We know that (see this wikipedia page) in the metric of Minkowski spacetime:
$$ds^2=(dt)^2-(dx)^2-(dy)^2-(dz)^2 \tag{1}$$
and we also know that in spherical coordinates this same metric becomes:
$$ds^2=(dt)^2-(dr)^2-r^2(d\theta)^2-r^2\sin^2{\theta}(d\phi)^2 \tag{2}$$
Let's prove this last statement:
We have that:
$$\begin{cases}t=t \\ x=r\sin{\theta}\cos{\phi} \\ y=r\sin{\theta}\sin{\phi} \\ z=r\cos{\theta}\end{cases}$$
we can think of $x,y,z$ as functions of $r,\theta,\phi$; so we get:
$$dx=\sin{\theta}\cos{\phi}dr+r\cos{\theta}\cos{\phi}d\theta-r\sin{\theta}\sin{\phi}d\phi$$
and so on for $dy,dz$; then we can square to get $(dx)^2,(dy)^2,(dz)^2$ written in terms of $(dr)^2,(d\theta)^2,(d\phi)^2$. Now we can put our findings back into equation (1) and if all goes right we should find equation (2).
However I didn't get to the end of this calculation because the algebra gradually becomes unbearable, especially when you get to the squaring part, where terms with mixed differentials start to pop out. However seems to me that this method should work out fine.
My questions are: This method will lead to the correct solution (2)? And even if this method is indeed correct: is there a better method to demonstrate (2) from (1)? Where better here means simply less algebra.
Best Answer
Yes! There's a way more simple method of converting the metric into spherical co-ordinates. In cartesian co-ordinates, the expression of the metric is of the form
$$\mathrm ds^2=-c^2\mathrm dt^2+(\text{infinitesimal displacement})^2\tag{1}$$
In cartesian co-ordinates,
$$\text{infinitesimal displacement}=\sqrt{\mathrm dx^2+\mathrm dy^2+\mathrm dz^2}$$
So now, our task is to find such an infinitesimal displacementin spherical co-ordinates. This is a purely mathematical task. Let's start of with a figure.
Image source
In the above image, all the three paths are mutually perpendicular/orthogonal, so the net displacement will just be the
$$\text{infinitesimal displacement}=\sqrt{(\text{path 1})^2+(\text{path 2})^2+(\text{path 3})^2}\tag{2}$$
But it's easy to see that
\begin{align} \text{path 1}&=r\mathrm\: d \theta\\ \text{path 2}&=r\sin \theta \: \mathrm d\phi\\ \text{path 3}&=\mathrm dr \end{align}
And voila, substitute the above expressions into equation $(2)$ and subsequently into equation $(1)$ to get the desired result.