As before, let $\vv=\dot{\rr}$ denote the velocity vectors of a family of geodesics, and $\uu=\rr'$ the separation vector between them. 1) Then along the geodesics we have \begin{align} \vv\,d\tau &= d\rr \\ v^k\,d\tau &= \sigma^k \end{align} and similar expressions hold along curves with tangent vector $\uu$, namely \begin{align} \uu\,ds &= d\rr \\ u^k\,ds &= \sigma^k \end{align}

Let $\ww=w^i\ee_i$ be a vector field. Then as usual we have \begin{equation} d\ww = \left( dw^i + \omega^i{}_j\,w^j \right)\,\ee_i = \left( dw^i + \Gamma^i{}_{jk}w^j\sigma^k \right)\,\ee_i \end{equation} where the $\Gamma^i{}_{jk}$ are the components of the connection 1-forms $\omega^i{}_j$. We can now compute \begin{align} \dot{\ww} &= \left( \dot{w}^k + \Gamma^k{}_{ij} w^i v^j \right) \,\ee_k \label{wdot} \\ \ww' &= \left( (w^k)' + \Gamma^k{}_{ij} w^i u^j \right) \,\ee_k \label{wprime} \end{align} Since $\vv$ corresponds to a geodesic, we have \begin{equation} \dot{v}^k + \Gamma^k{}_{ij} v^i v^j = 0 \label{vgeod} \end{equation} and by assumption we have \begin{equation} \dot{\uu} = \vv' \label{udot} \end{equation} so that \begin{equation} \dot{u}^k - (v^k)' = \Gamma^k{}_{ij} (v^i u^j - u^i v^j) \label{uvcomm} \end{equation}

We now calculate $\ddot{\uu}=(\vv')\Dot{}$. Using (\ref{wdot}) and (\ref{wprime}), we have \begin{equation} \vv' = \left( (v^k)' + \Gamma^k{}_{ij} v^i u^j \right) \,\ee_k \end{equation} and then \begin{align} (\vv')\Dot{} &= \left[ \left( (v^k)' + \Gamma^k{}_{ij} v^i u^j \right)\Dot + \Gamma^k{}_{ij} \left( (v^i)' + \Gamma^i{}_{mn} v^m u^n \right) v^j \right]\,\ee_k \nonumber\\ &= \left[ (\dot{v}^k)' + \dot{\Gamma}^k{}_{ij} v^i u^j + \Gamma^k{}_{ij} \left( \dot{v}^i u^j + v^i \dot{u}^j \right) \right.\nonumber\\ & \qquad\Bigl. + \Gamma^k{}_{ij}v^j \left( (v^i)' + \Gamma^i{}_{mn} v^m u^n \right) \Bigr]\,\ee_k \end{align} Inserting (\ref{vgeod}) and (\ref{uvcomm}) leads to \begin{align} (\vv')\Dot{} &= \left[ -(\Gamma^k{}_{ij})' v^i v^j - \Gamma^k{}_{ij} \left( (v^i)' v^j + v^i (v^j)' \right) + \dot{\Gamma}^k{}_{ij} v^i u^j \right.\nonumber\\ & \qquad\Bigl. + \Gamma^k{}_{ij} \left( \dot{v}^i u^j + v^i \dot{u}^j \right) + \Gamma^k{}_{ij}v^j \left( (v^i)' + \Gamma^i{}_{mn} v^m u^n \right) \Bigr]\,\ee_k \nonumber\\ &= \left[ -(\Gamma^k{}_{ij})' v^i v^j + \dot{\Gamma}^k{}_{ij} v^i u^j + \Gamma^k{}_{ij} \Gamma^j{}_{mn} v^i (v^m u^n - u^m v^n) \right.\nonumber\\ & \qquad\Bigl. + \Gamma^k{}_{ij} \Gamma^i{}_{mn} v^m (v^j u^n - u^j v^n ) \Bigr]\,\ee_k \nonumber\\ &= \left[ -(\Gamma^k{}_{ij})' v^i v^j + \dot{\Gamma}^k{}_{ij} v^i u^j \right.\nonumber\\ & \qquad\Bigl. + \left( \Gamma^k{}_{ij}\Gamma^j{}_{mn} + \Gamma^k{}_{jm}\Gamma^j{}_{in} \right) v^i (v^m u^n - u^m v^n ) \Bigr]\,\ee_k \label{uddot} \end{align} The first thing to note about the factor in square brackets is that it depends only on $\uu$ and $\vv$, not on their derivatives. To recognize this factor, we begin by computing the curvature 2-forms.

Taking the exterior derivative of \begin{equation} \omega^i{}_j = \Gamma^i{}_{jk}\,\sigma^k \end{equation} we obtain \begin{align} \Omega^k{}_i &= d\omega^k{}_i + \omega^k{}_j\wedge\omega^j{}_i \nonumber\\ &= d\Gamma^k{}_{ij}\wedge\sigma^j + \Gamma^k{}_{ij}\,d\sigma^j + \omega^k{}_j\wedge\omega^j{}_i \nonumber\\ &= d\Gamma^k{}_{ij}\wedge\sigma^j - \Gamma^k{}_{ij}\,\omega^j{}_m\wedge\sigma^m + \omega^k{}_j\wedge\omega^j{}_i \nonumber\\ &= d\Gamma^k{}_{ij}\wedge\sigma^j - \Gamma^k{}_{ij}\Gamma^j{}_{mn}\,\sigma^n\wedge\sigma^m + \Gamma^k{}_{jm}\Gamma^j{}_{in}\,\sigma^m\wedge\sigma^n \end{align} We are interested in the coefficient of $ds\wedge d\tau$, so we can assume without loss of generality that only $\tau$ and $s$ are changing, so that \begin{equation} df = \dot{f}\,d\tau + f'\,ds \end{equation} and any 2-form must be proportional to $ds\wedge d\tau$. Thus, \begin{align} \Omega^k{}_i &= \left(\dot{\Gamma}^k{}_{ij}d\tau+(\Gamma^k{}_{ij})'ds\right)\wedge\sigma^j + \left(\Gamma^k{}_{jm}\Gamma^j{}_{in} + \Gamma^k{}_{ij}\Gamma^j{}_{mn}\right) \sigma^m\wedge\sigma^n \nonumber\\ &= \left[ -\dot{\Gamma}^k{}_{ij} u^j + (\Gamma^k{}_{ij})' v^j + \left(\Gamma^k{}_{jm}\Gamma^j{}_{in} + \Gamma^k{}_{ij}\Gamma^j{}_{mn}\right) (u^mv^n - v^m u^n) \right] \nonumber\\ & \qquad ds\wedge d\tau \nonumber\\ &= \frac12 \,R^k{}_{imn}\,\sigma^m\wedge\sigma^n = \frac12 \,R^k{}_{imn}\,(u^mv^n-v^mu^n)\,ds\wedge d\tau \nonumber\\ &= R^k{}_{imn}\,u^mv^n\,ds\wedge d\tau \label{devcur} \end{align}

Comparing (\ref{udot}), (\ref{uddot}), and (\ref{devcur}) we finally conclude that \begin{equation} \ddot{\uu} = - R^k{}_{imn} v^iu^mv^n \ee_k \end{equation} as claimed. This expression for the relative acceleration of nearby geodesics ($\vv$) in the direction of $\uu$ is called the equation of geodesic deviation.