Suppose you have a smooth spacelike hypersurface $\Sigma$ in some spacetime (four-dimensional Lorentzian manifold). Let $\gamma$ be a timelike geodesic meeting $\Sigma$ orthogonally and let $p$ be a point on $\gamma$. Then $p$ is said to be conjugate to $\Sigma$ iff there exists a Jacobi field $J$ orthogonal to the tangent vector $\gamma'$, which is zero at $p$, nonzero on $\Sigma$, and comes from a variation of $\gamma$ by geodesics intersecting $\Sigma$ orthogonally.

Why is this the definition? Why not say that $p$ is conjugate to $\Sigma$ iff there exists a Jacobi field $J$ which is zero at $p$ and zero on $\Sigma$ (and thus $J$ is orthogonal everywhere to $\gamma$), the same definition as conjugacy along a geodesic? If you do that, you cannot ask of the variation to come from geodesics intersecting $\Sigma$ (if $q$ is the intersection of $\gamma$ and $\Sigma$, then the orthogonal space to the tangent space at $q$ of $\Sigma$, i.e. $(T_q\Sigma)^{\perp}$, is one-dimensional, so any geodesic intersecting $\Sigma$ orthogonally at $q$ must be a reparametrization of $\gamma$), but why does one care about intersecting orthogonally?

EDIT: If you take $\theta$ to be the expansion of the timelike geodesics through $p$, then $p$ is conjugate to $q$ along $\gamma$ iff $\theta$ goes to infinity at $q$.

I would like to see a proof for that in the hypersurface case: $p$ is conjugate to $\Sigma$ along $\gamma$ iff $\theta$, the expansion of the timelike geodesics orthogonal to $\Sigma$, goes to infinity at $p$. In particular, I do not know how to impose the condition that the Jacobi field $J$ arises from a variation of geodesics orthogonal to $\Sigma$.