Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 316 / 4

The variables $y_j=I_j{e}^{i{\Omega }_j}$ are complex inclinations, despite the question's description as eccentricities. Diagonalize the Laplace-Lagrange inclination matrix: let real eigenvectors ${\mathbf{v}}_k$ satisfy $B{\mathbf{v}}_k={\lambda }_k{\mathbf{v}}_k$, and form $V=({\mathbf{v}}_1,{\mathbf{v}}_2,{\mathbf{v}}_3)$. The physical matrix is diagonalizable by positive diagonal symmetrization of a matrix, as shown below.

The matrix exponential solution isWrite $c_k=I_k{e}^{i{\gamma }_k}$ with $I_k\ge 0$, and define $I_{jk}=I_k{v}_{jk}$. This givesThe eigenvalues determine nodal precession frequencies, the eigenvectors determine relative inclinations within each secular eigenmode, and the initial complex inclinations determine their amplitudes and phases. The real $I_{jk}$ can have either sign, encoding opposite nodal phases. The zero-frequency mode is a common fixed tilt.

Let $a=B_{12}$, $b=B_{13}$, $c=B_{21}$, $d=B_{23}$, $e=B_{31}$, and $f=B_{32}$. Zero row sums put $B$ in the form used in the spectrum of a three-dimensional matrix with zero row sums. In particular, $B(1,1,1{)}^T=0$. DefineExpanding the characteristic polynomial gives $\det (\lambda I-B)=\lambda ({\lambda }^2+S\lambda +T)$, soFor the physical coefficients, $L_j{B}_{jk}=L_k{B}_{kj}$ with $L_j=M_j\sqrt{G{M}_{\star }{a}_j}$. Thus $W=\mathrm{diag}(L_j)$ gives positive diagonal symmetrization of a matrix, andConsequently the eigenvalues are real and nonpositive, and $B$ has a complete eigenvector basis. This justifies the mode expansion in part (i).

For small $\alpha$, expand the defining integrand of the Laplace coefficient:HenceThe constant term integrates to zero and ${\int }_0^{2\pi }{\cos }^{2}xdx=\pi$, givingThere is no quadratic term: its factors are combinations of $\cos x$ and ${\cos }^{3}x$, which both integrate to zero over a period. The linear approximation is used for the distant planet, not for the closely spaced inner pair.

Keep the inner-pair Laplace coefficient unexpanded because $\alpha =a_1/a_2$ is close to one. Put $\ell =L_1/L_2$ andUsing the small-argument Laplace coefficient for the distant planet and $n_2/n_1={\alpha }^{3/2}$ givesThe reverse-coupling expressions follow from the angular momentum identity $L_j{B}_{jk}=L_k{B}_{kj}$. If desired, $L_1/L_2=(M_1/M_2)\sqrt{\alpha }$, $L_1/L_3=(M_1/M_3)\sqrt{\beta }$, and $L_2/L_3=(M_2/M_3)\sqrt{\beta /\alpha }$ express all ratios in the supplied quantities. The diagonal entries are $B_{11}=-(B_{12}+B_{13})$, $B_{22}=-(B_{21}+B_{23})$, and $B_{33}=-(B_{31}+B_{32})$.

For comparable masses and closely spaced inner orbits, $\alpha \simeq 1$, $\ell \sim 1$, and $L_1/L_3,L_2/L_3=O(\sqrt{\beta })$. ThusThe distant couplings $B_{13},B_{23}$ scale as ${\beta }^3$ relative to a fixed inner architecture, and the reverse couplings have an additional $\sqrt{\beta }$ suppression.

Ignoring the distant planet at first, the inner two-by-two block has eigenvalues zero and $-(B_{12}+B_{21})$. Therefore its fast differential secular eigenmode hasFor the slow mode, strong inner coupling makes $y_1\simeq y_2=y_{\mathrm{in}}$. Add the first two evolution equations weighted by angular momenta and divide by $L_{12}=L_1+L_2$. The rigid inner pair in hierarchical secular inclination dynamics has equationswithThe two-plane matrix has eigenvalues zero and $-c(1+L_{12}/L_3)$. The architecture implies $L_3\gg L_{12}$ for comparable masses, so its leading slow frequency isThe neglected factor $1+L_{12}/L_3$ gives a relative correction of order $\sqrt{\beta }$; imperfect locking of the inner pair supplies higher-order corrections in its weak coupling to the outer planet.

Initially the two inner orbital normals coincide, so their combined angular momentum is $L_{12}=L_1+L_2$ along one direction. Use the outer orbital normal as the vertical axis. The inner pair's vector then has perpendicular component $L_{12}\sin \Delta I$ and parallel component $L_{12}\cos \Delta I$. Adding the outer angular momentum givesThe invariable plane is perpendicular to this total vector. Thus the outer orbit's orbital inclination to that plane equals the angle between its own normal and the total normal. The inclination to the invariable plane of two orbital planes isThis uses the small, prograde inclinations for which the denominator is positive; in general use the corresponding two-argument arctangent to choose the quadrant. If $A\gg 1$, $I_3\simeq \Delta I/(A+1)$ for small $\Delta I$: the distant planet carries most of the angular momentum and lies close to the invariable plane.