In a <coordinate basis>, parallel transport is
$$
\frac{ds^\mu}{d\tau}+\Gamma^\mu_{\rho\nu}\frac{dx^\rho}{d\tau}s^\nu=0.
$$
Along the circular orbit, $dr=d\theta=0$ and $d\phi/dt=\Omega$. Dividing by $dt/d\tau$ gives
$$
\boxed{\frac{ds^\mu}{dt}+\Gamma^\mu_{t\nu}s^\nu
+\Omega\Gamma^\mu_{\phi\nu}s^\nu=0}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 309 / 1 / b / Solution

The <retarded and advanced null coordinates> invert to
$$
t=\frac{u+v}{2},\qquad z=\frac{v-u}{2}.
$$
Thus
$$
-dt^2+dz^2=-du\,dv
$$
and the <Minkowski metric> becomes
$$
\boxed{ds^2=-du\,dv+dx^2+dy^2.}
$$

The <coordinate basis> transforms by the chain rule:
$$
\boxed{
\partial_t=\partial_u+\partial_v,\qquad
\partial_z=-\partial_u+\partial_v,
}
$$
or equivalently
$$
\boxed{
\partial_u=\frac12(\partial_t-\partial_z),\qquad
\partial_v=\frac12(\partial_t+\partial_z).
}
$$
In the $(z,t)$ diagram, $\partial_t$ points vertically upward and $\partial_z$ horizontally right. The vector $\partial_u$ points along the future-left null ray and $\partial_v$ along the future-right null ray; their factors of one half affect length in the coordinate drawing but not direction.

Using these derivative relations, the <Minkowski wave operator in null coordinates> is
$$
\boxed{
\Box=-\partial_t^2+\partial_z^2+\partial_x^2+\partial_y^2
=-4\partial_u\partial_v+\partial_x^2+\partial_y^2.
}
$$

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 357 / 3 / a / Solution

{title2=$\partial/\partial x^\mu$}

Local coordinates $x^\mu$ induce the coordinate-basis tangent vectors $\partial_\mu=\partial/\partial x^\mu$. Under a coordinate change they transform by the chain rule.


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