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Kinetic energy is the energy associated with motion. For a nonrelativistic point mass of mass $m>0$, measured in a specified inertial frame,Here $\mathbf{v}$ is velocity and $v$ is speed; $T$ is another common symbol for kinetic energy. The dot product makes $K$ nonnegative regardless of the direction of velocity. At fixed mass, doubling speed quadruples kinetic energy: a $2\mathrm{kg}$ point mass at $2m/s$ has $4\mathrm{J}$, whereas at $4m/s$ it has $16\mathrm{J}$. The SI units are $\mathrm{kg}{\mathrm{m}}^2/{\mathrm{s}}^2=\mathrm{J}$. Different reference frames can assign different velocities and therefore different kinetic energies. ^[ref]

A short outline explains why kinetic energy has this formula. In a nonrelativistic inertial frame, constant mass and Newton's second law give, using the chain rule,Integrating over time proves the work-energy theorem, $W_{\mathrm{net}}=K_f-K_i$. The net force is the sum of all forces, and $W_{\mathrm{net}}$ is their total work, not merely the work of a chosen driving force. This route to final speed is useful when the forces as functions of position are known but the full motion as a function of time would be harder to find. ^[ref]

In the running example, a $2\mathrm{kg}$ point mass starts at rest and receives a constant net $3\mathrm{N}$ force through $4\mathrm{m}$. Its net work is $12\mathrm{J}$, soAt any intermediate distance $x\ge 0$, $K(x)=(3\mathrm{N})x$ and $v(x)=\sqrt{2Fx/m}$: kinetic energy is a linear function of distance, while speed is proportional to the square root of distance. If a net opposing $3\mathrm{N}$ braking force subsequently acts for $4\mathrm{m}$, it does $-12\mathrm{J}$ of work and brings this point mass to rest. Negative net work decreases kinetic energy; it does not make kinetic energy negative. Beyond the stopping point the motion must be reconsidered, since the point mass cannot keep moving forward with a negative $K$.

Kinetic energy also makes the exchange with potential energy intuitive: a falling body's loss of gravitational potential energy can appear as increased kinetic energy. The formula above applies when speed is small compared with the speed of light; at higher speeds, special relativity changes the expression. ^[ref]

A description of how mass is assigned to positions, expressed for example by point masses, a mass density, or a measure. Writing $dm$ allows centre of mass and moment of inertia formulas to cover both discrete and continuous cases.

The moment of inertia about a specified axis measures how a mass distribution resists changes of angular velocity about that axis. For point masses $m_i$ at perpendicular distances $r_{\perp i}$, or a continuous distribution with mass density $\rho$,Its SI unit is $\mathrm{kg}{\mathrm{m}}^2$. The distance is to the entire axis, not to one point on it: moving a point mass along the axis leaves its contribution unchanged. The moment of inertia depends on both the distribution and the chosen axis; it is not determined by total mass alone. About an axis fixed in an inertial frame, a rigid body with larger moment of inertia has smaller angular acceleration under the same net axial torque, and greater rotational kinetic energy at the same angular speed.^[ref]

For a simple model, put four equal point masses of total mass $M$ on light rigid spokes at radius $a$. Their moment of inertia is $M{a}^2$. Doubling $a$ quadruples the moment of inertia without changing $M$. The illustrated placements $a=R/3$ and $a=R$ give $M{R}^2/9$ and $M{R}^2$. Actual spokes and finite-sized weights contribute their own moments of inertia; the point mass model neglects them. Repositioning weights during rotation also breaks the fixed-distribution assumption, so it requires a separate work and angular momentum calculation.

