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Use metric signature $+---$ and the parity matrix $P^{\mu }{}_{\nu }=\mathrm{diag}(1,-1,-1,-1)$. If a Dirac field transforms as ${\psi }^{\prime }(x)={\eta }_P{S}_P\psi (x_P)$, the chain rule changes the spatial derivatives' signs. Covariance of its Dirac equation therefore requiresThe Clifford algebra gives precisely these identities for $S_P={\gamma }^0$. In an irreducible Dirac representation any other solution differs by a scalar phase, because its ratio with ${\gamma }^0$ commutes with all gamma matrices. With $|{\eta }_P|=1$, the Dirac mass bilinear and kinetic term consequently obeyThe factor ${\gamma }^0$ compensates the sign reversal of spatial derivatives. A transformation containing only the argument change would not preserve the massive Dirac equation. This establishes parity symmetry of the free or vector-coupled Dirac theory; it does not make a chiral weak interaction parity symmetric.

For charge conjugation, transpose the adjoint Dirac equation to obtain $(i{\gamma }^{\mu T}{\partial }_{\mu }+m){\bar{\psi }}^T=0$. Multiplying by $C$ gives the original Dirac equation for ${\psi }^c=C{\bar{\psi }}^T$ ifOne may choose $C$ unitary; in four dimensions a conventional choice in a standard Dirac or chiral basis is $C=i{\gamma }^2{\gamma }^0$, with $C^T=-C$. Its overall phase and ${\eta }_C$ do not affect the following bilinear transformation. Reordering the Grassmann fermion fields gives, for a color matrix $X$ commuting with the gamma matrices,Thus a vector color current changes sign and transposes its color generator under charge conjugation.

Write the Hermitian color connection as ${\mathcal{A}}_{\mu }=A_{\mu }^a{T}^a$. The interaction coming from $D_{\mu }={\partial }_{\mu }+ig{\mathcal{A}}_{\mu }$ is $-g\bar{\psi }{\gamma }^{\mu }{\mathcal{A}}_{\mu }\psi$. Its charge-conjugation symmetry requires ${\mathcal{A}}_{\mu }^C=-{\mathcal{A}}_{\mu }^T$. Parity acts on the Lorentz index as on a vector connection, so combining the two givesIn components, ${\mathcal{A}}_0\mapsto -{\mathcal{A}}_0^T(x_P)$ and ${\mathcal{A}}_i\mapsto +{\mathcal{A}}_i^T(x_P)$. The transpose is a color-space transpose, independent of the spinor matrix $C$. Intrinsic fermion phases cancel between a field and its adjoint.

Define the matrix gauge field strength unambiguously byThe derivative terms transform with the expected parity factors and a minus transpose. For the commutator, the crucial identity is $[X^T,Y^T]=-[X,Y{]}^T$, so the nonlinear term has the same transformation as the derivative terms. This is the CP transformation of non-Abelian field strength:For example, ${\mathcal{F}}_{0i}\mapsto +{\mathcal{F}}_{0i}^T(x_P)$ while ${\mathcal{F}}_{ij}\mapsto -{\mathcal{F}}_{ij}^T(x_P)$. With $\mathrm{tr}(T^a{T}^b)=\frac{1}{2}{\delta }^{ab}$, the color sum in the theta operator is twice a trace of two field strengths. The two charge-conjugation minus signs cancel, and transposition reverses the trace factors without changing their trace. The four parity matrices contracted with the Levi-Civita symbol contribute $\det P=-1$. HenceThe theta operator is CP odd; a generic fixed nonzero coefficient violates CP. Setting the coefficient to zero removes this term. Although the density is locally a total derivative, nontrivial gauge topology means that it need not be irrelevant to the quantum theory. If a properly normalized theta angle is identified periodically, invariance at special values equivalent to their negatives is a separate global question; it does not change the CP odd transformation of the local operator.

A stable symmetry-breaking minimum requires $\lambda >0$ and $m^2<0$. Put $v^2=-m^2/\lambda >0$. The common scalar potential can then be written, up to a constant, asIts minima form the sphere $\Phi \cdot \Phi =v^2$. Choose the vacuum direction ${\Phi }_0=(0,0,v{)}^T$. If instead $m^2>0$ with positive $\lambda$, the minimum is at the origin and the assumed nonzero vacuum expectation value is not a stable vacuum of these models.

