So, from Gron (p. 89), using the convention that spacelike intervals squared are positive, in a rotating reference frame with cylindrical coordinates given by:
(t,r,\theta,z)
The line element is:
ds^2 = -\gamma^{-2} c^2 dt^2 + dr^2 + 2 r^2 \omega dt d\theta + r^2 d\theta^2 + dz^2
where
\gamma = (1 - r^2 \omega^2/c^2)^{-1/2}
And the metric tensor is:
\mathbf g =<br />
\left(<br />
\begin{array}{cccc}<br />
-\gamma ^{-2} c^2 & 0 & r^2 \omega & 0 \\<br />
0 & 1 & 0 & 0 \\<br />
r^2 \omega & 0 & r^2 & 0 \\<br />
0 & 0 & 0 & 1<br />
\end{array}<br />
\right)<br />
NB \gamma is given by Gron as part of the line element and metric for the rotating frame in equations 5.3-5.5, and is only equal to 1 for the special case of \omega=0.
Finally, some of the Christoffel symbols in the rotating reference frame are non-zero (Gron p. 149). Specifically:
\Gamma^{r}_{tt}=-\omega^2r
\Gamma^{r}_{\theta \theta}=-r
\Gamma^{r}_{\theta t}=\Gamma^{r}_{t \theta}=-\omega r
\Gamma^{\theta}_{rt}=\Gamma^{\theta}_{tr}=\omega/r
\Gamma^{\theta}_{\theta r}=\Gamma^{\theta}_{r \theta}=1/r
Now, the worldline of a particle starting on the x-axis at t=0 and undergoing uniform circular motion at angular velocity \omega in the x-y plane in an inertial frame is given by the following expression in the rotating frame:
\mathbf X = (t,r_0,0,0)
From this we can derive the four-velocity in the rotating frame as follows:
\mathbf U = \frac{d \mathbf X}{d \tau} = i c \frac{d \mathbf X}{ds} = i c \frac{d \mathbf X}{dt} \frac{dt}{ds} = i c \; (1,0,0,0) \; \frac{1}{\sqrt{-\gamma^{-2} c^2}} = (\gamma,0,0,0)
The norm of the four-velocity is given by:
||\mathbf U||^2=U_{\mu} U^{\mu}= g_{\mu\nu} U^{\nu} U^{\mu} = -c^2
So this agrees with my previous results so far as expected since the norm is a frame invariant quantity.
Now we can derive the four-acceleration in the rotating frame as follows:
A^{\mu}=\frac{DU^{\mu}}{d\tau}=\frac{dU^{\mu}}{d\tau}+\Gamma^{\mu}_{\nu\lambda}U^{\nu}U^{\lambda}
\frac{d \mathbf U}{d\tau}= i c\frac{d \mathbf U}{ds}= i c\frac{d \mathbf U}{dt}\frac{dt}{ds}= i c \; (0,0,0,0) \; \frac{dt}{ds}=(0,0,0,0)
There is only one non-zero component of:
\Gamma^{\mu}_{\nu\lambda}U^{\nu}U^{\lambda}=\Gamma^{r}_{tt}U^{t}U^{t}=-\gamma^2 r \omega^2
So, substituting back in we obtain the four-acceleration in the rotating frame:
\mathbf A = (0,-\gamma^2 r \omega^2,0,0)
The norm of the four-acceleration, which is equal to the magnitude of the proper acceleration, is given by:
||\mathbf A||^2=A_{\mu} A^{\mu}= g_{\mu\nu} A^{\nu} A^{\mu} = \gamma^4 r^2 \omega^4
So this also agrees with my previous results as expected since the norm is a frame invariant quantity.
In summary, if you use four-vectors it does not matter which frame you do the calculations in, they will all agree on the norms. The magnitude of the proper acceleration, which is equal to the norm of the four-acceleration, is a frame-invariant quantity, and it is given by the above expression. The quantity starthaus derived is a frame-variant coordinate acceleration, not the frame-invariant proper acceleration.