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Let the uniform mass density be $\rho =3M/(4\pi a^3)$. In spherical polar coordinates about the rotation axis, the perpendicular distance is $r\sin \theta$ and the volume element is $r^2\sin \theta drd\theta d\varphi$. The moment of inertia is thereforeIn the given rigid body kinetic energy decomposition, $\frac{1}{2}M|\dot{\mathbf{R}}{|}^2$ is the translational kinetic energy of the whole mass moving at its center of mass velocity. The term $\frac{1}{2}I{\omega }^2$ is the rotational kinetic energy about the center of mass. The mixed term vanishes because the mass-weighted relative positions sum to zero.

For rolling without slipping, the point of contact has zero instantaneous velocity relative to the stationary surface. Its rotational velocity is opposite the center of mass velocity and has magnitude $a|\omega |$. Hence $V=a|\omega |$. Substituting this and the moment of inertia of a uniform solid sphere givesThe rotational axis is through the center of mass and parallel to the surface, perpendicular to the direction of rolling.

Write $M=M_1+M_2$, $a=a_1+a_2$, and $\mu =M_1{M}_2/M$, the reduced mass. In the center of mass frame, $a_1=a{M}_2/M$ and $a_2=a{M}_1/M$. Both components of the circular orbit have the same angular velocity $\Omega$. Thus their total orbital angular momentum isUsing Kepler's third law, ${\Omega }^2=GM/a^3$, and the orbital period $P=2\pi /\Omega$, gives the circular-binary orbital angular momentum

Take ${\dot{M}}_1<0$, so ${\dot{M}}_2=-f{\dot{M}}_1$ and $\dot{M}=(1-f){\dot{M}}_1$. A parcel in an isotropic stellar wind has, on average, the donor star's orbital velocity. Its wind velocity relative to the donor star averages to zero, so its mean specific orbital angular momentum about the center of mass is $j_1=a_1^2\Omega$. The escaping mass per unit time is $-(1-f){\dot{M}}_1$. With negligible stellar spin, no additional wind torque, and internal redistribution of the angular momentum of retained matter, donor-wind angular-momentum loss therefore givesIsotropy is in the donor star's frame: it does not make the escaping orbital angular momentum vanish. This is also different from isotropic re-emission from a binary star, where matter escapes from the accretor.

Logarithmically differentiate the expression for $J$ along a slowly evolving sequence of circular orbits:Substitution of the donor-wind angular-momentum loss and mass rates givesFor constant $f$, the last expression is $-3fd\log M_1/dt-3d\log M_2/dt-2d\log M/dt$. Hence the period invariant for constant-fraction donor-wind mass loss isFor a time-dependent $f$, the differential equation still holds, but this integrated power law does not. At $f=1$ it reduces to the period-product invariant for conservative mass transfer; at $f=0$, $M_2$ is constant and $P{M}^2$ is constant, as in Jeans-mode mass loss.

Set the binary mass ratio $q=M_1/M_2$. Since $a^3=GM{P}^2/(4{\pi }^2)$, the Kepler third law and the preceding orbital period derivative implyThe specified approximation to the Roche lobe gives $\log R_L=\log (0.46)+\log a+\frac{1}{3}(\log M_1-\log M)$. Consequently the donor-wind Roche-lobe response isThis is a local Roche-lobe radius response exponent, so it remains valid instantaneously even if $f$ varies.

The stellar radius response exponent of $R_1\propto M_1^{-n}$ is ${\zeta }_{\ast }=-n$. Maintaining Roche-lobe overflow in exact contact requires ${\dot{R}}_1/R_1={\dot{R}}_L/R_L$, yieldingFor a precise feasibility of donor-wind binary contact test, put ${\zeta }_0=\frac{1}{3}-\frac{4q}{3(1+q)}$ and ${\zeta }_1=2q-\frac{5}{3}$. The Roche-lobe radius response exponent is ${\zeta }_L=(1-f){\zeta }_0+f{\zeta }_1$. Thus a permitted contact fraction exists exactly whenUnless $A={\zeta }_1-{\zeta }_0=2q+4q/[3(1+q)]-2$ vanishes, the required fraction is $f=(-n-{\zeta }_0)/A$. At $q=(\sqrt{10}-1)/3$, both limiting Roche-lobe radius response exponents coincide: contact is possible for any $f$ only if $n=5/3-2q$, and for no $f$ otherwise.

