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A change $\lambda \mapsto A\lambda +B$ rescales a geodesic tangent and all its linear Killing constants by $1/A$. A null norm remains zero and cannot fix $A$, unlike proper-time normalization of a material particle. Ratios of the constants, such as the photon impact parameter $b=|h|/(ck)$ in Schwarzschild spacetime, are independent of the affine scale. A physically normalized four-momentum fixes the photon energy and hence selects a tangent scale.

For a nearly radial Kepler orbit, $q\simeq h^2/(2GM)$, so modest changes in specific angular momentum can strongly change pericentre. A planetary torque acting for a reference orbital time gives $\delta h\sim (m_p/M)\sqrt{GM{a}_p}$. Coherent kicks change $q$ by order itself after $N\sim (M/m_p)\sqrt{q/a_p}$ orbital-scale encounters, whereas uncorrelated kicks require an order-$N^2$ random-walk count. Both estimates are conditional on encounter geometry: a close encounter with a smaller impact parameter can give a much larger kick.

A spherical surface with areal radius $R>3m$, where $m=GM/c^2$, has apparent angular radius $\psi$ for a static exterior observer at $r_0>R$. Its grazing null geodesic has impact parameter $b=R/\sqrt{1-2m/R}$. Projecting the photon tangent into the observer's local rest frame gives $\sin \psi =b\sqrt{1-2m/r_0}/r_0$, yielding the displayed result. The condition $R>3m$ ensures the grazing ray escapes, since the surface is outside the photon sphere. For $R=4m$, the distant image radius is $\sqrt{2}$ times the flat-space prediction.

For unit mass, incoming speed $v>0$, impact parameter $b>0$, and potential energy $\mp \kappa /r$, put $q=\kappa /v^2>0$. At closest approach, conservation of energy and conservation of angular momentum give the displayed radii and tangential speeds $u_{\mp }=bv/p_{\mp }$. Consequently $p_{-}{p}_{+}=b^2$ and $u_{-}{u}_{+}=v^2$. These relations compare equal force magnitudes. The attractive radial case $b=0$ is a singular collision, so the speed-product statement does not extend to that endpoint.

For an azimuthally symmetric classical scattering map from impact parameter $b$ to scattering angle $\theta$,When several impact parameters produce the same angle, their nonnegative contributions are summed.

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.

For a trajectory with incoming and outgoing straight asymptotes, the deflection angle is the angle between its limiting incoming and outgoing velocity directions. For an attractive hyperbolic orbit of eccentricity $e$ and impact parameter $d$, $\sin (\delta /2)=1/e$ and $\tan (\delta /2)=GM/(d{v}^2)$, where $v$ is the incoming speed at infinity. The incoming velocity points opposite to the incoming radial direction; ignoring that reversal gives the wrong angle.

For an unbound equatorial Schwarzschild spacetime geodesic, let $E_{\infty }$ and $L_{\infty }$ be its conserved energy and angular momentum measured relative to infinity. The ratio $J=c^2{L}_{\infty }/E_{\infty }$ has dimensions of area divided by time. For a massive test particle, $E_{\infty }=m{c}^2{\gamma }_{\infty }$ and $L_{\infty }=m\ell$, so $J=\ell /{\gamma }_{\infty }=b{v}_{\infty }$, where $\ell$ is specific angular momentum and $b$ the impact parameter. For a photon, $E_{\infty }=c{p}_{\infty }$ and $L_{\infty }=b{p}_{\infty }$, giving $J=bc$ without assigning a finite Lorentz factor or a proper time to the photon. This ratio removes the arbitrary normalization of a null affine parameter.

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

For $f=1-2M/r$ and squared areal radius $C=r(r-b)$, an equatorial null geodesic obeys ${\dot{r}}^2=E^2[1-B^{2}f/C]$, with impact parameter $B=L/E$. At $b=2M$, $f/C=1/r^2$ in the exterior. Its supremum is $1/(4M^2)$ at the null singular boundary, rather than at a regular exterior photon sphere. Rays with $B>2M$ turn at $r=B$; rays with $B<2M$ reach the boundary. Assuming those rays are absorbed gives the displayed geometric-optics cross-section despite zero limiting horizon area.