In the ungauged theory, the internal symmetry is global $O(3)$; its connected part is $SO(3)$. Constant orthogonal transformations preserve both the kinetic term and the potential. The chosen vacuum is invariant under $O(2)$ acting on its first two components, so the continuous breaking is $SO(3)\to SO(2)$ with two broken generators. The vacuum orientations are genuinely different degenerate global vacua.

The global triplet scalar symmetry breaking is visible directly in the fluctuation masses. Write $\Phi =({\pi }_1,{\pi }_2,v+h{)}^T$. The potential Hessian isThe quadratic Lagrangian is thereforeThere is one radial scalar of mass $m_h^2=2\lambda v^2=-2m^2$ and two massless Goldstone bosons. The two Goldstone modes describe motion tangential to the vacuum sphere. They remain physical, since a global rotation has only constant parameters and cannot remove arbitrary spacetime-dependent fluctuations. All three original real-scalar degrees of freedom remain. Interactions persist: the shifted potential is $V-V_{\min }=\frac{\lambda }{4}(2vh+h^2+{\pi }_1^2+{\pi }_2^2{)}^2$, containing cubic and quartic terms. Masslessness of the angular modes is protected by the exact continuous symmetry through Goldstone's theorem.

In the gauged theory, the three-component real field is the vector representation of local $SO(3)$, equivalently the Adjoint representation of $SU(2)$. The center of $SU(2)$ acts trivially on this field, so both descriptions have the same local field content and perturbative spectrum. Take $(t^a{)}_{bc}=-i{ϵ}^{abc}$, for which $[t^a,t^b]=i{ϵ}^{abc}{t}^c$. The covariant kinetic term and the gauge kinetic term are invariant under local transformations with the associated connection transformation. The scalar inversion $\Phi \mapsto -\Phi$, with the gauge field unchanged, is also a discrete invariance of the displayed action.

The vacuum leaves rotations about its third axis unbroken. The residual gauge group is $SO(2)$, or the corresponding $U(1)$ subgroup in the $SU(2)$ description. Expanding the covariant kinetic term about ${\Phi }_0$ givesThus the gauge-boson mass matrix is $g^2{v}^2\mathrm{diag}(1,1,0)$. The third generator annihilates the vacuum, while the first two do not. The masses are $m_{B^1}=m_{B^2}=gv$, $m_{B^3}=0$, and $m_h=\sqrt{2\lambda }v$.

The quadratic derivative mixing displays what happens to the Goldstone fields. With the generator convention just chosen,Local rotations can set the two angular fields to zero in unitary gauge. Then $\Phi =(0,0,v+h{)}^T$ andThe two Goldstone modes supply the longitudinal polarizations of the two massive vector bosons. They do not survive as additional physical massless scalars. The radial scalar is still physical, and $B^3$ retains only two transverse polarizations. This adjoint triplet Higgs spectrum realizes the Higgs mechanism; the gauge group is only partially broken, so one massless gauge boson remains.

The degree-of-freedom check isIn contrast, the global theory has just the three scalar modes, including its two physical Goldstone bosons. A vacuum expectation value in the local theory is a choice of gauge-fixed description: the direction can be changed by gauge transformations and is not itself an observable. Gauge-independent masses and the physical degree count express the actual effect. Both theories retain Lorentz symmetry in the constant scalar vacuum.

The flavor content is $K^0=d\bar{s}$ and ${\bar{K}}^0=s\bar{d}$. A charged-current box diagram changes strangeness by two units. The two internal quark lines can contain any up-type flavors $u_i,u_j\in \{u,c,t\}$, connected by two charged $W$ propagators. One allowed topology is shown below; its crossed counterpart also contributes to the full mixing amplitude.

Each charged-current vertex contains the appropriate CKM matrix element. The flavor sum has combinations ${\lambda }_i{\lambda }_j$, with ${\lambda }_i=V_{is}^{\ast }{V}_{id}$, multiplying mass-dependent loop functions. CKM matrix unitarity gives ${\sum }_i{\lambda }_i=0$, illustrating the GIM cancellation of flavor-independent loop terms. The diagram is at order $g^4$ and second order in the weak interaction.

