### Space-time in special relativity

For any space, if you know the distances between all pairs of neighbouring points, then you can reconstruct the whole geometry of that space. This is a fundamental principle of geometry, and is especially relevant to Riemannian geometry of general relativity.

In space-time, distance is replaced by proper time, but the principle still holds, and is encapsulated by the identity $$\mathrm{d}\tau^2 = \mathrm{d}t^2 - \mathrm{d}x^2 - \mathrm{d}y^2 - \mathrm{d}z^2$$ or, in index notation, $$\mathrm{d}\tau^2 = \eta_{\mu \nu} \mathrm{d}X^\mu \mathrm{d}X^\nu$$ where $$\eta_{\mu \nu} \rightarrow \begin{bmatrix}1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1\end{bmatrix}$$ are the components of the special-relativistic space-time metric.

Nb. We could use this statement to introduce special relativity axiomatically. We could show that the set of transformations that preserve proper time are in fact the Lorentz transformations, and this in turn would lead to the statement that the laws of physics are the same in all reference frames.

All of the components of the space-time metric of special relativity are constants. This is the defining characteristic of a flat space (or space-time).

More precisely, we say that a space (or space-time) is **flat** if there exists at least one set of coordinates in which the metric has constant components.

So, the fact that we can find a set of coordinates for space-time in special relativity in which the metric is constant means that space-time in special relativity is flat.

### Two-dimensional space

For simplicity, let's just consider distance, $\mathrm{d}s$, in two-dimensional space, rather than proper time, $\mathrm{d}\tau$, in four-dimensional space-time.

Many different sets of coordinates can be applied to normal two-dimensional space.

The usual Cartesian coordinates $(x, y)$ are characterised by the simplest metric of all, $$\mathrm{d}s^2 = \mathrm{d}x^2 + \mathrm{d}y^2$$

The metric is preserved under rotations, so we can write $$\mathrm{d}s^2 = \mathrm{d}{x'}^2 + \mathrm{d}{y'}^2$$

We can write the distance in component form, $$\mathrm{d}s^2 = \delta_{i j} \mathrm{d}X^i \mathrm{d}X^j$$ where the metric is just the two-dimensional *Kronecker delta*, $$\delta_{i j} \rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

Trivially, since this metric has constant components, we can say that normal two-dimensional space is flat, but we could consider other sets of coordinates for this space.

**Non-uniform scales**

For example, we could consider coordinates with different units on the $x,y$-axes.

Figure 8.2 has *metres* on the $x$-axis and *centimetres* on the $y$-axis.

We couldn't use the usual metric straightaway - we would need to convert both components into the same units first.

If we wanted the distance in *metres*, say, then we would have to substitute $$\mathrm{d}y \rightarrow 100 \mathrm{d}y$$ in the distance formula, $$\mathrm{d}s^2 = \mathrm{d}x^2 + 100^2 \mathrm{d}y^2$$

Note that the associated metric, (which, in its most general form, is given the symbol $g$) still has constant components, $$g_{i j} \rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 100^2 \end{bmatrix}$$

**Polars**

Again, the standard distance formula cannot be applied for the coordinates, $(r, \theta)$, since the second component is angular displacement, rather than linear displacement.

This time, however, the factor that scales the angular displacement, $\mathrm{d} \theta$, in the distance formula is $r$ and *not* constant, since it depends on how far away from the origin we are.

The distance is given by $$\mathrm{d}s^2 = \mathrm{d}r^2 + r^2 \mathrm{d} \theta^2$$ and so the associated metric for polar coordinates is $$g_{i j} \rightarrow \begin{bmatrix} 1 & 0 \\ 0 & r^2 \end{bmatrix}$$

Now, we might conclude - wrongly - that since the components of this particular metric are not all constant, then polar coordinates must therefore be describing a space that isn't flat. But, as we mentioned above, in order to determine whether or not a space is flat, we need only find one set of coordinates in which the metric is constant and this is true for Cartesian coordinates, hence the space is flat.