For a test mass $M$ and isotropic background particles of mass $m$, the hyperbolic two-body problem uses the relative deflection scale $b_{90}=G(M+m)/w^2$. The test body's longitudinal velocity change in one encounter is $-2mw/[(M+m)(1+b^2/b_{90}^2)]$. Integrating the encounter rate over impact parameter gives a logarithm; replacing its velocity-dependent value by a common Coulomb logarithm in stellar dynamics gives the displayed systematic drag. The approximation $M+m\simeq M$ requires $M\gg m$, not merely $M>m$. Velocity diffusion and exact cutoff dependence are distinct effects omitted by this leading-log drag model.

For a fast ion traversing a fixed material thickness with only a small fractional energy loss, the dipole excitation probability per atomic encounter scales as $Z^2/v^2$ at fixed impact parameter. Summing transferred excitation energies and integrating the encounter measure $2\pi bdb$ gives the leading electronic stopping power scaling, with slower logarithmic dependence from the allowed excitation energies and impact-parameter range.

At fixed ion kinetic energy $E=M{v}^2/2$, a heavier isotope is slower, so the leading loss scales as $Z^{2}M/E$. At fixed velocity, isotopes of the same element instead have approximately the same electronic loss; plotting against energy per nucleon removes the leading isotope mass dependence. The charge should be the effective projectile charge if its bound electrons screen the nucleus. At lower speeds, excitation thresholds, slow-passage suppression, charge exchange and nuclear energy transfer invalidate a simple extrapolation of the high-speed law.

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.

A Schwarzschild static observer at areal radius $R$ emits an outward null geodesic at a small angle $\psi$ to the radial direction. Its impact parameter is $b=R\sin \psi /\sqrt{f}$, and the accumulated azimuth to infinity is ${\int }_R^{\infty }bdr/[r^2\sqrt{1-f(r)b^2/r^2}]=b/R+O(b^3/R^3)$. Thus weak curvature magnifies a narrow outward beam angle by $1/\sqrt{f(R)}$, rather than producing the full infinity-to-infinity deflection $4GM/(b{c}^2)$.

The additional invariant is the squared norm of the affine tangent. Metric compatibility and the geodesic equation make it constant, and a photon has the null conditionOn the equatorial plane this givesand the conserved quantities of part (a) yieldFor a nonradial ray $\ell \ne 0$, put $u=1/r$. Sincethe orbit equation becomesHere $b=|\ell |/(c\epsilon )$ is the positive impact parameter for a future-directed ray arriving from infinity. A strictly radial ray has $\ell =0$ and must be described by the radial equation rather than by $\varphi$ as parameter.

Let $f(r)=1-2m/r$ and work in the exterior $r>2m$, with $G=c=1$. The affinely parametrized geodesic Lagrangian isIts polar-angle equation isThe initial data $\theta =\pi /2$, $\dot{\theta }=0$ therefore remain in that plane. Rotational symmetry allows the axes to be chosen so that any nonradial initial trajectory has those data: its conserved angular-momentum direction fixes an orbital plane through the centre. Radial trajectories can also be placed in an equatorial plane.

The coordinates $t,\varphi$ are cyclic in $L$, so their canonical momenta are constant:The corresponding Killing vector fields are ${\partial }_t$ and ${\partial }_{\varphi }$. With the displayed signature, $g({\partial }_t,\dot{x})=E$ and $g({\partial }_{\varphi },\dot{x})=-J$, explaining the sign in the second canonical momentum. Thus $E$ is conserved energy per unit rest mass for a massive particle parametrized by proper time, and $J$ is its axial angular momentum per unit rest mass. For a photon, affine rescaling rescales both constants; after the tangent is normalized to the physical four-momentum they represent its energy at infinity and axial angular momentum. Their ratio $b=J/E$ is the signed impact parameter.

In the equatorial plane, the photon satisfies the null geodesic constraintwhich givesFor $J\ne 0$, put $u=m/r$ and $\alpha =Em/J$. Thenand consequentlyThe $J=0$ radial case is not represented by $\varphi$ as an orbit parameter.

