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The relativistic orbit-shape equation can be derived without dividing by $\dot{r}$ at a turning point. Substitute the norm constraint into the radial geodesic equation to eliminate $\dot{t}$ and ${\dot{r}}^2$. Using $f^{\prime }=2\mu /r^2$ givesFor a massive particle with proper time as parameter, $\kappa =c^2$. Put $u=1/r$, and use primes in this paragraph for derivatives with respect to $\varphi$. The constant specific angular momentum gives $\dot{\varphi }=h{u}^2$, so the chain rule yieldsSubstituting these expressions into the radial equation and dividing by $h^2{u}^2$ for a nonradial orbit givesThe extra $3\mu u^2$ term is responsible for Schwarzschild perihelion precession. It vanishes in the Newtonian weak-field limit $\mu /r\to 0$. The derivation uses a massive particle; for a null geodesic, $\kappa =0$ and the constant $GM/h^2$ on the right is absent.

Use the $+---$ metric signature and abbreviate $f(r)=1-2\mu /r$, with $\mu =GM/c^2$ a length. Dots initially mean differentiation with respect to an affine parameter $\lambda$. The geodesic Lagrangian in the equatorial plane isThe planarity of geodesics in spherical symmetry permits this plane to be chosen. More explicitly, the full angular Euler-Lagrange equation is $\ddot{\theta }+2\dot{r}\dot{\theta }/r-\sin \theta \cos \theta {\dot{\varphi }}^2=0$, so initial data $\theta =\pi /2$, $\dot{\theta }=0$ preserve it.

Both $t$ and $\varphi$ are cyclic coordinates. Their Euler-Lagrange equations give conserved conjugate momenta $p_t=c^{2}f\dot{t}$ and $p_{\varphi }=-r^2\dot{\varphi }$. Defining constants $k,h$ therefore givesThe corresponding second-order geodesic equations areand variation of $r$ givesThe geodesic Lagrangian has no explicit $\lambda$ dependence. Its conserved Lagrangian energy function is ${\dot{x}}^{\alpha }\partial \mathcal{L}/\partial {\dot{x}}^{\alpha }-\mathcal{L}=\mathcal{L}$, since it is quadratic in the tangent components. Thus twice this function is a constant:This is the constant squared tangent norm of an affinely parametrized geodesic.

For a massive particle choose $\lambda =\tau$, its proper time, and obtain $\kappa =c^2$. For a photon, the tangent is null and $\kappa =0$. A general rescaling of a massive particle's affine parameter rescales a positive $\kappa$; the numerical value $c^2$ requires the proper time choice. With that choice, $h=L_z/m$ is signed specific angular momentum, of units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$, and $k=E/(m{c}^2)$ is dimensionless conserved specific Killing energy. The time coordinate is normalized to proper time at infinity, so this $E$ is the particle's conserved energy referred to infinity. A photon's constants scale with the normalization of its null tangent; it has no rest frame or proper time parametrization.

Substitution of $\dot{t}=k/f$ and $\dot{\varphi }=h/r^2$ into the norm gives the general radial first integralFor a massive particle this becomesAdopting the usual half-normalization for the effective potential for timelike Schwarzschild geodesics, defineAll terms have units ${\mathrm{m}}^2{\mathrm{s}}^{-2}$, or energy per unit mass. The first two terms are the Newtonian gravitational and centrifugal contributions; the last is the relativistic correction.

A circular orbit requires both $\dot{r}=0$ and $V_{\mathrm{eff}}^{\prime }(R)=0$. Differentiating at fixed $h$ givesThe radial first integral also givesFor the future-directed circular orbit at $R=4\mu$, these constants are $|h|=4\mu c$, $k=1$, and $f=1/2$. ThereforeSince Schwarzschild time agrees with the clock time of a static observer at infinity, the intended infinity-referred tangential orbital speed isThis uses the circumferential areal radius $R$ with the clock at infinity; it is not a local measurement made at the orbit. A local static observer's clock instead has $d{\tau }_{\mathrm{stat}}=\sqrt{f}dt$ and measures $v_{\mathrm{local}}=v_{\infty }/\sqrt{f}=\boxed{c/\sqrt{2}}$. These two operationally different speeds explain the significance of the phrase “at infinity”.

