The separation vector *is* a Jacobi field because it obeys the Jacobi equation.

Here I will derive geodesic deviation from scratch because I find MTW's derivation hard to follow. (Much like everything else in that book.)

**Definition 1.** Consider a family of timelike geodesics, having the property that in a sufficiently small open region of the Lorentz manifold $(M,g)$ precisely one geodesic passes through every point. Such a collection is called a *congruence*. The tangent field to this set of curves, parameterized by proper time $s$, is denoted by $u$ and is normalized as $\langle u,u\rangle=-1$.

Let $\gamma(t)$ be some curve transversal to the congruence. This means that the tangent $\dot\gamma$ is never parallel to $u$ in the region under consideration. Imagine that every point on the curve $\gamma(t)$ moves a distance $s$ along the geodesics which passes through that point. Let the resulting point be $H(s,t)$. This defines a map $H:O\longrightarrow M$ for some open $O\subset \mathbb{R}^2$. For each $t$, the curve $s\longmapsto H(s,t)$ is a timelike geodesic with tangent vectors $u(H(s,t))$. We say that $u$ defines a vector field $u\circ H$ along which the map $H$ is *tangential*, i.e. of the form
$$u\circ H=H_*\circ\frac{\partial}{\partial s}\quad\text{or}\quad u\big(H(s,t)\big)=T_{(s,t)}H\cdot\frac{\partial}{\partial s}$$
In other words, $u$ and $\partial/\partial s$ are $H$-related. We shall use for $u\circ H$ the same letter $u$. Let $v$ be the tangential vector field along $H$ belonging to $\partial/\partial t$
$$v=H_*\circ\frac{\partial }{\partial t}$$
The vectors $v(s,t)$ are tangent to the curves $t\longmapsto H(s,t)$ ($s$ fixed), and represent the separation of points which are moved with the same proper time along the neighboring geodesics of the congruence (beginning at arbitrary starting points). This is shown in the following diagram (taken from Straumann, *General Relativity* [2013]):

Since $\partial/\partial s$ and $\partial/\partial t$ commute, the tangential fields $u$ and $v$ also commute. To see this, recall that if the vectors of two Lie brackets are $H$-related, then the Lie brackets themselves are $H$-related.

To get the distance between curves, we create a projection operator $id+u\otimes u$ onto the subspace of the tangent space orthogonal to $u$. Thus the relevant infinitesimal separation vector $n$ is
$$
n=v+\langle v,u\rangle u
$$
Note that $\langle n,u\rangle=\langle v,u\rangle-\langle v,u\rangle=0$, so $n$ is indeed perpendicular to $u$. We now show that $n$ is also Lie transported. We have
\begin{align*}
\mathcal{L}_un&=[u,n]=[u,v]+[u,\langle v,u\rangle u]=\big(u\langle v,u\rangle\big)u\\
u\langle v,u\rangle&=\frac{\partial}{\partial s}\langle v,u\rangle
\end{align*}
The normalization condition $\langle u,u\rangle=-1$ implies
$$0=\frac{\partial}{\partial t}\langle u,u\rangle=2\langle \nabla_v u,u\rangle$$
Furthermore, since $\nabla_vu=\nabla_uv$, it follows that
$$\frac{\partial}{\partial s}\langle v,u\rangle=\langle \nabla_uu,v\rangle+\langle u,\nabla_uv\rangle=\langle u,\nabla_vu\rangle=0$$
This shows that indeed
$$\tag{1}
\mathcal{L}_un=0
$$

Next, we consider
$$\nabla^2_uv=\nabla_u\nabla_uv=\nabla_u\nabla_vu=[\nabla_u,\nabla_v]u$$
However, the Riemann tensor for three vectors $X,Y,Z$ is $R(X,Y)Z=[\nabla_X,\nabla_Y]Z-\nabla_{[X,Y]}Z$. Due to $[u,v]=0$, we obtain
$$\nabla^2_uv=R(u,v)u$$
This is called the *Jacobi equation* for the field $v$.

