I'm given a conformally flat metric $$g_{\mu\nu} =\Omega(x)\eta_{\mu\nu}$$, and asked to show that for timelike geodesics the geodesic equation takes the form: $$\frac{\partial}{\partial \tau} (\Omega \frac{\partial x^{\mu}}{\partial \tau}) +\eta ^{\mu \lambda} \partial_{\lambda} (log(\sqrt{\Omega}))$$
I almost have the result but am off by an extra factor of $\Omega$.
The Christoffel symbols evaluate to: $$\frac{1}{2} \Omega^{-1} \eta^{\rho m} (\frac{\partial \Omega}{\partial x^{\nu}} \eta_{m \mu} +\frac{\partial \Omega}{\partial x^{\mu}} \eta_{m \nu} -\eta_{\mu \nu} \partial_m \Omega)$$, and carrying out the summations over the dummy index m give: $$\frac{1}{2} \Omega^{-1} (\frac{\partial \Omega}{\partial x^{\nu}} \delta_{\mu}^{\rho} +\frac{\partial \Omega}{\partial x^{\mu}} \delta_{\nu}^{\rho} -\eta_{\mu \nu}\eta^{\rho m} \partial_m \Omega)$$So plugging this result into the geodesic equation:$$ \frac{\partial^{2} x^{\mu}}{\partial \tau^{2}} + \Gamma_{\alpha \beta}^{\mu} \frac{dx^{\alpha}}{d \tau}\frac{dx^{\beta}}{d \tau} =0$$ gives: $$ \frac{\partial^{2} x^{\mu}}{\partial \tau^{2}} + \frac{1}{2} \Omega^{-1} (\frac{\partial \Omega}{\partial x^{\beta}} \delta_{\alpha}^{\mu} +\frac{\partial \Omega}{\partial x^{\alpha}} \delta_{\beta}^{\mu} -\eta_{\alpha \beta}\eta^{\mu m} \partial_m \Omega) \frac{dx^{\alpha}}{d \tau}\frac{dx^{\beta}}{d \tau} =0$$ and evaulating further $$ \frac{\partial^{2} x^{\mu}}{\partial \tau^{2}} + \frac{1}{2} \Omega^{-1} (\frac{\partial \Omega}{\partial x^{\beta}}\frac{dx^{\beta}}{d \tau} \frac{dx^{\mu}}{d \tau} +\frac{\partial \Omega}{\partial x^{\alpha}} \frac{dx^{\alpha}}{d \tau} \frac{dx^{\mu}}{d \tau} -\eta^{\mu \lambda} \partial_{\lambda} \Omega \eta_{\alpha \beta}\frac{dx^{\alpha}}{d \tau}\frac{dx^{\beta}}{d \tau}) =0$$ Noting that: $$\frac{\partial \Omega}{\partial x^{\beta}}\frac{dx^{\beta}}{d \tau} = \frac{\partial \Omega}{\partial \tau}, \eta_{\alpha \beta}\frac{dx^{\alpha}}{d \tau}\frac{dx^{\beta}}{d \tau} =-1$$ I get: $$\frac{\partial^{2} x^{\mu}}{\partial \tau^{2}} + \frac{1}{2} \Omega^{-1}(2\frac{\partial \Omega}{\partial \tau}\frac{dx^{\mu}}{d \tau} + \eta^{\mu \lambda} \partial_{\lambda} \Omega) =0$$ which is rearranged to: $$\frac{\partial^{2} x^{\mu}}{\partial \tau^{2}} + \Omega^{-1}(\frac{dx^{\mu}}{d\tau}\frac{\partial \Omega}{\partial \tau}) + \eta ^{\mu \lambda} \partial_{\lambda} (log(\sqrt{\Omega})) =0$$ Which is indeed off by a factor of $\Omega$ from the desired expression. How do I fix this?
It seems about fine until you put $$ \eta_{\mu\nu} \dot{x}^{\mu}\dot{x}^{\nu} = -1 $$ which is incorrect since your actual metric is $g_{\mu\nu} = \Omega\eta_{\mu\nu}$. So the timelike geodesic should satisfy $g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu} = -1$ instead. Hence after your line $$ \frac{\partial^{2} x^{\mu}}{\partial \tau^{2}} + \frac{1}{2} \Omega^{-1} (\frac{\partial \Omega}{\partial x^{\beta}}\frac{dx^{\beta}}{d \tau} \frac{dx^{\mu}}{d \tau} +\frac{\partial \Omega}{\partial x^{\alpha}} \frac{dx^{\alpha}}{d \tau} \frac{dx^{\mu}}{d \tau} -\eta^{\mu \lambda} \partial_{\lambda} \Omega \, \color{red}{\eta_{\alpha \beta}\frac{dx^{\alpha}}{d \tau}\frac{dx^{\beta}}{d \tau}}) =0,$$ it should follows that $$ \ddot{x}^{\mu} +\frac{1}{2} \Omega^{-1} \big( 2\dot{\Omega} \dot{x}^{\mu} - \eta^{\mu\lambda} \partial_{\lambda}\Omega \, \color{red}{(-\Omega^{-1})} \big) = 0. $$ And therefore the result follows.