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Let $f(\mathbf{p})d^{3}p$ be particle number per unit volume in a momentum element, including the spin-state count in $f$. This is the local momentum part of a phase-space distribution function, normalized by the number density $n=\int f(\mathbf{p})d^{3}p$. For an isotropic distribution with particle energy $E(p)=\sqrt{m^2{c}^4+p^2{c}^2}$, the kinetic pressure of an isotropic gas is the average momentum flux:The kinetic energy density, explicitly excluding rest mass, isFor $p\ll mc$, $E-m{c}^2=p^2/(2m)$ and $\mathbf{v}=\mathbf{p}/m$, giving $P=2u_{\mathrm{kin}}/3$. For $p\gg mc$, $E\simeq pc$ and $v\simeq c$, giving $P\simeq u_{\mathrm{kin}}/3$. ThusNeither relation assumes a Maxwellian distribution. At intermediate momenta neither constant ratio is exact, and anisotropic distributions require a pressure tensor rather than this scalar pressure.

For a classical Maxwell-Boltzmann distribution, $f=n(2\pi m{k}_{B}T{)}^{-3/2}\exp [-p^2/(2m{k}_{B}T)]$. Three Gaussian component integrals give $⟨p^2⟩=3m{k}_{B}T$, so the ideal gas hasIn a fully ionized mixture, sum over independent species to obtain $P_{\mathrm{gas}}=\rho k_{B}T/({\mu }_{\mathrm{mol}}{m}_u)$, where the mean molecular weight counts ions and free Electrons. This excludes partial-ionization and interaction corrections. A dilute classical ultrarelativistic gas still has $P=n{k}_{B}T$ but $u_{\mathrm{kin}}\simeq 3n{k}_{B}T$; its momentum distribution is proportional to $e^{-pc/(k_{B}T)}$ instead of the nonrelativistic Gaussian.

For a thermal photon gas, the two polarization states and zero photon chemical potential give the Planck photon distributionUse $x=pc/(k_{B}T)$ and ${\int }_0^{\infty }{x}^3/(e^x-1)dx={\pi }^4/15$. The radiation constant and photon equation of state are thereforeLikewise $n_{\gamma }=2\zeta (3)(k_{B}T/\hslash c{)}^3/{\pi }^2$. Unlike a gas with a fixed particle number, photons do not have pressure proportional to baryonic mass density.

For fully degenerate Electrons, the Fermi-Dirac distribution becomes a filled momentum sphere with two spin states. Counting them gives the Fermi momentumHere ${\mu }_e$ is the mean molecular weight per electron. The equation of state of a cold electron gas follows by integrating momentum flux up to $p_F$:Its two limits areThe corresponding kinetic energy densities are $3n_e[p_F^2/(2m_e)]/5$ and $3n_e{p}_{F}c/4$, respectively. Thus in the high-density relativistic limit the Electron pressure scales as ${\rho }^{4/3}$ and is nearly independent of temperature. This is the ideal noninteracting, fixed-composition Electron result, not a universal equation of state at nuclear densities where captures, interactions and the composition change.

The boundaries on a stellar equation-of-state regime diagram concern the Electron component. Define the electron relativistic density threshold and thermal electron relativistic threshold byA degenerate Electron gas is nonrelativistic well below ${\rho }_{\ast }$ and ultrarelativistic well above it: this is an approximately vertical division on a log-density plot. A nondegenerate Electron gas instead becomes thermally relativistic near $T_{\ast }$: this is an approximately horizontal division. The regimes have broad crossovers rather than a discontinuity at either line.

The exact electron Fermi temperature, subtracting Electron rest energy, isFor Electrons with negligible thermal pairs, strong degeneracy requires $T\ll T_F$, whereas $T\gg T_F$ gives a nondegenerate gas. In the pair-rich regime the actual Electron and Positron distributions must instead be treated with their chemical potentials, as discussed below. The two limiting degeneracy boundaries areThese slopes $2/3$ and $1/3$ explain the bent degeneracy boundary on the logarithmic graph. A thermal-wavelength test $n_e[2\pi {\hslash }^2/(m_e{k}_{B}T){]}^{3/2}\sim 2$ gives the same nonrelativistic density/temperature scaling, with an order-one definition of the crossover.

