Equatorial null orbit in Schwarzschild spacetime

For inverse radius $y=1/r$ and azimuth $\varphi$, a Schwarzschild null geodesic obeyswhere $b=L/E$ is the impact parameter.

For a captured nonradial null geodesic with $u=1/r$, the large-$u$ orbit equation is $(du/d\varphi {)}^2\sim 2\mu u^3$. Integration on the inward branch givesThe curvature singularity is reached at finite azimuth. This does not prevent a near-critical ray from winding repeatedly near the photon sphere before its final plunge. A radial ray has constant azimuth, so this asymptotic coordinate parametrization is inapplicable to it.

For nonzero angular constant $h$, the interior null radial equation gives ${\dot{r}}^2\sim 2\mu h^2/r^3$. Its inward branch integrates to $r^{5/2}\sim \frac{5}{2}\sqrt{2\mu }|h|({\lambda }_{\ast }-\lambda )$. Thus the singularity is at finite affine parameter and $r\propto ({\lambda }_{\ast }-\lambda {)}^{2/5}$. The zero-angular-momentum case has a different affine scaling.

For an equatorial null geodesic in Schwarzschild spacetime, let $k=(1-2\mu /r)dt/d\lambda$ and $h=r^{2}d\varphi /d\lambda$ be the conserved quantities for an affine parameter $\lambda$. The null metric condition gives ${\dot{r}}^2/2+V_{\mathrm{eff}}=c^2{k}^2/2$. For $h\ne 0$, the effective potential is zero at the Schwarzschild event horizon, positive outside, and tends to zero at infinity. Its unique exterior maximum is at $r=3\mu$, with value $h^2/(54{\mu }^2)$ and second derivative $-h^2/(81{\mu }^4)$. This maximum gives the unstable photon sphere.

The impact parameter of an incoming ray is $b=|h|/(ck)$ for $k>0$. Capture requires its radial energy to exceed the maximum, namely $b<3\sqrt{3}\mu$. At equality the incoming ray approaches the circular orbit asymptotically; it does not cross the horizon. For $h=0$, the potential is identically zero and the ingoing radial ray is captured.

At the critical impact parameter $b=3\sqrt{3}M$, the null orbit equation factors as $(u^{\prime }{)}^2=(u-1/3{)}^2(2u+1/3)$ for $u=M/r$. One oriented family has the displayed solution. As $\varphi \to -\infty$, positive $A$ approaches the photon sphere from larger radii and negative $A$ from smaller radii; $A=0$ is the circular orbit. The physical domain must retain $r>0$ and the positive square root in its implicit parametrization.

A circular null geodesic remains at constant areal radius. It lies at a stationary point of the null radial effective potential and may be stable or unstable according to the sign of the second derivative there.

For $0<a<M$, the two exterior equatorial circular null geodesics of the Kerr metric have radii solving the signed equation above. The plus sign gives the prograde branch and the minus sign gives the retrograde branch. To derive it put $b=L/E$ in the double-root conditions of the equatorial Kerr null geodesic radial equation: $r^2=(b^2-a^2)/3$ and $r^3=M(b-a{)}^2$. Both exterior radii exceed $r_{+}=M+\sqrt{M^2-a^2}$ and approach $3M$ when $a\to 0$. The horizon generators are a separate case with $E=L=0$; dividing by $E$ excludes them. Constant-radius null directions at other exterior radii generally have nonzero acceleration and are not geodesics.

A photon sphere is a spherical set of circular null geodesics. In Schwarzschild spacetime it occurs at $r=3M$ and is unstable.

At $2\mu <r\le 3\mu$, measure the emission angle $\alpha$ from the outward radial direction in a static observer's orthonormal tetrad. Local photon energy is $E_{\mathrm{loc}}=E_{\infty }/\sqrt{1-2\mu /r}$, and its tangential momentum is $E_{\mathrm{loc}}\sin \alpha /c$. Thus its impact parameter is $b=r\sin \alpha /\sqrt{1-2\mu /r}$. Combining this with the null effective potential in Schwarzschild spacetime gives escape precisely for outward directions $0\le \alpha <{\alpha }_{\mathrm{crit}}$; the limiting ray approaches the photon sphere and does not escape.

As $r\downarrow 2\mu$ the cone shrinks to zero opening. A static observer does not exist on the horizon itself, and no future-directed ray starting on the Schwarzschild event horizon reaches exterior infinity. At $r=3\mu$, the critical half-angle is $\pi /2$; the exactly tangential critical ray stays on the unstable circular orbit. Escape cones for a moving observer are changed by relativistic aberration.

For a Reissner-Nordstrom black hole with $|Q|\le M$, the outer unstable circular null geodesic has radius $r_{\mathrm{ph}}=(3M+\sqrt{9M^2-8Q^2})/2$. Maximizing $f/r^2$ for $f=1-2M/r+Q^2/r^2$ gives the critical impact parameter $b_c^2=r_{\mathrm{ph}}^2/f(r_{\mathrm{ph}})=2r_{\mathrm{ph}}^3/(r_{\mathrm{ph}}-M)$. The geometrical optics capture cross-section is $\pi b_c^2$, reducing to $27\pi M^2$ in Schwarzschild spacetime and $16\pi M^2$ in the extremal charged case. Finite-wavelength absorption need not equal this geometric limit.

// Target: general-relativity.bigb