The moment of inertia of a uniform disc follows by dividing it into concentric annuli: $dm=(2M/R^2)rdr$ and $I={\int }_0^R{r}^2(2M/R^2)rdr=M{R}^2/2$. For an ideal thin ring every mass element is at $R$, so $I=M{R}^2$. A ring of finite radial thickness instead uses the moment of inertia of a uniform annulus, $I=M(R^2+b^2)/2$, with inner radius $b$. A moment of inertia is thus a compact summary of radial placement rather than a complete description of shape. For several joined pieces, their moments of inertia add only when each is calculated about the same axis.^[ref]

A physical pendulum makes the moment of inertia measurable through timing. Martín Monteiro, Cecilia Stari, Cecilia Cabeza and Arturo C. Martí used a metal bar with suspension holes $1.0\mathrm{cm}$ apart and an attached smartphone. They report bar length $1.199\mathrm{m}$, bar mass $0.2518\mathrm{kg}$ and phone mass $0.1311\mathrm{kg}$. Changing the suspension hole changes the pivot axis. Their readouts were phone gyroscope angular velocity records, analyzed by autocorrelation or a fit to a sinusoid, and an independent photogate; the logged records for the sinusoid fits lasted $10\mathrm{s}$. They observed agreement between the three timing methods and the geometry-based model, including a minimum of oscillation period as the suspension point varied. These are reported apparatus details and observations, not measurements made for this article.^[ref]

To infer a moment of inertia from such a measurement, determine the total mass $M_{\mathrm{tot}}$ including every attached part, locate the assembly's centre of mass $G$, and measure its distance $h>0$ below the horizontal pivot $O$. Record successive crossings in the same direction, or the elapsed time $t_N$ for $N$ complete oscillations, giving oscillation period $P=t_N/N$. The physical pendulum model with a fixed pivot, a uniform gravitational field, negligible friction and the small-angle approximation givesThe crossing times, masses and geometric distances are the measured inputs; $I_{O,\mathrm{inferred}}$ is an inference conditional on this model, while an integral over the assumed shape supplies an independent model prediction. The parallel axis theorem converts the inferred pivot moment of inertia to one about a parallel axis through $G$. The controls and interpretation of that comparison are described there. An axial measurement determines one scalar moment of inertia, not the complete inertia tensor needed for general rotation.

For a nonrelativistic point mass of fixed mass $m>0$, measured in an inertial frame, its momentum isMomentum is a vector pointing along the velocity, with International System of Units unit $\mathrm{kg}m/s$. Doubling either mass or velocity doubles momentum. In one dimension, a chosen positive direction makes $p=mv$ a signed quantity: a $2\mathrm{kg}$ body moving right at $3m/s$ has $p=+6\mathrm{kg}m/s$; moving left at the same speed gives $p=-6\mathrm{kg}m/s$. ^[ref]

For a collection, total momentum is the vector sum $\mathbf{P}={\sum }_i{m}_i{\mathbf{v}}_i$. Two equal masses moving with opposite velocities can have zero total momentum while both have positive kinetic energy. Direction is therefore essential when using momentum conservation; adding speeds instead of signed velocities gives the wrong balance.

Newton's second law connects the net force on one fixed-mass point mass to its momentum through ${\mathbf{F}}_{\mathrm{net}}=d\mathbf{p}/dt$. Integrating that change gives impulse. Here $\mathbf{p}$ denotes mechanical momentum; when canonical momentum differs from mass times velocity, the distinction in mechanical momentum below applies. The nonrelativistic formula also has a limited regime: special relativity is needed at speeds comparable with the speed of light.

Momentum conservation states that total momentum stays constant when the net force exerted from outside the chosen collection is zero. Use a closed collection, meaning the same fixed-mass point masses are followed throughout, and an inertial frame. ThenTo see why, sum Newton's second law over the members. If ${\mathbf{F}}_{ij}$ is the force on member $i$ from member $j$, Newton's third law assumes ${\mathbf{F}}_{ij}=-{\mathbf{F}}_{ji}$. Each pair cancels in the total:The internal forces act on different bodies and can change their individual momenta; their contributions cancel only in the collection's sum. An isolated system has no external interaction, but balanced external forces also suffice for momentum conservation. A closed system merely excluding matter exchange need not have conserved momentum. ^[ref]

The usefulness is that the detailed internal forces during a collision need not be known. Both an elastic collision and a perfectly inelastic collision obey the same isolated momentum balance; kinetic energy conservation is an additional condition for the elastic collision. If the external impulse is small but nonzero, the total momentum change is correspondingly small. For a variable-mass system exchanging matter, the entering or leaving momentum must also be included.

