---
abstract: 'The Lidov-Kozai mechanism allows a body to periodically exchange its eccentricity with inclination. It was first discussed in the framework of the quadrupolar secular restricted three-body problem, where the massless particle is the inner body, and later extended to the quadrupolar secular nonrestricted three body problem [@1969CeMec...1..200H; @1976CeMec..13..471L; @1994CeMDA..58..245F]. In this paper, we propose a different point of view on the problem by looking first at the restricted problem where the massless particle is the outer body. In this situation, equilibria at high mutual inclination appear [[@2006PhyD..213...15P]]{}, which correspond to the population of stable particles that @2008MNRAS.390.1377V [@2009MNRAS.394.1721V] find in stable, high inclination circumbinary orbits around one of the components of the quadruple star HD 98800. We provide [a simple analytical framework using a vectorial formalism for these situations. We also look]{} at the evolution of these high inclination equilibria in the non restricted case.'
author:
- |
    F. Farago[^1] and J. Laskar\
    ASD, IMCCE-CNRS UMR8028, Observatoire de Paris, UPMC, 77 avenue Denfert-Rochereau, 75014 Paris, France
bibliography:
- 'FL2009.bib'
title: 'High inclination orbits in the secular quadrupolar three-body problem'
---

celestial mechanics – planetary systems – methods: analytical – methods: $N$-body simulations.

Introduction {#se:intro}
============

As it is known, the secular three-body problem after node reduction has two degrees of freedom (e. g. [@1905QB351.P73......; @2002CeMDA..84..283M]). However, due to what @1976CeMec..13..471L called a [*happy coincidence*]{}, this problem is integrable when it is expanded up to order two in the ratio of semi-major axes, i.e. at the quadrupolar approximation. Indeed, the argument of perihelion of the outer body does not explicitly appear in the quadrupolar expansion of the secular problem, thus giving one more integral of motion linked to the eccentricity of the outer body.

The limiting case where the inner body has no mass has been extensively studied . We will call this problem the [*inner restricted problem*]{}, while the converse case where the two inner bodies are massive and the outer body is massless will be called the [*outer restricted problem*]{}. In the inner restricted case, the conservation of the normal component of the angular momentum enables the inner particle to periodically exchange its eccentricity with inclination (the so-called Lidov-Kozai mechanism). The inner restricted model is well suited when the inner body has a small mass with respect to the other two. However, when looking at higher mass ratios, for example in triple star systems, this is no longer justified.

Since the Hamiltonian of the quadrupolar problem of three masses is very similar to that of the inner restricted problem when it is written in elliptic variables, the study of the massive problem has mainly focused on the dynamics of the two inner bodies [@1969CeMec...1..200H; @1976CeMec..13..471L; @1994CeMDA..58..245F]. These previous works completely classified the different dynamical regimes and bifurcations, using the equations of motion of the inner binary.

There is however another limit-case to the massive problem, which is the outer restricted problem. [@2006PhyD..213...15P have studied this case and discussed the existence and stability of equilibria in the non-averaged system using the framework of KAM theory. We give here a very simple model of the outer restricted case which provides an alternate formulation of these previous results and which is directly usable in an astronomical context. We also fully describe the possible motions of the bodies and give an analytical expression of their frequencies.]{} We use this model to explain the results of @2008MNRAS.390.1377V [@2009MNRAS.394.1721V], who find populations of particles at very high inclinations around one of the components of the double-binary star HD 98800, which are stable even under the perturbation of the other component. We then look at the quadrupolar problem of three masses from the perspective of the outer restricted problem [and show how the inner and outer restricted cases are related to the general case]{}. Vectorial methods as developed by @2006Icar..185..312B [@2009Icar..201..750B; @2009AJ....137.3706T] are extremely well suited for this approach.

Secular outer restricted problem {#se:rtbp}
================================

Derivation of the Hamiltonian
-----------------------------

We consider here the case of a massless particle orbiting a central binary object. We do not restrict ourselves with respect to inclinations or eccentricities. The components of the binary have masses $m_0$ and $m_1$, the binary’s total mass is ${\mathcal{M}_{01}}=m_0+m_1$ and its reduced mass is ${\beta_1}=m_0m_1/(m_0+m_1)$. The two massive components have barycentric positions ${\mathbf{u}}_0\ \text{and}\ {\mathbf{u}}_1$. We also denote $\delta=m_0/{\mathcal{M}_{01}}$ and ${{\mathbf{r}}_1}={\mathbf{u}}_1-{\mathbf{u}}_0$, and ${\mathbf{r}}_2$ is the position of the outer particle relatively to the barycentre of the inner binary. Using these notations, the particle has the following Hamiltonian:

$$H = \frac{\tilde{{\mathbf{r}}}^2_2}{2}-G\left(\frac{m_0}{|{\mathbf{r}}_2-{\mathbf{u}}_0|}+\frac{m_1}{|{\mathbf{r}}_2-{\mathbf{u}}_1|}\right)\ ,$$

where $\tilde{{\mathbf{r}}}_2=\dot{{\mathbf{r}}}_2$ is the canonical momentum associated to the position of the massless particle, ${\mathbf{r}}_2$. Since ${\mathbf{u}}_0=-(1-\delta){{\mathbf{r}}_1}$ and ${\mathbf{u}}_1=\delta{{\mathbf{r}}_1}$, we can rewrite the Hamiltonian as:

$$H=\frac{\tilde{{\mathbf{r}}}^2_2}{2}-G\left(\frac{m_0}{|{\mathbf{r}}_2+(1-\delta){{\mathbf{r}}_1}|}
 +\frac{m_1}{|{\mathbf{r}}_2-\delta{{\mathbf{r}}_1}|}\right)\ .$$

We now suppose that ${r_1}\ll r_2$ and expand the Hamiltonian to order 2 in $r_1/r_2$:

$$\label{eq:hamqp}
 H = \left(\frac{\tilde{{\mathbf{r}}}^2_2}{2}-\frac{G{\mathcal{M}_{01}}}{r_2}\right)
 -\frac{G{\beta_1}}{2r_2^3}\left(3\frac{{\mathopen ( {\mathbf{r}}_2 . {{\mathbf{r}}_1}\mathclose )}^2}{r_2^2}-{r_1}^2\right)\ .$$

The first term is the Keplerian energy of the particle interacting with the binary, seen as a point mass ${\mathcal{M}_{01}}$. It is equal to $-G{\mathcal{M}_{01}}/2a_2$, where $a_2$ is the semi major axis of the particle.

Since we are interested in the secular behaviour of the particle, we average this quadrupolar Hamiltonian over the mean anomalies of the binary (${M_1}$) and of the particle ($M_2$). In order to do this, we first introduce four unit vectors: $({\mathbf{i}},{\mathbf{j}},{\mathbf{k}})$ are bound to the orbit of the binary, remain constant, and will provide a natural reference frame; ${\mathbf{w}}$ is bound to the orbit of the particle and will vary. More precisely, ${\mathbf{i}}$ points in the direction of the perihelion of the binary, ${\mathbf{k}}$ is colinear to the angular momentum of the binary, and ${\mathbf{j}}={{\mathbf{k}} \wedge {\mathbf{i}}}$; the last vector ${\mathbf{w}}$ is colinear to the angular momentum of the massless particle.

We can then compute the following averaged quantities, where quantities indexed with $1$ relate to the binary, quantities with index $2$ relate to the particle, and ${\mathbf{u}}$ is an arbitrary fixed vector (see for instance the appendix of @2006Icar..185..312B):

$$\begin{aligned}
 {\left\langle {r_1}^2 \right\rangle}_{{M_1}} & = & {a_1}^2\left(1+\frac{3}{2}{e_1}^2\right)\ ; \\
 {\left\langle {\mathopen ( {{\mathbf{r}}_1}. {\mathbf{r}}_2 \mathclose )}^2 \right\rangle}_{{M_1}} & = & \frac{{a_1}^2}{2}(r_2^2-{\mathopen ( {\mathbf{k}} . {\mathbf{r}}_2 \mathclose )}^2)\nonumber\\
    &&+\frac{{a_1}^2{e_1}^2}{2}(4{\mathopen ( {\mathbf{i}} . {\mathbf{r}}_2 \mathclose )}^2 - {\mathopen ( {\mathbf{j}} . {\mathbf{r}}_2 \mathclose )}^2)\ ; \\
 {\left\langle \frac{1}{r_2^3} \right\rangle}_{M_2} & = & \frac{1}{a_2^3(1-e_2^2)^{3/2}}\ ; \\
 {\left\langle \frac{{\mathopen ( {\mathbf{r}}_2 . {\mathbf{u}} \mathclose )}^2}{r_2^5} \right\rangle}_{M_2} & = &
    \frac{u^2-{\mathopen ( {\mathbf{w}} . {\mathbf{u}} \mathclose )}^2}{2a_2^3(1-e_2^2)^{3/2}}\ .\end{aligned}$$

The substitution of these expressions in yields:

$$\begin{gathered}
\label{eq:hmoy}
{\left\langle H \right\rangle}_{{M_1},M_2} = -\frac{G{\mathcal{M}_{01}}}{2a_2} -\frac{3}{8}\frac{G{\beta_1}{a_1}^2}{a_2^3(1-e_2^2)^{3/2}}\times\\
\left[\left({e_1}^2-\frac{1}{3}\right)
+{\mathopen ( {\mathbf{k}} . {\mathbf{w}} \mathclose )}^2-{e_1}^2(4{\mathopen ( {\mathbf{i}} . {\mathbf{w}} \mathclose )}^2-{\mathopen ( {\mathbf{j}} . {\mathbf{w}} \mathclose )}^2)\right]\end{gathered}$$

Since the particle has no mass, the only variable element of the binary is its mean anomaly ${M_1}$. After averaging over this angle, it is no longer present in the Hamiltonian. After averaging over the mean anomaly of the particle, its semi major axis $a_2$ becomes constant. Moreover, ${\mathbf{w}}=\sin i_2 \sin\Omega_2\ {\mathbf{i}} -
\sin i_2 \cos\Omega_2\ {\mathbf{j}} + \cos i_2\ {\mathbf{k}},$ so the argument of pericentre $\omega_2$ of the particle does not appear in the averaged Hamiltonian. Hence, at the quadrupolar order, the conjugate momentum associated to $\omega_2$, i.e. the norm of the angular momentum of the particle $G_2=\sqrt{G{\mathcal{M}_{01}}a_2(1-e_2^2)}$, is constant. Therefore the eccentricity $e_2$ of the particle is constant. This fact is a feature of the quadrupolar expansion, not a property of the restricted problem. As such it remains true when the outer body has a non-zero mass (see section \[se:ftbp\]). This is the [*happy coincidence*]{} that @1976CeMec..13..471L noted. Finally, only one degree of freedom remains, related to the couple $(i_2,\Omega_2)$.

