= Solution
With $u=t-z$, the quadratic <geodesic Lagrangian> is
$$
L=\frac12(-\dot t^2+\dot x^2+\dot y^2+\dot z^2)
+xyA(u)\dot u^2.
$$
Its transverse <Euler-Lagrange equations> are
$$
\boxed{\ddot x=yA(u)\dot u^2,
\qquad \ddot y=xA(u)\dot u^2}.
$$
The longitudinal equations are
$$
\boxed{\ddot t=\ddot z
=2A(u)(\dot x,y+x\dot y)\dot u+xyA'(u)\dot u^2},
$$
after using the difference of those same equations. In particular,
$$
\ddot u=\ddot t-\ddot z=0,
\qquad
\boxed{\dot t-\dot z=\dot u=\text{constant}}.
$$
This conserved quantity also follows from the <Killing vector field> $\partial_t+\partial_z$ and the <geodesic conserved quantity from a Killing vector>.
Back to article page