In a translation-invariant field theory, integrating over the spacetime zero mode enforces momentum conservation at interactions and scattering amplitudes.

The Newtonian gravitational field is the gravitational force per unit mass of a test particle. For a point mass $M$ at the origin,The test particle's mass cancels: different small masses have the same gravitational acceleration at the same position. In SI, the Newtonian gravitational field has units $\mathrm{N}\mathrm{k}{\mathrm{g}}^{-1}=\mathrm{m}{\mathrm{s}}^{-2}$. The same exterior expression holds for a spherically symmetric source, by the spherical shell theorem.

Choose a reference radius $r_0>0$, set $g_0=GM/r_0^2$, and use nondimensionalization with $z=r/r_0$. Then $g/g_0=z^{-2}$: at $z=1,2,4$, the magnitudes are $1,1/4,1/16$. This makes the spatial variation visible without specifying a planet. ^[ref]

The Newtonian gravitational potential $\Phi$ is potential energy per unit mass of a test particle. Its gradient determines the Newtonian gravitational field:In SI, $U$ has units $\mathrm{J}$, while $\Phi$ has units $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}={\mathrm{m}}^2{\mathrm{s}}^{-2}$; its gradient has units of acceleration, not force. Adding a constant to $\Phi$ leaves the Newtonian gravitational field unchanged. For an isolated point mass, the convenient reference is $\Phi (\infty )=0$. ^[ref]

Increasing the radius from $r_0$ to $r_1>r_0$ raises the Newtonian gravitational potential energy:An external agent supplies this positive work if the initial and final kinetic energies agree; gravity does work $-\Delta U$. For $r_1=2r_0$, the required work is $GMm/(2r_0)$. ^[ref]

In an inertial frame, a body with zero net external force remains at rest or moves with constant velocity. For a point mass of constant positive mass, write the net external force as the vector sum of all external forces on that body:Here ${\mathbf{r}}_0$ and ${\mathbf{v}}_0$ are the position and velocity at time $t_0$. The zero net external force must hold throughout the interval, not merely at one instant. Both the magnitude and direction of the velocity stay fixed: constant speed along a curved trajectory still involves acceleration. Rest is the case ${\mathbf{v}}_0=0$. For an extended closed system, the analogous statement concerns its center of mass; zero net external force need not prevent its parts from moving relative to one another.

The short mathematical argument is to use Newton's second law with $m>0$, then take the integral of $d\mathbf{v}/dt=0$ and of $d\mathbf{r}/dt={\mathbf{v}}_0$. This gives the two constants fixed by the initial position and velocity. The physical content also includes the existence of inertial frames in which these relations apply; an arbitrary accelerating reference frame need not show constant velocity for a body with zero actual net external force.

The useful question is therefore which forces change the motion, rather than which force keeps it going. A free-body diagram may show several nonzero forces whose vector sum is zero. Equal opposing forces on the same body can balance; a third-law interaction pair acts on different bodies and cannot be cancelled within that one body's diagram.

In 1687, Isaac Newton stated Law I in Philosophiae Naturalis Principia Mathematica: a body's rest or uniform straight motion persists unless impressed forces compel a change. The following paragraph identifies air resistance and gravity as influences on the trajectory of a thrown body. The vector equations above are a modern expression of the law, not a quotation of his notation.^[ref]

Newton's law of universal gravitation gives the attractive force between two point masses. With source mass $M$ at the origin and relative position $\mathbf{r}$ of mass $m$, write $r=|\mathbf{r}|>0$:Here $G$ is the gravitational constant. The minus sign points toward $M$; the magnitude follows an inverse-square law, so doubling the separation quarters the force. The opposite force on $M$ satisfies Newton's third law. ^[ref]

Both masses accelerate. Applying Newton's second law to each in the isolated two-body problem givesThe relative standard gravitational parameter includes both masses; replacing it by $GM$ requires the test particle approximation $m\ll M$.