Let $\Theta$ be the antiunitary CPT operator. It interchanges the neutral-kaon flavor states up to phases. For a Hermitian Hamiltonian invariant under CPT,since the diagonal expectation is real. Therefore CPT gives $R_{11}=R_{22}$. More generally an effective decay Hamiltonian has $R=M-i\Gamma /2$, where both $M$ and $\Gamma$ are Hermitian. CPT gives $M_{11}=M_{22}$ and ${\Gamma }_{11}={\Gamma }_{22}$, hence the same equality of the complex diagonal entries. It does not require $R_{12}=R_{21}$; that is a CP condition. Nor is $R_{21}=R_{12}^{\ast }$ true for the full decay matrix in general.

Choose the CP convention $CP|K^0⟩=-|{\bar{K}}^0⟩$ and $CP|{\bar{K}}^0⟩=-|K^0⟩$. Then CP acts as $-{\sigma }_x$ on the flavor basis. Invariance of $H^{\prime }$ means $R={\sigma }_{x}R{\sigma }_x$, and thereforeWith a different flavor-state phase convention this relation carries the corresponding phase factors; equality is the relation in the convention adopted here. For a Hermitian mass matrix it makes the off-diagonal element real.

For the requested production-time mass eigenstates, discard the absorptive part. From now on $R$ denotes the Hermitian part $(R_{\mathrm{original}}+R_{\mathrm{original}}^{†})/2$, so its entries have the formIf the original matrix already was Hermitian no replacement is needed. The neutral-kaon mass matrix diagonalization gives eigenvalues $r_0\pm |z|$, since $(r_0-\lambda {)}^2-|z{|}^2=0$. Write $z=r{e}^{i\phi }$, $r=|z|>0$. Use the kaon mixing square-root branch conventioncontinuously from the CP-conserving convention in which $z>0$. Independent unrelated root signs would interchange the eigenvalue labels. Direct multiplication gives $R(a,-b{)}^T=(r_0-r)(a,-b{)}^T$ and $R(a,b{)}^T=(r_0+r)(a,b{)}^T$. Assigning the larger mass to $K_L^0$ giveswith mass shifts $r_0-r$ and $r_0+r$ respectively. A common strong-interaction mass may simply be added to both. Their norms are one and their inner product is zero. A Hermitian mass-only calculation determines the lighter and heavier combinations; it does not determine their different lifetimes.

In the CP-conserving limit, the chosen lighter and heavier combinations areThey are CP eigenstates with eigenvalues $+1$ and $-1$ in the convention above. The sign of the real off-diagonal entry and the flavor-state convention are chosen together so that the stated $K_1^0=K_S^0$ limit is the lower-mass state.

Changing to this CP basis givesConsequently, provided $a+b\ne 0$, defineAbsorbing the common phase of $a+b$ into the definitions of the mass eigenstates and normalizing now yieldsFor this Hermitian approximation, $ϵ=i\tan (\phi /2)$ is purely imaginary, so $⟨K_S^0|K_L^0⟩=(ϵ+{ϵ}^{\ast })/(1+|ϵ{|}^2)=0$. The phase convention matters for the kaon CP mixing parameter. Physical kaon decay eigenstates instead diagonalize the generally non-Hermitian matrix $M-i\Gamma /2$, for which the mixing parameter can have a real part and the eigenstates need not be orthogonal. The formula here is the requested mass-only result, not a calculation of that full decay dynamics.

Put $V=|v_0|>0$, take $q>0$ without loss of generality, and write $s=R^2+z^2/q^2$. The Newtonian gravitational potential is singular at the origin; all pointwise claims below concern $s>0$. The zero of the Newtonian gravitational potential is the one in the paper, with an implicit fixed length unit inside the logarithm.