There is a circular photon orbit at $r=3m$, but it is unstable. The radial effective potential isA circular orbit requires $V^{\prime }(r)=0$ and $E^2=V(r)$, givingIts second derivative isIt is a maximum, rather than a minimum, so radial perturbations depart from the orbit. Equivalently the second-order orbit equation is $u^{\prime \prime }=3u^2-u$. Linearizing about $u_c=1/3$ gives $\delta u^{\prime \prime }=\delta u$, with modes $e^{\varphi }$ and $e^{-\varphi }$. The existence of a growing mode directly contradicts the printed stability claim. Exact initial data can remain circular, but that does not make the orbit stable.

The supplied implicit expression describes the critical-energy family with ${\alpha }^2=1/27$, as fixed by the circular orbit. DefineRationalizing once gives $F=(\sqrt{3}-s)/(2(\sqrt{3}+s))$ andDifferentiating $F(u)=A{e}^{\varphi }$ gives $F^{\prime }{u}^{\prime }=F$, soThis verifies the first-order equation for every allowed integration constant $A$ on its real branch. For $A=0$ the derivative argument is replaced by direct substitution: $u=1/3$ is constant and solves the critical equation. The displayed family does not solve the equation at an independently arbitrary value of $\alpha$.

For $A>0$, write $q=A{e}^{\varphi }$. On the physical exterior branch,As $\varphi \to -\infty$, $q\to 0^{+}$ andThus for $A=0$ the photon remains exactly on the circular orbit. For $A>0$ it approaches $r=3m$ from outside through indefinitely many revolutions in the negative-$\varphi$ limit. With increasing $\varphi$, this same branch peels away outward and reaches infinity when $q=1-\sqrt{3}/2$. Reversing the direction of motion gives an incoming critical ray asymptotically approaching the photon circle. These are separatrix trajectories, not evidence of stability.

The potential maximum and the outward-growing radial separation display the same instability in two complementary descriptions.

The geodetic precession per orbit isThe orbital position and four-velocity repeat after a full orbit, so the direction to distant Earth in the instantaneous rest frame repeats. The freely transported gyroscope does not: its beam axis has turned through $\Delta \psi$ relative to the original aiming direction. This applies to any initial in-plane spin direction by linearity of parallel transport, not only the radial initial condition used in part (b). The beam's angular diameter is $1^{\circ }$, giving a half-angle of $0.5^{\circ }$. Thus Earth lies outside the beam after one orbit; comparing the offset with the diameter would already show a miss, but the correct acceptance angle is the half-angle.

The initial radial ray must be outward from the black hole for a direct path to Earth. A precisely inward radial null geodesic falls into the black hole and cannot provide the assumed initial communication. For outgoing near-radial Schwarzschild light deflection, the angular correction is especially small. If a static observer at $R$ measures a small emission angle $\psi$ from the outward radial direction, its impact parameter is $b=R\sin \psi /\sqrt{f(R)}$. The outgoing null geodesic givesRelative to flat propagation, curvature changes the small pointing angle and beam radius by the fraction $f(R{)}^{-1/2}-1\simeq 0.00251$. At an offset of $1.35^{\circ }$ this is only about $0.0034^{\circ }$; the beam's $0.5^{\circ }$ radius changes by about $0.0013^{\circ }$. The exactly radial central ray has zero angular deflection. This is not the geometry of the usual $4GM/(b{c}^2)$ scattering formula, which assumes a ray coming from infinity, passing the black hole, and returning to infinity.

Relativistic aberration also distinguishes a coordinate-radial light ray from a beam aimed radially in the moving crew's frame. Here $v/c=1/\sqrt{398}\simeq 0.0501$. Successful initial aiming includes that fixed correction; the same correction applies at the repeated orbital position. More precisely, an outgoing static-radial ray initially has crew-frame angle ${\psi }_0^{\prime }=-\arcsin (v/c)$ from the radial axis. The local small-angle aberration factor about that aim is $d\psi /d{\psi }^{\prime }=\gamma =\sqrt{398/397}$; including the outgoing propagation factor gives $\gamma /\sqrt{f}=1/q\simeq 1.00377$. Thus the asymptotic miss is about $1.358^{\circ }$ and the beam half-angle about $0.502^{\circ }$, with the same conclusion. The small distortion of the outward beam cannot correct the much larger accumulated geodetic precession.