The radiation-to-gas pressure boundary in the nondegenerate fully ionized regime isRadiation dominates above this line. Once Electrons are strongly degenerate, compare radiation pressure with $P_e(\rho )$ instead of continuing the ideal-gas comparison into that regime. The radiation-to-degeneracy pressure boundary is $T=[3P_e(\rho )/a_{\mathrm{rad}}{]}^{1/4}$, with logarithmic slopes $5/12$ and $1/3$ in the nonrelativistic and ultrarelativistic limits. Degeneracy of the Electrons and dominance of their pressure are distinct criteria.

For the pair curve, distinguish baryonic net Electrons from thermally created Electrons and Positrons. Chemical equilibrium with photons requires opposite Electron/Positron chemical potentials when the one-particle energy includes rest energy, as in the distributions below. At low temperature and low degeneracy the zero-chemical-potential density per charge species isCharge neutrality gives $n_{-}-n_{+}=n_b=\rho /({\mu }_e{m}_u)$, while the nondegenerate equilibrium product is $n_{-}{n}_{+}=n_0^2$. HencePairs become important when $n_0$ is comparable with or exceeds $n_b$, approximately the electron-positron thermal pair abundance curve ${\rho }_{\mathrm{pair}}(T)={\mu }_e{m}_u{n}_0(T)$. Below its density at a fixed temperature, pairs dominate over the net charge Electrons. The exponential makes the low-temperature portion steep in a log-log plot. For ${\mu }_e=2$, the full zero-potential integral gives approximate pair-marker densities $1.80\times 10^3\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$ at $10^9\mathrm{K}$ and $7.52\times 10^5\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$ at $3\times 10^9\mathrm{K}$. At relativistic temperature use the full zero-chemical-potential Fermi-Dirac distribution instead; it gives $n_0\to [3\zeta (3)/(2{\pi }^2)](k_{B}T/\hslash c{)}^3$, so the high-temperature pair curve approaches slope three in log-density versus log-temperature. Strong net-electron degeneracy suppresses Positrons and requires the full chemical-potential-dependent distribution.

The original diagram uses fully ionized helium, ${\mu }_e=2$ and ${\mu }_{\mathrm{mol}}=4/3$, only to set numerical locations. It plots the full $T_F$ curve and the zero-chemical-potential pair integral, rather than extending their asymptotes into the crossover. The pair boundary is deliberately an approximate abundance marker, not a phase transition; at high temperature the nonrelativistic ion approximation and fixed-composition model also have limits. For $T\gg T_{\ast }$ with abundant nondegenerate pairs, the combined Electron/Positron energy density is $7a_{\mathrm{rad}}{T}^4/4$, so photons plus pairs have $u=11a_{\mathrm{rad}}{T}^4/4$ and $P=u/3$. Real stellar matter adds Coulomb effects, partial ionization, nuclear reactions and, at sufficiently high density, nuclear-matter physics beyond the ideal regime map.

Assume a time-independent spherical relative potential $\psi (r)$, with acceleration $\nabla \psi$. The relative energy and specific angular momentum areAlong an orbit, $\dot{E}=\nabla \psi \cdot \mathbf{v}-\mathbf{v}\cdot \nabla \psi =0$ and $\dot{\mathbf{\ell }}=\mathbf{r}\times \nabla \psi =0$, since a spherical gradient is radial. Thus both $E$ and $L$ are integrals of motion. Choose the escape level $\psi =0$ and take the bound galactic distribution function to vanish for $E\le 0$.

For an isotropic $f(E)$, integrate over a spherical coordinate system in velocity space:Differentiating gives ${\rho }^{\prime }(\psi )=2\pi \sqrt{2}{\int }_0^{\psi }f(E)/\sqrt{\psi -E}dE$. Convolve once more with the reciprocal-square-root kernel. Reversing the order of integration, under the usual integrability assumptions, usesTherefore ${\int }_0^{E_0}{\rho }^{\prime }(\psi )/\sqrt{E_0-\psi }d\psi =2\sqrt{2}{\pi }^2{\int }_0^{E_0}f(E)dE$. Differentiating proves Eddington inversion:Here ${\pi }^2$ is outside the square root, as in the original PDF; the TeX transcription wrongly places it inside. The assumptions include a locally integrable bound distribution and sufficient density regularity, with no extra unbound or boundary population. An inverted expression must additionally be nonnegative to be physical. Since the velocity measure and $f$ are invariant under all velocity rotations, $⟨v_r^2⟩=⟨v_{\theta }^2⟩=⟨v_{\varphi }^2⟩=⟨v^2⟩/3$ whenever these moments exist; mixed moments vanish.