If we drop the constant terms in , and introduce the mean motion $n_1$ of the binary into the Hamiltonian ($n_1^2{a_1}^3=G{\mathcal{M}_{01}}$), we get the following expression[^2] [(see also eq. 10 in [@2006PhyD..213...15P])]{}:

$${\left\langle H \right\rangle} = -\frac{\alpha G_2}{2}
\left[{\mathopen ( {\mathbf{k}} . {\mathbf{w}} \mathclose )}^2-{e_1}^2(4{\mathopen ( {\mathbf{i}} . {\mathbf{w}} \mathclose )}^2-{\mathopen ( {\mathbf{j}} . {\mathbf{w}} \mathclose )}^2) \right]\ ,$$

where $$\alpha = \frac{3}{4}n_1 \left(\frac{{a_1}}{a_2}\right)^{7/2} \frac{{\beta_1}}{{\mathcal{M}_{01}}}\frac{1}{(1-e_2^2)^2}\ .$$

This Hamiltonian can be rewritten in a very compact form as: $${\left\langle H \right\rangle} = -\frac{1}{2}\phantom{.}^t{\mathbf{w}}.{\mathbf{T}}.{\mathbf{w}}\ ,\ \text{where:}$$ $${\mathbf{T}} = \alpha G_2\left(\begin{array}{ccc}
                -4{e_1}^2 & 0 & 0 \\
                0 & {e_1}^2 & 0 \\
                0 & 0 & 1
               \end{array}\right)\ .$$

We also give the expression of the Hamiltonian using the inclination and the node of the particle:

$$\label{eq:HrestIO}
 {\left\langle H \right\rangle}=-\frac{\alpha G_2}{4}\left[2\cos^2i_2 - {e_1}^2\sin^2i_2\left(3-5\cos2\Omega_2\right)\right]\ .$$

Equations of motion {#se:restreqns}
-------------------

As discussed in [@2006Icar..185..312B], the equations of motion for ${\mathbf{w}}$ are simply obtained by:

$$\dot{{\mathbf{w}}} = \frac{1}{G_2}\nabla_{{\mathbf{w}}}{\left\langle H \right\rangle}\wedge{\mathbf{w}}\ .$$

After computing the gradient, we find:

$$\dot{{\mathbf{w}}} = -\alpha
  \left[(\mathbf{k}.\mathbf{w})(\mathbf{k}\wedge\mathbf{w})
       -{e_1}^2(4(\mathbf{i}.\mathbf{w})(\mathbf{i}\wedge\mathbf{w})
              -(\mathbf{j}.\mathbf{w})(\mathbf{j}\wedge\mathbf{w}))\right]\ .$$

If we note $x=(\mathbf{i}.\mathbf{w})$, $y=(\mathbf{j}.\mathbf{w})$, and $z=(\mathbf{k}.\mathbf{w})$, we get the following system for $(x,y,z)$:

$$\begin{aligned}
\dot{x} & = & \alpha(1-{e_1}^2)yz\ ; \label{eq:xdot}\\
\dot{y} & = & -\alpha(1+4{e_1}^2)zx\ ; \\
\dot{z} & = & 5\alpha {e_1}^2 xy\ .\label{eq:zdot}\end{aligned}$$

In these variables, the fact that ${\mathbf{w}}$ is a unit vector and the energy integral translate into the following equalities:

$$\begin{aligned}
 x^2+y^2+z^2 & = & 1\ ; \label{eq:wsphere}\\
 z^2 - {e_1}^2(4x^2-y^2) & = & h = \text{Cst}\ . \label{eq:Hsurf}\end{aligned}$$

The system of three differential equations – has thus two independent first integrals and is as such integrable. It is also straightforward from these two relations that $$\label{eq:hineg}
    -4{e_1}^2\leqslant h\leqslant 1\ .$$

Motion of a massless body around a circular binary
--------------------------------------------------

In the case of a circular binary, the Hamiltonian and the equations of motion greatly simplify[^3]:

$${\left\langle H \right\rangle} = -\frac{\alpha G_2}{2}{\mathopen ( {\mathbf{k}} . {\mathbf{w}} \mathclose )}^2\ ,$$

$$\dot{{\mathbf{w}}} = -\alpha
  (\mathbf{k}.\mathbf{w})(\mathbf{k}\wedge\mathbf{w})$$

The scalar product $(\mathbf{k}.\mathbf{w})=\cos i_2$ remains constant, and the nodes of the orbit of the particle simply precess around the angular momentum of the binary, with a constant precession rate: $$\dot{\Omega}=-\alpha\cos i_2 =
 -\frac{3}{4}n_1 \left(\frac{{a_1}}{a_2}\right)^{7/2} \frac{{\beta_1}}{{\mathcal{M}_{01}}}\frac{\cos i_2}{(1-e_2^2)^2}$$

This precession is equivalent to the precession generated by the quadrupolar potential of a circular and homogeneous ring of mass $\beta_1$ and of radius $a_1$ [following an idea which can be traced back to Gauss (see [@2009MNRAS.394.1085T] and references therein)]{}.

Motion of a massless body around an elliptic binary
---------------------------------------------------

### Qualitative overview

When the binary is elliptic, the situation changes and cannot be explained any longer by the quadrupolar torque of a circular ring. If we substitute $z^2$ in using , we get: $$\begin{aligned}
\label{eq:hinterw}
 (1+4{e_1}^2)x^2+(1-{e_1}^2)y^2 &=1-h\geqslant 0\ ,\\
  x^2+y^2+z^2 &= 1\ .\end{aligned}$$

The intersections of the energy surfaces and the normalized angular momentum sphere of the particle can thus be seen as the intersections of elliptic cylinders with the unit sphere. For a given value of the energy $h$, the extremity of the unit angular momentum vector of the particle ${\mathbf{w}}$ will move on the intersection of the corresponding cylinder with the unit sphere. Figures \[fig:Hxy\].a and c show these intersections for different values of the energy as dotted lines drawn on the unit sphere, in two situations where the binary has an eccentricity of $0.5$ and $0.2$ respectively. The three axes correspond to the scalar products $x$, $y$ and $z$ that are defined in section \[se:restreqns\].

![\[fig:Hxy\]Intersections of the energy surfaces and the unit angular momentum sphere (a) and its projection in the $(x,y)$ plane (b) for ${e_1}=0.5$. Intersections of the energy surfaces and the unit angular momentum sphere (c) and its projection in the $(x,y)$ plane (d) for ${e_1}=0.2$.](FL2009-fig1){width="\columnwidth"}

There are four visible kinds of trajectories: closed trajectories around the two poles of the sphere $(x,y,z)=(0,0,\pm 1)$, and closed trajectories around the points $(x,y,z)=(\pm 1,0,0)$.

When the extremity of the angular momentum of the particle ${\mathbf{w}}$ follows a trajectory around the north pole, it means that it precesses around the angular momentum of the binary ${\mathbf{k}}$ with an inclination that is strictly inferior to $90^\circ$: in this case, the orbital motion of the particle is prograde relatively to the orbital motion of the binary.

When the extremity of the angular momentum of the particle ${\mathbf{w}}$ follows a trajectory around the south pole, it means that it precesses around the opposite of the angular momentum of the binary, $-{\mathbf{k}}$, with an inclination that is strictly superior to $90^\circ$: in this case, the orbital motion of the particle is retrograde relatively to the orbital motion of the binary.

When the extremity of the angular momentum of the particle ${\mathbf{w}}$ follows a trajectory around one of the two points $(x,y,z)=(\pm 1,0,0)$, it precesses around the direction of the perihelion of the binary or the opposite of this direction. In this case, the inclination oscillates around $\pm 90^\circ$.

### Frequencies

The frequencies of these motions can be found analytically. Indeed, using equation , we see that $x$ and $y$ are on ellipses or arcs of ellipses bounded by the unit circle (figures \[fig:Hxy\].b and d show respectively the cases where ${e_1}=0.5$ and $0.2$). Thus, there is an angle $\phi$ such that:

$$\begin{aligned}
 x & = & \sqrt{\frac{1-h}{1+4{e_1}^2}} \cos\phi\ , \label{eq:xell}\\
 y & = & \sqrt{\frac{1-h}{1-{e_1}^2}} \sin \phi\ .\end{aligned}$$

Using , we can then express $z^2$ as:

$$\label{eq:z2phi}
 z^2 = \frac{h+4{e_1}^2}{1+4{e_1}^2} - \frac{5{e_1}^2}{1+4{e_1}^2}\frac{1-h}{1-{e_1}^2} \sin^2\phi\ .$$

There are two opposite values of $z$ for each $\phi$, reflecting the symmetry of the system with respect to the orbital plane of the binary. If we use expression in combination with equation , we obtain the following equation for $\dot{\phi}$:

$$\begin{aligned}
 \dot{\phi}&=&\mp\alpha\sqrt{(1-{e_1}^2)(h+4{e_1}^2)}
\sqrt{1-k^2\sin^2\phi}\ , \label{eq:dotphi}\\\
k^2&=&\frac{5{e_1}^2}{1-{e_1}^2}\frac{1-h}{h+4{e_1}^2}\ .\end{aligned}$$

The constant $k^2$ is positive because of relation . The value $k^2=1$ defines a limit between two dynamical regimes. If $k^2<1$, or equivalently if $h>{e_1}^2$, $\dot{\phi}$ never vanishes and the projection of ${\mathbf{w}}$ on the orbital plane of the binary moves along the full ellipse . In this case, ${\mathbf{w}}$ precesses around the angular momentum of the binary, ${\mathbf{k}}$. If $z>0$ the mutual inclination of the two orbits is always less than $90^\circ$ so the orbital motion of the particle is prograde; conversely, if $z<0$ the mutual inclination of the two orbits is always superior to $90^\circ$ so the orbital motion of the particle is retrograde.

If $k^2>1$ (or $h<{e_1}^2$), then $\dot{\phi}$ vanishes for $\phi_0=\pm \arcsin(1/k)$, changes its sign (which is accompanied by a change of sign in the $z$ variable), and the angle $\phi$ librates between $-\phi_0$ and $+\phi_0$. Thus, the projection of ${\mathbf{w}}$ on the orbital plane of the binary is bounded by the unit circle to stay on an arc of ellipse . In this case, ${\mathbf{w}}$ precesses around the direction of perihelion of the binary, so that both the inclination and the node of the particle librate around $\pm 90^\circ$.

In both cases, the period of the motion can be calculated with a simple quadrature using equation :

$$\label{eq:Tpart}
 T=\frac{16}{3n_1}\frac{{\mathcal{M}_{01}}}{{\beta_1}}\left(\frac{a_2}{{a_1}}\right)^{7/2}
 \frac{K(k^2)(1-e_2^2)^2}{\sqrt{(1-{e_1}^2)(h+4{e_1}^2)}}\ ,$$

where $K(k^2)$ is the elliptic integral of the first kind defined by: $$K(k^2) = \left\{\begin{array}{cc}
                  \int_0^{\pi/2}\frac{d\phi}{\sqrt{1-k^2\sin^2\phi}} & \text{if}\ k^2<1\\
                  \int_0^{\phi_0}\frac{d\phi}{\sqrt{1-k^2\sin^2\phi}} & \text{if}\ k^2>1
                 \end{array}\right.\ .$$

The last case where $k^2=1$ (or $h={e_1}^2$) corresponds to the trajectories that separate the previous two types. They link the points $(x,y,z)=(0,\pm 1,0)$, and the associated period is infinite. In the projection on the $(x,y)$ plane, these separatrices form the ellipse which is tangent to the unit circle. Since all trajectories that are inside this ellipse correspond to the precession of ${\mathbf{w}}$ around ${\mathbf{k}}$, the width $\Delta x_{\text{sep}}$ of the separating ellipse in the $(x,y)$ plane gives an indication on the proportion of such trajectories. Using equation and the fact that $h={e_1}^2$ on the separatrix, we get:

$$\Delta x_{\text{sep}} = 2\sqrt{\frac{1-{e_1}^2}{1+4{e_1}^2}}$$

Therefore, when the inner binary is circular, this width is equal to $2$, the full width of the unit circle, and the only possible motion is precession of ${\mathbf{w}}$ around $\pm{\mathbf{k}}$. When the eccentricity of the binary increases, the width of the separatrix decreases to zero, which is a limit case since it can only be reached for a value of the binary’s eccentricity equal to 1. The precession motions of ${\mathbf{w}}$ around $\pm{\mathbf{i}}$ thus become predominant when the eccentricity of the binary grows.

Comparison with numerical studies
---------------------------------

![Energy levels of the Hamiltonian in the $(i_2\cos\Omega_2,i_2\sin\Omega_2)$ plane for values of the eccentricity of the binary ${e_1}=0$ (a), ${e_1}=0.1$ (b), ${e_1}=0.79$ (c), ${e_1}=0.9$ (d).\[fig:Heb\]](FL2009-fig2){width="\columnwidth"}

In [@2009MNRAS.394.1721V], the authors investigate a family of particles at high inclinations around the binary HD 98800 Ba-Bb, which remain stable even under the perturbation of an outer third stellar companion. They isolate a nodal precession imposed by the inner binary as the stabilizing mechanism working against the destabilizing Kozai perturbations of the outer companion. They run simulations of test particles orbiting the binary HD 98800 Ba-Bb using non secular equations. They observe the libration islands around $i_2=\pm90^\circ$ and $\Omega_2=\pm90^\circ$ that we discussed in the previous section. As they show their results in the $(i_2\cos\Omega_2,i_2\sin\Omega_2)$ plane, we plotted the energy levels of the outer restricted Hamiltonian using these same coordinates for an easier comparison. Figure \[fig:Heb\] shows these levels for different values of the eccentricity. The c. panel in particular uses the same value for the eccentricity of the binary (${e_1}=0.79$) as figures 4 and 5 of [@2009MNRAS.394.1721V].

