Begin by letting the standard Minkowski metric in ##(t,x,y,z)## be
$$
ds^2 = c^2\,dt^2 \;-\; dx^2 \;-\; dy^2 \;-\; dz^2.
$$
Define a vector field ##X## on ##\mathbb{R}^4## by
$$
X = \omega\,\bigl(y\,\partial_x \;-\; x\,\partial_y\bigr),
$$
where ##\omega## is a constant. Observe that ##X## generates rotations in the ##(x,y)##-plane at angular speed ##\omega##. For each fixed ##t## and ##z##, let ##\bigl(x(\sigma), y(\sigma)\bigr)## satisfy the flow equation
$$
\frac{d}{d\sigma}
\begin{pmatrix} x(\sigma) \\[4pt] y(\sigma) \end{pmatrix}
=
\omega
\begin{pmatrix} y(\sigma) \\[2pt] -\;x(\sigma) \end{pmatrix},
\quad
\bigl(x(0), y(0)\bigr) = (x_0, y_0).
$$
Solving this gives
$$
\begin{pmatrix} x(\sigma) \\[3pt] y(\sigma) \end{pmatrix}
=
\begin{pmatrix}
x_0 \cos(\omega \sigma) \;+\; y_0 \sin(\omega \sigma)\\[4pt]
-\;x_0 \sin(\omega \sigma) \;+\; y_0 \cos(\omega \sigma)
\end{pmatrix}.
$$
Define a diffeomorphism ##\Phi:\mathbb{R}^4\to\mathbb{R}^4## by letting ##\Phi(t,x,y,z)## be the point
$$
\bigl(t, x(t), y(t), z\bigr)
\quad
\text{where}
\quad
\begin{pmatrix}x(t)\[2pt]y(t)\end{pmatrix}
=
\begin{pmatrix}x\[2pt]y\end{pmatrix}
\ast
\text{the above flow from } \sigma=0 \text{ to } \sigma=t.
$$
Hence, for each fixed ##(x,y)##, we follow the integral curve of ##X## up to the “time” ##\sigma = t##. Explicitly, if ##\Phi(t,x,y,z) = (t', x', y', z')##, then
$$
t' = t,
\quad
\begin{pmatrix}x'\[3pt]y'\end{pmatrix}
=
\begin{pmatrix}
x\cos(\omega t) \;+\; y\sin(\omega t)\[3pt]
-\;x\sin(\omega t) \;+\; y\cos(\omega t)
\end{pmatrix},
\quad
z' = z.
$$
Label the new coordinates by ##(t', x', y', z')##. In these coordinates, write the differentials
$$
dt' = dt,
\quad
dx' = \frac{\partial x'}{\partial t}\,dt + \frac{\partial x'}{\partial x}\,dx + \frac{\partial x'}{\partial y}\,dy,
$$
$$
dy' = \frac{\partial y'}{\partial t}\,dt + \frac{\partial y'}{\partial x}\,dx + \frac{\partial y'}{\partial y}\,dy,
\quad
dz' = dz.
$$
Substitute into
$$
ds^2 = c^2\,dt^2 \;-\; dx^2 \;-\; dy^2 \;-\; dz^2.
$$
For instance,
$$
dx = \bigl[\cos(\omega t)\,dx' \;-\;\sin(\omega t)\,dy'\bigr]
$$
plus an extra piece from the partial derivatives with respect to ##t##. Carefully expand:
$$
dx = \cos(\omega t)\,dx' + \sin(\omega t)\,dy'
\quad
\text{only if } dt=0,
$$
but here ##t' = t## implies there is also
$$
\frac{\partial x}{\partial t} = -\,\omega\,y(\sigma)\ldots
$$
Writing each of ##dx## and ##dy## in terms of ##dt', dx', dy'## leads to cross-terms of the form ##dt'\,d\phi'##. After all substitutions, one finds that the metric in ##(t',x',y',z')## picks up off-diagonal terms and changes in the coefficient of ##dt'^2##. In cylindrical-like form (setting ##r' = \sqrt{x'^2 + y'^2}##, ##\phi'## its polar angle), the resulting line element becomes
$$
ds^2
=
\bigl(c^2 - \omega^2\,r'^2\bigr)\,dt'^2
\;-\;2\,\omega\,r'^2\,dt'\,d\phi'
\;-\;dr'^2
\;-\;r'^2\,d\phi'^2
\;-\;dz'^2.
$$
Symbolically,
$$
ds^2
=
c^2\,dt'^2
\;-\;
(\text{some expression})\,dt'\,d\phi'
\;-\;\dots
$$
where ##\phi'## is the azimuthal angle around the ##(x',y')##-plane. The key observation is that ##\partial/\partial t'## in the primed system is not globally orthogonal to the spatial directions, which reveals that a particle’s proper time ##\tau## is *not* simply ##\tau = t'\sqrt{1 - v^2/c^2}##. Instead, we obtain
$$
d\tau^2
=
-\frac{1}{c^2}\,g_{\mu\nu}\,dx'^\mu\,dx'^\nu,
$$
where ##g_{\mu\nu}## is the induced metric from the above transformation. If the particle remains “at rest” with ##(x',y')## fixed (so ##dr'=0## and ##d\phi'=0##), then
$$
ds^2
=
\bigl(c^2 - \omega^2\,r'^2\bigr)\,dt'^2,
\quad
d\tau
=
\sqrt{1 - \frac{\omega^2\,r'^2}{c^2}}\;dt',
$$
proving there is a time-dilation factor at radius ##r'##. In general, a trajectory at rest in the rotating chart is actually undergoing centripetal acceleration in the original inertial frame, and the integral
$$
\tau
=
\int
\sqrt{
-\,g_{\mu\nu}\,
\frac{dx'^\mu}{d\lambda}\,
\frac{dx'^\nu}{d\lambda}
}
\,d\lambda
$$
will differ from
##\int dt'##, showing that time coordinate labels in a rotating chart are not identical to the proper time of rotating observers. The non-orthogonality of the time coordinate to spatial directions is made explicit by the off-diagonal components ##g_{t'\phi'} \neq 0##. This geometric fact ensures that we cannot simply take the usual inertial-frame time-dilation factor ##\sqrt{1 - v^2/c^2}## to compute elapsed proper time. Instead, all such relativistic effects emerge from the full metric tensor, which carries the imprint of rotation through these mixing terms, even though we began in flat Minkowski space.