For the half-anisotropic distribution function, write the polar velocity angle from the radial direction as $\eta$. Then $L=rv\sin \eta$, andThe apparent $1/L$ singularity is integrable because it cancels the factor $v\sin \eta$ in the velocity measure. Integrating $\eta \in [0,\pi ]$ and $\phi \in [0,2\pi ]$ givesThe density relation requires $r\rho \to 0$ at the escape boundary and $f_E\ge 0$ for a physical distribution. Consider a monotonically decreasing $\psi$. The largest allowed radius at fixed binding energy is the radial-orbit limit $r_E$ with $\psi (r_E)=E$; a nonzero-$L$ orbit normally has a smaller apocenter because of its centrifugal term. Changing from $\psi$ to $r$ givesThe numerator in the original PDF instead uses $\rho +d\rho /d\psi$. That is a genuine printed error: the chain rule for $d(r\rho )/d\psi$ requires $\rho +rd\rho /dr$. It is not legitimate to prove the printed expression as written. A concrete counterexample is the unit-parameter Hernquist model at $r=1$: ${\psi }^{\prime }=-1/4$, ${\rho }^{\prime }=-(5/2)\rho$ and $d\rho /d\psi =10\rho$. The printed quotient is $-44\rho$, whereas the correct $g$ is $6\rho >0$. The former would give a negative distribution.

The angular weight after cancellation is uniform in $\eta$, so the averages of ${\cos }^2\eta$ and ${\sin }^2\eta$ are both $1/2$. The azimuthal angle splits the tangential term equally. Hence $⟨v_r^2⟩=⟨v_{\theta }^2+v_{\varphi }^2⟩$ and each individual tangential moment is half the radial one. The velocity-anisotropy parameter is consequentlyThis is a radially biased constant-anisotropy distribution function, not an isotropic one or purely radial motion. It does not by itself prove dynamical stability; it preferentially weights small angular momentum while remaining integrable.

For the Hernquist model, use the Poisson equation for Newtonian gravity with the relative-potential sign, ${\nabla }^2\psi =-4\pi G\rho$. Since ${\psi }^{\prime }=-GM/(r+a{)}^2$,Its enclosed mass is $M{r}^2/(r+a{)}^2$, which tends to $M$. Expressing the augmented density in the relative potential gives $r\rho =a{\psi }^3/(2\pi G^3{M}^2)$. The preceding derivative therefore giveswith zero bound-model population outside the allowed energy domain. It is nonnegative and reproduces the density by direct integration. The Hernquist model has a central density cusp, so the formulas for the local $1/L$ distribution are interpreted at $r>0$. As a further check, the radial second moment obtained from the same integral is $\psi /4$ and each tangential second moment is $\psi /8$, consistent with $\beta =1/2$.

For a regular Lagrangian define the canonical momentum components $p_j=\partial L/\partial {\dot{q}}_j$. Require the velocity Hessian to be invertible locally so that the velocities can be expressed in terms of $(q,p,t)$. The Legendre transform in mechanics definesThe explicit time argument cannot generally be omitted when the given Lagrangian depends on time. Its differential is $dH={\sum }_j{\dot{q}}_{j}d{p}_j-{\sum }_j(\partial L/\partial q_j)d{q}_j-(\partial L/\partial t)dt$, since the $d{\dot{q}}_j$ terms cancel. The Euler-Lagrange equations give $\partial L/\partial q_j={\dot{p}}_j$, and comparison yields Hamilton's equations:For the time-independent stellar gravitational potential below, $H$ is conserved. Its sign convention differs from the binding energy in question 2: here $E=H=v^2/2-\psi$ is the ordinary specific orbital energy.

The prescribed nondegenerate prolate spheroidal coordinates require $0<\alpha <\beta$; equality collapses the coordinate family. Their physical velocity components are $v_j=h_j{\dot{q}}_j$, and $h_{\varphi }=R$. Per unit stellar mass the Lagrangian is $L=\frac{1}{2}{\sum }_j{h}_j^2{\dot{q}}_j^2+\psi$, so $p_j=h_j^2{\dot{q}}_j=h_j{v}_j$. Its Hamiltonian isThis uses acceleration $+\nabla \psi$, as specified, rather than importing the opposite gravitational potential convention.