Historically, Newton's Book III, Proposition VII argues that gravitational attraction applies to all bodies and scales with their masses; its second corollary states the inverse-square law for individual particles. ^[ref]

In an inertial frame, a point mass of constant mass $m>0$ obeysThe net force is the sum of all forces acting on that body; momentum, velocity and acceleration refer to the same reference frame. A force changes velocity, rather than sustaining it: zero net force implies constant velocity, including a nonzero one. Doubling the net force at fixed mass doubles acceleration; doubling mass at fixed net force halves it. ^[ref]

For example, an otherwise balanced block of mass $0.5\mathrm{kg}$ receives a constant horizontal net force $2\mathrm{N}$ for $0.5\mathrm{s}$. Its acceleration is $4m/{\mathrm{s}}^2$, so its velocity changes by $2m/s$. Equivalently, the impulse is $F\Delta t=1\mathrm{N}\mathrm{s}$, and $\Delta v=\Delta p/m$. Starting from rest, its displacement during the pulse is $a(\Delta t{)}^2/2=0.5\mathrm{m}$. After the pulse, zero net force leaves it moving at constant velocity.

Applying the law therefore requires both a specification of the net force and initial conditions. In the harmonic oscillator example, the force is $-kx$, so Newton's second law becomes a differential equation for position. In an accelerating reference frame, the measured relative acceleration includes a contribution from the frame's own motion; the equation of motion in a rotating frame illustrates the extra terms required.

Constant mass is essential to the displayed $m\mathbf{a}$ form. A variable-mass system exchanging matter also exchanges momentum; its balance includes the entering or leaving matter's velocity. Simply differentiating $m\mathbf{v}$ and equating it to external force on that open body can omit this flux. The existing variable-mass system equation specifies the needed correction.

In the 1687 Principia, Newton's second law was stated in terms of change of motion proportional to impressed force, directed along that force. Its original statement is verbal; the derivative equation above is modern notation. ^[ref]

A point mass constrained to the fixed graph $y=y(x)$ has velocity $(\dot{x},y^{\prime }\dot{x})$. Under uniform downward gravity, its Lagrangian is $L=m(1+y^{\prime 2}){\dot{x}}^2/2-mgy$, so its canonical momentum is $p_x=m(1+y^{\prime 2})\dot{x}$, not the horizontal mechanical momentum. The Euler-Lagrange equation gives the equation in the title, and the conserved energy is $m(1+y^{\prime 2}){\dot{x}}^2/2+mgy$. A straight segment with constant slope $s$ has constant horizontal acceleration $-gs/(1+s^2)$; zero curvature does not imply zero acceleration. A cart held to a track satisfies this mechanical constraint only while the assumed contact or retaining mechanism remains effective.

Each simple pendulum moves in the vertical plane containing the pivots. Ignoring contact between the ideal rods and point masses, each rod can turn through a full circle. Thus the configuration space is the torusChoose generalized coordinates ${\theta }_1,{\theta }_2$, measured from the downward vertical, positive towards the right and understood modulo $2\pi$. These angles are local coordinates on the torus, rather than a single globally one-to-one real coordinate chart. With upward Cartesian coordinate $y$, the bob positions areTheir velocities satisfy $|{\dot{\mathbf{r}}}_j{|}^2=l^2{\dot{\theta }}_j^2$, so kinetic energy from Cartesian position vectors givesThe spring length is the Euclidean distanceTaking gravitational potential energy zero at pivot height gives $V_g=-mgl(\cos {\theta }_1+\cos {\theta }_2)$. The spring potential energy is $V_s=k(d-r{)}^2/2$, since $r$ is its natural length. Consequently the Lagrangian isIn the paper's notation $t({\theta }_1,{\theta }_2)=V_s$; this function is distinct from the time variable. This is the exact Lagrangian of two spring-coupled pendulums.