Verrier and Evans notice no apparent structure in the dynamics of the $(e_2,\omega_2)$ couple apart from the circulation of the perihelion. This is in agreement with the fact that the particle’s eccentricity is constant at the quadrupolar approximation.

The authors also suggest that the projection of the angular momentum of test particles along the line of apses of the binary may be an integral of motion. From the results of the previous section, it is straightforward to see that the projection $x$ of the angular momentum of test particles along the line of apses of the binary is not constant. It varies with an amplitude that decreases to $0$ when the inclination of the particle approaches $\pm 90^\circ$, which can be misleading when looking at numerical results for highly inclined particles. However, the norm of the angular momentum of the test particles is an integral of the secular motion.

![\[fig:period\] Dependence of the period with respect to the eccentricity of the binary, in normalized units. The full line corresponds to the calculated period, while the dashed line corresponds to the power-law fit given by @2009MNRAS.394.1721V. We used a least squares method to fit the relative position of the two curves.](FL2009-fig3){width="\columnwidth"}

Finally, the authors give a power-law fit of the period of the libration of the node with respect to three parameters: the eccentricity of the binary, the ratio of the semimajor axes $a_2/a_1$, and the mass ratio of the binary, $\delta$. Their power-law is fitted using particles with fixed inclinations ($85^\circ$). They give in their equation (5): $$T\propto e_1^{-1.1}\delta^{-0.8}\left(\frac{a_2}{a_1}\right)^{3.37}\ .$$

By rewriting the mass dependences of equation , we get the following analytical dependence with respect to the mass ratio and the semi-major axis of the binary: $$T\propto\left(\frac{a_2}{a_1}\right)^{3.5}(\delta(1-\delta))^{-1}\ .$$

These two exponents compare very well with the fitted power law, in spite of the differences between the two models. The dependency with respect to $e_1$ is rather complex in equation , and it is best compared in figure \[fig:period\].

The grid of initial conditions for the particles in [@2009MNRAS.394.1721V] extends however from 3 to 10 AU for a binary separation of 1 AU, so the quadrupolar approximation may not be sufficient to fully describe the motion of the particles with the lowest semi major axes. In particular, Verrier and Evans state that some low inclination particles show large eccentricity variations and even instability. This could be due to a low initial semi-major axis and to resonances that are eliminated in our secular model by the averaging over the mean anomalies.

Problem of three massive bodies {#se:ftbp}
===============================

As we already stated, the quadrupolar secular three-body problem is still integrable when all the bodies have positive masses. As such, it is possible to show how the outer restricted problem we discussed in the previous section relates to the general case, and to the inner restricted case studied by @1962AJ.....67..591K and . We will first express the Hamiltonian of the secular quadrupolar problem using the same vectorial method as in the previous section in order to focus on the relative movements of the orbits. In their studies of the secular quadrupolar problem, @1976CeMec..13..471L and @1994CeMDA..58..245F have shown that this problem depends on two parameters. We will then point out which regions of parameter space are topologically equivalent to the outer restricted case, and which regions correspond to the inner restricted case, in order to show the continuity that exists between both situations.

Hamiltonian
-----------

Let us consider three masses $m_0$, $m_1$ and $m_2$, this time with $m_2\neq0$. We note the barycentric coordinates and impulsions $({\mathbf{u}}_i,{\mathbf{\tilde{u}}}_i)_{i=0,1,2}$. As in the previous section, we suppose that the two bodies of indices 0 and 1 form a binary and that the distance of the third body to this binary is much larger than the separation of the binary. We still note $\delta =
m_0/(m_0+m_1)$. We first perform a canonical change of variables to Jacobi coordinates,

$$\begin{aligned}
 {\mathbf{r}}_0 &={\mathbf{u}}_0 &\tilde{{\mathbf{r}}}_0 &= \tilde{{\mathbf{u}}}_0+\tilde{{\mathbf{u}}}_1+\tilde{{\mathbf{u}}}_2 ={\mathbf{0}} \\
 {\mathbf{r}}_1 &={\mathbf{u}}_1-{\mathbf{u}}_0 &\tilde{{\mathbf{r}}}_1 &= {\mathbf{\tilde{u}}}_1 + (1-\delta)\tilde{{\mathbf{u}}}_2 \\
 {\mathbf{r}}_2 &={\mathbf{u}}_2-\delta{\mathbf{u}}_0-(1-\delta){\mathbf{u}}_1 & \tilde{{\mathbf{r}}}_2 &= {\mathbf{\tilde{u}}}_2\end{aligned}$$

In these coordinates, the Hamiltonian of the three bodies is [@1989LaskarGoutelas]: $$\begin{gathered}
 H=\left(\frac{{\tilde{\mathbf{r}}}_1^2}{2\beta_1}-\frac{\mu_1\beta_1}{r_1}\right) + \frac{{\tilde{\mathbf{r}}}_2^2}{2\beta_2}\\
   -Gm_2\left(\frac{m_0}{|{\mathbf{r}}_2+(1-\delta){\mathbf{r}}_1|}+\frac{m_1}{|{\mathbf{r}}_2-\delta{\mathbf{r}}_1|}\right),\end{gathered}$$

where $\beta_1^{-1}=m_0^{-1}+m_1^{-1}$, $\beta_2^{-1}=(m_0+m_1)^{-1}+m_2^{-1}$, $\mu_1=G(m_0+m_1)$ and $\mu_2=G(m_0+m_1+m_2)$.

Using the fact that $r_1\ll r_2$, we expand the Hamiltonian to order two in $r_1/r_2$ as in the previous section: $$\begin{gathered}
 H=\left(\frac{{\tilde{\mathbf{r}}}_1^2}{2\beta_1}-\frac{\mu_1\beta_1}{r_1}\right)+
 \left(\frac{{\tilde{\mathbf{r}}}_2^2}{2\beta_2}-\frac{\mu_2\beta_2}{r_2}\right)\\
 -G\frac{\beta_1m_2}{2r_2^3}\left(3\frac{({\mathbf{r}}_1.{\mathbf{r}}_2)^2}{r_2^2}-r_1^2\right)\ .\end{gathered}$$

The first two terms are Keplerian energies and are equal respectively to $-\mu_1\beta_1/2a_1$ and $-\mu_2\beta_2/2a_2$, where $a_1$ and $a_2$ are the semi major axes of the inner and the outer body in our system of coordinates.

We now average over the two mean anomalies $M_1$ and $M_2$ in order to get the secular part of the Hamiltonian. We will define 4 unit vectors which are analogous to the 4 vectors we used in the first section: $({\mathbf{i}}_1,{\mathbf{j}}_1,{\mathbf{k}}_1)$ are tied to the orbit of the inner binary; ${\mathbf{i}}_1$ points in the direction of the perihelion of the inner binary, ${\mathbf{k}}_1$ points in the direction of its angular momentum, and ${\mathbf{j}}_1={\mathbf{k}}_1\wedge{\mathbf{i}}_1$. The last vector ${\mathbf{k}}_2$ is colinear to the angular momentum of the outer body. In this section, the vectors tied to the orbit of the inner binary will no longer have fixed directions.

Using the same averaging formulae as in the previous section and using the fact that $({\mathbf{i}}_1.{\mathbf{k}}_2)^2+
({\mathbf{j}}_1.{\mathbf{k}}_2)^2+({\mathbf{k}}_1.{\mathbf{k}}_2)^2={\mathbf{k}}_2^2=1$, we can write:

$$\begin{gathered}
\label{eq:tmpavg2}
  {\left\langle H \right\rangle}_{M_1,M_2}=-\frac{\mu_1\beta_1}{2a_1}-\frac{\mu_2\beta_2}{2a_2}
  -\frac{3}{8}\frac{Gm_2\beta_1}{(1-e_2^2)^{3/2}}\frac{a_1^2}{a_2^3}\\
  \times\left[-\frac{1}{3}+2e_1^2+
  (1-e_1^2){\mathopen ( {\mathbf{k}}_1 . {\mathbf{k}}_2 \mathclose )}^2-5e_1^2{\mathopen ( {\mathbf{i}}_1 . {\mathbf{k}}_2 \mathclose )}^2\right]\ .\end{gathered}$$

After averaging over the two mean anomalies, the semi-major axes are constant. There are thus four degrees of freedom in the system, associated to the two eccentricities and the two inclinations. As we explained in the previous section, the argument of perihelion of the outer body does not appear in the quadrupolar expansion, and thus the norm of the angular momentum of the outer body $G_2=\beta_2\sqrt{\mu_2a_2(1-e_2^2)}$, is constant. This implies that its eccentricity $e_2$ is constant. Using the reduction of the nodes would leave only one degree of freedom in the reduced Hamiltonian, associated to the couple $(e_1,\omega_1)$. The full reduction of the Hamiltonian and its expression in elliptical variables is the approach that has been used widely, since it yields a very similar Hamiltonian function as in the inner restricted problem [@1969CeMec...1..200H; @1976CeMec..13..471L; @1994CeMDA..58..245F].

We want however to look at the motion of the nodes, or equivalently the motion of the vector ${\mathbf{k}}_2$ in the moving frame $({\mathbf{i}}_1,{\mathbf{j}}_1,{\mathbf{k}}_1)$ of the orbit of the second body.

In order to easily compute the equations of motion, we introduce two vectors associated to the orbit of the binary that are colinear to ${\mathbf{i}}_1$ and ${\mathbf{k}}_1$, and include in their norm the eccentricity of the binary, as in [@2009AJ....137.3706T]: $$\begin{aligned}
 {\mathcal{K}}_1 &= \sqrt{1-e_1^2}{\mathbf{k}}_1\ , &{\mathcal{I}}_1 &=e_1 {\mathbf{i}}_1\ .\end{aligned}$$

If we drop all the constant terms in equation \[eq:tmpavg2\] and use the above vectors, we get:

$${\left\langle H \right\rangle}_{M_1,M_2}=-\frac{\alpha'G_2}{2}\left[2
 {\mathcal{I}}_1^2+({\mathcal{K}}_1.{\mathbf{k}}_2)^2
                  -5({\mathcal{I}}_1.{\mathbf{k}}_2)^2\right]\ ,$$

where $$\alpha' = \frac{3}{4}n_1\left(\frac{a_1}{a_2}\right)^{7/2}\frac{\beta_1}{{\mathcal{M}_{01}}}\frac{1}{(1-e_2^2)^2}
        \sqrt{1+\frac{m_2}{{\mathcal{M}_{01}}}}\ ,$$ and ${\mathcal{M}_{01}},n_1$ are defined as in section \[se:rtbp\].

Equations of motion {#equations-of-motion}
-------------------

The components of ${\mathcal{K}}_1,{\mathcal{I}}_1$ and ${\mathbf{k}}_2$ have the following Poisson brackets[^4] [@2005BorisovMamaev; @2006Icar..185..312B]:

$$\begin{aligned}
 \{{\mathcal{K}}_{1i} ,{\mathcal{K}}_{1j} \} &= -\frac{\epsilon_{ijk}}{\Lambda_1} {\mathcal{K}}_{1k}\,, &
           \{k_{2i} ,k_{2j} \} &= -\frac{\epsilon_{ijk}}{G_2} k_{2k}\,, \\
 \{{\mathcal{I}}_{1i} ,{\mathcal{I}}_{1j} \} &= -\frac{\epsilon_{ijk}}{\Lambda_1} {\mathcal{K}}_{1k}\,, &
           \{{\mathcal{K}}_{1i} ,{\mathcal{I}}_{1j} \} &= -\frac{\epsilon_{ijk}}{\Lambda_1} {\mathcal{I}}_{1k}\,,\end{aligned}$$

where $\Lambda_1=\beta_1\sqrt{\mu_1a_1}$ and $\epsilon_{ijk}$ is the Levi-Civita symbol[^5].

