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For an axial Killing vector $\xi$ and zero cosmological constant, set $F^{ab}={\nabla }^b{\xi }^a$. The Killing equation and Ricci identity give ${\nabla }_b{F}^{ab}=-R^a{}_b{\xi }^b$. In a vacuum region this vanishes, so Stokes theorem makes the Komar angular momentum equal on homologous enclosing surfaces. The region may contain gravitational waves with nonzero Weyl curvature: Ricci flatness, rather than complete flatness, is what removes the flux. This proves absence of axial angular-momentum transport by exactly axisymmetric gravitational radiation, without assuming that a radiating spacetime is stationary.

An exact axial Killing vector field makes Komar angular momentum independent of the surrounding vacuum surface. Therefore gravitational waves in a configuration with exact axisymmetry carry no net angular momentum about that symmetry axis, although they can carry energy. This is specific to axial angular momentum; it is not an absence of radiation.

For a rigid body rotating around its third principal axis, the anisotropic part of its second mass moment tensor oscillates at angular frequency $2\Omega$. The quadrupole formula gives the displayed luminosity. Equal transverse principal moments remove this rotating quadrupole and its leading gravitational wave emission.

For either tensor helicity, neutrino free streaming produces the memory equationThe neutrino tensor free-streaming kernel encodes the angular propagation of the perturbation. Energy transferred to neutrino directional anisotropy damps the oscillation amplitude of a gravitational wave after horizon entry. This effect was calculated by Weinberg.

For a slowly evolving circular binary star with separation $a$, reduced mass $\mu$ and orbital angular frequency $\Omega$, the leading gravitational wave losses satisfy $\dot{E}=\Omega \dot{J}$. One direct check uses the standard quadrupole angular-momentum loss lawwhere $Q$ is the mass quadrupole moment and ${ϵ}_{ijk}$ is the Levi-Civita symbol. For $\mathbf{r}=a(\cos \Omega t,\sin \Omega t,0)$, put $A=\mu a^2$; then $Q_{xx}=A(1/6+\cos (2\Omega t)/2)$, $Q_{yy}=A(1/6-\cos (2\Omega t)/2)$, $Q_{xy}=A\sin (2\Omega t)/2$, and $Q_{zz}=-A/3$. Substitution gives ${\dot{J}}_z=-32G{A}^2{\Omega }^5/(5c^5)$ and the quadrupole formula gives $\dot{E}=-32G{A}^2{\Omega }^6/(5c^5)$. Their ratio proves the balance relation. The same relation follows from the rotating wave's angular harmonic number two and frequency $2\Omega$. The additional angular-momentum flux law, not the energy flux alone, is needed to justify preservation of secular circularity.

The gravitational-wave memory effect is a lasting change in the relative configuration of freely falling test bodies after a transient gravitational wave has passed.

For Roche-lobe overflow, $R_2=R_L$. Cubing the Roche lobe approximation and using the donor's mass-radius relation givesKepler's third law therefore yields the Roche-lobe period-mass relation for a linear donor radius law:The component distances from the centre of mass are $a_1=a{M}_2/M$ and $a_2=a{M}_1/M$. Summing their angular momenta gives the circular-binary orbital angular momentumDuring conservative binary mass transfer, $\dot{M}=0$ and ${\dot{M}}_1=-{\dot{M}}_2$. Logarithmic differentiation at constant angular momentum gives $\dot{a}/a=-2(1-q){\dot{M}}_2/M_2$. Thus the Roche-lobe radius response exponent givesTheir difference is $d\log (R_L/R_2)/dt=(2q-8/3){\dot{M}}_2/M_2$. Since the donor loses mass, this is positive for $q<4/3$: the donor shrinks relative to its Roche lobe, so transfer shuts off without an external driver. The Roche lobe itself expands only for $q<5/6$; for $5/6<q<4/3$ it shrinks more slowly than the donor.

Gravitational-wave emission from a binary system provides such a driver: the orbit emits gravitational waves carrying energy and angular momentum, and the shrinking Roche lobe can maintain contact. Allowing $\dot{J}<0$, the binary mass-transfer contact equation becomesIt gives a negative transfer rate for $q<4/3$.

