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For two uniform angular motions with $P_b<P_c$, an inner exoplanet transit at zero and next outer exoplanet transit at $\tau$, the next conjunction solves $(n_b-n_c)t+n_c\tau =2\pi$. Hence its longitude modulo $2\pi$ isFor an eccentric outer body this is only an approximation. Let $f_L,M_L$ be its line-of-sight anomalies and ${\Delta }_e(t)=f(t)-M(t)-(f_L-M_L)$. Exact true alignment instead requires $(n_b-n_c)t+n_c\tau -{\Delta }_e(t)=2\pi$. The correction follows directly from uniform mean anomaly and nonuniform true anomaly.

For bound Kepler orbits, the Delaunay variables obey $0<G<L$, $-G<H<G$, and three periodic angles. A canonical transformation preserves phase space volume. At fixed $L$, integrating $H$ supplies the measure $2GdG$. Since $G=L\sqrt{1-e^2}$, the fraction with orbital eccentricity less than $e$ is $L^{-2}{\int }_{L\sqrt{1-e^2}}^{L}2GdG=e^2$. Thus the probability density function is $2e$, independently of any normalizable energy-only phase-space distribution function. A Boltzmann distribution is unnecessary. Even the time-dependent density $F(E(L))[1+\eta \cos (l-n(L)t)]$, $|\eta |<1$, has this same marginal because integrating the mean anomaly removes its oscillatory part; an eccentricity marginal alone does not imply equilibrium.

For a bound noncollision Kepler orbit with gravitational parameter $\mu$, the canonical actions per unit mass are $\mathcal{L}=\sqrt{\mu a}$, $\mathcal{G}=\mathcal{L}\sqrt{1-e^2}$ and $\mathcal{H}=\mathcal{G}\cos i$. Their angles are mean anomaly $\ell$, argument of periapsis $\omega$ and longitude of ascending node $\Omega$. The Hamiltonian is $-{\mu }^2/(2{\mathcal{L}}^2)$, so only $\ell$ advances in the unperturbed problem. In the prograde spherical-action chart, $J_r=\mathcal{L}-\mathcal{G}$, $J_{\theta }=\mathcal{G}-\mathcal{H}$, $J_{\varphi }=\mathcal{H}$; the type-two canonical transformation is generated by $F_2=(\mathcal{L}-\mathcal{G})w_r+(\mathcal{G}-\mathcal{H})w_{\theta }+\mathcal{H}{w}_{\varphi }$. Retrograde spherical actions have $J_{\theta }=\mathcal{G}-|\mathcal{H}|$, so the absolute value cannot be dropped globally. Circular and coplanar limits are singular angle charts, not singular orbits.

For two circular Kepler orbits with mutual inclination $I_m$, independently average the stellar quadrupole disturbing function over both mean anomalies. A circular ring with unit normal $\mathbf{k}$ has $⟨{\hat{r}}_i{\hat{r}}_j⟩=({\delta }_{ij}-k_i{k}_j)/2$. Contracting the two ring tensors gives $⟨{\cos }^2\psi ⟩=[1+(\mathbf{k}\cdot {\mathbf{k}}_p{)}^2]/4$, proving the displayed exact angular factor. For small orbital inclinations relative to a common plane, $I_m^2=I^2+I_p^2-2I{I}_p\cos ({\Omega }_p-\Omega )+O(I^4)$. Thus the secular angular factor is $1/4-3(I^2+I_p^2)/8+3I{I}_p\cos ({\Omega }_p-\Omega )/4$. This potential generates the slow stellar contribution to satellite nodal precession.

Take an illustrative elliptic Kepler orbit with semi-major axis $a=1$, orbital eccentricity $e=0.6$ and gravitational parameter $\mu =1$, in consistent dimensionless units. Its semi-minor axis is $b=a\sqrt{1-e^2}=0.8$, mean motion is $n=\sqrt{\mu /a^3}=1$, and specific angular momentum magnitude is $h=\sqrt{\mu a(1-e^2)}=0.8$. About the attracting focus,The two sectors below each span $\Delta t=0.5$, giving $\Delta A=0.2$. Near periapsis the short radius sweeps a wide angle; near apoapsis the long radius sweeps a narrow angle. Constant areal velocity therefore does not require constant angular velocity or constant speed.