For the Stäckel potential, use the Hamilton-Jacobi equation with principal function $S=W_{\lambda }(\lambda )+W_{\mu }(\mu )+l\varphi -Et$. Axisymmetry makes $p_{\varphi }=l=R{v}_{\varphi }$ constant. Write $d=\lambda -\mu$ and $A(\tau )=(\tau +\alpha )(\tau +\beta )$. Multiplying $H=E$ by $d$ and usingseparates it as $F(\lambda )-E\lambda =F(\mu )-E\mu$, where $F(\tau )=2A(\tau )(W_{\tau }^{\prime }{)}^2-(\alpha -\beta )l^2/[2(\tau +\alpha )]-G(\tau )$. Each side depends on its own coordinate only and therefore equals a constant, chosen as $-I$. Here $G(\tau )$ is the generating function of the gravitational potential, not the gravitational constant. HenceThis proves the conserved axisymmetric Stäckel third integral. Eliminating $E$ between the two expressions and using $p_j=h_j{v}_j$ givesAt regular generic phase-space points the three separation constants $E,l,I$ are independent; special circular or degenerate orbits can reduce their local independence. Thus the model is an integrable Hamiltonian system.

For the intended regular, phase-mixed steady model, Jeans theorem permits a galactic distribution function $f=F(E,l,I)$. At any fixed point, $E$ and $I$ depend on $v_{\lambda }$ and $v_{\mu }$ only through their squares, and $l$ depends only on $v_{\varphi }$. Thus $f$ is even separately in $v_{\lambda }$ and $v_{\mu }$, whether or not it is even in $v_{\varphi }$. Odd integration proves velocity-ellipsoid alignment in a Stäckel potential:The argument allows azimuthal streaming: it does not require zero mean $v_{\varphi }$. The means in the two meridional directions vanish, so the off-diagonal velocity covariance also vanishes. Second moments must exist for this conclusion.

There is a hypothesis behind the printed universal wording. On a generic nonresonant regular invariant torus, a continuous steady distribution is constant along a dense orbit and hence depends only on the torus integrals; this justifies the phase-mixed form. Bare stationarity alone is not sufficient in a degenerate resonant gravitational potential with additional integrals. For example $H=\frac{1}{2}(v^2+{\omega }^2{r}^2)$ is a Stäckel Hamiltonian: choose $G(\tau )=-\frac{1}{2}{\omega }^2[{\tau }^2+(\alpha +\beta )\tau ]$, up to a constant vacuum shift. The additional conserved quantity $K_{xy}=v_x{v}_y+{\omega }^{2}xy$ gives a positive normalizable steady tracer $f\propto \exp (-aH+ϵK_{xy})$ for $a>|ϵ|>0$. Its local $⟨v_x{v}_y⟩=ϵ/(a^2-{ϵ}^2)\ne 0$. At azimuth zero and a regular point where the $\lambda$ direction has a radial component, this gives nonzero $⟨v_{\lambda }{v}_{\varphi }⟩$. This tracer need not itself source the harmonic gravitational potential. It exhibits why the stronger, ordinary phase-mixed three-integral assumption is necessary instead of claiming that every imaginable steady distribution has the asserted alignment.

The geometry follows by multiplying the coordinate-defining equation and evaluating at $\tau =-\alpha ,-\beta$:Constant $\lambda$ surfaces are prolate spheroids, with equatorial semiaxis $\sqrt{\lambda +\alpha }$ and polar semiaxis $\sqrt{\lambda +\beta }$. Constant interior $\mu$ surfaces are two-sheeted hyperboloids. Their common foci lie at $z=\pm \sqrt{\beta -\alpha }$. The orbit volume in an axisymmetric Stäckel potential is determined byIts allowed coordinate intervals have turning boundaries or coordinate endpoints. A typical bound orbit with $l\ne 0$ oscillates between inner and outer spheroids, oscillates vertically between hyperboloidal limits, and circulates about the symmetry axis with $\dot{\varphi }=l/R^2$. The equatorial fold $\mu =-\beta$ joins the positive and negative $z$ branches; it is not necessarily a physical turning point in $z$.