The equations of motion for the three vectors are thus: $$\begin{aligned}
 \dot{{\mathcal{K}}}_1 & = & - \frac{1}{\Lambda_1} \left( {\mathcal{K}}_1 \wedge \nabla_{{\mathcal{K}}_1} H
                      + {\mathcal{I}}_1 \wedge \nabla_{{\mathcal{I}}_1} H\right)\ , \\
 \dot{{\mathcal{I}}}_1 & = & - \frac{1}{\Lambda_1} \left( {\mathcal{I}}_1 \wedge \nabla_{{\mathcal{K}}_1} H
                      + {\mathcal{K}}_1 \wedge \nabla_{{\mathcal{I}}_1} H\right)\ , \\
 \dot{{\mathbf{k}}}_2 & = & - \frac{1}{G_2} {\mathbf{k}}_2 \wedge \nabla_{{\mathbf{k}}_2} H\ .\end{aligned}$$

In order to look at the motion of the vector ${\mathbf{k}}_2$ in the moving frame $({\mathbf{i}}_1,{\mathbf{j}}_1,{\mathbf{k}}_1)$ of the orbit of the second body, we use as @2006Icar..185..312B the above system to derive equations for $x=({\mathbf{k}}_2.{\mathbf{i}}_1)$, $y=({\mathbf{k}}_2.{\mathbf{j}}_1)$, $z=({\mathbf{k}}_2.{\mathbf{k}}_1)$ and $e_1$. Indeed, $x=({\mathbf{k}}_2.{\mathcal{I}}_1)/|{\mathcal{I}}_1|$, $z=({\mathbf{k}}_2.{\mathcal{K}}_1)/|{\mathcal{K}}_1|$, $e_1=|{\mathcal{I}}_1|$, and $y$ is obtained using the identity $x^2+y^2+z^2=1$:

$$\begin{aligned}
 \dot{x} & =  \alpha'(1-e_1^2)yz + \alpha'\frac{G_2}{\Lambda_1}\sqrt{1-e_1^2}y(2-5x^2)
 \label{eq:fullmotion1}\\
 \dot{y} & =  -\alpha'(1+4e_1^2)xz  \nonumber \\
 &\phantom{=\dots}-\alpha'\frac{G_2}{\Lambda_1}\frac{x}{\sqrt{1-e_1^2}}[(1-e_1^2)(2-5x^2)+5e_1^2z^2] \\
 \dot{z} & =  5\alpha'e_1^2xy +\alpha'\frac{G_2}{\Lambda_1}\frac{5e_1^2}{\sqrt{1-e_1^2}}xyz \\
 \dot{e}_1 &=  \alpha'\frac{G_2}{\Lambda_1}5e_1\sqrt{1-e_1^2}xy \label{eq:fullmotion4}\end{aligned}$$

The equations for $x,y$ and $z$ contain two terms: the first one is identical to the outer restricted system, and the second one includes the motion of the reference frame $({\mathbf{i}}_1,{\mathbf{j}}_1,{\mathbf{k}}_1)$ induced by the interaction with the third body. Note that when $\Lambda_1$ is very large compared to $G_2$ so that we can assume that $G_2/\Lambda_1$ is equal to zero, which corresponds to the case where $m_2\ll m_0\text{ and }m_1$, the above system is identical to the outer restricted system –.

The conservation of the total angular momentum ${\mathbf{C}}={\mathbf{G}}_1+{\mathbf{G}}_2$ introduces the two main parameters of the problem. Indeed, $$\Lambda_1^2(1-e_1^2) + G_2^2 + 2 \Lambda_1\sqrt{1-e_1^2} G_2 z = C^2\ .$$

We note $\gamma=C/\Lambda_1$, $\gamma_2=G_2/\Lambda_1$. The above expression of the norm of the total angular momentum can be rewritten as a second degree equation giving $\sqrt{1-e_1^2}$ as a function of $z$ using the two parameters $\gamma$ and $\gamma_2$: $$\label{eq:redangmom}
 (1-e_1)^2 + 2\gamma_2z\sqrt{1-e_1^2} + \gamma_2^2-\gamma^2 = 0\ .$$

The Hamiltonian can then be rewritten as: $${\left\langle H \right\rangle}=-\frac{1}{2}\alpha'\Lambda_1\gamma_2[z^2+e_1^2(2-z^2-5x^2)]\ .$$

The inequalities $-1\leqslant z\leqslant 1$ and $0\leqslant e_1 < 1$ give the boundaries of the parameter space and the range of possible values of $e_1$ for any given couple of parameters[^6] $(\gamma,\gamma_2)$: $$\begin{aligned}
    |\gamma-\gamma_2|&\leqslant 1\ , \label{eq:paramspace}\\
    |\gamma-\gamma_2|&\leqslant \sqrt{1-e_1^2} \leqslant \min[\gamma+\gamma_2,1] \label{eq:allowedeta}\ . \end{aligned}$$

With these notations, the outer restricted problem of section \[se:rtbp\] corresponds to the limit where $\gamma_2=0$, and in this case $e_1=\sqrt{1-\gamma^2}$ is constant as we saw. Note that when $\gamma_2>\gamma$, we have $G_2>C$, so this part of the parameter space contains only retrograde motions. Our aim in this paper is to show the continuity between the outer restricted case we studied in section \[se:rtbp\], and the inner restricted case that was investigated by @1962AJ.....67..591K and . Both these problems lie in the region of parameter space where $\gamma>\gamma_2$ so we will restrict our study to this case[^7].

In our case where $\gamma>\gamma_2$, there is only one acceptable root to [equation ]{}, which is: $$\sqrt{1-e^2_1}=-\gamma_2z+\sqrt{(\gamma_2z)^2+\gamma^2-\gamma_2^2}\ .$$

This relation implies that $e_1$ is a growing function of $z$. Note that $z=\cos i_2$, where $i_2$ is the inclination of the outer body in the reference frame of the inner binary. As such, coplanar prograde motions ($z=1$) will always occur for the maximal value of the eccentricity of the inner binary: $$e_{1,\max}=\sqrt{1-(\gamma-\gamma_2)^2}\ .$$

Conversely, low eccentricities for the binary will be associated to lower values of $z$, and thus higher inclinations. Relation implies that the inner binary can only have a circular motion if $\gamma+\gamma_2\geqslant 1$. In this case, coplanar retrograde motion ($z=-1$) is not allowed, and the lowest value of $z$ is: $$\label{eq:z0}
   z_0 = \cos i_{2,\max} = \frac{\gamma^2-\gamma_2^2-1}{2\gamma_2}\ .$$

When $\gamma+\gamma_2<1$ however, coplanar retrograde motion ($z=-1$) is possible and the associated value of the eccentricity of the binary is: $$e_{1,\min}=\sqrt{1-(\gamma+\gamma_2)^2}\ .$$

![\[fig:paramspace\] Parameter Space. The dark gray areas are excluded by equation , the light gray area correspond to the part of parameter space corresponding to $\gamma_2\geqslant\gamma$ which we do not study. The dotted line $\gamma+\gamma_2=1$ separates the zone where there can be coplanar retrograde motion associated to a minimum eccentricity for the inner binary that is strictly higher than $0$ (below the dotted line) and the zone where the inner binary can be circular but the inclination is bounded (see section \[sse:eqcirc\]). In zone O, the problem is topologically equivalent to the outer restricted problem. In zones I and I’ it is topologically equivalent to the inner restricted problem, with zone I being equivalent to situations above the critical inclination and zone I’ being equivalent to situations under the critical inclination. The letters a–f correspond to the values of the parameters used to plot the corresponding panels in figures \[fig:IOPlane\] and \[fig:Spheres\].](FL2009-fig4){width="\columnwidth"}

Dynamical regimes {#se:dynreg}
-----------------

![\[fig:IOPlane\] Trajectories in the $(i_2\cos\Omega_2,i_2\sin\Omega_2)$ plane for different values of the parameters. [See section \[se:dynreg\] for a detailed discussion and appendix \[se:fixp\] for calculations.]{} a: $(\gamma,\gamma_2)=(0.8,0)$ outer restricted case with $e_1=0.6$; b: $(\gamma,\gamma_2)=(0.8,0.25)$; c: $\gamma=0.8,\,
\gamma_2^2=(1/3)(1-\gamma^2)$; d: $(\gamma,\gamma_2)=(0.8,0.4)$; e: $(\gamma,\gamma_2)=(1.08,0.4)$; f: $(\gamma,\gamma_2)=(1.28,0.4)$.](FL2009-fig5){width="\columnwidth"}

![\[fig:Spheres\] Trajectories on the unit angular momentum sphere for different values of the parameters. [See section \[se:dynreg\] for a detailed discussion and appendix \[se:fixp\] for calculations.]{} a: $(\gamma,\gamma_2)=(0.8,0)$ outer restricted case with $e_1=0.6$; b: $(\gamma,\gamma_2)=(0.8,0.25)$; c: $\gamma=0.8,\,
\gamma_2^2=(1/3)(1-\gamma^2)$; d: $(\gamma,\gamma_2)=(0.8,0.4)$; e: $(\gamma,\gamma_2)=(1.08,0.4)$; f: $(\gamma,\gamma_2)=(1.28,0.4)$.](FL2009-fig6){width="\columnwidth"}

In appendix \[se:fixp\], we briefly derive in the framework of the present study the fixed points of the system and the boundaries of the dynamical regimes in parameter space that are given in [@1994CeMDA..58..245F]. [The fixed points are named as follow: the north pole is called $N$, and the south pole $S$; linearly stable fixed points are named $E$, as elliptic, and linearly unstable points are named $H$, as hyperbolic; finally, signs are placed as indices to refer to the symmetry of the problem with respect to the two planes $x=0$ and $y=0$.]{} There are three dynamical regimes in the region of parameter space we study.

In region O of figure \[fig:paramspace\], the parameter $\gamma_2=G_2/\Lambda_1$ is small (less than 1/2). This is the case in particular when the mass ratio $m_2/m_1$ is small. [Moreover, $\gamma^2+3\gamma_2^2<1$.]{} The phase space is topologically equivalent to the outer restricted problem of section \[se:rtbp\]. The north pole, which corresponds to coplanar prograde motion with maximal eccentricity for the binary is linearly stable. There are two additional stable fixed points $\text{E}_{\pm}$ in the plane $y=0$ [(see section \[sse:eqy0\])]{}. They belong to the same family as the fixed points $y=z=0$, $x=\pm 1$ of the outer restricted problem that are responsible for the stable high [inclination]{} orbits observed by @2009MNRAS.394.1721V. When $\gamma+\gamma_2\leqslant1$, the south pole which corresponds to coplanar retrograde motion with minimal eccentricity for the binary, is also [linearly]{} stable. [The a panels of figures \[fig:IOPlane\] and \[fig:Spheres\] provide a visualisation of the topology of this case.[^8]]{} When $\gamma+\gamma_2>1$, the south pole is no longer accessible as stated in the previous section. It [is however replaced by a stable trajectory]{} at a [maximal]{} inclination given by equation , as can be seen on panel b [of figures \[fig:IOPlane\] and \[fig:Spheres\]]{}. This trajectory corresponds to a circular inner binary [(see section \[sse:eqcirc\])]{}. Finally, there are two unstable points $\text{H}_{\pm}$ in the $x=0$ plane that belong to the same family as the unstable points of the outer restricted problem $x=z=0$, $y=\pm 1$ [(see section \[sse:eqx0\])]{}.

[Panels c of figures \[fig:IOPlane\] and \[fig:Spheres\] show the limiting case between regions O and I. On this boundary, $\gamma^2+3\gamma_2^2=1$. The two unstable points $H_{\pm}$ are now located on the boundary of the accessible part of the sphere.]{}

[Regions I and I’ of figure \[fig:paramspace\] are both in the part of the parameter space defined by $\gamma^2+3\gamma_2^2>1$. In this zone, the problem becomes topologically equivalent to the inner restricted problem studied by and @1962AJ.....67..591K. In the inner restricted case, there is a critical value of the inclination ($\cos i_2=\sqrt{3/5}$) under which a circular inner binary is always linearly stable, and above which a circular inner binary is always linearly unstable, giving rise to Kozai cycles.]{}

In region I of figure \[fig:paramspace\], the dynamical regime is topologically equivalent to the inner restricted problem in the case where the inclination is superior to the critical value. The limit trajectory $z=z_0$ which corresponds to a circular inner binary becomes [linearly]{} unstable. However, the north pole and the two fixed points $\text{E}_{\pm}$ are still stable. In the inner restricted phase space, when the inclination is superior to the critical value, there are two possible behaviours for the periastron of the inner particle: it can either circulate, or librate around $\pm 90^{\circ}$. In our representation, the circulation case corresponds to trajectories around the north pole, and the libration islands correspond to the two fixed points $\text{E}_{\pm}$. This is shown in panels d and e in figures \[fig:IOPlane\] and \[fig:Spheres\].

