Orthogonal Jacobi Fields Remain Orthogonal Let $\gamma(t)$ be a geodesic and suppose $\left<J(0), \gamma'(0)\right> = \left<J'(0), \gamma'(0)\right> = 0$ where $J'(0)$ indicates the covariant derivative of $J$ along $\gamma$. There exists a unique Jacobi field $J(t)$ with initial conditions $J(0), J'(0)$. How can I see that $J(t)$ and $\frac{DJ}{dt}$ remain perpendicular to $\gamma(t)$ for all $t$? 

Since $\langle J(0), \gamma'(0) \rangle =0$,  it suffices to show that $\frac{d}{dt}\langle J(t), \gamma'(t) \rangle =0$ for all $t$.
To do this, first compute
$\frac{d}{dt}\langle J(t), \gamma'(t) \rangle = 
\langle J'(t), \gamma'(t) \rangle + \langle J'(t), \gamma''(t) \rangle =
\langle J'(t), \gamma'(t) \rangle,  
$
where we've used compatibility of the Levi-Civita connection with the metric and the fact that $\gamma$ is a geodesic.
Now, since $\langle J'(0), \gamma'(0) \rangle=0$, it suffices to show that
$\frac{d}{dt}\langle J'(t), \gamma'(t) \rangle =0$ for all $t$.
To this end, compute 
$\frac{d}{dt}\langle J'(t), \gamma'(t) \rangle = \langle J''(t), \gamma'(t) \rangle = \langle R(\gamma'(t), J(t))\gamma'(t),\gamma'(t)\rangle=0$,
where we've used that $J$ is a Jacobi field and that the Riemann tensor is antisymmetric in its last two entries. 