The effective-potential stability of a circular orbit follows fromConsequently the $R=4\mu$ orbit exists but is unstable to radial perturbations. The original plot below shows its position at a maximum, for $h=4\mu c$.

For a weak-field elliptic orbit, the relativistic correction makes the direction of closest approach advance after each radial cycle. The derivations under i and ii below giveper orbit to first order, where $a_{\mathrm{orb}}$ is the semi-major axis and $e$ the orbital eccentricity. For Mercury, the additional Schwarzschild perihelion precession is approximately 43 arcseconds per century, over and above the much larger Newtonian precession produced primarily by other planets. It is the unexplained residual after those effects have been included that provides the classic quantitative test of general relativity; one should not identify the whole observed precession with this relativistic correction.^[ref]

The labels i and ii in the original paper introduce optional orbit-equation aids, rather than separate mandatory subquestions. They are retained here for derivations of both aids and of the perihelion result.

The specific angular momentum of an elliptic Kepler orbit is $h=\sqrt{\mu a(1-e^2)}$. At periapsis, $r=a(1-e)$, soThe mean motion understates instantaneous angular speed near periapsis. A particle in an exterior mean-motion resonance can briefly overtake the planet in angle despite completing full orbits more slowly.

If the relative acceleration is $\ddot{\mathbf{r}}=-GM\mathbf{r}/r^3+\mathbf{f}$, with constant masses, then the specific orbital energy $\epsilon$, specific angular momentum $\mathbf{h}$ and eccentricity vector obeyThe inverse-square-force terms cancel in each derivative. These identities express perturbing work and torque, and track the instantaneous change of the ellipse's shape and orientation.

The plunging region begins near the innermost stable circular orbit, where nearly circular motion becomes unstable and radial inflow becomes rapid. If the inflow time is shorter than the stress-transport time, the gas approximately conserves its specific angular momentum there.

For an attractive inverse-square force, $r=p/(1+e\cos f)$, where $f$ is true anomaly, $e$ is orbital eccentricity, and $p=h^2/\mu$ is the semilatus rectum. The specific angular momentum relation $\dot{f}=h/r^2$ converts this geometric equation into an angular evolution law.

At fixed specific angular momentum $j$, perturb the effective potential $\Phi (r)+j^2/(2r^2)$ about a circular orbit. Linear radial motion has $\ddot{x}+{\kappa }^{2}x=0$, where ${\kappa }^2={\Phi }^{\prime \prime }+3{\Phi }^{\prime }/r=r^{-3}d(j_c^2)/dr$. Positive ${\kappa }^2$ gives stable oscillations; axial symmetry alone does not force that sign.

On a nonradial zero-energy Kepler orbit, conserved specific angular momentum $h$ gives periapsis $r_p=h^2/(2GM)$ and ${\dot{r}}^2=2GM(r-r_p)/r^2$. Integrating with $s=\sqrt{r-r_p}$ yields $t-t_p=\pm \sqrt{2/(GM)}(r+2r_p)\sqrt{r-r_p}/3$. Squaring gives the displayed version of the Barker equation, valid on both incoming and outgoing branches. The time origin $t_p$ is passage through periapsis, not an arbitrary point on the orbit.

For inviscid cylindrical rotation without a magnetic field, an axisymmetric radial interchange conserves each parcel's specific angular momentum $l=r^2\Omega$. Its restoring coefficient is ${\kappa }^2=r^{-3}d(l^2)/dr$. Increasing $l^2$ gives centrifugal stability; decreasing $l^2$ gives an exponentially growing interchange mode when radial displacement is allowed by the boundaries. A vanishing derivative is marginal, not a strictly restoring condition.