In region I’ of figure \[fig:paramspace\], the dynamical regime is topologically equivalent to the inner restricted problem in the case where the inclination is inferior to the critical value. Only one stable fixed point remains, on the north pole of the sphere, associated to prograde coplanar motion. This is shown in panel f in figures \[fig:IOPlane\] and \[fig:Spheres\].

In both regions I and I’, the parameter $\gamma_2=G_2/\Lambda_1$ can take higher values. This is in particular true when the mass ratio $m_2/m_1$ increases.

The curve between regions I and I’ is linked to the critical inclination that is defined in the inner restricted case. Indeed, along that curve, given by equation \[eq:critlim\], we have the following limits when $\gamma \rightarrow \infty$: $$\begin{aligned}
    \frac{\gamma_2}{\gamma} &\rightarrow 1\, , & \gamma_2-\gamma \rightarrow -\sqrt{\frac{3}{5}}\ . \end{aligned}$$

When $G_2$ is very large compared to $G_1$, we can make the following first order expansion:

$$\begin{aligned}
 \gamma_2-\gamma &=\frac{G_2-C}{\Lambda_1} \\
  &=\frac{G_2-\sqrt{G_2^2+G_1^2+2({\mathbf{G}}_2.{\mathbf{G}}_1)}}{\Lambda_1} \\
  &\approx \frac{G_2 - G_2(1+ ({\mathbf{G}}_2.{\mathbf{G}}_1)/G_2^2)}{\Lambda_1} \\
  &\approx - \frac{({\mathbf{G}}_2.{\mathbf{G}}_1)}{\Lambda_1G_2} \\
  &\approx -z \sqrt{1-e_1^2} \ .\end{aligned}$$

As such, we see that along the border between regions I and I’, when $\gamma$ and $\gamma_2$ both tend to infinity, we have the relation

$$z \sqrt{1-e_1^2} \approx \sqrt{\frac{3}{5}}\ .$$

Recall that $z=\cos i_2$, where $i_2$ is the inclination of the outer orbit in the reference frame of the inner orbit. Thus, the inclination of the inner orbit relatively to the outer orbit is $i_1=-i_2$, and the above equation becomes:

$$\cos i_1 \sqrt{1-e_1^2} \approx \sqrt{\frac{3}{5}}\ .$$

This relation is precisely the one giving the critical value of the normal component of the angular momentum of the inner body in the inner restricted problem.

Conclusion {#conclusion .unnumbered}
==========

We first studied the case of a massless particle orbiting a binary at a long distance, and, in the secular and quadrupolar approximations, gave a full analytical description of the motion along with the expression of the period of the secular motion. When the inner binary is circular, only nodal precession takes place. However, when the binary is elliptic, libration islands appear at high inclinations, and these islands grow bigger when the eccentricity of the binary rises. @2008MNRAS.390.1377V [@2009MNRAS.394.1721V] observe a similar nodal libration in their study of the stability of particle populations in the quadruple stellar system HD 98800, and we showed that the analytical framework that we derived for the outer restricted problem is well suited to explain the results of Verrier and Evans.

The quadrupolar secular three-body problem is still integrable when all the bodies have positive masses [@1969CeMec...1..200H; @1976CeMec..13..471L; @1994CeMDA..58..245F]. Using a vectorial formalism as [@2006Icar..185..312B; @2009Icar..201..750B; @2009AJ....137.3706T], we looked at this problem from the point of view of the outer restricted case. We showed how the outer restricted problem relates to the general case, and to the inner restricted case studied by @1962AJ.....67..591K and : when the mass of the outer body is small enough compared to the mass of the inner body, the general case behaves similarly to the outer restricted problem. When the mass of the outer body increases enough, the general case behaves like the inner restricted problem. We gave an expression of the boundary between these two regimes.

The outer restricted problem and its generalization to the non restricted case provide an interesting starting point in the study of circumbinary planetary systems, such as the one discovered recently around the eclipsing sdB+M system HW Virginis [@2009AJ....137.3181L]. In this system, the inner binary is very tight with a period of 2.8 hr, while the proposed planetary companions have periods of 9.1 yr and 15.8 yr, so the quadrupolar expansion is fully justified. Another field of application of the outer restricted problem is the study of the motion of stars orbiting around binary black holes [@2008AJ....135.2398M; @2009ApJ...692.1075G; @2009ApJ...693L..35M].

Fixed points and bifurcations {#se:fixp}
=============================

The fixed points and boundaries presented in section \[se:dynreg\] have already been studied by @1976CeMec..13..471L and @1994CeMDA..58..245F. We briefly present here their derivation in the framework of the present formalism. With the notations of section \[se:ftbp\], we will limit ourselves to $\gamma>\gamma_2$.

Poles of the sphere, $x=y=0$ {#sse:eqpol}
----------------------------

This case corresponds to case 1 in section 5 of [@1994CeMDA..58..245F]. Note that their sphere is constructed using the eccentricity and perihelion of the [inner binary]{}, and is thus different from our angular momentum sphere.

For all values of the parameters in the domain we study, the north pole $z=1$, which corresponds to coplanar prograde motion, is a fixed point of the system. The associated eccentricity of the inner binary is: $$e_{1,\max}=\sqrt{1-(\gamma-\gamma_2)^2}\ .$$ It is the maximal value of the eccentricity. This fixed point is always linearly stable. It is noted N in figures \[fig:IOPlane\] and \[fig:Spheres\]. Figure \[fig:IOPlane\] shows the lines of equal energy in the $(i_2\cos\Omega_2,i_2\sin\Omega_2)$ plane, and figure \[fig:Spheres\] shows these lines plotted on the sphere of unit angular momentum of the outer body ${\mathbf{k}}_2^2=1$.

When $\gamma+\gamma_2<1$ (under the dotted line in figure \[fig:paramspace\]), the south pole $z=-1$ (noted S in the following figures), which corresponds to coplanar retrograde motion, is also a linearly stable fixed point of the system. The eccentricity of the inner binary is minimal and equal to: $$e_{1,\min}=\sqrt{1-(\gamma+\gamma_2)^2}\ .$$

Note that in this region of parameter space, the inner binary cannot be circular.

When $\gamma+\gamma_2\geqslant1$ (above the dotted line in figure \[fig:paramspace\]), the minimal eccentricity [of]{} the binary is $0$ as deduced from . The south pole $z=-1$ does not correspond to a real value of the eccentricity in this case. This limit however is not a bifurcation strictly speaking. When crossing it, the stable south pole of the sphere [is replaced by a stable trajectory at maximal inclination]{}.

Circular Trajectories for the inner binary {#sse:eqcirc}
------------------------------------------

In the region of parameter space where circular trajectories exist for the binary (above the dotted line in figure \[fig:paramspace\]), the value of $z$ which corresponds to such trajectories is minimal and equal to: $$z_0= \frac{\gamma^2-\gamma_2^2-1}{2\gamma_2}\ .$$

The equations of motion on the small circle of the sphere $z=z_0$ are: $$\begin{aligned}
 \dot{x}_{\phantom{1}} & = & \phantom{-} \alpha' y(z_0 + \gamma_2 (2-5x^2))\ , \label{eq:circmotionx}\\
 \dot{y}_{\phantom{1}} & = & - \alpha'x(z_0 +\gamma_2 (2-5x^2))\ , \label{eq:circmotiony}\\
 \dot{z}_{\phantom{1}} & = & \phantom{-} 0\ , \\
 \dot{e}_1 &= &\phantom{-}  0\ .\end{aligned}$$

The right hand sides of equations and vanish for a certain value of $x$ equal to: $$x_0^2 = \frac{\gamma^2+3\gamma_2^2-1}{10\gamma_2^2}\ .$$

The curve $\gamma^2+3\gamma_2^2=1$ separates in figure \[fig:paramspace\] the regions noted O and I. We can distinguish three cases:

1.  $\gamma^2+3\gamma_2^2<1$. In region O of figure \[fig:paramspace\], $x_0^2<0$ so there is no fixed point on the circle $z=z_0$. As such, this circle is a trajectory for which the inner binary is circular and the outer orbit precesses at a fixed inclination given by $i_{2,\max}=\text{ArcCos}\,z_0$. Moreover, this trajectory is linearly stable.

2.  $\gamma^2+3\gamma_2^2=1$. There are two fixed points on the circle $z=z_0$ at the coordinates $(x=0,y=\pm\sqrt{1-z_0^2})$.

3.  $\gamma^2+3\gamma_2^2> 1$. In this case, we must also check that $y_0^2=1-x_0^2-z_0^2\geqslant 0$. The limit case where there is equality yields: $$5 \gamma_2^4 - (4 + 10 \gamma^2)\gamma_2^2 +  (5 \gamma^4  - 8 \gamma^2 + 3) = 0 \ .$$

This boundary limits the regions I and I’ in figure \[fig:paramspace\]. When solving the above equation for $\gamma_2^2$ and selecting only the relevant solution satisfying $\gamma>\gamma_2\, , \gamma+\gamma_2\geqslant
1$, we obtain a solution that corresponds to equation 44 in section 5.2 of [@1994CeMDA..58..245F] and that can be written using our notations as: $$\label{eq:critlim}
 \gamma_2^2 = \frac{2+5\gamma^2-\sqrt{60\gamma^2-11}}{5}\, , \quad \gamma+\gamma_2\geqslant 1\ .$$

In region I, $y_0^2>0$ so there are four fixed points on the circle $z=z_0$, at the coordinates $(\pm x_0, \pm
y_0)$. They are noted $\text{H}_{\pm\pm}$, in panels d and e of figures \[fig:IOPlane\] and \[fig:Spheres\]. Moreover, the trajectories that correspond to circular binaries are unstable in this zone. In region I’ however, $y_0^2<0$ so we are again in a region of parameter space where there are no fixed points on the circle $z=z_0$, and the trajectories associated to circular binaries are again stable.

The $x=0$ plane {#sse:eqx0}
---------------

When $x=0$, the only non trivial equation remaining in system – is $\dot{x}=0$. Looking for a fixed point different from $x=y=0$, we have to solve $$z\sqrt{1-e_1^2}+2\gamma_2=0\ ,$$

which after using relation yields:

$$\begin{aligned}
 e_1 & = &\sqrt{1-\gamma^2-3\gamma_2^2}\ ,\\
x & = & 0 \ , \\
y & = & \pm\frac{\sqrt{\gamma^2-\gamma_2^2}}{\sqrt{\gamma^2+3\gamma_2^2}}\ ,\\
z & = & -\frac{2\gamma_2}{\sqrt{\gamma^2+3\gamma_2^2}}\ .\end{aligned}$$

We thus have two symmetric fixed points in the plane $x=0$. They are noted $\text{H}_{\pm}$ in figures \[fig:IOPlane\] and \[fig:Spheres\]. For these fixed points to exist, the associated eccentricity must be a real number. As such, their domain of existence is the region noted O in figure \[fig:paramspace\]. This is case 2.1 in section 5 of [@1994CeMDA..58..245F].

These two fixed points are linearly unstable in their domain of existence. Note that in the outer restricted problem ($\gamma_2=0$), these fixed points are simply $x=z=0, y=\pm1$.

The $y=0$ plane {#sse:eqy0}
---------------

When $y=0$, the only non trivial equation we must solve is $\dot{y}=0$. Here again, we look for another fixed point than $x=y=0$, thus we have to solve:

$$(1+4e_1^2)z +\frac{\gamma_2}{\sqrt{1-e_1^2}}[(1-e_1^2)(2-5x^2)+5e_1^2z^2]=0\ .$$

Substituting $1-z^2$ in place of $x^2$ and then $\sqrt{1-e_1^2}$ in place of $z$ using , we get: $$(1-e_1^2)^3 - \left(\gamma^2+\frac{1}{2}\gamma_2^2+\frac{5}{8}\right)(1-e_1^2)^2+\frac{5}{8}(\gamma^2-\gamma_2^2)^2=0$$

This equation is the same as equation number 40 in [@1994CeMDA..58..245F]. In our region of parameter space, there is at most one root which satisfies to the constraint . The curve separating the zone where there is one solution and the zone where there is no solution corresponds to the [case where the limit value $e_1=0$ is a solution]{}, and coincides with the boundary between regions I and I’ in figure \[fig:paramspace\] which is given by equation .