For the classical nova, assume isotropic re-emission from a binary star: ejecta carry the accretor's specific angular momentum $j_1=a_1^2\Omega$. The eruption lasts many orbital periods, so the orbit responds approximately adiabatically and remains nearly circular; an instantaneous asymmetric kick would be a different model. To first order,Differentiating the circular-binary orbital angular momentum formula during the ejection givesThe accompanying Roche lobe change is thereforeThe donor does not change mass in the eruption and moves inside its expanded Roche lobe: this is nova-induced binary detachment.

Put $\Gamma =-\dot{J}>0$ for the constant external loss. During detachment the component masses are fixed, so the Roche lobe shrinks at fractional rate $2\Gamma /J$. Closing the fractional gap $4\delta m/(3M)$ takes $t_d=2J\delta m/(3M\Gamma )$. During the subsequent semidetached binary phase, the binary mass-transfer contact equation gives $|{\dot{M}}_2|=3M_2\Gamma /[J(4-3q)]$, so accumulating the next eruption's mass takes $t_s=J(4-3q)\delta m/(3M_2\Gamma )$. ConsequentlyBoth times are evaluated to first order in $\delta m/M$, with the same equilibrium donor mass-radius relation; changes of $q,J$ and the transfer rate within one cycle affect higher orders.

General relativity models spacetime by a smooth four-dimensional Lorentzian manifold whose metric tensor governs clock readings, causal cones and free-particle motion. The Einstein equivalence principle motivates local special relativity and universal geodesic free fall for test bodies; it does not by itself select a unique gravitational field equation. We assume a Levi-Civita connection, a metric tensor as the gravitational field with no additional dynamical gravitational variables, and matter with a covariantly conserved stress-energy tensor. Units have $c=1$.

The existence of local inertial frames is a local statement. At any regular event choose an orthonormal frame and Riemann normal coordinates, so that $g_{ab}={\eta }_{ab}$ and ${\Gamma }^a{}_{bc}=0$ at the event, with metric signature $(+---)$. Metric compatibility then gives ${\partial }_c{g}_{ab}=0$ there. The geodesic equation at that point is the special-relativistic inertial equation. Along a timelike geodesic, a parallel-transported orthonormal frame gives Fermi normal coordinates, eliminating the connection along the central worldline. These constructions do not in general eliminate second derivatives of the metric tensor in a neighborhood. The Riemann curvature tensor is built from the derivatives of the connection coefficients as well as their quadratic terms; nonzero Riemann curvature tensor components can remain when all connection coefficients vanish at a chosen point. Conversely, Polar coordinates in Minkowski spacetime can have nonzero connection coefficients but zero Riemann curvature tensor. Connection coefficients alone are therefore not an invariant measure of gravity.

For the tensorial content, consider a two-parameter family of geodesics parametrized by an affine parameter, with tangent $U$ and connecting field $J$. Their coordinate construction gives $[U,J]=0$, and torsion-freeness gives ${\nabla }_{U}J={\nabla }_{J}U$. Since ${\nabla }_{U}U=0$, commuting derivatives in the convention of Question 1 yields geodesic deviation:The left side is covariant relative acceleration, rather than the second coordinate derivative of an arbitrary separation. In an orthonormal frame in free fall with $U=(1,0,0,0)$, its spatial components are $D^2{J}^i/d{\tau }^2=-R^i{}_{00j}{J}^j$. This is a measurable tidal force, and a coordinate transformation cannot make a nonzero Riemann curvature tensor disappear. It is the distinction between gravity at one event and the behavior of an extended freely falling laboratory.