// Target: astrophysics.bigb

A planar orbit whose radius oscillates as its polar angle increases, producing successive lobes around the force center. For nonconstant $r$ with fundamental angular radial period $2\pi /\nu$ and nonzero specific angular momentum, the full position and velocity repeat precisely when $\nu$ is a rational number, provided the radius is bounded above and away from zero. If $\nu =p/q$ in lowest terms, $p$ radial periods advance the angle by $2\pi q$; $dt=r^{2}d\theta /h$ then gives a finite time period. This proves closure rather than merely a spatial self-intersection.

Let $\Psi (r)>0$ be the relative potential of a spherical stellar system, $\mathcal{E}=\Psi -v^2/2$ the specific binding energy, and $L=rv\sin \vartheta$ the specific angular momentum. Consider the galactic distribution functionwith $f=0$ for $\mathcal{E}\le 0$ and $f_0>0$. Velocity integration gives the mass densityFor this ordinary locally integrable distribution, the conditions are $\beta <1$ and $n>-1$. They ensure convergence at purely radial directions and at the escape-energy boundary, respectively.^[ref]

To derive the coefficient, write the velocity-volume element as $2\pi v^2\sin \vartheta dvd\vartheta$. The angular integral is $\sqrt{\pi }\Gamma (1-\beta )/\Gamma (3/2-\beta )$. In the radial integral set $x=v^2/(2\Psi )$, givingMultiplication proves the result using the gamma function normalization.

The angular moments give velocity-anisotropy parameter $\beta$, explaining the name constant-anisotropy distribution function. As a check, $\beta =0$, $n=7/2$ gives an isotropic distribution with $\rho \propto {\Psi }^5$, as in the Plummer model. The formula converts a phase-space ansatz into a density-potential relation; a self-consistent stellar model must additionally solve the Poisson equation for Newtonian gravity.

The distance from a focus to a conic section along a line perpendicular to its symmetry axis. In focus-centred polar coordinates, an ellipse may be written $r(1-e\cos \theta )=\ell$, where $e$ is its eccentricity of a conic.

The distance from a focus of a conic section to the curve measured perpendicular to its principal axis. In the focus-based polar coordinates equation $r=p/(1+e\cos \varphi )$ it is the length $p$, where $e$ is the eccentricity. For an ellipse, $p=a(1-e^2)=b^2/a$. For a Newtonian Kepler orbit, $p=j^2/(GM)$ with $j$ the specific angular momentum.

Let a weak tangential deceleration $D>0$ act on a test body moving along a slowly evolving family of circular orbits. Write its circular speed as $V_c(r)>0$ and its specific angular momentum as $j=r{V}_c(r)$. The specific torque is $\dot{j}=-rD$. Defining $q=d\log V_c/d\log r$ givesThis derivation follows angular-momentum balance; simply setting ${\dot{V}}_c=-D$ would omit the angular momentum changed by radial migration.

An adiabatic inward drift along radially stable circular orbits requires $1+q>0$. Indeed, with $\Omega =V_c/r$, the radial epicyclic frequency satisfiesFor this stable branch, the positive local timescale defined as $r/(2|\dot{r}|)$ is $(1+q)V_c/(2D)$. It is $V_c/D$ in a uniform-mass density core with $q=1$, and $V_c/(2D)$ for a flat galaxy rotation curve with $q=0$.

At $q=-1$, the circular-orbit angular momentum is locally independent of radius, so division by $1+q$ fails. For $q<-1$, the algebraic drift formula points outward, but those circular orbits are radially unstable; it is not a stable slow-inspiral solution. The quasistatic approximation also requires the drift time to exceed an orbital time. Finally, this local timescale is not automatically the remaining time to the center, which requires integrating the radius-dependent drift.

