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Let $f(r)=1-2M/r$, with $M>0$, and let a dot denote differentiation with respect to an affine parameter $\lambda$. The geodesic Lagrangian is one half the squared tangent norm. Its Euler–Lagrange equations givewhere $f^{\prime }=df/dr$ in these formulas only. The constants from time translation and axial rotation areThey are the Killing energy and axial angular momentum per chosen affine normalization. A rescaling of $\lambda$ rescales both, while their ratio is invariant. More generally the rotational Killing fields conserve the full angular-momentum vector: differentiating any Killing contraction along an affine geodesic gives $k^a{k}^b{\nabla }_{(a}{\xi }_{b)}=0$. Thus its direction can be chosen as the polar axis.

Equivalently rotate the initial position and angular tangent into one plane through the centre. The angular geodesic equation then has initial data $\theta =\pi /2$, $\dot{\theta }=0$, for which $\theta =\pi /2$ solves the equation identically. Uniqueness keeps the geodesic in that plane. A radial geodesic has no distinguished orbital plane but can be placed in any equatorial plane. Hence no physical restriction is imposed by setting $\theta =\pi /2$.

Write $L=L_z$ in that plane. For a null geodesic the tangent norm vanishes, givingThe full radial equation also gives $\ddot{r}=L^2(r-3M)/r^4$. These equations are used away from the coordinate singularity $r=2M$; the geodesics themselves extend across the horizon in regular coordinates. For radial rays $L=0$, $\dot{r}=\pm E$ and azimuth is not an orbit parameter.

For the nonradial case $L\ne 0$, set $u=M/r$ and now use a prime for $d/d\varphi$. Since $u^{\prime }=-M\dot{r}/L$, the null first integral becomesHere $b$ is the signed impact parameter; its magnitude is the ordinary positive impact parameter for an exterior future-directed ray. The second-order orbit equation is $u^{\prime \prime }+u=3u^2$. It follows from the radial geodesic equation and therefore holds even where $u^{\prime }=0$; differentiation of the squared first integral alone would not establish a circular solution at a turning point.

A finite circular photon orbit has $u^{\prime }=u^{\prime \prime }=0$ and nonzero $u$. Hence $u=1/3$, while the first integral fixes ${\alpha }^2=1/27$:The null radial potential $L^{2}f/r^2$ has second derivative $-2L^2/(81M^4)$ there, so the photon sphere is unstable. The formal root $u=0$ lies at infinite radius, not a finite orbit. Radial null generators of the event horizon have fixed angles and are not circular photon orbits.

At the critical impact parameter the first integral factors asChoose the oriented branch $u^{\prime }=(u-1/3)\sqrt{1+6u}/\sqrt{3}$. Put $w=\sqrt{1+6u}$; then $w^{\prime }=(w^2-3)/(2\sqrt{3})$. The functionsatisfies $dQ/d\varphi =Q$ by direct differentiation. Integrating givesThis establishes the displayed family as actual critical-impact-parameter geodesics. Reversing the azimuthal orientation gives the complementary $A{e}^{-\varphi }$ parametrization.

To expose the geometry, put $x=A{e}^{\varphi }$. Solving for the positive square root givesFor physical $r>0$ this branch uses $-1/2<x<(2-\sqrt{3})/2$, with the upper endpoint corresponding to infinite radius. As $\varphi \to -\infty$,Thus the critical photon orbits of Schwarzschild spacetime wind around the photon circle infinitely many times with an exponentially shrinking radial displacement. For $A>0$ the approach is from outside $r=3M$; for $A<0$ it is from inside; for $A=0$ the orbit is the circle itself. Increasing $\varphi$ along the chosen orientation sends the positive branch outward toward infinity and the negative branch inward, through the horizon toward the singularity. The limiting description at $\varphi \to -\infty$ concerns the asymptotic photon sphere rather than a finite-angle radial turning point.