For three incommensurate frequencies, the regular trajectory densely samples its projected allowed volume, with nonuniform occupation determined by its speeds. It does not follow one coordinate surface, and it need not close. Resonant orbits may be closed or lie in a smaller projected set; $l=0$ confines azimuthal motion to a meridional plane (with the usual axis coordinate qualification), and circular or equatorial limits are also lower-dimensional. Smooth Stäckel separation removes the generic chaotic behavior of nonintegrable potentials; it does not assert that every orbit fills a three-dimensional volume.

For Roche-lobe overflow in the stipulated stellar thermal equilibrium, $R_2=R_L$. Using the mass-radius relation and Kepler's third law,The total mass cancels, givingHere $R_{\odot }$ is the solar radius. This is the Roche-lobe-filling period-density relation specialized to $R_2\propto M_2$.

Neglecting spin, the angular momentum of a circular orbit is $J=M_1{M}_2\sqrt{Ga/M}$. Between classical novae the transfer conserves total mass, so $\dot{M}=0$ and ${\dot{M}}_1=-{\dot{M}}_2$. Its logarithmic derivative givesIn particular, for conservation of angular momentum,The source PDF has $\dot{J}=0$ here; the TeX's $J=0$ is a transcription defect. For $q>4/3$, the Roche-lobe radius response exponent exceeds the equilibrium stellar radius response exponent $1$. Since ${\dot{M}}_2<0$, the Roche lobe then shrinks faster than the donor star; the overfill increases and drives more transfer, giving positive feedback. Equality at $q=4/3$ is marginal in this linear response test.

Two qualifications matter. First, $q>4/3$ is a formal extrapolation outside the stated $q<1$ range, and the supplied Roche lobe approximation need not remain accurate there. Second, true dynamical stability of binary mass transfer compares ${\zeta }_L$ with the adiabatic stellar radius response exponent ${\zeta }_{\mathrm{ad}}$, not ${\zeta }_{\mathrm{eq}}=1$. The stellar thermal equilibrium argument supplies the displayed feedback threshold under the imposed radius law, rather than a necessary and sufficient physical dynamical threshold. For example, the polytropic mass-radius relation for a fully convective adiabatic stellar polytrope with index $3/2$ gives ${\zeta }_{\mathrm{ad}}=-1/3$; within this same Roche lobe approximation its dynamical threshold is $q=2/3$. Thus $q<4/3$ by itself does not guarantee dynamical stability.

Gravitational-wave emission from a binary system removes orbital energy and angular momentum. Magnetic braking of a binary star provides another important loss: a magnetized stellar wind carries away donor spin, while tidal locking couples that spin to the orbit. Tides alone redistribute angular momentum and are not an external sink. Matter expelled from the system can also carry orbital angular momentum, although appreciable continuous mass loss would require nonconservative modifications of the preceding equations. Setting ${\dot{R}}_L/R_L={\dot{R}}_2/R_2$ gives the binary mass-transfer contact equationFor $q<4/3$, a negative external $\dot{J}$ therefore sustains negative ${\dot{M}}_2$ along the stipulated stellar thermal equilibrium contact sequence, provided the system is otherwise stable and can remain thermally relaxed.

During the classical nova, the ejecta carry the white dwarf's specific angular momentum, not zero angular momentum. With ${\Omega }^2=GM/a^3$ and the accretor's distance from the centre of mass $a_1=a{M}_2/M$, isotropic escape givesThe slow ejection relative to the orbital period permits the adiabatic circular orbit approximation; the donor star does not transfer appreciable additional mass during it. Take $\delta M_1=-\delta m$, $\delta M_2=0$ and $\delta M=-\delta m$. Differentiating $J=M_1{M}_2\sqrt{Ga/M}$ givesSince $M/M_1=1+q$, this yieldsThese are first-order formulae with errors of order $(\delta m/M{)}^2$. The orbital widening agrees with Jeans-mode mass loss. The donor star's radius is unchanged to this order by the assumed ejection, so its Roche lobe expands away from it, causing nova-induced binary detachment. This idealization neglects changes in the donor from irradiation or interaction with ejecta; those are not supplied in the model.

Let $\Gamma =-\dot{J}>0$ be the fixed external loss rate. During the detached binary phase both component masses are fixed, and ${\dot{R}}_L/R_L=2\dot{J}/J=-2\Gamma /J$. Closing the fractional gap $4\delta m/(3M)$ takesIn the subsequent semidetached binary, the binary mass-transfer contact equation gives $-{\dot{M}}_2=3M_2\Gamma /[J(4-3q)]$. Accumulating a fresh layer of mass $\delta m$ therefore takesThroughout these first-order estimates $J,M,q$ can be evaluated at the start of the cycle: their fractional changes during it are of order $\delta m/M$. The limit $\Gamma =0$ is excluded, since no finite reconnection time follows without an external shrinkage mechanism.