To compute the sectors without mistaking equal angles for equal times, parametrize the ellipse by eccentric anomaly $E$:This follows by integrating $\frac{1}{2}(xdy-ydx)$. Kepler's equation gives $E-e\sin E=M=n(t-t_{\mathrm{p}})$, with mean anomaly $M$ measured from periapsis time $t_{\mathrm{p}}$. Consequently $\Delta A=abn\Delta t/2=h\Delta t/2$. The displayed sectors use mean anomaly intervals $[-0.25,0.25]$ and $[\pi -0.25,\pi +0.25]$, rather than equal eccentric anomaly increments.

An exoplanet passes in front of its host star as seen by an observer and blocks part of the stellar flux. Exoplanet transit photometry infers geometry and radii from the resulting light curve. A true alignment involves the true anomaly, not generally the mean anomaly.

For an elliptic Kepler orbit, Kepler's equation relates mean anomaly $M$ and eccentric anomaly $E$ by

The relative displacement of a Kepler orbit sweeps equal areas about the attracting focus of a conic in equal times. A short interval near periapsis covers a wide angle at small separation; the same interval near apoapsis covers a narrow angle at large separation. The swept areas are equal even though the travelled distances are different. In particular, the sectors have curved ellipse boundaries: triangles formed by joining their endpoints only approximate the swept areas for short intervals.

Let $h=|\mathbf{r}\times \dot{\mathbf{r}}|$ be the relative specific angular momentum. The areal velocity isA central force exerts zero torque about its centre, so conservation of angular momentum makes this rate constant. This explanation also applies to other central forces; the ellipse shape specifically comes from Newtonian gravity.

In the $a=1$, $e=0.6$, $\mu =1$ example, specific angular momentum is $h=\sqrt{\mu a(1-e^2)}=\sqrt{0.64}=0.8$, giving areal velocity $0.4$. At periapsis and apoapsis, the velocity is perpendicular to the relative displacement, so $h=rv$. The endpoint speeds are consequentlyThe fourfold speed difference is consistent with the fourfold separation difference. Away from these endpoints, $h/r$ gives only the transverse component of velocity; the full speed follows from the vis-viva equation.

To actually place equal-time markers, use Kepler's equation. Write mean anomaly as $\mathcal{M}=nt$, with $t=0$ at periapsis and mean motion $n=\sqrt{\mu /a^3}=1$. The different symbol $\mathcal{M}$ avoids confusing this angle with the mass $M$. Solve for eccentric anomaly $E$ in radians, then compute the relative position:This timing procedure is explained in NASA educational archive, How Orbital Motion is Calculated.

For the first interval $0\le t\le \pi /6$, the final eccentric anomaly is $E\approx 1.041495$. For the equal interval $\pi \le t\le 7\pi /6$ starting at apoapsis, it ends at $E\approx 3.471065$. Both swept areas are $\Delta A=(h/2)\Delta t=0.4\pi /6=\pi /15\approx 0.209440$, one twelfth of the entire ellipse's area $0.8\pi$. The diagram computes its markers from Kepler's equation, making visible how uniform time steps become uneven spatial steps.

Uniform mean anomaly implies a line probability inversely proportional to the orbital speed. For small orbital eccentricity and true anomaly $f$, it is proportional to $1-e\cos f$.

Use $\mu =\gamma m$ for the gravitational parameter, to distinguish it from the Delaunay action called $G$. The displayed actions are per unit reduced mass. Accordingly use specific canonical momenta; with physical relative momenta, all three actions acquire a common reduced-mass factor, which leaves the eccentricity fraction below unchanged.

The canonical transformation. In spherical coordinates, separation of the Kepler Hamilton-Jacobi equation givesA local type-two generating function, continued across turning points with the appropriate momentum branches, isThe old momenta are $\partial W/\partial (r,\theta ,\varphi )$ and the new angles are $l={\partial }_{L}W$, $g={\partial }_{G}W$, $h={\partial }_{H}W$. This defines a canonical transformation, since its generating differential is $dW=p_{r}dr+p_{\theta }d\theta +p_{\varphi }d\varphi +ldL+gdG+hdH$. With standard choices of angle origins, the resulting Delaunay variables arewith $l$ the mean anomaly, $g$ the argument of periapsis, and $h$ the longitude of ascending node. In particular, the signed $H$ includes retrograde orbits; the squared equation in the PDF must not be read as $H\ge 0$.