In a static Newtonian limit, $g_{00}=1+2\Phi$ and ${\Gamma }^i{}_{00}={\partial }_i\Phi$, so the test-body acceleration is $-\nabla \Phi$. The curvature-sign convention in Killing derivative identities gives $R^i{}_{00j}\simeq {\partial }_i{\partial }_j\Phi$. Hence relative acceleration is the negative Hessian of the gravitational potential. For the Newtonian potential of a point mass $\Phi =-GM/r$, the Hessian has radial eigenvalue $-2GM/r^3$ and two tangential eigenvalues $GM/r^3$: radial separations stretch and tangential ones compress. Outside the source the Ricci tensor can vanish while the Riemann curvature tensor remains nonzero. Its vacuum tidal part is encoded by the Weyl tensor; vacuum does not mean flatness. Gravitational waves likewise carry Riemann curvature tensor disturbances without a local material source.

To obtain the field equations, make the additional dynamical assumptions explicit: locality and diffeomorphism invariance of general relativity; a symmetric metric tensor field equation using at most second derivatives of the metric tensor; a geometrical left-hand side linear in the Riemann curvature tensor with constant coefficients; and compatibility with local matter conservation. These restrictions exclude higher-curvature or extra-field alternatives. Under them the available rank-two curvature contractions give the ansatzThe contracted Bianchi identity and metric compatibility giveFor this to vanish identically for arbitrary metric tensors, $b=-a/2$. Assuming $a\ne 0$ and absorbing its normalization into the coupling givesThe permitted constant term is the cosmological constant. Its value is not fixed by the equivalence principle. Conservation of $T$ follows for generally covariant matter on its equations of motion: varying the matter action under a compactly supported infinitesimal diffeomorphism, then integrating its term $T^{ab}{\nabla }_a{\xi }_b$ by parts, gives ${\nabla }_a{T}^{ab}=0$. The divergence-free Einstein tensor is consistent with this identity.

The coupling sign must be fixed in the actual curvature-sign convention in Killing derivative identities, with $R_{ab}=R^c{}_{acb}$. For the static weak-field approximation, $R_{00}\simeq -\Delta \Phi$. Set $\Lambda =0$ for the local asymptotically flat Newtonian comparison. In four dimensions the trace equation gives $R=-\kappa T$, so $R_{00}=\kappa (T_{00}-\frac{1}{2}{g}_{00}T)\simeq \kappa \rho /2$ for slowly moving matter. Matching $\Delta \Phi =4\pi G\rho$ forces $\kappa =-8\pi G$. Thus, with this convention,Under the opposite curvature-sign convention in Killing derivative identities the coupling sign reverses, as in Einstein-equation coupling under reversed curvature. This sign statement is compatible with the supplied linearized handout. Local inertial frames remove the connection at an event; tidal Riemann curvature tensor effects remain tensorial, and the field equations require dynamical assumptions beyond the equivalence principle.

Let ${\gamma }_{\alpha \beta }={\delta }_{\alpha \beta }$ be the positive Euclidean spatial background metric. Write a tensor-sector metric perturbation as $h_{00}=h_{0\alpha }=0$, $h_{\alpha \beta }=-2E_{\alpha \beta }$, with ${\gamma }^{\alpha \beta }{E}_{\alpha \beta }=0$ and ${\partial }^{\alpha }{E}_{\alpha \beta }=0$. Spatial indices here are raised with $\gamma$, and $\square ={\partial }_t^2-\Delta$.

Substitute the linearized Levi-Civita connection into the cover's curvature definition. To first order,In the transverse-traceless tensor sector $h=0$ and ${\partial }^c{h}_{ac}=0$, so $\delta C=0$ andThe tensor part of the Linearized Einstein equations therefore givesIn a calculation strictly confined to the tensor sector, the source on this equation means its tensor projection; it is already trace-free there.