Set $\mathbf{r}={\mathbf{x}}_1-{\mathbf{x}}_2$, $M=M_1+M_2$ and $\mu =M_1{M}_2/M$, the reduced mass. Newton's law of universal gravitation gives ${\ddot{\mathbf{x}}}_1=-G{M}_2\mathbf{r}/r^3$ and ${\ddot{\mathbf{x}}}_2=G{M}_1\mathbf{r}/r^3$. Subtraction establishesFor fixed masses, the orbital energy and angular momentum areTaking derivatives gives $\dot{E}=\mu \mathbf{v}\cdot (\ddot{\mathbf{r}}+GM\mathbf{r}/r^3)=0$ and $\dot{\mathbf{J}}=\mu \mathbf{r}\times \ddot{\mathbf{r}}=0$. These are conservation of energy and conservation of angular momentum in the unperturbed Kepler orbit. To see directly that the orbital eccentricity is fixed, put $\mathbf{h}=\mathbf{r}\times \mathbf{v}$ and $\mathbf{n}=\mathbf{r}/r$. The eccentricity vector isSince $\dot{\mathbf{h}}=0$, the cross product identity $\mathbf{r}\times (\mathbf{r}\times \mathbf{v})=\mathbf{r}(\mathbf{r}\cdot \mathbf{v})-r^2\mathbf{v}$ gives $\ddot{\mathbf{r}}\times \mathbf{h}=GM\dot{\mathbf{n}}$, so $\dot{\mathbf{e}}=0$. Its magnitude $e$ is the orbital eccentricity. Equivalently,Thus the semi-major axis and orbital eccentricity of a bound noncollision Kepler orbit are constant, including $e=0$.

Take the centre of mass as origin. The two positions are ${\mathbf{x}}_1=(M_2/M)\mathbf{r}$ and ${\mathbf{x}}_2=-(M_1/M)\mathbf{r}$. For $x\gg r$, a Taylor expansion givesThe mass dipole vanishes because ${\sum }_A{M}_A{\mathbf{x}}_A=0$, and the second mass moment tensor is ${\sum }_A{M}_A{x}_{Ai}{x}_{Aj}=\mu r_i{r}_j$. The gravitational quadrupole potential of a point-mass binary is consequentlyBoth printed tensors are trace-free. Contracting them with the Kronecker delta summed over three spatial dimensions givesThis proves the requested expression through second order. In particular the printed convention is $q_{ij}=\frac{3}{2}{Q}_{ij}$, where $Q_{ij}=\mu (r_i{r}_j-r^2{\delta }_{ij}/3)$ is the standard mass quadrupole moment; the radiation coefficient must be adjusted accordingly. No mass dipole term may be retained in centre of mass coordinates.

To evaluate the quadrupole formula without assuming a circular orbit, put $k=GM$, $u=\dot{r}=\mathbf{n}\cdot \mathbf{v}$ and ${\mathbf{v}}_{\perp }=\mathbf{v}-u\mathbf{n}$. Differentiating the gravitational acceleration gives the kinematic jerkFor $S_{ij}=r_i{r}_j$, the product rule then givesInserting these into the printed mass quadrupole moment convention yieldsThe two tensors in brackets are orthogonal in their tensor contraction, because $\mathbf{n}\cdot {\mathbf{v}}_{\perp }=0$. Their squared norms are $6$ and $2v_{\perp }^2$, respectively. HenceThe instantaneous quadrupole luminosity of a Kepler binary is thereforeThe radial derivative $\dot{r}$ is not the vector speed. The bracket equals $v_{\perp }^2+u^2/12$, so the radiated luminosity is nonnegative for every instantaneous velocity.

For a slowly evolving circular orbit, there is no preferred orbital phase at which to excite a persistent eccentricity vector. More precisely, the rotating mass quadrupole moment emits at twice the orbital frequency, with angular harmonic number two; its gravitational-wave energy and angular-momentum balance is $\dot{E}=\Omega \dot{J}$. The circular Kepler orbit sequence has $J=\mu \sqrt{GMa}$ and $dE/dJ=\Omega$, so this loss is tangent to the sequence. At leading adiabatic order it preserves zero secular orbital eccentricity. This is a circular gravitational-wave inspiral, with a small radial drift rather than an exactly fixed-radius Newtonian circle. The energy flux alone would not establish circularity without this symmetry and angular momentum balance.

