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Let $\mathbf{r}={\mathbf{r}}_2-{\mathbf{r}}_1$ and $\mathbf{v}=\dot{\mathbf{r}}$. Relative to the centre of mass, the two positions are ${\mathbf{r}}_1=-(M_2/M)\mathbf{r}$ and ${\mathbf{r}}_2=(M_1/M)\mathbf{r}$. Their velocities have the same mass factors. Adding their kinetic energies and angular momenta gives, with reduced mass $\mu =M_1{M}_2/M$,These are the internal orbital quantities, excluding any uniform centre-of-mass motion. Newton's law of universal gravitation givesFor constant masses, differentiating $E$ makes the gravitational work cancel the derivative of the potential energy, so $\dot{E}=0$. Also $\dot{\mathbf{h}}=\mathbf{r}\times \ddot{\mathbf{r}}=0$, since the force is central; hence conservation of angular momentum gives $\dot{\mathbf{J}}=0$.

The eccentricity vector, the normalized attractive Laplace-Runge-Lenz vector, isUsing $\dot{\hat{\mathbf{r}}}=\mathbf{v}/r-\mathbf{r}(\mathbf{r}\cdot \mathbf{v})/r^3$ andshows that its derivative also vanishes. Adding a relative perturbing acceleration $\mathbf{f}$ changes these cancellations only by the perturbing terms. The perturbed Kepler-orbit conservation laws are therefore

For the distant Newtonian gravitational potential, expand each source term about the binary's centre of mass:The dipole vanishes because ${\sum }_i{M}_i{\mathbf{r}}_i=0$. With $\mathbf{a}=\mathbf{r}$, the quadratic mass moment is ${\sum }_i{M}_i{\mathbf{r}}_i{\mathbf{r}}_i=\mu \mathbf{aa}$, so the quadrupole potential of a circular binary isOn the inner binary's fast timescale, the distant companion's position is approximately fixed. Averaging over the inner circular orbit in the common plane gives $⟨{\cos }^2\theta ⟩=1/2$ and henceThis averaged potential function is independent of both time and azimuth. Consequently the companion's energy and axial angular momentum are conserved in the averaged problem. The exact motion has small fast variations; the statement is a secular approximation for a hierarchical system away from an orbital resonance.

For a circular outer orbit, force balance givesso the circumbinary orbital-period correction isConservative binary mass transfer leaves $M$ unchanged, so it does not change the leading monopole force. It changes $\mu$ and $a$, and therefore the quadrupole correction. If the inner orbital angular momentum is conserved, $J_{\mathrm{in}}=\mu \sqrt{GMa}$ gives $a\propto {\mu }^{-2}$ and $Q=\mu a^2\propto {\mu }^{-3}$. The correction is thus largest as the binary becomes strongly unequal in mass and widens, within the hierarchy $a\ll x$.

One must also allow the outer radius to respond. For slow axisymmetric evolution its specific angular momentum $h_3$ remains constant, while its energy need not: the averaged potential now depends on time. A nearly circular outer orbit has $h_3^2=GMx+3GQ/(4x)$ and $P_3=2\pi x^2/h_3$. At fixed $M,h_3$, differentiation givesThus mass exchange inside the binary can produce a small measurable outer-period change even without total mass loss. Nonconservative mass loss additionally changes the leading monopole term.

A binary star of total mass $M$, reduced mass $\mu$ and separation vector $\mathbf{a}$ has far-field potential functionwhere $\theta$ is the angle between $\mathbf{x}$ and $\mathbf{a}$. The dipole term vanishes in centre of mass coordinates. Averaging a circular coplanar binary over its fast phase gives $\overline{\Phi }=-GM/x-G\mu a^2/(4x^3)$.

A Schrodinger operator is a differential operator of the formwhere $\Delta$ is the Laplace operator and $V$ is a potential function. Its eigenvalue problem is the Time-independent Schrodinger equation.