# Tricks to speed up calculation of Christoffel symbols

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!

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!

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.