This is sort of a follow-up to the question I asked here: Confusion about timelike spatial coordinates
The important context is that we imagine a metric that, as $t\rightarrow\infty$, approaches the following:
$$ds^2=\text{d}t^2−\text{d}x^2$$
This is a metric in which the spatial coordinate $x$ is timelike, and it allows for closed timelike curves.
Mathematically, it is always possible to find a set of coordinates in which the metric locally looks like the Minkowski metric $\eta_{\mu\nu}$. What I don't understand is why this is physically acceptable in a situation like this.
From what I've learned in Sean M. Carroll's text, the transformation law for the metric at a point $p$ in spacetime is given by $(g_{\hat{\mu}\hat{\nu}})_p=(\frac{\partial x^{\mu}}{\partial x^{\hat{\mu}}}\frac{\partial x^{\nu}}{\partial x^{\hat{\nu}}})_p(g_{\mu\nu})_p$. The Minkowski metric is given by $ds^2=-\text{d}t^2+\text{d}x^2$. If I've done my math correctly (I may have flipped the signs), the necessary derivatives at this point are $$\frac{\partial t}{\partial \hat{t}}=\frac{\partial x}{\partial \hat{x}}=0$$ $$\frac{\partial x}{\partial \hat{t}}=1$$ $$\frac{\partial t}{\partial \hat{x}}=-1$$
But then it would appear that (to first order) we have set $x=\hat{t}$ and $t=-\hat{x}$. In other words, to transfer to a local Minkowski frame, we need to reassign space and time. Why are we even allowed to do this? Does the Einstein Equivalence Principle require it for some reason?
