densities relative to the critical density $\rho_{c,0} = 3H_0^2/(8\pi G)$, and $$ \Omega_{K,0} = 1 - \Omega_{R,0} - \Omega_{M,0} - \Omega_{\Lambda,0}. $$ Observations indicate that $\Omega_{R,0}\approx 0$, $\Omega_{K,0}\approx 0$, $\Omega_{M,0}\approx 0.3$ and $\Omega_{\Lambda,0}\approx 0.7$.

You can solve this equation to calculate $t(a)$ (see this post) and invert it to get $a(t)$. As you can see, this is not simply a linear function of $t$, but a much more complicated function.

Answer 2

For any other readers of this post, let me quickly say what the scale factor is. The scale factor, $a(t)$, is a term in Cosmology that in layman's terms represents the expansion rate of the universe. The question, therefore, is asking why the expansion of the universe is not written as being linear with respect to time.

There are a few reasons why $a(t)\neq t$.

First (and if you ask a mathematician, most importantly), we have evidence that the expansion rate of the universe is accelerating. This means that $\ddot a(t)=\frac{d^2 a(t)}{d t^2}$ must be positive, which implies that $\dot a(t)=\frac{d a(t)}{d t}$ must be a function of $t$ (because of integrals and whatnot), which further means that $a(t)\neq t$ because if it did, its first derivative would be a constant and its second derivative would be zero.

If you were to ask an astronomer or a cosmologist, however, they might give you reason the second: Based on patterns of redshifts, other empirical evidence, and some firmly grounded theories, we have strong reason to believe that the universe is expanding in a way that is asymptotically converging toward an exponential function of $t$. Thus, the scale factor would have to be approaching an exponential function over time.

If you want to read more and perhaps learn about some of the reasoning behind the information in my answer, I suggest starting with the Wikipedia page and exploring its linked pages and references. It is all great material.

Answer 3

The expansion model implied by this question is known as the "coasting universe".

Triggered mostly by the observed coincidence that the gravitational radius $R_0$ of the universe at current time $t_0$ equals $c t_0$, once in a while research papers appear that try to fit cosmological observations to this model. However, the observed acceleration of the cosmological expansion renders such exercises futile. (See also: this question.)

Answer 4

A smart-aleck answer to this would be to ask "what do you mean by $t$?"

In general relativity, we have general coordinate invariance, so we are always free to rescale the time coordinate to a new coordinate $\tau$ by an arbitrary mapping $\tau = f(t)$, so long as $f(t)$ is an invertible function. Since, for all solutions of the Robertson-Walker metric, the metric is piecewise invertible, it is always possible, in a finite neighborhood of a point $t$, to choose $f(t) = a(t)$, and then, your coordinate time IS the cosmological scale factor. Of course, the price we pay for this is having the metric go from it's nice form:

$$ds^{s} = - dt^{2} + a(t)^{2} d{\vec r}^{3}$$

(where $d{\vec r}^{3}$ is the spatial metric) to the "uglier" form:

$$ds^{s} = - \frac{1}{{\dot a}^{2}}d\tau^{2} + \tau^{2} d{\vec r}^{3}$$

where ${\dot a}$ is the derivative of $a$ fed through the inverse of $a$ (so that it is a function of $\tau$ rather than $t$ now).

If by "$t$", you mean "the proper time of a stationary observer at $r=0$", see the other answers.

Answer 5

Because it doesn't match scientists' theory on the rate of expansion of the universe. It would indicate that the universe is less than 7000 years old which conflicts directly with their theoretical universal age of 14 billion years.