Metric coefficients in rotating coordinates Let $(t,x,y,z)$ be the standard coordinates on $\mathbb{R}^4$ and consider the Minkowski metric 
$$ds^2 = -dt^2+dx^2+dy^2+dz^2.$$ 
I am trying to compute the metric coefficients under the change of coordinates given by:
$t' = t$
$x' = \sqrt{x^2+y^2}\cos(\omega t + \phi)$
$y' = \sqrt{x^2+y^2}\sin(\omega t + \phi)$
$z' = z$
where $\phi = \tan \frac{y}{x}$. I know one way to do this is to compute write each of the original coordinate maps as a composition of these new coordinate maps and then take derivatives to express $(\partial_{t'},\partial_{x'},\partial_{y'},\partial_{z'})$ in terms of $(\partial_t,\partial_x,\partial_y,\partial_z)$. This is rather cumbersome though and it seems like there should be a more elegant way to do this. Can anyone suggest a cleaner way to transform the metric?
 A: 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.
A: This answer is essentially a combination of @Muphrid's answer, the comments given there,
together with some significant simplifications of my own.
The Minkowski metric rewritten in cylindrical cooordinates is
$$ds^2=-dt^2+dr^2+r^2d\phi^2+dz^2. \tag{1}$$
Your coordinate transformation expressed in cylindrical coordinates simply is
$$\begin{align}
t'&=t \\
r'&=r \\
\phi'&=\phi+\omega t \\
z'&=z
\end{align} \tag{2}$$
Combining (1) and (2) you can immediately write down
the metric in primed cylindrical coordinates $(t',r',\phi',z')$.
There is no need to fiddle with the derivatives
$(\partial_{t'},\partial_{r'},\partial_{\phi'},\partial_{z'})$
and $(\partial_{t},\partial_{r},\partial_{\phi},\partial_{z})$.
$$\begin{align}
ds^2&=-dt'^2+dr'^2+r'^2(d\phi'-\omega\ dt')^2+dz'^2 \\
&=(-1+r'^2\omega^2)dt'^2+dr'^2+r'^2d\phi'^2+dz'^2-2r'^2\omega\ d\phi'dt'
\end{align} \tag{3}$$
Transforming from cylindrical coordinates $(t',r',\phi',z')$
back to cartesian coordinates $(t',x',y',z')$ you finally get
(I leave out the calculational details here)
$$ds^2=(-1+\omega^2(x'^2+y'^2))dt'^2+dx'^2+dy'^2+ dz'^2 + 2\omega(y'dx'-x'dy')dt'. \tag{4}$$
The resulting metric (4) is the Minkowski metric augmented
with two additional terms. The one additional term
($\propto dt'^2$) gives rise to the centrifugal force,
the other ($\propto dt'$) to the Coriolis force.
A: Hints:


*

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

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

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

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

*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,
$$ \frac{\partial}{\partial t^{\prime}}~=~\frac{\partial}{\partial t}
+ \omega\left( y\frac{\partial}{\partial x}
-x\frac{\partial}{\partial y}\right), \qquad 
\frac{\partial}{\partial u^{\prime}}
~=~e^{-i\omega t}\frac{\partial}{\partial u}. $$
