Minkowski metric: Why does it look like it does? [duplicate]

Question

I have been searching for why would we even start with Minkowski spacetime metric as being written as:

$$ds^2=-dt^2+dx^2+dy^2+dz^2.$$

No really, so why would we have a negative sign for temporal side and positive for spatial side (or vice versa). How did Minkowki along with Einstein reach this metric?

Few of the answers I got were:

Because Einstein wanted preserve the fact that speed light is constant so he reached this... but I never understood this reasoning.

Another one

time being a new coordinate is treated as $it$ and thus we get the interval of $\sqrt{(it)^2+x^2+y^2+z^2}$ the same way we write an interval in Euclidean space as $\sqrt{(x^2+y^2+z^2}$).

I persoanlly haven't gotten very convinced by the second answer. So, if someone can explain the first answer (mathematically maybe), or suggest another reasoning I would be very grateful.

Answer 1

Thanks to Pythagoras, we know Euclidean space satisfies $\text{d}s_E^2=\text{d}x^2+\text{d}y^2+\text{d}z^2=\text{d}r^2+r^2\text{d}\theta^2+r^2\sin^2\theta\text{d}\phi^2$, where the last result is obtained by the chain rule. Clearly, any manifold with local coordinates satisfies a result of the form $\text{d}s^2=\sum_{\mu\nu}g_{\mu\nu}\text{d}x^\mu\text{d}x^\nu$, where the matrix representation of the metric tensor $g_{\mu\nu}$ depends on the choice of coordinate system. This matrix has to be symmetric and invertible for manipulations such as $\text{d}x_\mu=\sum_\nu g_{\mu\nu}\text{d}x^\nu$. With a suitable coordinate system, we can diagonalise the matrix and its diagonal entries will be its eigenvalues, which will be nonzero. The crucial question is which are positive and which are negative, and these details are called the signature of the metric.

A non-Newtonian theory introduces time into $\text{d}s^2$. Working with a coordinate system that diagonalises the metric tensor, "flat" spacetime has constant eigenvalues which can be scaled to $\pm 1$. Thus $\text{d}s^2=\text{d}s_E^2\pm\text{d}t^2$. If we choose a $+$ sign, we just have Euclidean space with another dimension. But if we choose a $-$ sign, the $\text{d}s^2$-preserving linear transformations of $\text{d}x^\mu$, which for Euclidean space would be rotations, become what are called Lorentz transformations. You can then check this gives the time dilation, Lorentz contraction, energy-momentum relation etc. that lectures often initially derive for motion along a straight line (which saves us two space coordinates).