Metric coefficients in rotating coordinates

Edit edit: as has been pointed out, I was incorrect to say $\partial_t = \partial_{t'}$ and so on. Serves me right for trying to look at it by inspection instead of being rigorous.

Nevertheless, I do think cylindrical coordinates simplifies the problem somewhat. Recall the cylindrical line element:

$$ds^2 = -dt^2 + dr^2 + r^2 \, d\phi^2 + dz^2$$

Now, there are several ways you can compute $\partial_{\mu'}$. In this case, it's simple enough to invert the coordinate system transformation, which naturally expresses the Jacobian terms in the correct manner. Note that $r' = r$ and $\phi' = \phi + \omega t$. We can then start reading off the transformations of the partial derivatives.

$$\begin{align*} \partial_{t'} &= \frac{\partial t}{\partial t'} \partial_t + \frac{\partial \phi}{\partial t'} \partial_\phi = \partial_t - \omega \partial_\phi \\ \partial_{r'} &= \frac{\partial r}{\partial r'} \partial_r = \partial_r \\ \partial_{\phi'} &= \frac{\partial \phi}{\partial \phi'} \partial_\phi = \partial_\phi \\ \partial_{z'} &= \frac{\partial z}{\partial z'} \partial_z = \partial_z\end{align*}$$

These give as a metric,

$$ds^2 = ({r'}^2 \omega^2 - 1) \, {dt'}^2 + {dr'}^2 + {r'}^2 \, {d\phi}' + {dz'}^2$$

There is only one derivative of any consequence to calculate. Moreover, in cylindrical coordinates, we can see clearly that there is some strange stuff going on at $r' = 1/\omega$.

Convert this back into your primed cartesian coordinates, and you're done.

Hints:

1. The coordinate $z^{\prime}=z$ is a passive spectator variable, so one may consider the reduced $2+1$ dimensional problem.

2. View the remaining two spatial coordinates as one complex coordinate, i.e., $$u~:=~x+iy, \qquad u^{\prime}~:=~x^{\prime}+iy^{\prime}. $$

3. The rotational transformation then simplifies to $$t^{\prime}=t, \qquad u^{\prime}~=~ u e^{i\omega t}. $$

4. Recall the chain rules $$ \frac{\partial}{\partial t} ~=~\frac{\partial t^{\prime}}{\partial t}\frac{\partial}{\partial t^{\prime}} +\frac{\partial x^{\prime}}{\partial t}\frac{\partial}{\partial x^{\prime}} +\frac{\partial y^{\prime}}{\partial t}\frac{\partial}{\partial y^{\prime}}, $$ $$ \frac{\partial}{\partial u} ~=~\frac{\partial t^{\prime}}{\partial u}\frac{\partial}{\partial t^{\prime}} +\frac{\partial u^{\prime}}{\partial u}\frac{\partial}{\partial u^{\prime}} +\frac{\partial \bar{u}^{\prime}}{\partial u}\frac{\partial}{\partial \bar{u}^{\prime}}. $$

5. Derive via the chain rule $$ \frac{\partial}{\partial t}~=~\frac{\partial}{\partial t^{\prime}} + \omega\left( x^{\prime}\frac{\partial}{\partial y^{\prime}} -y^{\prime}\frac{\partial}{\partial x^{\prime}}\right), \qquad \frac{\partial}{\partial u} ~=~e^{i\omega t}\frac{\partial}{\partial u^{\prime}}, $$ or conversely, $$ \frac{\partial}{\partial t^{\prime}}~=~\frac{\partial}{\partial t} + \omega\left( y\frac{\partial}{\partial x} -x\frac{\partial}{\partial y}\right), \qquad \frac{\partial}{\partial u^{\prime}} ~=~e^{-i\omega t}\frac{\partial}{\partial u}. $$