The Poisson equation for Newtonian gravity in cylindrical coordinates givesThe two contributions to the Laplacian areConsequently the mass density of the axisymmetric logarithmic gravitational potential isThe coefficient of $R^2$ is positive. The coefficient of $z^2$ determines the density positivity for an axisymmetric logarithmic potential: strict positivity at every noncentral point requires $q^2>1/2$ and $v_0\ne 0$. Nonnegativity permits $q^2=1/2$, where the mass density vanishes on the noncentral symmetry axis. For $q^2<1/2$ it is negative near that axis. Either sign of a nonzero real $q$ gives the same model, and $q=0$ is undefined. Setting $v_0=0$ yields vacuum away from the singular origin, not strictly positive mass density. The $r^{-2}$ central cusp is locally integrable and contains no hidden point mass: the Newtonian gravitational field flux through a shrinking sphere is of order its radius.

An equatorial circular orbit requires inward gravitational acceleration equal to $v_c^2/R$. Thus the circular speed isThis is a flat galaxy rotation curve. The singular center does not define an additional circular orbit at $R=0$.

The specific orbital energy $E=\frac{1}{2}(v_R^2+v_{\varphi }^2+v_z^2)+\Phi$ is conserved because the Newtonian gravitational potential is time independent. The axial specific angular momentum $L_z=R{v}_{\varphi }$ is conserved because the Newtonian gravitational potential is axisymmetric. Hence the Jeans theorem, or directly $dF(E,L_z^2)/dt=0$ in the Collisionless Boltzmann equation, permits a stationary two-integral galactic distribution function. Using $L_z^2$ makes this galactic distribution function even under reversal of the azimuthal velocity, so it describes a nonstreaming population. A general axisymmetric population can have an odd part in $L_z$, or dependence on a third integral of motion; neither is required for this construction.

At each position define $\rho ⟨g⟩=\int gF{d}^{3}v$. The two-integral galactic distribution function is invariant under swapping $v_R$ and $v_z$, since only their sum of squares occurs in $E$. It is also even in each velocity component. Swapping integration variables proves meridional velocity isotropy of a two-integral distribution, while reflecting one variable makes each mixed integrand odd. ThereforeAll mean velocities vanish, so these are also the corresponding statements for the velocity ellipsoid and its centered covariance tensor. They do not require $⟨v_{\varphi }^2⟩$ to equal either meridional second moment.

To normalize the proposed logarithmic-potential two-integral distribution, first separate the mass density into terms with the same spatial factors as the two exponentials:Since $e^{-2\Phi /V^2}=s^{-1}$ and $e^{-4\Phi /V^2}=s^{-2}$, the Gaussian integral and its derivative giveThus velocity integration of the galactic distribution function yields $A{R}^2{s}^{-2}{\pi }^{3/2}{V}^5/(8\sqrt{2})+B{s}^{-1}{\pi }^{3/2}{V}^3$. Matching the independent spatial factors determinesThis verifies both the stationary Collisionless Boltzmann equation and the required Poisson equation for Newtonian gravity. A shift $\Phi \mapsto \Phi +C$ changes the normalizations to $A{e}^{4C/V^2}$ and $B{e}^{2C/V^2}$; the physical mass density is unchanged.

There is an important distinction between positive mass density and a nonnegative galactic distribution function. For $1/2\le q^2\le 1$, both coefficients above are nonnegative. For a prolate model $q^2>1$, $A<0$ and one must test the sum rather than reject the model just because one term is negative. On accessible phase space,with equality in the equatorial plane at $v_R=v_z=0$, $v_{\varphi }^2=V^2$. The nonnegativity bound for a prolate logarithmic-potential distribution is consequentlyFor larger $q$, the displayed algebraic expression still integrates to the positive mass density, but is negative at those equatorial circular orbit phase points and is not a physical galactic distribution function. This extra restriction is distinct from the earlier density-only answer.

For an axisymmetric Newtonian gravitational potential, the vacuum Laplace equation in cylindrical coordinates is $R^{-1}{\partial }_R(R{\partial }_R\Phi )+{\partial }_z^2\Phi =0$. Use separation of variables, $\Phi =X(R)Z(z)$. Dividing by $XZ$ and setting the separated constant to $k^2$, with $k>0$, givesThe first equation has solutions $e^{\pm kz}$; with $u=kR$, the second is the order-zero Bessel differential equation. Regularity at the axis selects the Bessel function of the first kind $J_0(kR)$, rather than the singular Bessel function of the second kind $Y_0$. ThusThere are also zero-separation-constant modes, and other boundary conditions can select other separated families. For a localized reflection-symmetric thin disk, decay away from the disk selects $e^{-k|z|}$ on the respective half-spaces.

