= Solution
Put $f=1-2\mu/r$ and let dots now denote an <affine parameter> <derivative>. The <null geodesic> <lagrangian> in the <orbital plane> is $L=(fc^2\dot t^2-\dot r^2/f-r^2\dot\phi^2)/2$. The time and azimuthal <Euler-Lagrange equations> integrate once; the null normalization gives
$$
f\dot t=k,\qquad r^2\dot\phi=h,\qquad
fc^2\dot t^2-\dot r^2/f-r^2\dot\phi^2=0.
$$
For completeness, the differential radial <geodesic equation> is
$$
\ddot r+\frac{c^2ff'}2\dot t^2-\frac{f'}{2f}\dot r^2
-fr\dot\phi^2=0;
$$
the remaining differential equations are $\ddot t+(f'/f)\dot t\dot r=0$ and $\ddot\phi+2\dot r\dot\phi/r=0$. Spherical symmetry permits the equatorial choice.
Eliminating $\dot t,\dot\phi$ yields $\dot r^2+h^2f/r^2=c^2k^2$. For $h\ne0$, divide by $h^2$:
$$
\boxed{\frac{\dot r^2}{h^2}+V_{\rm eff}(r)=\frac1{b^2},\qquad
V_{\rm eff}=\frac1{r^2}-\frac{2\mu}{r^3},\qquad
b=\frac{|h|}{c|k|}.}
$$
The <impact parameter> is unchanged by a common rescaling of the <affine parameter>. At infinity it is $c$ times <angular momentum> divided by <energy>. Radial <photons> have $h=0$ and obey the undivided equation $\dot r^2=c^2k^2$.
The <effective potential> is zero at $r=2\mu$, positive outside, and tends to zero as $r\to\infty$. Its <derivative> is $V'_{\rm eff}=2(3\mu-r)/r^4$. Thus
$$
\boxed{r_{\rm circ}=3\mu,\quad
V_{\rm eff}(3\mu)=\frac1{27\mu^2},\quad
b_{\rm crit}=3\sqrt3\,\mu.}
$$
The <circular null geodesic> lies at a maximum, since $V''_{\rm eff}(3\mu)=-2/(81\mu^4)<0$. It is \b[unstable]: a small radial displacement away from the maximum permits motion away from the orbit. For incoming <photons> from infinity in the exterior of a <black hole>, an outer <turning point> exists when $b>b_{\rm crit}$; for $b<b_{\rm crit}$ the <photon> can cross the <event horizon>. For a material central body, the <vacuum> circular orbit is relevant only if $3\mu$ lies outside its surface.
\Image[/past-exam-of-the-relativity-course-of-the-university-of-cambridge/2012/paper-1-photon-potential.png]
{title=Photon potential and horizon-penetrating light cones}
For the new coordinates, set $q=2\mu/r$, so $f=1-q$. The <geodesic Lagrangian> is $L'=[fc^2\dot t'^2-2qc\dot t'\dot r-(1+q)\dot r^2-r^2\dot\phi^2]/2$. Cyclic-coordinate conservation and the null condition give the required analogous equations:
$$
\boxed{\begin{aligned}
f\dot t'-\frac q c\dot r&=k,& r^2\dot\phi&=h,\\
fc^2\dot t'^2-2qc\dot t'\dot r
-(1+q)\dot r^2-r^2\dot\phi^2&=0.
\end{aligned}}
$$
Here $k$ denotes the same conserved time-translation quantity as before. The transformation is $t'=t+(2\mu/c)\log|r/(2\mu)-1|$. In particular, the displayed time coordinate belongs to <Ingoing Kerr-Schild coordinates for Schwarzschild spacetime>; the usual advanced null coordinate of <ingoing Eddington-Finkelstein coordinates> is $v=t'+r/c$. Using a <metric tensor> with a vanishing $dr^2$ coefficient while retaining the name $t'$ would therefore solve a different coordinate representation.
An elimination that stays finite at the horizon is especially simple:
$$
\begin{aligned}
c^2k^2-\dot r^2
&=(fc\dot t'-q\dot r)^2-\dot r^2\\
&=f\left[fc^2\dot t'^2-2qc\dot t'\dot r
-(1+q)\dot r^2\right]\\
&=f r^2\dot\phi^2=\frac{fh^2}{r^2}.
\end{aligned}
$$
The middle equality uses $q^2+f(1+q)=1$. Hence \b[the radial <energy> equation and $V_{\rm eff}(r)$ are unchanged], including the horizon limit. The radial <metric tensor> block has <determinant> $-c^2$ and is regular at $r=2\mu$. Its radial null directions are $dr/(c\,dt')=-1$ and $dr/(c\,dt')=f/(1+q)$, as shown in the original schematic.
This <coordinate invariance of the Schwarzschild photon radial potential> follows because the transformation changes the time slicing but preserves the <areal radius> and the stationary <Killing vector field> $\partial_{t'}=\partial_t$. Consequently the conserved <Killing energy>, <angular momentum> and <impact parameter> are the same. <photon> capture, the circular orbit and the radial <turning points> cannot depend on coordinate labels; coordinate velocities can.
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