The sign of the overfilling change resolves the failure of contact:If ${\zeta }_{\ast }>\max ({\zeta }_0,{\zeta }_1)$, mass loss makes the donor star underfill its Roche lobe: the system detaches and contact-driven transfer stops. If ${\zeta }_{\ast }<\min ({\zeta }_0,{\zeta }_1)$, mass loss increases the overfilling: transfer is destabilized, and rapid transfer or a common envelope may result. Calling this a failure of dynamical stability of binary mass transfer specifically requires $-n$ to be the adiabatic stellar radius response exponent; a thermal or equilibrium response concerns a different timescale. Additional angular momentum losses or intrinsic stellar expansion can change these outcomes by changing the contact equation.

In the center of mass frame, the velocities are ${\mathbf{v}}_1=(M_2/M)\mathbf{v}$ and ${\mathbf{v}}_2=-(M_1/M)\mathbf{v}$. Hence the kinetic energy is $T=\frac{1}{2}\mu v^2$, while the Newtonian gravitational potential energy is $U=-G{M}_1{M}_2/r=-GM\mu /r$. The two-body orbital energy is thereforeThe relative equation of Newtonian gravity is $\ddot{\mathbf{r}}=-GM\mathbf{r}/r^3$. It is a central force, so conservation of angular momentum confines the motion to a plane and preserves the specific angular momentum $h=|\mathbf{r}\times \mathbf{v}|=r^2\dot{\theta }$.

For completeness, derive the polar equation of a Kepler orbit. Let $w=1/r$ and use primes for $\theta$ derivatives. Then $\dot{r}=-h{w}^{\prime }$ and $\ddot{r}-r{\dot{\theta }}^2=-h^2{w}^2(w^{\prime \prime }+w)$. The radial equation gives the Binet equationChoose the angular origin at closest approach. Its solution is $w=(GM/h^2)(1+e\cos \theta )$, soFor the ellipse, the orbital eccentricity satisfies $0\le e<1$. Its extreme separations are $r_p=l/(1+e)$ and $r_a=l/(1-e)$. The semi-major axis is half their sum, giving $a=l/(1-e^2)$ and thus $l=a(1-e^2)$. The true anomaly $\theta$ runs through a full $2\pi$; the endpoints are the same position, and for a circular Kepler orbit the angular origin is arbitrary.

The radial and transverse relative speeds are $\dot{r}=(GM/h)e\sin \theta$ and $r\dot{\theta }=h/r$. Substituting these into the specific orbital energy givesTherefore the conserved orbital energy isThe negative specific orbital energy and fixed angular momentum characterize the bound Kepler orbit; neither should be assumed unchanged through a mass-ejecting explosion.

Initially the circular Kepler orbit has $r=a$ and $v^2=GM/a$. Treat the supernova kick in a binary star as impulsive: the relative position does not change, the companion's velocity is unchanged during the impulse, and the new neutron star receives $\mathbf{u}$ with $|\mathbf{u}|=\alpha v$. Thus ${\mathbf{v}}^{\prime }=\mathbf{v}+\mathbf{u}$, and with $\psi$ the angle between these vectors before the kick,Write $M^{\prime }=M_1^{\prime }+M_2$, $\beta =M^{\prime }/M$ and ${\mu }^{\prime }=M_1^{\prime }{M}_2/M^{\prime }$. The post-explosion specific orbital energy isUsing ${\epsilon }^{\prime }=-G{M}^{\prime }/(2a^{\prime })$ givesHere $a^{\prime }>0$ for a bound Kepler orbit. For a hyperbolic Kepler orbit, this formula uses the signed energy parameter $a^{\prime }<0$, rather than the positive geometric magnitude of the hyperbola's semi-major axis. At the parabolic Kepler orbit boundary, $1/a^{\prime }=0$.