Apply the divergence theorem to the Poisson equation for Newtonian gravity across a narrow pillbox enclosing disk area. Writing the volume mass density as $\Sigma (R)\delta (z)$ givesReflection symmetry makes ${\Phi }_z(R,0^{+})=2\pi G\Sigma (R)$. A superposition of the regular vacuum modes therefore satisfiesThe order-zero Hankel transform and its inversion now determine the Hankel representation of thin-disk gravity:The sign is fixed by the positive derivative of an attractive Newtonian gravitational potential above the disk. For ordinary finite disks the representation uses the isolated boundary condition, fixing any extra vacuum field or additive constant. Infinite scale-free disks require suitable regularization of the potential itself.

Equatorial circular orbit balance is $v_c^2=R{\Phi }_R(R,0)$. Using $J_0^{\prime }(u)=-J_1(u)$ in the Hankel representation of thin-disk gravity givesDifferentiation under the integral is justified by convergence for regular models, or by first working off the disk and taking the limiting radial field.

Write $C={\Sigma }_0{R}_0$. For the Mestel disk, the supplied Bessel function integral gives $S(k)=-2\pi GC/k$ for $k>0$. The circular speed is thenMeanwhile the interior mass is $M(R)=2\pi {\int }_0^R(C/s)sds=2\pi CR$. ConsequentlyThe Mestel disk has a flat galaxy rotation curve, despite its nonspherical geometry.

Its infinite total mass prevents setting the Newtonian gravitational potential to zero at infinity. Indeed the integral for $\Phi$ diverges at $k=0$ because $S(k)\sim 1/k$. Subtracting the potential at a reference point removes this additive divergence. As an explicit check, the Mestel disk potential-density pair isIt is harmonic for $z\ne 0$, its vertical derivative jump is $4\pi GC/R$, and $R{\Phi }_R(R,0)=2\pi GC$. Equivalently, an Abel regularization of an oscillatory integral with damping factor $e^{-ak}$ in the radial-field integral gives $R{\int }_0^{\infty }{e}^{-ak}{J}_1(kR)dk=1-a/\sqrt{a^2+R^2}$, which tends to one as $a\downarrow 0$. The force is well-defined although the unreferenced potential is not.

The equality with $GM(R)/R$ is special to the Mestel disk, not a thin-disk shell theorem. The spherical shell theorem normally makes that formula possible, but does not apply to an astrophysical disk. Exterior annuli exert outward radial gravitational acceleration inside their radii, while interior annuli do not act exactly as point masses at the origin. For this particular scale-free surface density, their combined contributions happen to produce the same result as the interior mass expression. Changing the radial surface density or truncating the disk generally destroys the equality: enclosed mass does not determine a disc rotation curve.

Let $M=M_1+M_2$ and $\alpha =M_2/M$, with $0<\alpha <1$. In the centre of mass frame the two primary distances are $r_1=\alpha R$ and $r_2=(1-\alpha )R$. Applying Newton's law of universal gravitation and circular centripetal acceleration to the first primary gives $M_1{\omega }^2\alpha R=G{M}_1{M}_2/R^2$, and henceThe same result follows from the second primary or from the relative two-body equation of motion.

A rotating-frame equilibrium has zero rotating-frame velocity, so its Coriolis force vanishes. Above the orbital plane both primaries have downward gravitational components; below it both have upward components. The centrifugal force has no $z$ component. More explicitly,which vanishes only at $z=0$ for finite nonsingular points. Thus all equilibria lie in the orbital plane.

The equation of motion in a rotating frame for the dwarf isOn the orbital plane the centrifugal contribution is $m{\omega }^2\mathbf{r}$. Thus the rotating-frame effective potential is $E=U-m{\omega }^2(x^2+y^2)/2$; equilibria are exactly its stationary points. The Coriolis force affects motion about those equilibria but not their positions. This is the circular restricted three-body problem, with the dwarf treated as a test mass.