A photon follows a null geodesic, whose spatial path approaches straight lines far from the deflecting mass. Extend the incoming asymptote forward: this is the reference path the photon would follow if the mass were removed while its incoming direction and transverse offset were held fixed. The impact parameter $b$ is the perpendicular distance from the mass to that incoming asymptote. It is not, in general, the distance of closest approach.

The positive deflection angle $\delta$ is the angle between the outgoing propagation direction and the continued incoming direction. The photon bends toward the mass, so a ray passing above it has a downward outgoing slope in the figure. In the weak-field approximation for an isolated spherical mass, Schwarzschild light deflection gives $\delta \simeq 4GM/(b{c}^2)$ for distant endpoints. This angle differs from the apparent image angle measured from the observer's line of sight to the mass.

By spherical symmetry, choose the orbital plane to be $\theta =\pi /2$. The geodesic Lagrangian of the Schwarzschild metric isFor a massive test particle, dots denote proper time derivatives and $2\mathcal{L}=-c^2$. The cyclic coordinates $t$ and $\varphi$ give, through the Euler-Lagrange equations, two constants:At infinity, $\mathcal{E}={\gamma }_{\infty }=1/\sqrt{1-v_{\infty }^2/c^2}$ is the Lorentz factor. The constant $\ell$ is the specific angular momentum, and the energy per unit rest mass is $E_{\infty }/m=c^2{\gamma }_{\infty }$. The normalization of the four-velocity givesUse $dr/d\varphi =\dot{r}{r}^2/\ell$ and define the energy-normalized angular momentum for a Schwarzschild geodesicThenFor an incoming massive test particle with impact parameter $b$, $L_{\infty }=m{\gamma }_{\infty }{v}_{\infty }b$, so $J=b{v}_{\infty }$. In particular, $J$ is not the same as the specific angular momentum $\ell$ unless ${\gamma }_{\infty }=1$.

A photon has no proper time parametrization and no finite Lorentz factor. For it, use an affine parameter in the same geodesic Lagrangian and impose $2\mathcal{L}=0$. The ratio $J=c^2{L}_{\infty }/E_{\infty }=bc$ remains well defined, and the null result isEquivalently this is the ultrarelativistic limit of the massive formula at fixed $J$, with $1/{\gamma }_{\infty }^2\to 0$. The printed massive formula needs this interpretation before it is applied to a photon.

Let $m_g=GM/c^2$, $u=1/r$, and use a prime for $d/d\varphi$. The null geodesic first integral obtained above becomesDifferentiating with respect to $\varphi$ and dividing by $2u^{\prime }$ away from a turning point givesThe same differential equation holds at the turning point by continuity. When $m_g=0$, the general solution is $u_0=C_1\cos \varphi +C_2\sin \varphi$. Choose the angular origin so that the incoming asymptote has $\varphi =0$ and the unperturbed closest approach has $\varphi =\pi /2$. The impact parameter then fixesThis is the straight reference path, not a curved orbit.

For perturbation theory with $m_g/b\ll 1$, write $u=u_0+u_1+O(m_g^2/b^3)$, where $u_1=O(m_g/b^2)$. Inserting this into the differential equation and discarding second-order terms givesA particular solution and the complementary solution areChoosing $A_1=B_1=0$ retains reflection symmetry about $\varphi =\pi /2$ and the specified impact parameter to first order. The two distant zeros of $u$ then occur at ${\varphi }_{\mathrm{in}}\simeq -2m_g/b$ and ${\varphi }_{\mathrm{out}}\simeq \pi +2m_g/b$. Their separation exceeds $\pi$ by $4m_g/b$, consistently giving the Schwarzschild light deflection $\delta \simeq 4GM/(b{c}^2)$.