### Metrics

Metrics in general relativity are rank-two tensors, denoted by $g_{\mu \nu}$, and, in general, are functions of the coordinates in which they defined, $$g_{\mu \nu} = g_{\mu \nu}\left(X^\gamma\right)$$ and in the most abstract terms, a **metric** allow us to completely reconstruct the space upon which it has been defined, since it gives us a formula for the *distance* between neighbouring points in that space.

In relativity, both general and special, the space we are considering is space-time. That is, it's actually the *proper time* rather than ‘normal’ spatial distance that defines the metric.

For special relativity, the space-time metric is especially simple, $$g_{\mu \nu} = \eta_{\mu \nu} \rightarrow \begin{bmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix}$$ but the four assumptions we made about the universe, namely that it is homogeneous, isotropic, flat and expanding over time, led us to the metric $$g_{\mu \nu} \rightarrow \begin{bmatrix}1&0&0&0\\0&-a(t)^2&0&0\\0&0&-a(t)^2&0\\0&0&0&-a(t)^2\end{bmatrix}$$ where the *scale factor*, $a(t)$, was an increasing function of time, as yet undefined.

It is somewhat by convention that we keep time component of the metric, $$g_{00} = 1 (\equiv c)$$ although we could consider other alternatives. For example, we don't expect the time component to change over time, but we could imagine that is a constant other than $1$. But, then, we could just absorb it into a different set of coordinates that did have $g_{00} = 1$

Similarly, we choose to retain the minus-sign in front of the spatial diagonal components, $$g_{mm} = -a(t)^2$$ instead of, for example, including it into the scale factor, $a(t)$.

### Curved three-dimensional spaces

Relativity in general deals with four-dimensional space-time. That is, one dimension of time and three dimensions of space, so when we talk about space being flat or curved, we are talking about a three-dimensional space.

Now, a three-dimensional sphere is the curved surface of a four-dimensional ball. It is impossible to visualise, but note that the mathematics of a 3-sphere is generalised from a 2-sphere, so for example we can write $$\begin{align*}2\text{-sphere} &\rightarrow \mathrm{d}x^2 + \mathrm{d}y^2 + \mathrm{d}z^2 = \mathrm{d}r^2\\3\text{-sphere} &\rightarrow \mathrm{d}x^2 + \mathrm{d}y^2 + \mathrm{d}z^2 + \mathrm{d}w^2 = \mathrm{d}r^2\end{align*}$$

So, we can use the 2-sphere to visualise a positively curved, homogeneous, isotropic 3-space, but we just have to remember that we are really dealing in three dimensions of space, not two. Similarly we can use a 2-saddle to visualise a negatively curved, homogeneous, isotropic 3-space.

Suppose soon after the big bang, the whole of the observable universe was crammed together into a 3-space of the order of $\text{cm}^3$, say. There is no sense in which any particle in that space can be considered to be a special location, or moving in a special direction, whether the space is flat or curved. And as this universe expands, the distance and relative speed between two particles will increase, whether the space is flat or curved.

For example, consider any two points on a small 2-sphere. If we allow the sphere to expand over time, then both the distance between the points and the speed with which they move apart will increase over time, but they will remain on the sphere. Thus, we can see that it is perfectly reasonable for us to have assumed that a space can be homogeneous, isotropic, expanding and *curved*, rather than flat.

In fact, the closed, finite, spherical universe was originally the most popular cosmological model, and is still possibly correct, but curvature is still undetectable in the observable universe (at large enough scales) and so the radius of curvature ($r$ as above) must be larger than the observable universe.

With that in mind, for our purposes we can assume that space is flat.