The kick-direction binary survival criterion is bound if $S<2\beta$ and unbound with positive asymptotic speed if $S>2\beta$. To prove the printed sufficient disruption condition, choose a perpendicular kick, $\cos \psi =0$: then $S=1+{\alpha }^2$, so $M^{\prime }<\frac{1}{2}(1+{\alpha }^2)M$ produces positive specific orbital energy. This is sufficient, not the sharp existence threshold. Sincethe complete kick-direction binary survival criterion isThe second inequality follows by taking a kick directly opposite to $\mathbf{v}$. All directions remain bound if $2\beta >(1+\alpha {)}^2$, while every direction has positive escape energy if $2\beta <(1-\alpha {)}^2$. For $\alpha >0$, the marginal direction obeys $\cos \psi =(2\beta -1-{\alpha }^2)/(2\alpha )$ when the right-hand side lies in $[-1,1]$. Equalities give marginal parabolic Kepler orbits for the relevant extreme direction. With zero kick, losing more than half the original total mass unbinds the circular Kepler orbit.

For an escaping pair, the Newtonian gravitational potential energy approaches zero at infinite separation. Conservation of the post-explosion two-body orbital energy then gives $\frac{1}{2}{\mu }^{\prime }{V}^2={\mu }^{\prime }{\epsilon }^{\prime }$, or the asymptotic relative speed of a disrupted binaryThis is the relative recession speed, not the velocity of the new center of mass or either star's individual velocity in the original inertial frame. A marginal parabolic Kepler orbit separates with $V=0$ at infinity.

Use a circular orbit, as in the stated Roche lobe approximation, with orbital angular speed $\Omega$ and negligible stellar spin. The center of mass distances are $a_1=a{M}_2/M$ and $a_2=a{M}_1/M$. Summing the two orbital angular momenta givesKepler's third law, ${\Omega }^2{a}^3=GM$, makes this $J=M_1{M}_2\sqrt{Ga/M}$. In conservative mass transfer, both $M$ and $J$ are fixed, while $d{M}_2=-d{M}_1$. ThusThe Roche-lobe radius response exponent is consequentlyDuring a dynamical mass-loss perturbation, the deep interior and luminosity do not have time to change. The supplied giant structure therefore has stellar radius response exponent ${\zeta }_{\ast }=-0.27$. Its fractional overfill changes by $d\ln (R/R_L)=({\zeta }_{\ast }-{\zeta }_L)d\ln M_1$. The binary mass ratio uses donor mass divided by accretor mass. Since $d\ln M_1<0$, self-limiting Roche-lobe overflow requires ${\zeta }_{\ast }>{\zeta }_L$. Hence the dynamical stability condition isAt equality the linear restoring response vanishes. This conservative mass-transfer critical mass ratio uses the given radius exponent, rather than imposing the different fully convective $-1/3$ approximation.

The initially more massive component normally evolves first. If it first overflowed only after acquiring the given giant response, it would have $q>1$ and the mass loss would be dynamically unstable: its radius grows while its Roche lobe initially contracts. Starting overflow earlier can avoid this outcome. A main sequence or early post-main-sequence donor star can have a radiative envelope with a stabilizing contraction response. Transfer then reverses the binary mass ratio before the donor develops the giant structure. This is the route to an Algol binary through Case A mass transfer during core hydrogen burning, or early case B mass transfer after core hydrogen exhaustion but before a deep giant envelope develops.

For the initial Roche-lobe overflow to occur before the base of the giant branch, its lobe must be smaller than the base-of-branch stellar radius. Combining $R_L=0.462a(M_1/M{)}^{1/3}$ with Kepler's third law cancels the companion dependence:Writing $m_1=M_1/M_{\odot }$, substitute the mass-radius relation to obtain the pre-giant period limit:We use the specified $0.8\mathrm{days}$ below. This is a pre-giant Roche-lobe-filling period limit, not a condition on the current stripped donor's radius.