For a generic compact source, trace-free is not the same as transverse-traceless. For example, the static compactly supported spatial stress $T^{ij}={\partial }^i{\partial }^j\chi -{\gamma }^{ij}\Delta \chi$, with smooth compactly supported $\chi$, is divergence-free, but its trace-free part has divergence $2{\partial }^j\Delta \chi /3$, generally nonzero. There is a precise way to reproduce the unprojected trace-free formula used in the question while retaining this distinction. In Lorenz gauge in linearized gravity, the trace-reversed metric perturbation ${\bar{h}}_{ab}=h_{ab}-{\eta }_{ab}h/2$ obeysDefine its spatial trace-free half-strain byThen its wave equation has exactly the displayed trace-free source $\delta {\hat{T}}^{\alpha \beta }=\delta T^{\alpha \beta }-{\gamma }^{\alpha \beta }\delta T^{\mu }{}_{\mu ,\mathrm{spatial}}/3$. The physical tensor radiation is obtained afterwards with the transverse-traceless projector. Without that projection, this $E$ is a Lorenz-gauge trace-free coefficient, not necessarily a transverse tensor mode. At leading far-zone order, the same directional projector is applied to both sides of the quadrupole field formula below.

The flat contracted Bianchi identity implies stress-energy conservation, ${\partial }_b\delta T^{ab}=0$. LetCompact support removes boundary terms. Differentiate and integrate by parts once:A second time derivative, another integration by parts and symmetry of the stress-energy tensor giveTake the spatial trace-free part. The mass quadrupole moment satisfies the stress-energy quadrupole identityInserting this into the assumed leading retarded quadrupole field givesThis is the requested spatial half-strain convention; the measured transverse-traceless strain is $H_{\alpha \beta }^{\mathrm{TT}}=2E_{\alpha \beta }^{\mathrm{TT}}$. The formula selects the retarded source field and does not include an independently added homogeneous gravitational wave. Dropping the source-dependent part of retardation also needs the long-wavelength source approximation, besides $R/r\ll 1$.

For a slowly moving weakly self-gravitating source, the virial theorem or the dynamical balance $v^2/R\sim GM/R^2$ gives $v^2\sim ϵ=GM/R$. A varying nonspherical mass quadrupole moment has characteristic size $M{R}^2$ and dynamical time $\tau \sim R/v$, so $\ddot{Q}\sim M{R}^2/{\tau }^2\sim M{v}^2$. Consequently the quadrupole strain scaling of a weak self-gravitating source isThis is a characteristic scale, not a nonzero lower bound: a stationary or exactly spherical source does not radiate. Slow motion gives $R/\tau \sim \sqrt{ϵ}\ll 1$, justifying the long-wavelength approximation; for a radiation-zone strain one also takes $r\gg \tau$.

Use metric compatibility and a torsion-free connection. The coordinate formula for the Lie derivative of the metric tensor isSubstituting ${\partial }_a{g}_{bc}={\Gamma }^d{}_{ab}{g}_{dc}+{\Gamma }^d{}_{ac}{g}_{bd}$ and collecting connection terms gives $({\mathcal{L}}_{\xi }g{)}_{ab}={\nabla }_a{\xi }_b+{\nabla }_b{\xi }_a$. Thus a Killing vector obeys ${\nabla }_{(a}{\xi }_{b)}=0$, and conversely this Killing equation makes its flow preserve the metric. Its contraction gives ${\nabla }_a{\xi }^a=0$.

With the curvature convention fixed in Q1, contract the vector Ricci identity to obtainThe second term vanishes because the Killing divergence is zero. Metric compatibility converts the first term to ${\nabla }^b{\nabla }_a{\xi }_b$, and the Killing equation changes this to $-{\nabla }^b{\nabla }_b{\xi }_a$. Therefore the second covariant derivative of a Killing vector satisfiesThis derivation fixes the sign by the stated Ricci identity rather than by an unspecified convention for curvature.

For the axial Killing vector set $F^{ab}={\nabla }^b{\xi }^a$. It is antisymmetric and its divergence is ${\nabla }_b{F}^{ab}=-R^a{}_b{\xi }^b$. Work in geometrized units $G=c=1$ and with zero cosmological constant, as required by the asymptotically flat setting. If a spatial shell $\Omega$ has oriented boundary $S_2-S_1$, the given form of Stokes theorem applied to the Komar angular momentum givesThe vacuum Einstein field equations give $R_{ab}=0$ everywhere in the shell outside the matter source. The shell can still contain gravitational waves, since those have nonzero Weyl curvature rather than a nonzero Ricci tensor. HenceThis is angular momentum conservation in an axisymmetric vacuum region. Moving the outer surface through the radiative region adds no angular momentum. In an exactly axisymmetric spacetime there is no outward transport of the axial angular momentum by the waves. Axisymmetry does not imply stationarity, so this conclusion does not exclude radiation of energy.