For bound noncollision Kepler orbits, the independent ranges areThe limiting circular, radial and coplanar cases sit on boundaries where some angle charts degenerate. They have zero measure in the generic phase-volume calculation. The ranges are constrained as above, rather than a Cartesian box with three unconstrained actions.

The eccentricity count. A canonical transformation preserves phase space volume, soIntegrating the three angles contributes $(2\pi {)}^3$, and integrating $H$ gives $2G$. At fixed $L$, the condition that the orbital eccentricity be less than $e_0$ is $G>L\sqrt{1-e_0^2}$. ThusAny energy weight cancels between these integrals. If the total population is finite, or an explicit finite action band is used,This is the Delaunay phase-volume proof of the thermal eccentricity distribution. The source's $e^2$ law is a cumulative count, not a differential eccentricity density. Also, pointwise finite $F(E)$ does not alone imply finite $N_{\mathrm{tot}}$: $F=1$ gives ${\int }_0^{\infty }{L}^{2}dL=\infty$. The finite normalization of an energy-only Kepler distribution is the additional condition needed for a global probability statement; the fraction at every fixed energy is unaffected.

Why the observation does not imply relaxation or equilibrium. No Boltzmann distribution was used: every normalizable energy-only phase-space distribution function has the same eccentricity marginal, irrespective of collisional thermalization. An even stronger counterexample is a genuinely time-dependent collisionless distributionKepler motion has $\dot{l}=\nu (L)$ and constant actions and other angles, so ${\partial }_{t}f+\nu {\partial }_{l}f=0$: this is an exact nonstationary solution of the Collisionless Boltzmann equation. Integrating $l$ removes the cosine and leaves precisely the same $e^2$ cumulative count at every time. An eccentricity marginal alone cannot establish thermal relaxation, a relaxed age, or even a stationary distribution.

Write ${\mu }_g=\gamma m$ for the gravitational parameter of the relative Kepler orbit, and use specific orbital energy and momenta per unit reduced mass. The Hamiltonian isIf physical momenta rather than these specific momenta are used, multiply the actions by the reduced mass. This only introduces a constant factor into the phase space measure and leaves the eccentricity fractions unchanged. The formulas in the question adopt the specific-action convention.

Here is an explicit local canonical transformation. A separated complete integral of the Hamilton-Jacobi equation isChoose branches along an orbit and take $p_j=\partial S/\partial q_j$, $l=\partial S/\partial L$, $g=\partial S/\partial G$, $h=\partial S/\partial H$. Thenwhose exterior derivative proves preservation of the canonical symplectic form. Thus these variables are canonical, and the absolute Jacobian determinant is one. The angular separation gives $G=|\mathbf{r}\times \mathbf{v}|$ and $H=G\cos i$, including the sign of the angular momentum along the reference axis.

For clarity, the radial action can be evaluated rather than assumed. Let $r=a(1-e\cos u)$, with $u$ the eccentric anomaly. Then $p_r=\sqrt{{\mu }_g/a}e\sin u/(1-e\cos u)$, andIn the integral, use $e^2{\sin }^{2}u/(1-e\cos u)=1+e\cos u-(1-e^2)/(1-e\cos u)$ and ${\int }_0^{2\pi }(1-e\cos u{)}^{-1}du=2\pi /\sqrt{1-e^2}$. The polar action is $J_{\theta }=G-|H|$, while the signed azimuthal action is $J_{\varphi }=H$. The separated angular coordinates have the usual Delaunay variables interpretation: $g$ is the argument of periapsis, and $h$ is the longitude of ascending node. In particular,with a suitable origin at periapsis, so $l$ is the mean anomaly. We have therefore recoverednot merely assigned names to the new coordinates.