At fixed total mass, Kepler's third law gives $P\propto a^{3/2}$, while fixed orbital angular momentum gives $a\propto (M_1{M}_2{)}^{-2}$. Therefore the conservative period invariant isTo find the smallest possible pre-giant period ratio, use solar-unit masses $m_1,m_2$ and fixed $m=m_1+m_2$. Let $K=P(m_1{m}_2{)}^3$. Along this period-product invariant for conservative mass transfer,The denominator has logarithmic derivative $5/m_1-3/(m-m_1)$, with a strictly negative derivative of its own. Its unique maximum, hence the unique minimum of $F$, occurs atSince $F$ diverges at either endpoint, this is the global minimum period-to-donor-mass ratio for conservative evolution.

For the quoted Algol binary, $K=3(1\times 3{)}^3=81\mathrm{days}$ and $m=4$. Choosing the minimizing progenitor gives $m_{1,i}=2.5$, $m_{2,i}=1.5$ andIts pre-giant upper limit is $0.8(2.5{)}^2=5\mathrm{days}$, comfortably above this initial period. Subsequent conservative mass transfer of $1.5M_{\odot }$ produces the stated masses and period. Thus the data permit the proposed early-overflow formation route.

For RT Lac, instead $K=5(0.5\times 1.5{)}^3=2.109375\mathrm{days}$ and $m=2$. At its most favorable conservative progenitor, $m_{1,i}=1.25$, $m_{2,i}=0.75$ andNo conservative progenitor meets the pre-giant overflow condition. An initially more massive giant donor also fails the dynamical-stability condition derived above. Within this stable Algol-type formation model, the current RT Lac system therefore requires nonconservative mass transfer.

A plausible history is substantial envelope loss from the system, through a stellar wind or escaping overflow, possibly with a common envelope episode. Escaping matter also carries orbital angular momentum, so the total-mass and period-product invariants no longer constrain its initial orbit. A more massive progenitor can then shed its envelope and reach the present low donor mass without requiring the excluded conservative sequence. The supplied final data do not determine a unique mass-loss or angular-momentum-loss history.

For constant component masses, the center of mass frame has ${\mathbf{r}}_1=(M_2/M)\mathbf{r}$ and ${\mathbf{r}}_2=-(M_1/M)\mathbf{r}$. Newton's law of universal gravitation gives the relative equation $\ddot{\mathbf{r}}=-GM\mathbf{r}/r^3$. Substitution in the component angular momenta yieldsHere $\mu =M_1{M}_2/M$ is the reduced mass, while $\mathbf{h}$ is specific angular momentum.

The printed total-energy expression lacks a factor of $\mu$ in its kinetic term. With the stated physical separation coordinate, the dimensionally consistent total two-body orbital energy isThe second expression is specific orbital energy; these two conventions must not be mixed. Its conservation follows directly:For the eccentricity vector, differentiate $GM\mathbf{e}=\dot{\mathbf{r}}\times \mathbf{h}-GM\hat{\mathbf{r}}$. Since $\mathbf{h}$ is constant,The cancellation is the vector triple product identity. Thus all three corrected Kepler integrals are conserved. Dotting the eccentricity relation with $\mathbf{r}$ also gives the useful orbit identity

For ballistic streamline focusing by a moving star, work in the star's rest frame and take the incoming gas to have velocity $v{\mathbf{e}}_x$, with downstream behind the star at $x>0$. A streamline with impact parameter $d>0$ approaches from $(-\infty ,d,0)$. Its upstream specific angular momentum and eccentricity vector areAt its downstream axis crossing, $\mathbf{r}=b{\mathbf{e}}_x$. The orbit identity gives $b=d^2{v}^2/(GM)-b$, hence the collision distance isAt that point, $h_z=b{u}_y=-dv$, while the $y$ component of $GM\mathbf{e}=\mathbf{u}\times \mathbf{h}-GM\hat{\mathbf{r}}$ gives $dv{u}_x=d{v}^2$. ThereforeA symmetric streamline with opposite impact parameter has the opposite transverse velocity. Their shock wave removes the opposing transverse motion while preserving the common downstream component. Thus the remaining velocity is $v{\mathbf{e}}_x$ in the star frame. In the frame where the undisturbed medium is stationary, adding the star's velocity gives zero velocity immediately after this idealized collision.