One alternative Lagrangian isIt corresponds to setting gravitational potential energy zero at the downward position. The constant disappears from every Euler-Lagrange equation. More generally, total-time-derivative invariance of a Lagrangian allows $\tilde{L}=L+dF({\theta }_1,{\theta }_2,t)/dt$ for any smooth $F$.

No. A motionless isolated point mass produces a static gravitational field. All its mass multipole moments are time independent, so there is no outgoing gravitational radiation. In the quadrupole formula, $h_{ij}^{\mathrm{TT}}\propto d^2{Q}_{ij}^{\mathrm{TT}}/d{t}^2=0$.

The three nearly horizontal segments have lengths $2a,a,a$. To first order in small transverse slopes, the upward force on the first point mass is $T[(y_2-y_1)/a-y_1/(2a)]$. The force on the second is $T[(y_1-y_2)/a-y_2/a]$. Applying Newton's second law to masses $m,2m$ therefore givesThe approximation treats the string as massless, retains the imposed equilibrium tension and neglects gravity, as stipulated.

In cylindrical coordinates the particle's radial and vertical equations are $\ddot{r}-r{\dot{\varphi }}^2=-{\Phi }_r$ and $\ddot{z}=-{\Phi }_z$. Axisymmetry gives conservation of angular momentum, with specific angular momentum $h=r^2\dot{\varphi }$. Eliminating the cyclic azimuthal motion therefore givesThus $\boxed{{\Phi }_1=\Phi +h^2/(2r^2)}$ governs the meridional motion. Its conserved energy is $\frac{1}{2}({\dot{r}}^2+{\dot{z}}^2)+{\Phi }_1$. The additional term is the centrifugal barrier appropriate to fixed angular momentum.

For a point mass, $\Phi =-GM/\sqrt{r^2+z^2}$. If $h\ne 0$, the vertical derivative vanishes only at $z=0$, and the radial stationary condition there gives $r_0=h^2/(GM)$. This is the sole stationary point with $r>0$, at energy $-GM/(2r_0)$. Its Hessian matrix isConsequently the local contours are closed, approximately circular curves about a strict minimum. Finite negative-energy contours above the minimum remain closed; the centrifugal barrier prevents approach to the axis and the negative energy prevents escape to infinity. The zero contour satisfies $z^2=4r^4/r_0^2-r^2$, while positive-energy contours can be unbounded. The left panel illustrates these features; portions of the outer contours extend outside the plotting window.

Each panel uses its own circular radius as length unit and $GM/r_0$ as potential unit. Linearizing near the minimum givesThe radial and vertical epicyclic frequencies both equal the circular orbital frequency. Circular Kepler orbits are stable to small meridional perturbations of a freely moving particle. With no dissipation the oscillations persist rather than decay. A change in h shifts the circular radius to a nearby orbit; azimuthal phase itself is neutral by rotational symmetry. If $h=0$, there is no stationary point or centrifugal barrier and this circular-orbit stability discussion does not apply.

The imposed constraint now fixes $\dot{\varphi }=\omega$, while a torque allows $h=r^2\omega$ to vary as the particle moves radially. The meridional equations become $\ddot{r}=-{\Phi }_r+{\omega }^{2}r$ and $\ddot{z}=-{\Phi }_z$. Hence $\boxed{{\Phi }_2=\Phi -\frac{1}{2}{\omega }^2{r}^2}$ is the appropriate effective potential. One must not substitute a variable h into the fixed-angular-momentum potential and then differentiate as though h were constant.

For a point mass and nonzero $\omega$, the stationary point is $(r_c,0)$ with $r_c=(GM/{\omega }^2{)}^{1/3}$. Its meridional Hessian matrix isIt is a saddle, at ${\Phi }_2=-3GM/(2r_c)$. Near it the contours are hyperbolas, with the saddle contour locally tangent to $z=\pm \sqrt{3}(r-r_c)$, as shown in the right panel. On the equatorial plane the potential tends downward both toward the point mass and at large cylindrical radius; at fixed r it increases as $|z|$ increases. The radial perturbation obeys $\delta \ddot{r}=3{\omega }^2\delta r$, while the vertical perturbation obeys $\delta \ddot{z}=-{\omega }^2\delta z$. Thus the radial motion has growing and decaying exponential solutions and the vertical motion is oscillatory. For $\omega =0$ no such stationary point exists.