We now show that $n$ also satisfies this equation. From (1) it follows that
\begin{equation}
\nabla_un=\nabla_uv+\big(u\langle v,u\rangle\big)u+\langle v,u\rangle \nabla_uu=\nabla_uv
\end{equation}
Furthermore,
\begin{equation}
R(u,n)u=R(u,v)u+\langle v,u\rangle R(u,u)u=R(u,v)u
\end{equation}
Putting this all together, we see that $n$ satisfies the Jacobi equation
\begin{equation}
\nabla^2_un=R(u,n)u
\end{equation}
The Jacobi field $n$ is everywhere perpendicular to $u$. In physics we call this the *equation for geodesic deviation*. For a given $u$ the right hand side defines at each point $p\in M$ a linear map $n\longmapsto R(u,n)u$ of the subspace of $T_pM$ perpendicular to $u$.

In more familiar notation, we have
$$\frac{D^2 n^a}{D\tau^2}=R^a_{\;bcd}u^b n^c u^d$$
So why even bother calling this equation a Jacobi equation? Perhaps there is some general result of Jacobi fields that we wish to apply to GR? This is indeed the case.

**Definition 2.** Let $\gamma$ be a geodesic. A pair of points $p$, $q\in\gamma$ are said to be *conjugate* if there exists a Jacobi field $n$ which is not identically zero but vanishes at both $p$ and $q$.

There exists a *very powerful* theorem regarding conjugate points (suitably generalized here to spacetime, but a similar version holds in positive-definite spaces):

**Theorem 1.** Let $\gamma$ be a smooth timelike curve connecting two points $p$, $q\in M$. Then the necessary and sufficient condition that $\gamma$ locally maximizes proper between $p$ and $q$ over smooth one parameter variations is that $\gamma$ be a geodesic with no point conjugate to $p$ between $p$ and $q$.

This theorem is the basis for the *singularity theorems* of GR. Since MTW doesn't discuss these theorems (AFAIK, haven't read the whole thing), they don't need to mention Jacobi fields or conjugate points. Indirectly, this property of Jacobi fields is used in the justification for Big Bang cosmology.

For more in-depth discussions, I refer you to

Wald, *General Relativity* (1984)

Hawking & Ellis, *The Large Scale Structure of Spacetime* (1973)

(in that order) and references therein.

## Best Answer

Assume, that the photon path is parametrized by an affine parameter. The affine parameter for the null geodesic is usually denoted with $\lambda$, so I'll use $\lambda$ instead of $\tau$. Then by velocity I'll mean 4-vector $u^{\alpha} = \frac{d x^\alpha}{d \lambda}$.

I'll use Wikipedia notation for the Schwarzschild metric, assuming in addition, that $c=1$. Particularly, signature will be $(+---)$. Given any initial conditions for a photon, one can always find a "plane", containing initial position and velocity. (By a plane here I mean a big circle on a sphere, parametrized by $\varphi$-$\theta$ coordinates.) Then you can use rotational symmetry to move the initial conditions in $\theta=\pi/2$ plane, so that initially photon is in this plane and $\theta$ component of its velocity is 0. Then photon will always have $u^\theta=0$ and $\theta=\pi/2$.

The Schwarzschild metric is invariant with respect to time translation and rotation. In the traditional language of general relativity, $\frac{\partial}{\partial t}$ and $\frac{\partial}{\partial\varphi}$ are killing fields. This implies, that $u_t$ and $u_\varphi$ are constant (but not $u^t$ and $u^\varphi$). Together with $u^\alpha u_\alpha = 0$ this gives 3 equations for 3 non-zero components of the 4-velocity, so you can find all of them (as a functions of $r$). In this case physicists say, that the equation of motion is integrable. Now you may want to find $r(\lambda)$, $t(\lambda)$, $\varphi(\lambda)$, starting from solving $\frac{dr}{d\lambda} = u^r(r(\lambda))$ as an ordinary differential equation on the function $r(\lambda)$.

There are great notes on solving the same problem for a massive particle by Christopher Hirata. You may also want to look at other sections of the lecture notes for the course he gave at Caltech in 2011-2012. You may also think about photon trajectory as a particle trajectory in the limit $\textrm{mass}\to0$.