The same initial distance can lead to strikingly different orbits. Let $s=v_0/\sqrt{\mu /r_0}\ge 1$ and launch tangentially in one orbital plane. Then periapsis is the launch point, specific angular momentum is $h=r_0{v}_0$, andwhere $f$ is true anomaly. These follow by combining the initial specific orbital energy with the orbital eccentricity relation and the Kepler orbit equation. With $\mu =8{\mathrm{m}}^3/{\mathrm{s}}^2$ and $r_0=2\mathrm{m}$, the initial speeds $2$, $\sqrt{6}$, $2\sqrt{2}$, and $2\sqrt{3}m/s$ give $e=0,1/2,1,2$: a circular orbit, an elliptic orbit with apoapsis distance $6\mathrm{m}$, a parabolic trajectory, and a hyperbolic Kepler orbit. The two open paths have asymptotic speeds zero and $2m/s$ respectively. Their different directions at large distance are part of the Kepler orbit, not an extra scalar escape threshold.

A thin galactic disk is a stellar component whose vertical extent is small compared with its radial size. Small vertical velocity dispersion and substantial specific angular momentum are characteristic. Its survival constrains cumulative galactic disk heating, but its stars need not all have the same formation epoch.

A stationary axisymmetric galactic distribution function depending on the specific orbital energy and axial specific angular momentum is a two-integral distribution. These quantities are conserved in a time-independent axisymmetric Newtonian gravitational potential, so the Collisionless Boltzmann equation holds by the chain rule. Even dependence on $L_z$ removes net azimuthal streaming velocity; an odd part can supply rotation without changing the mass density.

For an instantaneous in-plane change of velocity of magnitude $\gamma v_K$ on a circular orbit, with angle $\theta$ measured from the forward tangential direction, the new specific orbital energy gives $a^{\prime }/a=(1-2\gamma \cos \theta -{\gamma }^2{)}^{-1}$. The specific angular momentum gives $e^{\prime 2}=1-(1-2\gamma \cos \theta -{\gamma }^2)(1+\gamma \cos \theta {)}^2$. The signed semi-major axis becomes negative for an unbound hyperbolic Kepler orbit.

A binary in which a black hole captures part of its companion's stellar wind. The captured gas can dissipate energy and, with sufficient specific angular momentum, form an accretion disk. This differs from gas overflowing a Roche lobe through the inner Lagrange point. The fraction captured, its angular momentum and changes in the wind affect the accretion flow and emitted X-rays.

Subtract local specific angular momentum times mass conservation from conservation of angular momentum including wind losses S,T. With outward radial mass flux $\mathcal{F}$ and outward viscous torque in an accretion disk $\mathcal{G}=-2\pi \bar{\nu }\Sigma r^3{\Omega }^{\prime }$, this gives $\mathcal{F}{h}^{\prime }+{\mathcal{G}}^{\prime }=2\pi r(Sh-T)$. For Keplerian rotation and $T=qSh$, substitution in mass conservation yields the displayed equation. A wind with q=1 removes local angular momentum but supplies no additional drift term; q>1 can drive accretion even with zero viscosity.

For a Keplerian disk, let $S$ be the total loss through both faces per unit projected area and $\dot{M}>0$ the inward accretion rate. Annular mass conservation gives ${\dot{M}}^{\prime }=2\pi RS$. A circular orbit has specific orbital energy $-GM/(2R)$, so a marginally escaping wind requires additional energy $GM/(2R)$ per unit lost mass. Equating this sink to a fraction $f$ of the two-face viscous dissipation $(9/4)\nu \Sigma {\Omega }^2$ gives the displayed relation. If the wind carries the local specific angular momentum $h=\sqrt{GMR}$, conservation of angular momentum gives $(\dot{M}h-\mathcal{G}{)}^{\prime }=h{\dot{M}}^{\prime }$, with viscous torque in an accretion disk $\mathcal{G}=3\pi \nu \Sigma h$. Eliminating the torque yields $R^2{\dot{M}}^{\prime \prime }+\frac{3}{2}R{\dot{M}}^{\prime }=\frac{3}{2}f\dot{M}$.