This saddle is the basis of magnetocentrifugal acceleration. A sufficiently strong, open poloidal magnetic field can enforce approximate corotation with its disc footpoint while allowing motion along the field. If its outward tangent makes angle $\vartheta$ with the vertical, a small distance l along it has $\delta r=l\sin \vartheta$, $\delta z=l\cos \vartheta$. Its leading potential change at a Keplerian rotation footpoint isThe thirty-degree magnetocentrifugal launching criterion follows: cold local launching has a downhill quadratic direction when $\boxed{\vartheta >30^{\circ }}$ from the vertical. Equality is marginal at quadratic order and depends on higher terms and field-line geometry. The field transmits the torque needed to increase the material's angular momentum; a freely moving particle has no such constraint and instead sees the stable minimum in part (a). The local criterion does not by itself prove global escape or determine the mass loading of a complete magnetohydrodynamic wind.

Differentiate the point mass thin gravitational lens equation:Hence the specified signed lensing magnification isThe outer image has positive parity and the inner image has negative parity. Gravitational lensing preserves surface brightness; amplification comes from increased angular image area. The observed flux magnifications are $|{\mu }_{+}|$ and $|{\mu }_{-}|$; a negative determinant reverses orientation, not flux. Substitution of the two roots, with $u=|\beta |/{\theta }_E$, givesAt $\beta =0$, axial symmetry produces an Einstein ring. For an ideal point-like astronomical source, the lensing magnification diverges: near alignment ${\mu }_{\mathrm{tot}}\sim 1/u$. Finite source size gives a finite observed flux and a ring of finite thickness.

For $0<\beta <{\theta }_E$, there are two images on opposite sides. Their separation is $\sqrt{{\beta }^2+4{\theta }_E^2}$, approaching $2{\theta }_E$ at alignment. For $\beta \ll {\theta }_E$, their positions are ${\theta }_{\pm }\simeq \pm {\theta }_E+\beta /2$ and ${\mu }_{\pm }\simeq 1/2\pm {\theta }_E/(2\beta )$: both images are bright, with the outer one slightly brighter. Moving away from alignment reduces the total amplification and makes the inner image relatively fainter.

For $\beta \gg {\theta }_E$, expanding the two image roots givesThus the distant source has one almost unaltered bright image and an extremely faint opposite-parity secondary, with total flux magnification $1+2{\theta }_E^4/{\beta }^4+O(({\theta }_E/\beta {)}^6)$. The secondary persists for every finite nonzero source offset in the point mass model.

Substituting $\eta =D_s\beta$ and $\xi =D_d\theta$ into the previous geometric relation verifiesFor a point mass gravitational lens, the given physical deflection yieldsAn aligned source has $\beta =0$, and therefore ${\theta }^2={\theta }_E^2$. Rotational symmetry turns these signed intersections into an Einstein ring, whose Einstein radius isFor an offset source, multiplying $\beta =\theta -{\theta }_E^2/\theta$ by $\theta$ gives a quadratic equation. Its two image positions areFor $\beta >0$, ${\theta }_{+}>0$ lies on the source side of the lens and ${\theta }_{-}<0$ lies on the opposite side.

All invariants in this part refer to the selected magnetic field line. Its squared poloidal Alfvén number isAt the Alfvén surface crossing it equals unity, so ${\rho }_a={\mu }_0{k}^2$ and $M_{Ap}^2=1/y$. Subtract the two azimuthal integrals in part (a) to findFinite azimuthal fields and velocity at $y=1$ require the Alfvén-surface regularity condition for an axisymmetric wind $\ell =\omega R_a^2$. HenceHere $v_{\varphi }$ denotes the azimuthal velocity relative to field-line rotation. For gravity of a point mass at $z=0$, $\Phi =-GM/R=-GM/(R_{a}x)$.

