# Calculate Ricci Tensor (Axial Symmetry)

I have a problem calculating a Ricci tensor.

My metric (Lorentz signature) is $\mathbf{g}=Xd\phi^2 + g_{ab}dx^adx^b$ where $X,g_{ab}$ don't depend on $\phi$, and $g_{ab}$ is a $2+1$ metric.

I define $\Phi = \frac{1}{2}\log(X)$, $\mathbf{R}_{\mu \nu}=0$ is the Ricci tensor corresponding to $\mathbf{g}$, $R_{ab} = -\mathbf{R}^{\phi}_{\;a\phi b}$ is the Ricci tensor corresponding to $g$. I am trying to prove $R_{ab} = D_a D_b \Phi + D_a \Phi D_b \Phi$. I am doing something wrong because there are some Christoffel coefficients I can't get out (Levi-Civita connection).

I have that $\mathbf{\Gamma}^\phi_{\phi b} = \partial_b \Phi$, $\mathbf{\Gamma}^a_{\phi\phi}=-\frac{1}{2}g^{as}\partial_s X$, and $\mathbf{\Gamma}^a_{bc} = \Gamma^a_{bc}$ ($a,b,c$ represent coordinates different from $\phi$). All the rest of the symbols are $0$.

Now $$R_{ab} = -\mathbf{R}^{\phi}_{\;a\phi b} = -\partial_\phi\mathbf{\Gamma}^\phi_{ab}+\partial_b\mathbf{\Gamma}^\phi_{a\phi}-\mathbf{\Gamma}^\nu_{ab}\mathbf{\Gamma}^\phi_{\nu\phi}+\mathbf{\Gamma}^\nu_{a\phi}\mathbf{\Gamma}^\phi_{\nu b}$$ from which $$R_{ab}=-0+\partial_b\partial_a\Phi- \partial_c\Phi \Gamma^c_{ab}+\partial_b\Phi\partial_a\Phi$$

I cannot, so far, remove $\partial_c\Phi\Gamma^c_{ab}$ from that equality. Am I doing something wrong? I am using G.Weinstein's paper "On Rotating Black Holes in Equilibrium in General Relativity" for reference, but trying to change to my notation, by the way.