To leading radiation order use $r=a$, $u=0$ and ${\mathbf{v}}^2=GM/a$ in the loss formula. Combining it with $E=-GM\mu /(2a)$ givesKeeping the fixed masses and leading quadrupole formula, integration givesThis formal coalescence time displays the strong $a_0^4$ dependence: gravitational-wave emission from a binary system matters far more for close binary stars than for wide ones. The orbital period decreases as $a^{3/2}$ and the gravitational-wave frequency increases, producing a chirp. A detached binary can be driven into Roche-lobe overflow; further evolution then depends on mass-transfer stability. A sufficiently close compact pair may merge. The weak-field, slow-motion, point-mass approximation ceases to apply before literal $a=0$; finite stellar radii or strong relativistic effects determine the final interaction.

The Dirichlet Laplacian eigenfunctions and positive Dirichlet Laplacian eigenvalues on the interval of length two areA Sturm-Liouville eigenfunction expansion gives $u={\sum }_j{q}_j(t){\varphi }_j$, with $q_j^{\prime \prime }+({\lambda }_j-\alpha )q_j=0$. Thus all normal modes oscillate precisely when $\alpha <{\lambda }_1$.

To justify a spatially uniform bound for arbitrary appropriate data, assume the usual wave equation energy class $u(0)\in H_0^1(-1,1)$ and $u_t(0)\in L^2(-1,1)$, or smoother compatible data. The energy method conservesThe sharp Poincare inequality is $\Vert u{\Vert }_{L^2}^2\le {\lambda }_1^{-1}\Vert u_x{\Vert }_{L^2}^2$. For $0\le \alpha <{\lambda }_1$, this makes $E$ coercive with constant $1-\alpha /{\lambda }_1$; for $\alpha <0$, coercivity is immediate. The one-dimensional Sobolev representative satisfies $|u(x,t)|\le \sqrt{2}\Vert u_x(t){\Vert }_{L^2}$ by the fundamental theorem of calculus and the Cauchy-Schwarz inequality. Hence the conserved energy bounds $u$ uniformly in both variables.

At $\alpha ={\lambda }_1$, the smooth solution $u=t{\varphi }_1$ is unbounded. At $\alpha >{\lambda }_1$, $u=e^{\sqrt{\alpha -{\lambda }_1}t}{\varphi }_1$ is unbounded. ThereforeThe bound depends on the initial data; it is not a single bound for all arbitrarily rescaled solutions. The energy-class hypothesis supplies pointwise meaning and excludes undefined rough-data interpretations.

The spatial shift in the PDF also requires a boundary value beyond the supplied right endpoint: if $m$ is the last interior index, the newest $m+2$ value is outside the interval. Thus the printed Dirichlet method has no specified boundary closure, and no unique stability range can be assigned to it as stated.

For completeness, the literal interior formula has an exact Von Neumann stability analysis on an infinite or periodic grid. Substitution of a Fourier mode $U_m^n={\xi }^n{e}^{im\theta }$ givesFor the root of a polynomial tending to one as $\theta \to 0$, expansion givesConsequently $\mu <3$ is unstable: sufficiently small nonzero frequencies have a root of a polynomial of modulus greater than one, and arbitrarily fine periodic grids contain such frequencies.

Conversely, if $\mu \ge 3$ and $s=\cos \theta$, thenFor $A=u+iv$ with $u\ge 3$,Since $|A|\ge 3$, the complex quadratic Schur criterion yields both roots of a polynomial in the unit disk, strictly inside except at $\theta =0$, where they are $1$ and $1/3$. Uniform stability follows directly from the BDF2 discrete energy identity: the modal recurrence is $3v^{n+2}-4v^{n+1}+v^n=-(A-3)v^{n+2}$, whose energy decreases because $\mathrm{Re}(A-3)\ge 0$. Summing the modal energies proves a bound independent of the grid and even of $\mu \ge 3$; no eigenvector-separation assumption is needed. This is the stability of a spatially shifted BDF2 stencil. Thus the literal periodic-grid result isThis cannot establish stability of an unspecified Dirichlet boundary condition closure. It also does not restore consistency of a numerical method.