Insert these quantities in the corotating energy integral with $w=0$, and divide by the gravitational energy scale $GM/R_a$. This gives the cold radial magnetohydrodynamic wind integral,withBoth are dimensionless. The apparent $0/0$ at $x=y=1$ must be evaluated on the smooth solution branch, not assigned an arbitrary value.

A constant-energy curve obeys $f_x+f_{y}dy/dx=0$. At a nondegenerate magnetosonic point, the vanishing coefficient of the mass density derivative requires simultaneous vanishing of its numerator:The curve must admit a real crossing slope. Expanding about such a point gives $f_{yy}{m}^2+2f_{xy}m+f_{xx}=0$, $m=dy/dx$, so a regular saddle point of a scalar function crossing has $f_{xy}^2-f_{xx}{f}_{yy}>0$. These are the form of the slow and fast regularity conditions in a warm wind, with the enthalpy contribution retained in $f$.

In the strictly cold approximation used to obtain the displayed $f$, the slow speed is zero. There is a cold-limit degeneracy of a slow magnetosonic point: a finite-speed interior slow crossing need not survive. Indeed,so this cold equation has no ordinary sub-Alfvénic slow critical point. A small retained enthalpy is needed to locate that point, which can approach the launching boundary as the sound speed tends to zero. The fast critical point still obeys the simultaneous conditions above wherever the reduced expression is nonsingular.

Treat the stars as point masses for the orbital motion, neglecting their spin angular momenta. Let $M=M_1+M_2$. Their distances from the centre of mass are $a_1=a{M}_2/M$ and $a_2=a{M}_1/M$. The sum of their orbital angular momenta isThus the circular-binary orbital angular momentum is

For conservative binary mass transfer, both $M$ and $J$ are fixed. In terms of the binary mass ratio $q=M_1/M_2$, $M_1=Mq/(1+q)$ and $M_2=M/(1+q)$, henceMass conservation also gives $d{M}_2=-d{M}_1$, soDifferentiating $J\propto M_1{M}_2{a}^{1/2}$ at fixed $M,J$ gives $d\log a/d\log M_1=2(q-1)$. Therefore the stipulated Roche lobe law has the Roche-lobe radius response exponent

Define the overfill by $\Delta =\log (R_1/R_L)$. On a time scale short compared with the slow change of $K$, the donor's response is $d\log R_1=\beta d\log M_1$. ThusDuring mass loss $d\log M_1<0$. If ${\zeta }_L>\beta$, mass loss increases the overfill and therefore drives more mass loss: this is positive feedback, giving more rapid Roche-lobe overflow. If ${\zeta }_L<\beta$, mass loss decreases the overfill and can balance the slow expansion due to stellar evolution. More explicitly, with $k_{\mathrm{ev}}=d\log K/dt>0$, contact requiresA slow negative transfer rate can satisfy this only on the stable side $\beta >{\zeta }_L$. The rapid-transfer condition is thereforeIts time scale depends on the physical donor response represented by $\beta$; it need not be a dynamical time scale if $\beta$ is a thermal-equilibrium mass-radius exponent.

For Roche-lobe overflow feedback with a constant wind torque factor, retain mass conservation to the stated approximation but allow the prescribed angular-momentum loss. Since $d\log J=fd\log M_1$, differentiation givesThe new Roche-lobe radius response exponent isThe same overfill calculation givesA loss of angular momentum has $f>0$ because ${\dot{M}}_1<0$, and it lowers the critical ratio by making the orbit contract more strongly.

For the specified solar-mass donor exponent $\beta =1/2$, the answer isWithout the wind torque this becomes $q>22/23\simeq 0.957$, so donors slightly less massive than their companions can already lie on the rapid-transfer side of this model. For $0<f<11/10$ the lower threshold is $(22-20f)/23$, and for $f\ge 11/10$ every positive $q$ satisfies the strict inequality. There is no finite upper bound on $q$ from this criterion. Since the paper does not specify $f$, the wind case has a conditional range rather than a unique numerical cutoff.