For nondegenerate bound Kepler orbits, the ranges areThe boundaries describe circular, coplanar or radial degeneracies, where some angles cease to be defined; they have zero measure in the present calculation. The actions are canonical coordinates with nested bounds, rather than three independently unrestricted Cartesian coordinates. The square of $H$ alone would miss the distinction between prograde and retrograde inclinations.

By the canonical transformation, the population measure becomesAt fixed $L$, the condition $e^{\prime }<e$ is $L\sqrt{1-e^2}<G<L$. Integrating all three angles and then $H$ givesConsequently the Delaunay phase-volume proof of the thermal eccentricity distribution yieldsThe PDF's $e^2$ dependence must be interpreted as a cumulative distribution function, not an eccentricity probability density function or the value of the original six-dimensional phase-space distribution function. A finite value of $F(E)$ at every energy also does not suffice to make the total population finite: $F=1$ on all bound energies gives the divergent integral ${\int }_0^{\infty }{L}^{2}dL$. The fractions are well defined at each fixed energy, over a finite allowed energy range, or for an energy distribution with a finite total integral.

A Boltzmann distribution is therefore unnecessary. Any normalizable energy-only distribution has the same eccentricity marginal, although such a distribution is a collisionless steady state by Jeans theorem. The marginal alone does not even prove stationarity: for $0<|\eta |<1$,is a positive time-dependent solution of the Collisionless Boltzmann equation and has exactly the same eccentricity marginal after integration over $l$. An observed $e^2$ cumulative law establishes neither thermal relaxation nor the age of the system or Universe.

For a bound Kepler orbit, the semimajor axis is half the long diameter of the ellipse; if $r_p,r_{mathrma}$ are its apsidal distances, $a=(r_p+r_{mathrma})/2$. The orbital eccentricity is $e=(r_{mathrma}-r_p)/(r_{mathrma}+r_p)$, with $0\le e<1$ and semiminor axis $b=a\sqrt{1-e^2}$. The gravitating centre is a focus, displaced by $ae$ from the ellipse's geometric centre.

Choose a reference plane, a positive normal and a fixed reference direction in that plane. The ascending node is where the orbit crosses toward positive normal displacement. The longitude of ascending node $\Omega$ is the reference-plane angle from the fixed direction to that node. The argument of periapsis $\omega$ is the angle from the ascending node to periapsis, measured in the orbital plane in the direction of motion. The orbital inclination $i$ is the angle between the reference normal and the orbital angular-momentum normal, with $0\le i\le \pi$. The diagram separates the ellipse's shape from these orientation angles.

Let $\mu =GM$ and use actions per unit test-particle mass. In the prograde spherical-action chart underlying the stated Hamiltonian, $J_{\varphi }=L_z$ and $J_{\theta }=L-L_z$, so $J_{\theta }+J_{\varphi }=L$. The radial action then satisfies $J_r+L=\sqrt{\mu a}$, consistently with $H_K=-\mu /(2a)$. DefineIf $(w_r,w_{\theta },w_{\varphi })$ are the old angles, a type-two generating function isDifferentiation with respect to the old angles recovers $J_r,J_{\theta },J_{\varphi }$, and differentiation with respect to the new actions gives the canonical transformationIts triangular integer action matrix has determinant one, so it preserves the angle periodicities. In the new coordinates,These are the Delaunay variables:With the usual choices of angle origins, $w_1=\ell$ is mean anomaly, $w_2=\omega$ is argument of periapsis, and $w_3=\Omega$ is longitude of ascending node. The generator $\mathcal{G}=L$ rotates the orbit within its plane, whereas $\mathcal{H}=L_z$ rotates that plane about the reference normal; this explains the angular identification. The fast angle obeys Kepler's equation $\ell =\mathcal{E}-e\sin \mathcal{E}$, where $\mathcal{E}$ is eccentric anomaly.