The printed energy unit has the wrong dimensions. The correct total energy unit is $mGM/R$, or $GM/R$ for specific energy; $GM/R^2$ is an acceleration. Dividing $E$ by $mGM/R$ and measuring lengths in units of $R$ gives the dimensionless effective potentialThe primary coordinates are $(-\alpha ,0)$ and $(1-\alpha ,0)$. Restricting to $y=0$ gives $F(x)=\mathcal{E}(x,0)$ andThe absolute values are essential; dropping them would reverse the attraction on the left of a primary. On each of the intervals separated by the two primaries, $F^{\prime }$ decreases strictly from $+\infty$ to $-\infty$. There is exactly one Collinear Lagrange point on each interval, a maximum along the $x$ direction.

For the inner Lagrange point, write $x=1-\alpha -\delta$ with $\delta >0$. For the L2 Lagrange point, write $x=1-\alpha +\delta$. The respective equations areBalancing the leading terms gives $\delta =(\alpha /3{)}^{1/3}+O({\alpha }^{2/3})$. Therefore the small-mass-ratio collinear Lagrange points obeyThe omitted $-\alpha$ displacement of the secondary is smaller than the stated error. The leading distance from the secondary is the dimensionless Hill radius.

For the L3 Lagrange point put $x=-1+c\alpha +O({\alpha }^2)$. It lies to the left of both primaries, soExpanding the two positive inverse-square factors gives $(1-\alpha )/(x+\alpha {)}^2=1+(2c+1)\alpha +O({\alpha }^2)$ and $\alpha /(x-1+\alpha {)}^2=\alpha /4+O({\alpha }^2)$. Thus $F^{\prime }=-(3c+5/4)\alpha +O({\alpha }^2)$, soAll three have $y=z=0$. The small-$\alpha$ equalities in the paper are asymptotic expressions, not exact coordinates for finite mass ratio.

For the remaining Lagrange points, introduce plane polar coordinates about the first primary: $x=-\alpha +s\cos \theta$, $y=s\sin \theta$. Its distance to the second primary is $d=(s^2+1-2s\cos \theta {)}^{1/2}$, andThe angular stationary condition is ${\partial }_{\theta }\mathcal{E}=\alpha s\sin \theta (d^{-3}-1)=0$. For an off-axis point $s>0$ and $\sin \theta \ne 0$, so $d=1$. The radial stationary condition becomesHence $s=1$, and $d=1$ implies $\cos \theta =1/2$. The Triangular Lagrange points are exactlyBoth triangles are equilateral. The collinear and off-axis calculations together exhaust the stationary points.

For the requested sketch, $\mathcal{E}$ tends to $-\infty$ at either primary and at large in-plane radius. Each Collinear Lagrange point is a saddle of the planar effective potential: its $x$ curvature is negative and its $y$ curvature is positive. To see the latter, let $d_1=|x+\alpha |$, $d_2=|x-1+\alpha |$ and $D=(1-\alpha )/d_1^3+\alpha /d_2^3$. At an inner point both distances are below one, so $D>1$. At an exterior stationary point, use barycentric positions $a_1=-\alpha$, $a_2=1-\alpha$ and weights $1-\alpha ,\alpha$: the equation implies $x(D-1)=\sum w_i{a}_i/|x-a_i{|}^3$. To the right, the weighted sum is positive because the positive primary is closer; to the left it is negative because the negative primary is closer. Thus $D>1$ there as well, and ${\mathcal{E}}_{yy}=D-1>0$.

At either Triangular Lagrange point, the planar Hessian isIts trace is $-3$ and determinant $27\alpha (1-\alpha )/4>0$, so these are local maxima. Their height is $\mathcal{E}=-3/2+\alpha (1-\alpha )/2$. The sketch below uses a finite mass ratio to separate the five points clearly; its singular wells are clipped only for display. These curvature labels concern the effective potential, not a stability test that ignores the Coriolis force.

