# FRW metric, convention misunderstanding?

1. Jul 12, 2015

### AstroPhysWhiz

So I have been following various derivations of the FRW metric and have a bit of confusion due to varying convention...

Would it be correct to say that curvature K can be expressed as both $$K = \frac{k}{a(t)^2}$$ and $$K = \frac{k}{R(t)^2}$$ where k is the curvature parameter?

If so, is it correct to say that the spatial line element for the k = 1 (closed) case may be expressed as

$$dl^2=\frac{dr^2}{1-\frac{r^2}{R(t)^2}}+ r^2d\Omega^2$$

If I then sub
$$r =R(t)\sin(\chi)$$,

using the fact that the full line element is
$$ds^2 = dt^2 - a(t)^2dl^2$$

I find
$$ds^2 = dt^2 - a(t)^2R(t)^2[d\chi^2 + \sin^2(\chi) d\Omega^2]$$

but the texts I have read state the metric to be

$$ds^2 = dt^2 - R(t)^2[d\chi^2 + \sin^2(\chi) d\Omega^2]$$

or

$$ds^2 = dt^2 - a(t)^2[d\chi^2 + \sin^2(\chi) d\Omega^2]$$

so I am clearly misunderstanding something with my extra factor, anyone able to clear things up for me?

Last edited by a moderator: Jul 12, 2015
2. Jul 12, 2015

### Chalnoth

You're mixing conventions.

First of all, a(t) and R(t) are, for the most part, just different names for the exact same parameter. They're both the scale factor.

There are also two different conventions for the value of the scale factor.

In one convention, we say that the scale factor today is equal to one. This makes $k$ into a floating-point value that is related to the current radius of curvature, $k = 1 / R_c^2$.

The other convention sets $k$ to be an integer value, with either $k = \{1, 0, -1\}$. With this convention, if $k$ is plus or minus one, then the scale factor is equal to the radius of curvature at that point in time. If $k$ is zero, then there is no radius of curvature and the scale factor takes on an arbitrary scaling.

Usually people use the variable $R(t)$ for the second convention, and the variable $a(t)$ for the first. But not all the time.