For retrograde orbits, signed $J_{\varphi }=L_z$ gives $J_{\theta }=L-|L_z|$ and the spherical Hamiltonian uses $J_r+J_{\theta }+|J_{\varphi }|$. The stated plus-sign form is therefore a prograde chart, not a global signed-action formula. The Delaunay variables still use $\mathcal{H}=L_z$; on the retrograde branch the generating function is $(\mathcal{L}-\mathcal{G})w_r+(\mathcal{G}+\mathcal{H})w_{\theta }+\mathcal{H}{w}_{\varphi }$. At $e=0$, periapsis is undefined, and at $i=0$ or $\pi$ the node is undefined: these are angle-chart degeneracies, not physical singularities.

For the perturbed planet orbit, the radial and vertical frequencies mean small oscillations about a circular equatorial orbit of cylindrical radius $R$. Write $r=\sqrt{R^2+z^2}$ and retain the stated quadrupole potential. On the equator,At fixed axial specific angular momentum $L_z$, the radial effective potential is $\varphi (R,0)+L_z^2/(2R^2)$. Its second derivative at circular balance $L_z=R^2{\Omega }_c$ gives the radial epicyclic frequencyReflection symmetry decouples vertical motion to first order. Expanding the potential in $z$ givesso the vertical epicyclic frequency isThe circular orbit is radially stable only for $R^2>3{\alpha }^2$. For $\alpha \ne 0$ in that domain, $\nu >{\Omega }_c>\kappa$. Vertical node crossings repeat at frequency $\nu$ while azimuth advances at ${\Omega }_c$, so the line of nodes has nodal precession rate ${\Omega }_c-\nu <0$. Radial periapsis passages repeat at $\kappa$, giving apsidal precession rate ${\Omega }_c-\kappa >0$. To leading quadrupole order, with $n=\sqrt{\mu /R^3}$,This is oblate-quadrupole epicyclic precession. Here $\varpi$ denotes the nearly equatorial longitude of periapsis, not $\omega$ alone.

The precession signs above have a small-eccentricity, small-inclination domain. They are not a universal claim for every inclined Kepler ellipse. To see the distinction with the same Delaunay variables, average the perturbation over an unperturbed orbit. Since $z/r=\sin i\sin (\omega +f)$ and $dt=r^{2}df/\mathcal{G}$, the term linear in $e\cos f$ integrates to zero and the ${\sin }^2(\omega +f)$ average in the $r^{-3}$ integral is $1/2$. ThusDifferentiating at fixed other actions yields the first-order secular ratesTheir nearly equatorial sum reproduces the prograde $\dot{\varpi }$. For a polar orbit, however, the node is fixed and $\dot{\omega }<0$. This supplies a concrete counterexample to extending the requested forward periapsis precession beyond the intended epicyclic regime.

For a circular orbit the origin of mean anomaly is arbitrary because there is no distinguished pericentre. Choose $M=0$ at the ascending node, so $M$ is also the argument of latitude. If a different periapsis convention is used, replace $M$ below by $M+\omega$. The orbital-frame rotation from inclination and node givesIn particular the crossing at $M=0$ has increasing $z$ for prograde $n>0$ and $0<I<\pi$, confirming the ascending node convention.

Put $ϵ=a/a_{\mathrm{pl}}$ and $c=\cos \psi =\hat{\mathbf{r}}\cdot {\hat{\mathbf{r}}}_{\star }$. The binomial series expansion of the direct disturbing function isThe indirect disturbing function subtracts $G{M}_{\star }ac/a_{\mathrm{pl}}^2$, cancelling the dipole term. The constant $G{M}_{\star }/a_{\mathrm{pl}}$ has no spatial derivative and does not perturb the particle. After omitting that constant, the leading stellar quadrupole disturbing function isThus the PDF's equality is understood modulo a dynamically irrelevant additive constant and higher multipoles.

Define $\ell =M+\Omega$ and let ${\ell }_{\star }$ be the star's orbital longitude as seen from the planet. Its orbital plane has $I_{\mathrm{pl}},{\Omega }_{\mathrm{pl}}$. The vector from planet to star is the negative of the planet's heliocentric vector: if ${\ell }_{\mathrm{pl}}$ describes that latter vector, use ${\ell }_{\star }={\ell }_{\mathrm{pl}}+\pi$. This sign matters in $\cos \psi$ but drops out of ${\cos }^2\psi$. Expanding the two rotated unit vectors givesFor example, the particle's horizontal correction is $-(I^2/2)\sin (\ell -\Omega )(-\sin \Omega ,\cos \Omega )$, while its vertical component is $I\sin (\ell -\Omega )$; their dot products with the corresponding stellar components produce these terms. No cubic terms occur because simultaneous reversal of both small orbital inclinations leaves the dot product unchanged.