For completeness the Einstein equations express the bulk current asOn an axisymmetric spacelike slice, $\xi$ is tangent to the slice and its contraction with $d^3{\Sigma }_a$ vanishes; there the trace contribution to the angular-momentum volume integral drops out. One must retain any inner surface charges if the integration region has inner boundaries. The vacuum shell argument itself does not require a matter-only representation of the total charge.

The current $Y$ is conserved on arbitrary slices, not just those to which $\xi$ is tangent. Indeed symmetry of the Ricci tensor and antisymmetry of ${\nabla }_a{\xi }_b$ giveThe last equality follows because a metric-preserving Killing flow also preserves the scalar curvature. Equivalently, the stress-energy conservation law and invariance of the stress trace give the same conserved current.

Let two spacelike hypersurfaces terminate on large enclosing spheres $S_1,S_2$ at the same asymptotic end. Join the spheres by a timelike tube lying wholly in the vacuum exterior. On this tube ${\nabla }_b{F}^{ab}=0$, so Stokes theorem gives zero difference of their Komar surface integrals, independently of the tube's shape. Alternatively, use the given Gauss theorem on a spacetime volume between the slices: ${\nabla }_a{Y}^a=0$, and the lateral flux is zero because $Y=0$ in the exterior vacuum tube. Taking the enclosing spheres to infinity with the usual asymptotic falloff therefore provesAll surface orientations in these equations use the same convention as the stated Komar integral. The argument concerns enclosing surfaces at the same asymptotic end, so it does not discard additional charges from other ends or inner boundaries.

Use $G=c=1$, select the retarded source field with no added homogeneous radiation, and work to leading order in the weak-field approximation and long-wavelength source approximation. The retarded fundamental solution of the Linearized Einstein equations in Lorenz gauge in linearized gravity isFor $r\gg d$, the denominator is $r$ to leading order. The delay is $t-r+\mathbf{n}\cdot {\mathbf{x}}^{\prime }+O(d^2/r)$, where $\mathbf{n}=\mathbf{x}/r$. In the usual slowly evolving source regime, its characteristic time $\tau$ obeys $d/\tau \ll 1$, so the source-size part of the delay is negligible:The stress-energy conservation equation ${\partial }_{\mu }{T}^{\mu \nu }=0$ is required at the approximation order used; it follows from the divergence of the gauge-fixed field equation. Compact support removes the integration by parts surface terms. Since $T^{00}=T_{00}$ and $T^{ij}=T_{ij}$ in this signature,Combining these identities gives the retarded quadrupole fieldFor an ordinary nonrelativistic bound source, $\tau \sim d/v$ makes $d/\tau \sim v\ll 1$. More generally the size relative to the variation timescale must also be small; speed alone does not exclude a rapidly varying small-amplitude motion.

The radiation boundary condition for linearized gravity is essential for a statement about the full field. Without it, the displayed implication is false: take $T_{\mu \nu }=0$ and add ${\bar{h}}_{11}=A\cos (k(t-z))$, ${\bar{h}}_{22}=-{\bar{h}}_{11}$, all other components zero. This weak plane gravitational wave satisfies both the homogeneous wave equation and Lorenz gauge in linearized gravity, while every $I_{ij}$ is zero. The proved formula is for the retarded source contribution, with the standard slow-source qualification. Its transverse–traceless projection gives the physical radiative strain.

At fixed masses, the equal-mass circular-binary quadrupole luminosity grows as $R^{-5}$. The binary's Newtonian binding energy is $-G{M}^2/(4R)$, so reducing the separation increases both the binding and the radiated power. Equivalently, its luminosity scaling iswhich makes the importance of orbital compactness explicit.