When there is a solution, the value of $e_1$ can be translated into a value of $z$ using . Since $y=0$, we get two values of $x=\pm \sqrt{1-z^2}$, and there are thus two symmetric fixed points on the sphere, which are both linearly stable. They are noted $\text{E}_{\pm}$ in figures \[fig:IOPlane\] and \[fig:Spheres\]. When $\gamma_2=0$, these fixed points become simply $y=z=0$, $x=\pm 1$, which are responsible of the stable orbits at high inclination as discussed in the previous sections.

[^1]: E-mail: farago@imcce.fr; laskar@imcce.fr

[^2]: We will from now write ${\left\langle H \right\rangle}$ for the averaged Hamiltonian, omitting the subscripts $M_1,M_2$.

[^3]: [The next non-zero term of the Hamiltonian which is the fourth order in $(a_1/a_2)$ plays an important part in the circular case as has been discussed in detail by @2006CeMDA..95...81P.]{}

[^4]: We use the following convention, where $p_i$ are momenta and $q_i$ positions: $\{f,g\}=\sum_i \left(\frac{\partial
    f}{\partial p_i}\frac{\partial g}{\partial q_i} -\frac{\partial f}{\partial q_i}\frac{\partial g}{\partial
    p_i}\right)$.

[^5]: $\epsilon_{ijk}=+1$ if $(i,j,k)$ is an even permutation of $(1,2,3)$, $\epsilon_{ijk}=-1$ is the permutation is odd, and $\epsilon_{ijk}=0$ in all other cases.

[^6]: The left part of the second inequality is strict if $\gamma=\gamma_2$.

[^7]: The other half of the parameter space ($\gamma\leqslant\gamma_2$) corresponds to retrograde motions which are of less physical interest and much more technical to study using our approach, in particular because equation does not have a unique solution in this case. The interested reader will find a complete discussion of this case in [@1976CeMec..13..471L; @1994CeMDA..58..245F].

[^8]: [Note that the south pole in figure \[fig:IOPlane\] a corresponds to the out-most trajectory; this is an artifact of the coordinate map $(i_2\cos\Omega_2,i_2\sin\Omega_2)$ which sends the south pole of the sphere onto the circle $i_2=180^{\circ}$.]{}


If a close loved one has passed away, you may find yourself responsible for making decisions about their cremation service, you may wonder how you should go about selecting an urn for their ashes. If so, ask yourself the following three questions during the decision making process.

How Much Money Do You Have To Spend On The Urn?

Before you start thinking about anything else, you should ask yourself how much money you have to spend on the urn. Knowing how much you have in the budget will guide the rest of your decisions.

For example, if you do not have much money to spend on the urn, you may need to go with a basic urn. These are sometimes included in the cremation fee. Or, you may be able to spend a little more to buy a more personalized one.

On the other hand, if you have a larger budget, you can pretty much have your pick of materials, colors, and designs.

What Type Of Urn Will Match The Deceased's Personality?

No matter what your budget, you want to try to find a type of urn that matches the deceased's personality. If your loved one was a simple, quiet person, muted colors and gentle lines may suit them.

However, if the deceased was more flamboyant, you could opt to go with a bold, ornate design accented with splashes of bright colors. Or, if they were a huge sports fan, you could even go with an urn that displays their favorite team's colors.

What Will You Do With The Urn After The Funeral Service?

Another thing you should consider when choosing your deceased loved one's urn is what you will be doing with it after the funeral service. This could also impact the design of the urn.

If you plan on placing the urn on a mantle or table, you may want to choose an urn that is personalized for the deceased as well as aesthetically pleasing. However, if you plan to bury the ashes or place them in a mausoleum, you may decide that you do not need to be as concerned about the long-term appearance.

After asking yourself the above questions, you should have enough basic information to start making an informed decision. Once you have an idea as to what type of urn you want for your loved one, make an appointment with the cremation service (think American Cremation Society) so they can direct you towards several options that may suit the dearly departed.

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About Me

A few weeks ago, one of my dear friends lost her long battle with cancer. I met this special lady at my church. She was warm, friendly, and compassionate. She always greeted me with a smile and a hug. A few weeks before she passed away, she planned her funeral. She didn’t want people to mourn her death at her funeral. Instead, she wanted them to celebrate her life. At this special service, a collection of candid photos was displayed on a large screen at the church. Friends and family members smiled as they viewed the snapshots and remembered this amazing woman. On this blog, you will discover how to plan a celebration funeral service for your deceased loved one.


Four piper

The term four piper is United States Navy terminology for classes of destroyers with four funnels. These include the  classes listed below:

These classes were built for use by the United States Navy during World War I and subsequently "moth-balled". Fifty were loaned, under Lend-Lease to the United Kingdom when convoy escorts were desperately needed in the early years of World War II.

External links
 Four Piper Destroyer Organization, 2005, "General Specifications"

Category:Ship types


What I Learned Working for Steve Jobs

The first time I met Steve Jobs, it was at his house for my first interview at Pixar, something I never thought would happen, not even in my wildest dreams. During our conversation, we talked about my experience and how it could apply to a role he was looking to fill—a liaison between Disney (where I worked for eight years) and the producers at Pixar Animation Studios.

After we talked about what the role would entail, I had one of the most surreal moments of my entire life: I said no to Steve Jobs. I wanted to work for Steve more than words could express, but I didn’t think putting someone between the producers and the distributors would ever work.

Steve ended up offering me a job running new business and marketing. I gave myself an Austin Powers-inspired title, “International Man of Marketing,” and dug into Pixar like a puppy in a playground. Although I thought I had successfully said no to that initial job offer, it became clear with every working day that my new role was turning into the one I’d rejected. What I didn’t know then is that Steve Jobs probably never really heard my no.

I began a professional descent that finally ended in a fateful meeting between Steve and me—a meeting that people who have worked for Steve know that most walk out of there having quit if they weren’t already fired.

I found it nearly impossible to stand up to his relentless and intimidating intellect. I respected Steve so much—basically worshiped the ground he worked on—so I figured my only move I had was to go with dignity. I did the honorable thing and resigned. That was one of the saddest days of my life.

I don’t regret any of the time I spent working for Steve—how could I? It was a privilege to be with him, even for a short and difficult time. Nearly two decades later, as the CEO of digital marketing firm Magnetic, I’m still learning things from my time working for him. Here are three:

1. Be clear and transparent.

While Steve was arguably the greatest marketer of our generation, sharing innovation with inspirational words to the masses, he wasn’t the best communicator with individuals.

Steve didn’t set defined expectations for me or other employees. He simply knew what he wanted. He wasn’t great at telling you what that was, but he knew it when he saw it, which was a major hurdle for any employee, at any level, to overcome.

Watching him operate made me recognize the importance of clarity and transparency with my team, and how imperative it is to set expectations. The more transparent I am about where we want to take the company, the clearer my team is about how to get there.

2. Build relationships at work.

The time we spend at work is also the time we spend with our work colleagues. One can’t exist without the other, so building relationships at work is the key to being successful—no matter rank or role. I didn’t do this well at Pixar—I focused on what Steve wanted at the expense of relationships with everyone else at the company.

When you make any decision that will affect the company, ensure that you help paint the picture and discuss with your team why they should want to be on board. If not, you could alienate your co-workers in the process and make progress very difficult.

3. Listen carefully when others speak.

Steve told me what he wanted when we first met, and I told him I didn’t want to do it. But when I think carefully about how everything played out, the role he first suggested was never very different from what I agreed to do. My excuse for not sticking to my guns is that Steve was one of the most persuasive people on the planet—but that’s not good enough. I should never have taken the job.

This lesson has led me to spend crazy amounts of time trying to figure out what people really want. I rarely try to change someone’s goals or opinions. No matter how driven, smart or successful your employee, if the person is not a good fit then you should end the relationship as quickly and as respectfully as possible.

When I walked away from Pixar, Steve said to me: “Life is long, and I’m sure our paths will cross again.” Sure enough, that was not the last time I talked to Steve Jobs.

One of the many things that makes Steve Jobs so inspirational is that he took his walk in the wilderness. He suffered setbacks, but he learned from them, matured and returned to even greater victories. The sweetness of the comeback is much richer and multifaceted than the first win. For this and many other reasons, Steve will always be a hero of mine.


Metro Manila (CNN Philippines) — The social media giant, Facebook, said that it has 47 million Filipinos on its network.

That is close to the 54 million registered to vote in the May 9 elections.

It's not hard to see how the social network can be a handy tool to reach voters and for voters to reach each other.

"Filipinos log on to Facebook daily because they want to know and learn more about the candidates that are running for office in these elections. But they're also debating the issues that are important to them on Facebook with their friends and learning more about those issues. And the candidates also want to reach out to the voters through Facebook," Elizabeth Hernandez, Facebook's head of policy for Asia, said.

The government is quick to learn handling this tool and now it is teaming up with Facebook for the polls.

Facebook will provide a means to crowdsource questions for the presidential and vice presidential debates. Users will be able to feed their questions and comments, which debate moderators will pose to the candidates during the live telecasts.

Also watch: Youth group launches mobile app on 2016 polls

On May 9, a "voter megaphone" will alert users that it's election day. After you finish voting, you have a quick way to tell all your friends that you've cast your ballot. That should encourage them to do the same.

Throughout the poll season, Facebook will keep tabs on political discussions and monitor trends and inform the Comelec what issues resonate with the public.

"We are eagerly looking forward to working with Facebook to ensure that the thoughts and opinions of Filipino voters remain an integral part not only of the debates but also of the elections. And with Facebook's help, we can see greater voter participation and meaningful engagement in the 2016 elections," Comission on Elections Chairman Andres Bautista said.

Partnership with Twitter, too

The Comelec is also partnering with Twitter, another social media giant.

Political analyst Dindo Manhit, a consultant in the Facebook partnership, said it could promote intelligent conversations on the elections.

"Mas mataas ang debate or substantive discussion sa Facebook so maaaring it can encourage deeper discussion sa ating eleksyon. So it could be an alternative venue by which people can discuss, people could answer, even our own candidates," Manhit said.

Trolling and cyber bullying are the risks in using social media in political discussions, but Facebook has a way to deal with that.

"When we look at a conversation, we don't measure whether it's positive or negative. We're just looking at the overall volume of conversation and discussions that people are having on Facebook. We have terms and conditions on what anybody can say on Facebook, whether it's yourself or a candidate," Facebook's Politics and Government Outreach Manager Katie Harbath said.

Election duty of public school teachers soon to be non-compulsory?

The Internet is changing the face of Philippine elections and the government trying to keep up and take advantage of the change.

It will again train public school teachers who will supervise the polls on how to operate vote counting machines with the help of Department of Science and Technology (DOST).

"The teachers of course have to be trained in respect of their functions and responsibilities and that's where the DOST comes in. So they will be certifying that the teachers who will be serving as BEI's have the requisite competence, skill and capacity to be able to serve as our Board of Election Inspectors," Bautista said.

Congress has passed a bill making election duty no longer compulsory for public school teachers. It also raises their pay from P4,500 to up to P7,000. The bill awaits President Benigno Aquino III's signature. The Comelec is confident some 300,000 teachers will volunteer for poll duty.


Q:

c# Get httponly cookie

How can i get a httponly cookie in a httpwebresponse ?
Habitually i use a CookieContainer to get the cookies in a httpwebresponse, but it doesnt work with httponly cookie.
Is there an other way to catch them ?