Put $k=p+1$, $\mu =G{M}_{\star }$ and $n_{\mathrm{pl}}=\sqrt{\mu /a_{\mathrm{pl}}^3}$. To leading order the mean-motion resonance fixes $a=a_{\mathrm{pl}}{k}^{-2/3}$. The specific angular momentum isThe asteroid's instantaneous angular speed is $h/r^2$. Thus the high-eccentricity angular-speed crossover satisfies $h/r_x^2=n_{\mathrm{pl}}$, givingHere $q\ll r_x\ll a$, so a parabolic Kepler orbit gives the local motion accurately. Its polar equation is $r=2q/(1+\cos f)$. Write $f_x=\pi -{ϵ}_x$. Since $1+\cos f_x\simeq {ϵ}_x^2/2$,This is the outward crossing; the inward crossing has true anomaly $-f_x$ modulo $2\pi$.

For the pericentre-to-crossover flight time use $D=\tan (f/2)$. The parabolic Kepler orbit has $r=q(1+D^2)$ and $dt=r^{2}df/h$, whence direct integration gives the Barker equationAt the crossover $D_x\simeq \sqrt{r_x/q}\gg 1$. Consequently the planet's angular displacement isThe error tends to zero as $q/a_{\mathrm{pl}}\to 0$ at fixed $k$; it includes the finite binding energy of the elliptic Kepler orbit as well as the subleading term in the Barker equation.

At the $j$th pericentre passage the asteroid's unwrapped mean longitude is $\lambda =\varpi +2\pi j$. The resonant argument therefore givesThus $\varphi /k$ specifies the planet's direction relative to the asteroid's longitude of pericentre when the asteroid passes pericentre, with $k$ branches separated by $2\pi /k$. It is not the instantaneous true-longitude separation throughout the orbit. At high orbital eccentricity the asteroid quickly sweeps to a direction nearly opposite pericentre, while the planet scarcely moves. During the long outer excursion the asteroid's direction changes slowly and the planet advances substantially. This makes an astronomical conjunction during the outer excursion much more dangerous than an astronomical conjunction close to pericentre.

At the following apocentre, the planet has advanced $\pi /k$. The asteroid's direction in the rotating reference frame is consequentlyThese $k$ apocentre directions are equally spaced. Phase protection of interior integer resonances places the planet in the largest gap between them, rather than at an apocentre direction. For $k=2$, $\varphi =\pi$ puts an apocentre toward the planet, whereas $\varphi =0$ puts the apocentres at $\pm \pi /2$. For $k=3$, $\varphi =0$ puts an apocentre toward the planet, whereas $\varphi =\pi$ puts the apocentres at $\pi /3,\pi ,5\pi /3$. Hence the geometrically favored resonant-argument libration centres in this regime of high orbital eccentricity areThis is an encounter-avoidance argument for likely stability, not a calculation of a resonant Hamiltonian or a claim that every orbit near either centre is stable at arbitrary orbital eccentricity.

The following rotating-frame drawing uses $a_{\mathrm{pl}}=1$, $a=3^{-2/3}$, $q=0.002$ and $\varphi =\pi$. It follows three successive Kepler orbits, which fill one planet orbital period and close the pattern. The blue segment is just the first outward half-orbit. The marked crossover is a maximum of its rotating-frame polar angle, since $d\psi /dt=h/r^2-n_{\mathrm{pl}}$ changes from positive to negative there. The planet stays at $(1,0)$; the three apocentres avoid that direction.

To estimate angular-momentum kicks to nearly radial orbits, note that $q\simeq h^2/(2\mu )$, so an order-one fractional change in $q$ requires $|\Delta h|$ of order $h$. A characteristic distant encounter at radii of order $a_{\mathrm{pl}}$ has tangential acceleration of order $G{M}_{\mathrm{pl}}/a_{\mathrm{pl}}^2$ and duration of order $n_{\mathrm{pl}}^{-1}$. Its torque per unit asteroid mass therefore producesWith the most favorable coherent signs, the crude encounter count isOrder-one factors depend on $p$ and encounter geometry. Uncorrelated signs instead give a random walk count of order $(M_{\star }/M_{\mathrm{pl}}{)}^{2}q/a_{\mathrm{pl}}$ when this is large. A librating phase-protected orbit can suppress the kicks even further.