Ordinary extended stars cannot remain separate at very small orbital radii: contact, mass transfer and tidal disruption intervene. A neutron star or black hole can remain a compact orbiting object down to separations of order a few gravitational radii, allowing high orbital speeds, rapidly changing mass quadrupole moments and strong gravitational waves. Thus compact, tightly bound binaries are especially efficient emitters. The Newtonian quadrupole formula explains the scaling; precision predictions near merger require relativistic dynamics, where that approximation itself ceases to be reliable.

Let $f=0.46$. At Roche-lobe overflow contact, the supplied equilibrium radius relation givesKepler's third law then impliesThus the Roche-lobe period-mass relation for a linear donor radius law isHere $R_{\odot }$ and $M_{\odot }$ denote the solar radius and solar mass. Cancellation of the total mass is the special Roche-lobe-filling period-density relation combined with $R_2\propto M_2$.

Between eruptions the conservative binary mass transfer has $M$ constant and ${\dot{M}}_1=-{\dot{M}}_2$. Neglecting spin, writeIts logarithmic derivative givesDifferentiating the Roche lobe relation consequently givesFor $\dot{J}=0$, the Roche-lobe radius response exponent is ${\zeta }_L=2q-5/3$. If the donor follows its specified equilibrium radius sequence, ${\dot{R}}_2/R_2={\dot{M}}_2/M_2$. Its logarithmic overflow therefore changes bySince ${\dot{M}}_2<0$, $q<4/3$ gives negative feedback in this equilibrium-response model, whereas $q>4/3$ gives increasing overflow. For $q>4/3$, loss of donor mass shrinks its Roche lobe faster than its modeled stellar radius, increasing transfer and hence amplifying the original disturbance. The $q>4/3$ comparison is a formal extension; the initial system is specified to have $q<1$.

The word dynamically needs a qualification. Thermal equilibrium specifies ${\zeta }_{\mathrm{eq}}=1$, but actual dynamical stability of binary mass transfer depends on the adiabatic stellar radius response exponent ${\zeta }_{\mathrm{ad}}$, which need not equal one. Stability after a rapid small mass loss requires ${\zeta }_{\mathrm{ad}}>{\zeta }_L$, because $d\log M_2<0$. The equilibrium relation alone does not supply this adiabatic response. As a formal counterexample to deriving a universal dynamical threshold from it, a radius model with a dimensionless entropy label $S$, $R(M_2,S)=C{M}_2^2{e}^S$, can have equilibrium $S_{\mathrm{eq}}=S_0-\log (M_2/M_{\ast })$, giving $R_{\mathrm{eq}}\propto M_2$ but ${\zeta }_{\mathrm{ad}}=2$ at fixed $S$. At $q=3/2$, ${\zeta }_L=4/3<2$, so that model is dynamically stable despite the extrapolated $q>4/3$ condition. This shows the missing response hypothesis, not a claim that this toy radius law describes the given low-mass star.

A more relevant illustration is an ideal fully convective adiabatic stellar polytrope of index $3/2$. At fixed entropy its polytropic mass-radius relation is $R\propto M_2^{-1/3}$, giving ${\zeta }_{\mathrm{ad}}=-1/3$ and a dynamical critical ratio $q=2/3$ under the same Roche lobe approximation. For example, $q=0.9<4/3$ can be unstable dynamically even though the equilibrium comparison has negative feedback. The requested $4/3$ threshold is the intended result with instantaneous response $R_2\propto M_2$; it is not a universal dynamical criterion. The subsequent cycle calculation uses the supplied equilibrium sequence and presumes a physically stable, sufficiently slow transfer regime.

Two important angular-momentum-loss mechanisms are gravitational-wave emission from a binary system and magnetic braking of a binary star. Gravitational waves directly remove orbital energy and angular momentum. A magnetized stellar wind removes donor spin angular momentum with a large magnetic lever arm; tidal synchronization replenishes that spin from the orbit, so the net effect is orbital angular-momentum loss. Treating the wind mass loss as small compared with the Roche-transfer flow is consistent with the conservative approximation used here; otherwise the mass and angular-momentum equations must both be modified. Neglecting the stored spin does not eliminate the torque transmitted through it.