The two mean longitudes are uniform for the circular Kepler orbits, so orbit averaging is uniform in both $\ell$ and ${\ell }_{\star }$. Set $c_0=\cos (\ell -{\ell }_{\star })$ and $c_2=\cos \psi -c_0$ through quadratic order. The needed averages follow by integrating sine and cosine products over $[0,2\pi {]}^2$:For the last identity, expand $c_0=\cos \ell \cos {\ell }_{\star }+\sin \ell \sin {\ell }_{\star }$ and factor the two integrations, each using $⟨{\sin }^2\ell ⟩=⟨{\cos }^2\ell ⟩=1/2$. Squaring $c_0+c_2$ now yieldswhere the error includes all fourth-order combinations of the two orbital inclinations. ThereforeMultiplying this by $G{M}_{\star }{a}^2/a_{\mathrm{pl}}^3$ gives the requested secular disturbing function. Equivalently, the exact angular factor for two circular rings is $(3{\cos }^2{I}_m-1)/8$, where $I_m$ is their mutual inclination; its small-angle expansion gives the same result. This is the double-averaged circular tidal potential.

For aligned equatorial and planetary orbital planes, set $I_{\mathrm{pl}}=0$. The secular stellar quadrupole disturbing function then has derivative ${\partial }_I{\overline{\mathcal{R}}}_{\star }=-3G{M}_{\star }{a}^{2}I/(4a_{\mathrm{pl}}^3)$. The supplied Lagrange planetary equations for nodal precession givesThe planetary quadrupole coefficient $J_2>0$ contributes ${\partial }_I{\overline{\mathcal{R}}}_{J_2}=-3G{M}_{\mathrm{pl}}{J}_2{R}_{\mathrm{pl}}^{2}I/(2a^3)$, henceBoth are retrograde nodal precession rates. Equating them cancels $n$ and gives the Laplace radiusThe planetary quadrupole dominates inside $a_L$ and the stellar tide outside it. These are small-inclination rates for planet-bound Kepler orbits in the hierarchical regime $a\ll a_{\mathrm{pl}}$. At exactly $I=0$, the node itself is undefined; the rates are the limiting precession frequencies of an infinitesimal tilt.

The printed line-density statement is incorrect for a phase-mixed orbit. A steady line density on a Kepler orbit is inversely proportional to speed: for a cross-sectional-area current $J_{\sigma }$, the area per unit arc length is $d\sigma /ds=J_{\sigma }/v$. Equivalently, phase mixing gives $d\sigma =({\sigma }_{\mathrm{tot}}/P)dt$, uniform in mean anomaly, where $P=2\pi \sqrt{a^3/\mu }$ is the orbital period.

For an optically thin population of blackbodies in radiative equilibrium, a fragment absorbs $\sigma L_{\star }/(4\pi r^2)$ and reradiates the same luminosity. Thus the fractional luminosity of a phase-mixed eccentric wire isThe specific angular momentum gives $dt/r^2=d\theta /h$, so the integral is $2\pi /h$. Since $Ph=2\pi a^2\sqrt{1-e^2}$,This derives the intended result after explicitly correcting the density to $1/v$.

The error is consequential. If one instead imposes the literal density $d\sigma /ds\propto v$, its time weighting is $v^{2}dt$. Using $⟨v^2⟩=\mu /a$, $⟨r^{-2}⟩=[a^2\sqrt{1-e^2}{]}^{-1}$ and $⟨r^{-3}⟩=[a^3(1-e^2{)}^{3/2}{]}^{-1}$ givesAt $e=1/2$ this is $5/3$ times the printed result. The inverse-radius averages can also be obtained directly with eccentric anomaly $E$, $r=a(1-e\cos E)$ and $dt=(1-e\cos E)dE/n$.