A:

Yes, it is possible to retrieve a HTTPOnly cookie, for instance from a client program using the "InternetGetCookieEx" function in the "Wininet.dll".
You must use PInvoke code like this :
/// <summary>
/// WinInet.dll wrapper
/// </summary>
internal static class CookieReader
{
    /// <summary>
    /// Enables the retrieval of cookies that are marked as "HTTPOnly".
    /// Do not use this flag if you expose a scriptable interface,
    /// because this has security implications. It is imperative that
    /// you use this flag only if you can guarantee that you will never
    /// expose the cookie to third-party code by way of an
    /// extensibility mechanism you provide.
    /// Version:  Requires Internet Explorer 8.0 or later.
    /// </summary>
    private const int INTERNET_COOKIE_HTTPONLY = 0x00002000;

    [DllImport("wininet.dll", SetLastError = true)]
    private static extern bool InternetGetCookieEx(
        string url,
        string cookieName,
        StringBuilder cookieData,
        ref int size,
        int flags,
        IntPtr pReserved);

    /// <summary>
    /// Returns cookie contents as a string
    /// </summary>
    /// <param name="url"></param>
    /// <returns></returns>
    public static string GetCookie(string url)
    {
        int size = 512;
        StringBuilder sb = new StringBuilder(size);
        if (!InternetGetCookieEx(url, null, sb, ref size, INTERNET_COOKIE_HTTPONLY, IntPtr.Zero))
        {
            if (size < 0)
            {
                return null;
            }
            sb = new StringBuilder(size);
            if (!InternetGetCookieEx(url, null, sb, ref size, INTERNET_COOKIE_HTTPONLY, IntPtr.Zero))
            {
                return null;
            }
        }
        return sb.ToString();
    }
}

The code is from MSDN.
I hope that helps !


Rapid expansion takes The Tinnitus Clinic across UK and Europe

Millions of people across the UK and Europe can now benefit from the very latest tinnitus treatments following a major expansion project.

The Tinnitus Clinic has opened at four new sites in just six months, marking a period of rapid growth for the business.

In 2018, clinics have launched in Newcastle, Salisbury, North Norfolk and South Wales in the UK. The first overseas licensed clinic has also been introduced in Gzira, Malta.

The Tinnitus Clinic is an independent audiological practice specialising in tinnitus treatment and management. It also operates at 12 other locations in the UK and Republic of Ireland.

Tinnitus is the sensation of hearing ringing, buzzing, hissing, chirping, whistling, or other sounds. The noise can be intermittent or continuous and can vary in loudness.

The Levo® Therapy System, Acute Tinnitus Therapy™, Acoustic CR® Neuromodulation, Tinnitus Desensitisation Therapy ™ and Hyperacusis Desensitisation are just some of the high-tech treatments available from The Tinnitus Clinic.

Audiology specialists also supply and fit hearing aids for tinnitus.

Carlo Laurenzi OBE, managing director at The Tinnitus Clinic, said: “2018 has been a fantastic year so far for The Tinnitus Clinic and we look forward to delivering further growth over the next six months.

“We’re proud to be bringing pioneering treatments, not found anywhere else in the UK and Europe, to millions of sufferers of this debilitating condition.

“Six million people alone in the UK live with tinnitus, and in some cases, the severity can lead to other problems such as anxiety, sleep deprivation and depression. We’re committed to ensuring as many sufferers as possible have access to treatment which can make a significant difference to their lives.”


Popful Mail Sound Box ’94

Overview

Popful Mail Sound Box ’94 comes on two CDs. Disc One is the Super Famicom version of the game’s score and Disc Two contains the music from the Mega CD and PC-Engine versions (that’s “Super Nintendo”, “Sega CD”, and “Turbo Grafx 16” respectively for the American gamers out there).At the end of Disc Two, there is a vocal version of Disc One’s ending theme, “Somebody Loves You”. It’s not the most spectacular Falcom soundtrack, but for what it is — a gamey, lighthearted album for an anime-inspired action RPG — it gets the job done. This is the kind of music you put in and simply drift into a good mood with. Three different versions are a nice touch, although they all have their strong points and weaknesses. Let’s take a closer look…

Body

Personally, I listen to disc 1 the most. The Super Nintendo has always had a great sound chip and it does the music for this game justice. It’s not as harsh on the ears as the CD versions and it comes out much better than, say, the SNES version of Ys III. All tracks loop once, giving plenty of time to get a feel for the music. Speaking of which, a lot of the music on this disc didn’t make the cut to the CD versions. Not a big loss given how well it turned out here. “Welcome Popful Mail” pretty much sets the tone for the entire soundtrack, with a bouncy feel. “Quick Tempered Spirit” sounds remarkably like the NES classic “Ice Climber,” with a lighthearted, jazzy sound. Tracks 10, 15, 19, 20 and 21 are the ones that correspond to the battle themes on Disc Two. I don’t know why they aren’t listed identically on the tracklist on the back. These are pretty much the meat of the disc, as the “Exploration” themes have a bit of a bland feel to them. Overall, a great CD.

I haven’t played the CD versions of the game, so I can’t say for sure if the tracks on Disc Two follow the redbook audio. I say this, because unlike other Falcom releases on CD that contain arrangements of the original music (the Ys games, most notably), these tracks play through once then fade out. At the very least, most redbook audio loops once or twice. These tracks give little to no variation to their Disc One counterparts, differing only in synth quality (the Mega CD tracks have more of a lighthearted feel, while the PC-Engine ones are closer to the typical Falcom rock). On top of that, the sound quality improvement isn’t that great. I don’t want to sound too hard on this disc because it has plenty of good tracks, but the disjointed listening experience puts a major damper on it.

For the most part, I prefer the PC-Engine versions over the Mega CD ones. “Opening” has a frantic around-and-around sound to it. The PC-Engine versions of “Volcanic Zone (Battle)” and “Geist Race” are excellent Falcom rock, and “Iceberg Zone (Battle)” has a great techno beat to it. Finally, the vocal version of “Somebody Loves You” is a great end for the disc. The soothing vocals lend themselves to another version of the song that is dead-on with its Disc One version. A little disappointing, but this disc has enough good points to warrant a few listens here and there. The liner notes are the icing on the cake. Inside are several black and white pictures of Popful’s adventures (there’s one that has her hugging a sword that costs 10,000 gold pieces, with the party completely broke. This was the one that really brought up the “Slayers” comparison). Very nice.

Summary

I have a soft spot for Popful Mail Sound Box ’94, but the main people I would recommend it to are those who have fond memories of playing the game. The Sega CD version miraculously made it to the US thanks to Working Designs. People not familiar with the game should listen to the samples and approach with caution. This isn’t Ys or Sorcerian music, so you might be a bit let down with the somewhat weak compositions. For those looking for a great Falcom album to add to their collection, try the Super Arrange Version of these tracks that is found on Falcom Special Box ’94, it’s a much better overall listening experience. Also, keep in mind that this pretty much the only source of the original music. The PC version on Falcom Special Box ’93 is mind-numbingly 8-bit and the rest of the Popful Mail CDs are drama albums. As for tracking it down, it came out in 1994 and most Falcom albums from that time period are a pain to track down. This set is no exception. Sold on eBay, it will probably hit between $70-$100. It’s cheaper on Yahoo! Japan Auctions, but still very infrequent. Falcom Special Box ’94 offers more bang for your buck, so I say go with that instead.

stars

1

3

b

8/1/2012

Popful Mail Sound Box ’94Andy Byus

Do you agree with the review and score? Let us know in the comments below!

3

Posted on August 1, 2012 by Andy Byus. Last modified on August 1, 2012.


productName: Peerio 2

description: Peerio desktop client

type: git

author: Technologies Peerio Inc. <support@peerio.com>

license: ISC

keychainService: PeerioMessenger

supportUser: support

feedbackUser: feedback


A1 motorway (Netherlands)

The A1 is a motorway in the Netherlands. The road connects the capital city of Amsterdam, near the interchange of Watergraafsmeer, with the German border, near Oldenzaal and Bad Bentheim, and the German Autobahn BAB 30. On its way, it crosses four provinces: North Holland, Utrecht, Gelderland and Overijssel.

European routes

The section of the road between its start at the interchange Watergraafsmeer, and the interchange of Hoevelaken near Amersfoort, is also a European route: the E231. This section of the A1 is the complete E231 route; the E231 does not consist of any other road or section.

Between the interchange Hoevelaken and the German border (and beyond, along the BAB 30), the European route E30 follows the A1 motorway. This European route is a so-called A-Class West-East European route, going from Cork in Ireland all the way to Omsk in Russia.

Exit list

References

External links

01
Category:Constituent roads of European route E30


Jeff Bezos told me that in theory, he's open to other device makers connecting to the Kindle store. If I were Sony, I woudl take him up on it --the Sony bookstore is no match for Amazon.com, and if Sony charged a lot less for a nice looking e-reader which connected to Amazon via computer, it might have a nice niche.

I say "bullshit", otherwise all the available Mobipocket-capable readers would be able to read AZW already.


Q:

Exercise 2.2.3 in Weibel Homological Algebra

Weibel Exercise 2.2.3 is as follows:
Let $P_\bullet$ be a complex of projectives with $P_i=0 \; \forall i<0$ then a map $\epsilon : P_0 \to M$ giving a resolution for $M$ is the same thing as a chain map $\epsilon : P_\bullet \to M_\bullet$, where $M_\bullet$ is the chain complex with $M$ concentrated in degree $0$.
But I am clearly misunderstanding this...
Clearly if we have such a map; $$ \to P_2 \xrightarrow{d_2} P_1 \xrightarrow{d_1} P_0 \xrightarrow{\epsilon_0} M \to 0$$
Making a projective resolution for $M$, then we have $ker(\epsilon_0)=Im(d_1)$ so commutative ladder diagram:
$$
\begin{array}{cccccc}
... \xrightarrow{d_2} & P_{1} & \xrightarrow{d_1} & P_0 & \xrightarrow{0} & 0 & \xrightarrow{0} ...\\
 & \big\downarrow & & \; \big\downarrow \epsilon & & \big\downarrow \\
... \xrightarrow{0} & 0 & \xrightarrow{0} & M & \xrightarrow{0} & 0 & \xrightarrow{0} ...
\end{array}
$$
.
However I cannot see how the two things are the same. We have no guarantee of exactness in the complex $P_\bullet$ for one. As well as no guarantee of equality $Im(d_1) = ker(\epsilon_0)$ and surjectivity of $\epsilon_0$.
Are them some missing conditions or have I misunderstood the statement?

A:

This is an error in Weibel, included in the list of errata here:
http://www.math.umd.edu/~jmr/602/bookerrors.pdf
"chain map" should be replaced by "quasi-isomorphism", which will fix the problem you observed.


// -----------------------------------------------------------------------
// Licensed to The .NET Foundation under one or more agreements.
// The .NET Foundation licenses this file to you under the MIT license.
// -----------------------------------------------------------------------

namespace Kerberos.NET.Entities
{
    public enum TransitedEncodingType
    {
        DomainX500Compress = 1
    }
}


Seattle area cyclists - help for a visitor

I'm visiting Seattle on vacation soon, and I have some questions about cycling in the city.

First, which bridges allow bicycles? I see there is a MUP on the I-90 bridge. Is that the only way to get from the Bellevue area to Seattle proper? It seems like there's plenty of cycling to be done, but that getting from point A to point B might be a bit more complicated.

Second, what are your suggestions for rides to "see the sights" in the city? I've never been to Seattle, and I really prefer to see new places by bike. I'm staying near Woodland park, and anticipate rides in the area of 25 miles or less, since I probably won't have a road bike (I'm either renting or borrowing a bike) and I suppose I should probably spend some of the vacation with the friends I'm going to see.

the ever shifting stable loaded with comfortable road bikes and city and winter bikes

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Hi, I'll pitch in. The only bridge that doesn't allow bikes is the 520 bridge across lake washington.

For rides in the city from Woodland park, 2 suggestions, ride west until you hit Puget Sound, then ride south across the ballard locks, then out to West Seattle for views of the city and the sound. Or, ride south until you hit Lake union, head east along the lake on the Burke-Gilman trail. ride out to the Red Hook Brewery, drink, then repeat the ride home.

Check out www.cascade.org for some rides you can join up with, and the city of Seattle has some great bicycling maps you can pick up for free at most bike shops. Closest shop from Woodland park is Gregg's Cycle on Green Lake.

I'm visiting Seattle on vacation soon, and I have some questions about cycling in the city.

First, which bridges allow bicycles? I see there is a MUP on the I-90 bridge. Is that the only way to get from the Bellevue area to Seattle proper? It seems like there's plenty of cycling to be done, but that getting from point A to point B might be a bit more complicated.

Second, what are your suggestions for rides to "see the sights" in the city? I've never been to Seattle, and I really prefer to see new places by bike. I'm staying near Woodland park, and anticipate rides in the area of 25 miles or less, since I probably won't have a road bike (I'm either renting or borrowing a bike) and I suppose I should probably spend some of the vacation with the friends I'm going to see.