Such losses shrink the Roche lobe and offset the donor's tendency to move inside it. Maintaining contact on the equilibrium sequence requires ${\dot{R}}_L/R_L={\dot{M}}_2/M_2$, hence the binary mass-transfer contact equation becomesFor $q<4/3$ and $\dot{J}<0$, this drives ${\dot{M}}_2<0$ and maintains the modeled semi-detached binary.

Now take $\delta m>0$ for the mass expelled from the accretor in one classical nova. The white dwarf loses $\delta M_1=-\delta m$, while the donor has $\delta M_2=0$ and the total mass change is $\delta M=-\delta m$. Isotropic ejection in the accretor rest frame supplies no mean kick and carries the accretor's specific orbital angular momentum. Its orbital radius is $a_1=a{M}_2/M$, soThis is the angular-momentum budget for isotropic re-emission from a binary star, without an additional spin or frictional torque on the ejecta. The general logarithmic change of $J=M_1{M}_2\sqrt{Ga/M}$ isUsing $M/M_1=1+q$ gives the first-order nova changesThe donor mass and its assumed equilibrium radius are unchanged during the eruption. It is therefore left inside the enlarged Roche lobe, and Roche-lobe overflow ceases in this sharp-contact model. An extended atmosphere, irradiation-induced donor expansion or additional ejecta torques would require a different treatment.

The stipulated circular orbit is an important idealization. Isotropic loss that is slow compared with the orbital period can preserve circularity. If it is truly impulsive compared with that period, an initially circular point-mass orbit instead acquires eccentricity. With unchanged velocity and position, its new specific orbital energy is ${\epsilon }^{\prime }=GM/(2a)-G(M-\delta m)/a=-G(M-2\delta m)/(2a)$ and its squared specific angular momentum stays $h^2=GMa$. The Kepler orbit energy and eccentricity relations then give $a^{\prime }=a(M-\delta m)/(M-2\delta m)$ and $e^{\prime }=\delta m/(M-\delta m)$. The latter still has the same first-order separation increase. The question assumes circularity; its first-order circular budget is consistent with adiabatic orbital mass loss or with subsequent circularization that changes the separation only at second order. Work symbolically with $\delta m$, since the printed numerical layer mass has no explicit unit.

Finally put ${\mathcal{L}}_J=-\dot{J}>0$, constant over both intervals as specified, and let $J,M,q$ denote their leading-order values over one small cycle. While detached, masses and donor radius are fixed, so ${\dot{R}}_L/R_L=2\dot{J}/J=-2{\mathcal{L}}_J/J$. The fractional gap $4\delta m/(3M)$ closes afterOnce contact resumes, the contact equation gives $-{\dot{M}}_2=3M_2{\mathcal{L}}_J/[J(4-3q)]$. Accumulation of the next layer takesDividing yields the detached-to-semidetached time ratioBoth times are first-order in $\delta m/M$. Corrections to either time are second-order in that small mass fraction, so the displayed ratio has relative corrections of order $\delta m/M$. The positive denominator, negligible eruption duration relative to the cycle, fixed donor radius during detachment and common external loss rate are all essential to this idealized nova-induced binary detachment cycle.

An isolated uncharged collapsing star initially has higher multipole moments and possibly time-dependent motion. The changing exterior emits gravitational waves, carrying away energy and nonspherical structure. Perturbations of the final black hole decay, so the late exterior is expected to approach a stationary spacetime.

Under the regularity, asymptotic flatness, and connected-horizon hypotheses of the black-hole uniqueness theorem, a stationary four-dimensional vacuum black hole is a Kerr black hole. Since the electric charge is zero, its intrinsic parameters areThe black-hole no-hair theorem expresses the loss of independently specifiable higher multipoles: those of the final Kerr black hole are determined by $M,J$. The direction of the rotation axis can be chosen by orienting the coordinates and is not an additional intrinsic parameter of the geometry.