In the mean-conjunction approximation set ${\lambda }_b={\lambda }_c=\Lambda$. The resonant conjunction geometry is thenThere are $q$ phase directions spaced by $2\pi /q$. The integer $p$ determines their temporal order: at exact resonance the synodic period is $P_{\mathrm{syn}}=(p+q)P_b/q=p{P}_c/q$, and the conjunction longitude advances by $2\pi p/q$ per conjunction. For coprime $p,q$ this visits all $q$ branches; resonant-argument libration broadens each direction by about its libration amplitude of a resonant argument divided by $q$.

Mean conjunction and actual alignment are different for an eccentric orbit. If $\Lambda$ is the true common longitude, write $x=\Lambda -{\varpi }_c=f_c$ and $y=M_c(f_c)$, with $M_c$ the mean anomaly. Because ${\lambda }_c={\varpi }_c+y$ while the circular planet has ${\lambda }_b=\Lambda$, the exact geometric relation isThe displayed equally spaced directions use $y=x+O(e_c)$, hence are leading-small-eccentricity geometry. With appreciable $e_c$, use Kepler's equation to locate true conjunctions; equating true and mean longitudes silently is not exact.

First adopt the uniform-angular-motion approximation implicit in the requested formula. Set the line of sight to longitude zero, the $b$ exoplanet transit to $t=0$, and the next $c$ exoplanet transit to $t=\tau$, with $0\le \tau <P_c$. Write $n_b=2\pi /P_b$, $n_c=2\pi /P_c$, with $P_c>P_b$. The next conjunction satisfiesThe common longitude is $n_{b}t$ modulo $2\pi$. Subtracting one $2\pi$ gives the conjunction longitude from a transit time lagSubstitution of the period ratio yieldsIt is exact in this uniform-angle model; no first-order expansion in $\delta$ is necessary here. For other choices of which exoplanet transit is used, choose the appropriate whole-turn branch.

For the finite-eccentricity orbit specified earlier, this is an approximation. Let $f_L$ and $M_L$ be the true anomaly and mean anomaly at the line of sight. The eccentric planet has $M_c(t)=M_L+n_c(t-\tau )$ and true angular advance $n_c(t-\tau )+{\Delta }_e(t)$, whereTrue conjunction therefore requires $(n_b-n_c)t+n_c\tau -{\Delta }_e(t)=2\pi$. The PDF omits this term, which is generally $O(e_c)$.

For a concrete counterexample take $p=q=1$, $\delta =0$, $e_c=1/5$, $\tau =P_b/4$, and a line of sight along $c$'s periapsis. The uniform model predicts $t=7P_b/4$ and longitude $3\pi /2$. At that time $M_c=3\pi /2$, but Kepler's equation gives $f_c\ne 3\pi /2$, so the planets are not truly aligned. Accurate finite-eccentricity conjunctions must be found from the corrected equation.

Measure angles from periapsis in the direction of motion. The true anomaly $f$ is the angle at the stellar focus between the pericentre ray and the star-particle radius vector. Draw the auxiliary circle of radius $a$ around the ellipse's center; the point on that circle with the same projected major-axis coordinate as the particle defines the eccentric anomaly $E$ at the center.

The mean anomaly $M=n(t-\tau )$ is a uniformly advancing time phase, with $M=0$ at pericentre passage $t=\tau$. It is the angle swept by a fictitious uniformly moving circular phase marker, not generally the actual position angle. The mean motion is the angular frequencyThe second formula is Kepler's equation, which converts the uniform time phase into geometric position.

Choose the common initial longitude of the particle and $G$ as zero. Let the fixed inertial longitude of periapsis be $\varpi$, positive ahead of that ray. At release $f_0=-\varpi$, but the mean anomaly isThereafter $M(t)=nt+M_0$, $f(t)=M(t)+2e\sin M(t)+O(e^2)$, and the particle's longitude is $\lambda (t)=\varpi +f(t)$. The reference ray has longitude ${\lambda }_G=n_{g}t$.

In the rotating reference frame, the radial $x$ axis points outward along the star-$G$ ray and the $y$ axis points forward. Define $\delta \lambda =\lambda -{\lambda }_G$. The exact positional relations are