The I-90 bridge is the best way to get between Bellevue and Seattle on a bicycle. You can't ride across on the SR520 bridge.
If you are near Woodland Park, go to either Aurora Bicycle Shop or Gregg's Greenlake Bicycle Shop and get the free Seattle Bicycling Guide Map and the King County Bicycle Map.
Seattle is very bicycle friendly. I have had very few problems with the drivers here, and have never been involved in any really serious incidents in 33 years of riding around the area. Believe it or not, the drivers are almost all mellow and patient! You can confidently ride anywhere in the city with the exception of a few major arterials.
Bellevue, Redmond, Kirkland, and the entire east side of Lake Washington are another matter. This is a much newer suburban area, very car-centric, and designed to actively discourage bicycling or walking. The drivers there are likely to find the presence of a bicyle as offensive, and they will let you know it. Also you can expect police harrassment, without regard to whether you are actually violating any laws. I would avoid riding on the east side, unless you are on one of the multi-use trails. The trails are fairly pleasant to ride although they can be congested on a nice day.
As for suggestions, ride the Burke Gilman trail out to Marymoor Velodrome to see the Friday night races ( round trip is about 40 miles ). Or ride along the Sound from Woodland Park, across the Chittenden Locks to Magnolia Avenue, Elliot Bay Trail, Alaskan Way, West Seattle Bridge, to the Alki Trail, and points south. Using your map, you can ride as far as south as Des Moines on roads that are very good for bicycling. Avoid Ambaum Blvd. between White Center and Burien if you go toward the south end.

Bellevue, Redmond, Kirkland, and the entire east side of Lake Washington are another matter. This is a much newer suburban area, very car-centric, and designed to actively discourage bicycling or walking. The drivers there are likely to find the presence of a bicyle as offensive, and they will let you know it. Also you can expect police harrassment, without regard to whether you are actually violating any laws. I would avoid riding on the east side, unless you are on one of the multi-use trails. The trails are fairly pleasant to ride although they can be congested on a nice day.

I live in Bellevue and the east side really isn't that bad. Bellevue is reasonably bike friendly although I certainly wouldn't go out of my way to bike here.

Here is a site with links to Seattle bike maps. Although the SR520 bridge is closed to bikes, buses frequently cross the bridge and most have bike racks. The bus bike racks in Seattle get lots of use so drivers are used to it. It's almost an oddity to see a bus without at least one bike on the rack on a nice day.

I live in Sammamish and regularly bike around the eastside. There are also some beautiful country roads to explore. There are plenty of cyclists about year-round so most drivers are used to seeing us out on the roads. I haven't had too many problems. Another area that's great to ride around is northeast of Everett and the Skagit Valley even further north. For a wonderful experience, take the ferry to Bainbridge Island or from Edmonds to Kingston and the Kitsap Penninsula. The rest of the San Juan chain is of course a cycling haven. Check out Orcas, Lopez and San Juan Island for some wonderful cycling. The ferry from Mukilteo to Whidbey Island is recommended.

I live in Sammamish and regularly bike around the eastside. There are also some beautiful country roads to explore. There are plenty of cyclists about year-round so most drivers are used to seeing us out on the roads. I haven't had too many problems. Another area that's great to ride around is northeast of Everett and the Skagit Valley even further north. For a wonderful experience, take the ferry to Bainbridge Island or from Edmonds to Kingston and the Kitsap Penninsula. The rest of the San Juan chain is of course a cycling haven. Check out Orcas, Lopez and San Juan Island for some wonderful cycling. The ferry from Mukilteo to Whidbey Island is recommended.

Perhaps I am being overly harsh about the area, but I have had bad experiences there and have avoided it for years. One thing I really don't like about the eastside is that it is laid out as hundreds of mega-malls and strip malls, interspersed with large suburban residential areas that are full of cul-de-sacs. The result is that most of the roads that go anywhere out of the immediate neighborhood are high-speed multilane arterials with no shoulder, not very suitable for biking. This is the same pattern you find in any modern suburban area in the US.

Where do you go to ride on the eastside ? The beautiful country roads sound interesting ! Do you know of a good, safe route to get from Seattle to the end of the Ironhorse Trail, for example ?

Where do you go to ride on the eastside ? The beautiful country roads sound interesting ! Do you know of a good, safe route to get from Seattle to the end of the Ironhorse Trail, for example ?

Best route would be to take the I-90 trail across the bridge into Bellevue. Cut through Mercer Slough Park and you will get dumped out in front of 128th. Then continue straight on across 128th. and pick up SE Eastgate. Take SE Eastgate Way to continue across 148th. You will eventually hit 35th. Pl. Hang a right and it will become 34th. Pl. Take it to 901/West Sammamish. where you will hang a right. Follow W. Samm towards Issaquah where it will become NW Sammamish Rd. and you will eventually hit the bottom of Lake Sammamish State Park. Cut east to stay on NW Samm Rd. and you will eventually hit East Sammamish. Hang a right and follow it for about 1/4 mile and then make a left to go up Black Nugget. Gear down low because this road hits 12%. Take Black Nugget till it ends at Issaquah-Pinelake and make a left. This will be about a 5% climb. The speed on that road is 35MPH but many drivers do go quite a bit faster so be careful. Most of them will give you room though. Take Issaquah Pinelake up to the next major light and continue on through. It will turn into Issaquah-Fall City. You will be greated by a bunch of 5% to 7% climbs along the way. You will eventually reach a fork where you can choose to continue on Issaquah-Fall City by hanging a right or continue on straight where the road turns into Duthie Hill. If you choose to go straight, you will eventually hit 202 in about 2 miles where you will hang a right towards Fall City. There's a high-speed winding descent that's fairly narrow. If you hang a right and go down Issaquah-Fall City, you will 202 about 2 miles further east. Then you take 202 past Fall City and towards Snoqualmie. Near the falls, you will find the entrance to Snoqualmie Valley Trail which you can take towards Rattlesnake Lake. Once at Rattlesnake Lake, you can pick up the John Wayne Pioneer Trail/Ironhorse Trail and take it towards the Snoqualmie Pass.

Yes, the eastside is chocked full of highspeed arterials that connect towns but they are fairly bike friendly. I have had no problems coexisting with 55MPH traffic on Hwy 202 for instance. Also, there are plenty of lower speed scenic routes. You just need to find them. The best route are the ones that follow along the terrain boundaries (ridgelines of the valleys, rivers, etc). Other good routes are the ones that parallel the major expressways just off to the side.

I would like to add that I have never been harrased by the law while riding in the Puget Sound area. I think you will find for the most part people are friendly toward bikes here. You can also go to the King Connty website for other ride options. Enjoy your stay here I think you will like it.


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package typings.devtoolsProtocol.anon

import typings.devtoolsProtocol.mod.Protocol.Debugger.RestartFrameRequest
import typings.devtoolsProtocol.mod.Protocol.Debugger.RestartFrameResponse
import scala.scalajs.js
import scala.scalajs.js.`|`
import scala.scalajs.js.annotation._

@js.native
trait ReturnTypeRestartFrameResponse extends js.Object {
  var paramsType: js.Array[RestartFrameRequest] = js.native
  var returnType: RestartFrameResponse = js.native
}

object ReturnTypeRestartFrameResponse {
  @scala.inline
  def apply(paramsType: js.Array[RestartFrameRequest], returnType: RestartFrameResponse): ReturnTypeRestartFrameResponse = {
    val __obj = js.Dynamic.literal(paramsType = paramsType.asInstanceOf[js.Any], returnType = returnType.asInstanceOf[js.Any])
    __obj.asInstanceOf[ReturnTypeRestartFrameResponse]
  }
  @scala.inline
  implicit class ReturnTypeRestartFrameResponseOps[Self <: ReturnTypeRestartFrameResponse] (val x: Self) extends AnyVal {
    @scala.inline
    def duplicate: Self = (js.Dynamic.global.Object.assign(js.Dynamic.literal(), x)).asInstanceOf[Self]
    @scala.inline
    def combineWith[Other <: js.Any](other: Other): Self with Other = (js.Dynamic.global.Object.assign(js.Dynamic.literal(), x, other.asInstanceOf[js.Any])).asInstanceOf[Self with Other]
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    }
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    def setParamsType(value: js.Array[RestartFrameRequest]): Self = this.set("paramsType", value.asInstanceOf[js.Any])
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  }

}


Common carriers, such as passenger airlines, bus lines, train lines, and passenger vehicles frequently include passenger seats. To increase a passenger's comfort and enjoyment, many passenger seats include movable arm rests, which may be raised to a stowed position to increase the amount of room available for a passenger, and may also be lowered to a deployed position to provide a location for a passenger to rest his or her arms, access certain amenities, or have a support for other objects, such as personal electronic devices, books, or other objects.
Conventionally, as illustrated in FIGS. 1-3, the raised and deployed positions of these arm rests have been controlled by up-stop mechanisms and down-stop mechanisms that contact the arm rest at the appropriate location along the rotational path of the arm rest to prevent further rotation in that direction. In most cases, the down-stop mechanism is located in a location that presents a pinching hazard when the arm rest is lowered to the deployed position. Thus, it may be desirable to provide a down-stop mechanism that does not pose a pinch hazard.


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1. Field of the Invention
The present invention relates to a heat processing apparatus and heat processing method for heating a substrate, which are used for manufacturing semiconductor devices while performing a coating/developing process on a substrate, such as a semiconductor wafer.
2. Description of the Related Art
In a photolithography step for semiconductor devices, a resist is applied onto a semiconductor wafer (which will be simply referred to as “wafer” hereinafter) to form a resist film. Then, the resist film is subjected to a light exposure process in accordance with a predetermined circuit pattern. Then, the light-exposed pattern thus formed is subjected to a developing process to form a circuit pattern on the resist film.
In a photolithography step of this type, various heat processes are performed, such as a heat process (pre-baking) performed subsequently to coating of a resist or a chemical solution for, e.g., a BARC, a heat process (post exposure baking) performed subsequently to light exposure, and a heat process (post baking) performed subsequently to development.
In general, these heat processes are performed by a heat processing apparatus (hot plate unit) provided with a heating plate (hot plate) heated by a heater. Conventionally, in a heat processing apparatus of this type, the following gas flow control is typically adopted to perform a heat process on a wafer as uniform as possible. Specifically, during the heat process, the heating plate is covered with a cover, in which gas flows are formed from around the heating plate toward the center, and gas is exhausted upward from the center of the cover.
In recent years, circuit patters of semiconductor devices show rapid progress in decreasing the line width, in decreasing the size, and in increasing the degree of integration, which requires the process uniformity of heat processes to be further improved. However, by use of the gas flow control described above, it has become difficult to attain desired process uniformity, particularly uniformity concerning gas flows relative to a wafer, which is necessary to uniformize critical dimensions (CD). Accordingly, it is required to provide a technique for performing a heat process with higher uniformity. As a technique for realizing a heat process with higher uniformity, Jpn. Pat. Appln. KOKAI Publication No. 2003-318091 discloses a technique for performing a heat process while supplying gas flows from above uniformly onto the entire surface of a substrate. According to this technique, since the gas flows uniformly strike the entire surface of the substrate, such as a wafer, it is possible to improve the process uniformity, particularly uniformity in the film thickness of a coating film.
However, devices for the next generation require very minute patterns with a CD value of 50 nm or less, in which even slight dust may cause a defect. Accordingly, it is necessary to swiftly remove sublimed substances and so forth generated by a heat process. In this respect, the technique disclosed in Jpn. Pat. Appln. KOKAI Publication No. 2003-318091 mentioned above cannot sufficiently remove sublimed substances and so forth, and thus it may deteriorate the yield of products. Particularly, such deterioration in the yield due to sublimed substances is easily caused in a coating film for argon fluoride (ArF) having a short wavelength. This coating film is used for further improving the resolving power of light exposure in accordance with a CD for the next generation.
Further, depending on coating films, heat processes of this kind include processes that should give weight to CD uniformity and processes that should give weight to an improvement in the product yield by removing substances, such as sublimed substances, that generate dust. However, the technique disclosed in Jpn. Pat. Appln. KOKAI Publication No. 2003-318091 mentioned above cannot address the latter type. Accordingly, there is a great demand for a heat processing technique that can address these two types.