# Trying to understand the normalisation of the scale factor to be 1 today

1. Aug 10, 2012

### Heldo Jelbar

Hello all! I'm trying to understand the standard normalisation of the scale factor to be set to 1 at today's time. Looking at the first Friedmann Equation for a spatially flat Robertson Walker metric with no cosmological constant gives

$$\frac{\dot{a}^2}{a^2} = \frac{8\pi G}{3}\rho$$

If we wanted to see how the density of the universe changed from the beginning of the matter dominated era to today, we would set

$$a(t) = t^{2/3}$$

This means that,

$$\frac{\dot{a}^2}{a^2} = \frac{4}{9t^2}$$

inserting this back into the Friedmann Equation, we get

$$\rho = \frac{1}{6\pi Gt^2}$$

So we see that in a expanding universe the density decreases as $1/t^2$, which is sensible. But my question is this: if we normalise the scale factor $a(t)$ such that $a(t_0) = 1$, where $t_0$ is today's time, then one way of doing this is to use units where $t_0 = 1$. This then would make $a(t_0) = 1$ straightforwardly for any power law expansion of scale factor. But normalising the scale factor in this way messes with the density time relation. As all times in the past have $t< 1$, a $1/t^2$ relation will actually show that the density is INCREASING in time as the universe expands, as $t$ is less than one before today. But this is no longer sensible.

So does anyone know how to correctly normalise the scale factor to avoid this issue? Any answers with their justifications would be great, and a reference to where I can read more about this would be even better! Many many thanks in advance!

2. Aug 10, 2012

### phyzguy

What are you talking about? If the density ~ 1/t^2, the density will always be decreasing as t increases, regardless of whether t>1 or t<1.

3. Aug 10, 2012

### Heldo Jelbar

Sorry, you're right. There was a deeper reason I was asking however. There seem to be two different ways in which the scale factor can be normalised, at least as far as I can see. Either you can:

1) Change your units of time such that $t_0 = 1$, and use $a(t) = t^{2/3}$

2) Normalise the scale factor by setting $a(t) = \left(\frac{t}{t_0}\right)^{2/3}$

I'm not sure which is the correct one. In the first case simply changing the units means that the scale factor has dimensions, time to the power 2/3. In the second case the scale factor is dimensionless. I think the correct answer is the second one, and I was trying to think of why the option 1) wouldn't be allowed, which made me make the mistake you pointed out. I think that the scale factor has to be dimensionless.

Does anyone know how exactly the scale factor is normalised? Is the normalisation factor as simple as in option 2), or are there more terms? Do you know where I can find out more? For instance, how is the value of $t_0$ calculated?

4. Aug 10, 2012

### phyzguy

Well, I'm not saying this is the only way to do it, but I think the convention is that the scale factor is defined to be a ratio between the proper distance at time t and the proper distance at time t0. As such it is dimensionless, and the scale factor at time t0 = 1.0. It is then arbitrary what time is defined as t0, but it is usually taken as t0 = today.

5. Aug 11, 2012

### Chalnoth

Well, setting $a(t_0) = 1$ is a trivial operation, because the scale factor has no units. However, setting $t_0 = 1$ is not a trivial operation, because $t_0$ has units. If you are using kilograms-meters-seconds units, for instance, performing the manipulations as you have above essentially ends up setting $t_0 = 1s$, not $t_0=1$. And arbitrarily setting the current age of the universe to one second is obviously wrong.

One simple way to take care of this would be to just put every time in the past in terms of $t_0$. This would be equivalent to Heldo Jelbar's second equation:

$$a(t) = \left({t \over t_0}\right)^{2 \over 3}$$