After transverse kinetic energy is dissipated, the remaining specific orbital energy is ${\epsilon }_{\mathrm{after}}=v^2/2-GM/b$. It is negative if $b<2GM/v^2$, or equivalently if the upstream impact parameter satisfies $d<d_{\mathrm{cap}}=2GM/v^2$. This is the post-shock capture criterion in ballistic accretion; bound axial streams can return to the star. The captured incident mass flux through the corresponding disk gives the ballistic Bondi--Hoyle--Lyttleton accretion rate:This is the pressureless, dissipative capture model, with the shock's transverse energy unavailable to unbind the wake.

For wind mass transfer in a binary star, steady isotropic donor mass loss gives the local wind mass density at the companion:A fast wind has $v_w\gg 2\pi a/P$ and a capture scale small compared with $a$, so its relative incident speed is $v_w$ to leading order. Apply the same Bondi--Hoyle--Lyttleton accretion rate with accretor mass $M_2$ to obtainFinally, Kepler's third law eliminates the unquoted separation, $a=[G(M_1+M_2)P^2/(4{\pi }^2){]}^{1/3}$, giving the rate in terms of the stated orbital period:Equivalently, the fast-wind accretion fraction in a circular binary is $[M_2/(M_1+M_2){]}^2(v_{\mathrm{orb}}/v_w{)}^4$, with $v_{\mathrm{orb}}=2\pi a/P$ the relative circular orbital speed. Its smallness is consistent with using an almost undisturbed isotropic stellar wind.

Let $\mathbf{r}$ be position relative to the center of mass, and let the new origin have vector $\mathbf{c}$ relative to it. The inertia tensor about this origin isExpanding and using $\int \mathbf{r}dm=0$ proves the tensor form of the parallel axis theorem:For a uniform cube of side $L$ and mass $M$, centered coordinate integrals give $I_0=(M{L}^2/6)I$. The displacement to a vertex is $(L/2,L/2,L/2)$ up to signs. Using axes along its incident edges givesThe principal moments of inertia are $\boxed{M{L}^2/6,11M{L}^2/12,11M{L}^2/12}$. The first principal axis is the body diagonal through the vertex; every direction perpendicular to it has the repeated moment.

First derive the needed gamma-matrix traces directly from the Clifford algebra. The cyclic property of the trace givesso $\mathrm{tr}({\gamma }^{\mu }{\gamma }^{\nu })=4{\eta }^{\mu \nu }$. For $T=\mathrm{tr}({\gamma }^{\mu }{\gamma }^{\nu }{\gamma }^{\rho }{\gamma }^{\sigma })$, move ${\gamma }^{\mu }$ successively through the other three matrices and cycle the final term back to the front:ThereforeThese are all the trace identities required.

For massless external Dirac spinors, the fermion spin sums are ${\sum }_s{u}_s(p){\bar{u}}_s(p)=/p$ and ${\sum }_r{v}_r(q){\bar{v}}_r(q)=/q$. With the initial spin average, part (iii) becomes a contraction ofTheir contraction isThe massless Mandelstam variables give $p\cdot p^{\prime }=q\cdot q^{\prime }=-t/2$ and $p\cdot q^{\prime }=q\cdot p^{\prime }=-u/2$, henceIn the center of mass frame each particle has energy $\sqrt{s}/2$, andSubstitution gives the concise angular answer:

Use the Lorentz-invariant phase-space measure from the original PDF, including the $1/(2E_{p^{\prime }})$ and $1/(2E_{q^{\prime }})$ factors missing from the local TeX. In the center of mass frame put $E=\sqrt{s}/2$ and $k=\sqrt{s-4m^2}/2$, assuming $s>4m^2$. The spatial Dirac delta distribution sets ${\mathbf{q}}^{\prime }=-{\mathbf{p}}^{\prime }$, leavingSince $d(2E_{k^{\prime }})/d{k}^{\prime }=2k^{\prime }/E_{k^{\prime }}$, the radial integral givesThe invariant flux factor is $2\sqrt{\kappa (s,m^2,m^2)}=2\sqrt{s(s-4m^2)}$, using the Källén function. Dividing the phase space by this flux yields $d\sigma /d\Omega =|\mathcal{M}{|}^2/(64{\pi }^{2}s)$, with the final-state labels retained as in the printed formula.

