Tricks to speed up calculation of Christoffel symbols A couple of months ago I was asked on an exam to calculate the Christoffel symbols of the following metric 
$$d s^{2}=\left[1-\omega^{2}\left(x^{2}+y^{2}\right)\right] d t^{2}+2 \omega y d t d x-2 \omega x d t d y-d x^{2}-d y^{2}-d z^{2}$$
given the inverse metric 
$$g^{\mu \nu}=\left(\begin{array}{cccc}
1 & \omega y & -\omega x & 0 \\
\omega y & -\left(1-\omega^{2} y^{2}\right) & -\omega^{2} x y & 0 \\
-\omega x & -\omega^{2} x y & -\left(1-\omega^{2} x^{2}\right) & 0 \\
0 & 0 & 0 & -1
\end{array}\right).$$
This seemed like a rather tedious task, so I assumed that there must be some trick to get this done faster. After some thinking I noticed the following things that might be useful


*

*The only components that depend on coordinates in the metric are $$g_{tt}= 1-\omega^2(x^2+y^2),\quad g_{tx}=\omega y,\quad g_{ty}=-\omega x.$$

*The metric is block diagonal $\{t,y,x\}$ and $\{z\}$, so we conclude that all the Christoffel symbols including $z$ vanish. 

*$\Gamma^\mu_{\alpha\beta}=\Gamma^\mu_{\beta\alpha}$

*There seems to be some kind of symmetry between the variables $x$ and $y$ in the metric that one might use, but I just couldn't figure out the transformation during the exam...


I wasn't able to come up with any more tricks, so I started calculating but it took me forever to get anywhere... I was wondering if I maybe had missed something... Are there any other tricks one can use to figure out the Christoffel symbols? Is this somehow "trivial"?
 A: An efficient way to compute the Christoffel symbols is to determine the geodesic equations for a metric from
$$\delta\int\frac{ds}{d\tau}d\tau=0$$
using the calculus of variations (with lots of integration by parts to turn $\delta\dot x$ into $\delta x$, etc.) and then read off the Christoffels by comparing the resulting equations to the general form of the geodesic equation,
$$\ddot{x}^\mu+\Gamma^\mu{}_{\alpha\beta}\dot{x}^\alpha\dot{x}^\beta=0.$$
No computer necessary!
A: I had to do this little while ago. I computed the spin connection rather than the Christoffels. It seemed easier. I would not have liked to do it in an exam! Here is my working: 
Rotating frame Born Metric is 
$$
d\tau^2 = (1-\Omega^2r^2)d{t'}^2 -2\Omega{ r}^2 d\theta dt' -d{r}^2-{r'}^2 d\theta^2 \\
= (1-|{\bf v}|^2)d{t'}^2 +2v_\xi d\xi dt'  +2v_\eta d\eta dt' -d\xi^2-d\eta^2 \\
= d{t'}^2- (d\xi- v_\xi dt')^2 -(d\eta-v_\eta dt')^2 
$$
where $v_\xi =\Omega \eta$, $v_\eta=-\Omega \xi$ is the local velocity of the inertial frame as seen from the rotating frame. 
In the $t,\xi,\eta$ basis
$$
g_{\mu\nu} = \left[\matrix{ 1-\Omega^2 r^2 & \Omega\eta&-\Omega \xi  \cr \Omega\eta  &-1& 0\cr -\Omega  \xi &0 & -1}\right]
$$
A Minkowski-orthonormal dreibein in the rotating coordinates is given by the  frame
$$
{\bf e}_t = \partial_t +\Omega  \eta \,\partial_\xi -\Omega \xi \,\partial_\eta,\\  
{\bf e}_\xi = \partial_\xi, \\  
{\bf e}_\eta = \partial_\eta,\nonumber
$$
and its dual  basis is 
$$
{\bf e}^{*t}= dt,\\
{\bf e}^{*\xi}= d\xi - \Omega \eta \,dt, \\
{\bf e}^{*\eta}= d\eta + \Omega \xi\, dt, 
$$
so
$$
d{\bf e}^{*t}= \phantom - 0,\\
d{\bf e}^{*\xi}= \phantom - \Omega \,dt\wedge d\eta\\
d{\bf e}^{*\eta}= -\Omega  dt\wedge \,d\xi.\nonumber
$$
We compare the torsion-free Cartan relation 
$$
d{\bf e}^{*a}+ {\boldsymbol \omega^a}_b \wedge {\bf e}^{*b}=0,
$$
and so read off that the only non-zero spin connection component is
$$
{\boldsymbol \omega}_{\xi\eta}= -{\boldsymbol  \omega}_{\eta\xi}=\Omega \, dt.
$$
It took me about an hour to be sure of that  had the vierbeins correct, so the Cartan relation, although easier than plugging into the Christoffel expression, is   not a very effective trick.
Later I  did the Christoffels, but I used a Mathematica package for them! 
A: 
I was wondering if I maybe had missed something ...

What OP seems to be missing and what other answers do not mention explicitly is that this metric is easily $(3+1)$ decomposable with the spatial geometry on each slice $t=t_0$ being Euclidean metric, while the shift between slices $t=t_0$ and $t=t_0+dt$ corresponds to pure rotation by an angle $\omega dt$ around the $z$–axis.
So, if OP's course has included discussion about such decomposition or possibly subjects like ADM formalism, then this should have been a hint about how to proceed here.
The 4D metric could be written as (Greek indices run from 0 to 3, Latin from 1 to 3):
$$
ds^2\equiv g_{\mu\nu}dx^\mu dx^\nu = -\gamma_{ij}(dx^i+N^i dt)(dx^j+N^j dt)+(N dt)^2,
$$
where  $N$ is the lapse function, $N^i$ is the shift vector  and $\gamma_{ij}$ is the 3D Euclidean spatial metric with following form:
$$
N=(g^{00})^{-\frac12}\equiv  1 , \qquad N^i = \epsilon ^{ijk}\omega^j x^k,\qquad \gamma_{ij}=\delta_{ij}.
$$
with Levi-Civita tensor defined by  $\epsilon^{123}=+1$ and $\omega^i=\omega \,\delta^i_3$.
Note, that the sign convention (mostly minus) used by OP is different from most GR textbooks (such as MTW) so one should check for possible sign errors I may have introduced when translating equations from MTW (§21.4).
It would be easier to calculate Christoffel symbols with all lower indices:
$$
Γ_{μ\,αβ}=\frac 12 (g_{βμ,α}+  g_{μα,β} - g_{αβ,μ}).
$$
Now we can proceed with the actual calculations. First the zero components:


*

*$Γ_{i\,jk}\equiv 0$, since the spatial metric is Euclidean. 

*$Γ_{0\,ij}\equiv 0$, since the shift vector field $N^i$ has zero shear.

*$Γ_{0\,00}\equiv 0$, since the metric does not depend on time.
Now the nonzero part:
$$Γ_{0\,0i}= -Γ_{i\,00}=\frac12 \partial _i (N_j N^j),\qquad
Γ_{i\,0j}=\frac12 (g_{0i,j}-g_{0j,i})=\epsilon_{ijk}\omega^k,$$
with nontrivial components:
$$
Γ_{0\,0x}=\omega^2 x, \qquad Γ_{0\,0y}= \omega^2 y, \qquad Γ_{x\,0y}=\omega, 
$$
while the rest of nonzero components are obtained by the symmetry relations.