This is a statement about the settled, isolated exterior in classical general relativity, conditional on settling and the hypotheses of the black-hole uniqueness theorem. It does not describe the entire radiating collapse spacetime with only two numbers, nor does it extend unchanged to additional long-range matter fields.

Put $A,B\in \{1,2\}$ and write $h_{AB}=H_{AB}{e}^{ik\cdot x}$, where $H_{11}=H_{+}$, $H_{22}=-H_{+}$ and $H_{12}=H_{21}=H_{\times }$. Physical fields are the real parts. Use $c=1$ here, and take the initial separation at equal time, so $L^0=0$. For propagation in the positive $x^3$ direction one may set $k_{\mu }=(-\omega ,0,0,\omega )$; only $k_0^2={\omega }^2$ enters the answer.

The linearized Levi-Civita connection in this transverse-traceless gauge has the potentially nonzero componentswith all others zero to this order. In particular ${\Gamma }^{\mu }{}_{00}=0$, so a particle initially at rest has fixed spatial coordinates in transverse-traceless gauge, and $t$ equals its proper time. Expanding the Riemann curvature tensor givesThe plus sign in the last expression agrees with $R(T,S)T$ in the geodesic deviation convention above.

For the separation measured in a parallel-propagated orthonormal frame, write $S^{\hat{\mu }}=L^{\hat{\mu }}+{\xi }^{\hat{\mu }}$. The frame removes the connection terms from the time derivative, so the geodesic deviation equation reduces to ${\ddot{\xi }}^{\hat{A}}=R^{\hat{A}}{}_{\hat{0}\hat{0}\hat{B}}{L}^{\hat{B}}={\ddot{h}}_{AB}{L}^{\hat{B}}/2$ at first order. The observable transverse relative acceleration isThese are the plus polarization and cross polarization tidal patterns; there is no longitudinal acceleration. The curvature is evaluated along the reference worldline, and replacing $S$ by $L$ on the right changes only second-order terms. This is the infinitesimal-separation approximation intrinsic to geodesic deviation.

There is an important component distinction if the symbols $L^{\mu }+{\xi }^{\mu }$ are instead interpreted literally in the transverse-traceless gauge coordinate basis. Expanding the left side, rather than replacing a covariant derivative by an ordinary derivative, givesThus in TT coordinatesWith zero initial coordinate displacement and velocity, ${\xi }_{\mathrm{coord}}=0$ throughout. This does not remove the physical gravitational wave strain. To first order an orthonormal coframe in spacetime along the reference worldline is ${\vartheta }^{\hat{A}}=({\delta }^A{}_B+h_{AB}/2)d{x}^B$; its dual frame is parallel propagated because ${\partial }_0({\delta }^A{}_B-h_{AB}/2)+{\Gamma }^A{}_{0B}=0$. Henceafter choosing the initial flat-frame reference. The TT coordinate separation and measured separation distinction accounts exactly for the apparently different accelerations. In a region initially free of the wave, the measured displacement is ${\xi }^{\hat{A}}=h_{AB}{L}^{\hat{B}}/2$ with the corresponding initial conditions.

Use linearization around Minkowski spacetime after constant rescaling of $x,y$: set $A=1+\epsilon a(u)$, $B=1+\epsilon b(u)$, $u=t-z$. To first order,Hence $s=a+b$ is affine in $u$, while $h=a-b$ is the freely propagating profile. The affine transverse trace in a diagonal linearized plane wave is pure gauge: it has zero linearized curvature and is locally a coordinate artifact. To see this without imposing extra boundary conditions, use a first-order linearized coordinate gauge transformation $h_{\mu \nu }\mapsto h_{\mu \nu }-{\partial }_{\mu }{\xi }_{\nu }-{\partial }_{\nu }{\xi }_{\mu }$, withThe $tx,ty,zx,zy$ changes cancel and the remaining time/longitudinal changes are proportional to $s^{\prime \prime }=0$. The transverse perturbation becomeswith all time and longitudinal components zero. It is a transverse-traceless gauge plane gravitational wave travelling in the positive $z$-direction, with plus polarization and no cross polarization in these transverse axes.