Let $x=\cos \theta$. Of the Mandelstam variables, $t$ isThe azimuthal integral contributes $2\pi$, and the angular endpoints are thereforeChanging variables gives the requested expression from the formula supplied in the paper:There is a normalization qualification: the displayed starting formula integrates over labeled final momenta and contains no $1/2!$. For the physical cross-section of two indistinguishable outgoing quanta of this real scalar field, the identical-particle factor in a final-state phase-space integral divides the full integral by two. With that convention $X=1/(32\pi )$; equivalently, integrate only one representative of each exchanged pair. The boxed value $1/(16\pi )$ follows the given formula exactly.

A Lagrange top is a rigid body that is symmetric about a principal axis, has a point on that axis fixed in space, and has its center of mass on the same axis while gravity acts uniformly. Here $I_1=I_2$ is the transverse principal moment of inertia about the fixed point, $I_3$ is the moment about the symmetry axis, $M$ is the total mass, and $l$ is the distance from the fixed point to the center of mass.

The Euler angles for a symmetric top use $\theta$ for inclination, $\varphi$ for precession, and $\psi$ for spin about the body axis. Since $\psi$ is a cyclic coordinate, its generalized momentumis conserved. The coordinate $\varphi$ is also cyclic, sois a second integral. Finally, the Lagrangian has no explicit time dependence, and conservation of energy from time-translation invariance gives the independent integral

For steady precession set $\theta$ constant and $\dot{\varphi }=\Omega$ constant. The $\theta$ Euler-Lagrange equation, with $0<\theta <\pi /2$, reduces after division by $\sin \theta$ toThis quadratic has a real precession rate precisely when its quadratic discriminant is nonnegative. Hence Steady precession of a Lagrange top is possible if and only if

Put $z=x_3$. The body is a solid of revolution with disk radiusIts volume and uniform mass density are

By the continuous center of mass formula, the distance of the center of mass from the pointed origin is

For a disk of radius $R$, $\int r^{2}dA=\pi R^4/2$. Therefore the axial moment isFor the transverse $x_1$-axis, $\int x_2^{2}dA=\pi R^4/4$, while every slice also contributes $z^2$ times its mass. HenceBy axial symmetry $I_2=I_1$. These values agree with the moment of inertia of a pointed cubic-profile solid of revolution.

Write the position of a mass element relative to the center of mass as $\mathbf{r}$, so its inertial-frame velocity is $\dot{\mathbf{X}}+\mathbf{\omega }\times \mathbf{r}$. The center-of-mass identity $\int \mathbf{r}dm=0$ removes all cross terms. Define the inertia tensor about the center of mass byThe total angular momentum about the origin and kinetic energy are then

Let $\rho (r)=C{r}^{-1}\sin (\pi r/R)$. Spherical symmetry makes $I=I_{0}1$. The mass normalization isso $C=M/(4R^2)$. A moment about any diameter isTherefore

Let $R$ be the center of mass and $g$ the uniform gravitational acceleration. The gravitational torque about $R$ is

Choose body-fixed principal axes of inertia. The inertial derivative of the angular momentum isFor torque-free motion $N=0$ and $L_i=I_i{\omega }_i$, giving the Euler equations for a torque-free rigid body

For a symmetric top, let $I_1=I_2=I$ and let $I_3$ be the axial moment. Then ${\omega }_3$ is constant andThus ${\omega }_1^2+{\omega }_2^2$ is constant and the transverse angular-velocity vector rotates uniformly about the body symmetry axis. Its rate is $|\Omega |$; it precesses in the positive axial sense when $(I_3-I){\omega }_3>0$ and in the opposite sense when this product is negative.