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Relative positions and velocities measured from galaxy $A$ are $r_B-r_A$ and $v_B-v_A$. Spatial homogeneity says that the same position-to-relative-velocity rule applies from every galaxy. Thus $v(r_B-r_A)=v(r_B-r_O)-v(r_A-r_O)$. Taking $r_O=0$ and putting $r_B=x+y$, $r_A=y$ gives additivity $v(x+y)=v(x)+v(y)$. For a continuous physical velocity field, additivity implies rational homogeneity and, by continuity, real homogeneity. Thereforefor a position-independent matrix $H$. Regularity is necessary to exclude pathological discontinuous additive functions.

Decompose the given matrix as $H=(5D/t)I+(D/t)W$, whereIts symmetric part is isotropic: no cosmological shear, expansion scalar $\nabla \cdot v=15D/t$, and linear scale factor proportional to $t^{5D}$ for $t>0$. Its antisymmetric part is rigid rotation with angular velocityThe vorticity is $2\mathbf{\omega }$. Comoving separations obeyThus all distances grow by the same power, volumes grow as $t^{15D}$, and the pattern rotates about the fixed axis $(1,-2,1)$ through angle $\sqrt{6}D\log (t/t_0)$. Homogeneity alone permits this rotation; full cosmological isotropy would not.

Use an Earth-centred rotating reference frame, with Earth's constant angular velocity vector $\mathbf{\omega }$. The vectors $\mathbf{r}$, $\mathbf{v}$, and $\mathbf{a}$ are the particle's position, velocity, and acceleration measured in that rotating frame. The vector $\mathbf{g}$ is the gravitational acceleration due to the actual gravitational force, expressed in the same axes; it excludes the separate centrifugal acceleration. The remaining term $-2\mathbf{\omega }\times \mathbf{v}$ is the Coriolis acceleration. A constant rotation axis and an unaccelerated Earth-centred origin account for the absence of an Euler term or a translational-origin term.

For the size estimates, Earth's angular speed isWith drop height $H=20\mathrm{m}$, unperturbed free fall gives the fall timeand final speed $gT=20{\mathrm{m}\mathrm{s}}^{-1}$. Thus the gravitational term has size $10{\mathrm{m}\mathrm{s}}^{-2}$, the Coriolis acceleration is at most about $2\omega gT\approx 2.8\times 10^{-3}{\mathrm{m}\mathrm{s}}^{-2}$, and the centrifugal acceleration is at most ${\omega }^2{R}_0\approx 2.9\times 10^{-2}{\mathrm{m}\mathrm{s}}^{-2}$. These are characteristic sizes $10^1$, $10^{-3}$, and $10^{-2}$ in acceleration units; geometric factors can reduce them, and the Coriolis term is zero at release.

For a suitable approximation to the rotational deflection of a falling body, take $\mathbf{g}$ constant over the short fall, start at rest relative to Earth, and use the zeroth-order velocity ${\mathbf{v}}^{(0)}(t)=\mathbf{g}t$ inside the small Coriolis term. In the centrifugal term replace $\mathbf{r}$ by ${\mathbf{R}}_0$, because $H/R_0\ll 1$. Integrating the resulting acceleration twice, with initial position ${\mathbf{r}}_{\mathrm{i}}$, givesSince ${\mathbf{r}}_{\mathrm{i}}+\mathbf{g}{T}^2/2={\mathbf{R}}_0$, evaluating at the unperturbed fall time yields the printed first-iterate estimate

The approximation keeps the leading Coriolis contribution and the leading centrifugal contribution enhanced by Earth's large radius. Feeding the Coriolis velocity correction back into the equation produces terms of size ${\omega }^{2}g{T}^4$, smaller than the retained centrifugal displacement by a factor of order $H/R_0$. Replacing $\mathbf{r}$ by ${\mathbf{R}}_0$ in that term has the same relative error. Here $\omega T\approx 1.4\times 10^{-4}$ and $H/R_0\approx 3.3\times 10^{-6}$, so both approximations are well controlled. Strictly, the displayed vector is the position at time $T$: its small upward component is removed by the slightly longer actual fall time. That arrival-time correction changes the retained horizontal displacements only at higher order. Thus the horizontal components of the printed estimate give the landing point on the locally flat surface; its vertical component should not be interpreted as a landing above the ground.

Let ${\hat{\mathbf{e}}}_E,{\hat{\mathbf{e}}}_N,{\hat{\mathbf{e}}}_U$ point east, north, and vertically upwards. They form a right-handed basis, with ${\hat{\mathbf{e}}}_N\times {\hat{\mathbf{e}}}_U={\hat{\mathbf{e}}}_E$. At latitude $\lambda$, in the spherical-Earth approximation,The required cross products areAt $\lambda =45^{\circ }$, the northerly and easterly displacements are thereforeThe negative northerly component means a southward displacement. With the supplied values, these are approximately $2.9\mathrm{cm}$ south and $1.3\mathrm{mm}$ east. The eastward effect comes from Coriolis acceleration, while the leading southward effect here comes from centrifugal acceleration.

The rotating-frame derivative formula is $(d\mathbf{A}/dt{)}_S=(d\mathbf{A}/dt{)}_{S^{\prime }}+\mathbf{\omega }\times \mathbf{A}$. Applying it twice to the position vector, with constant angular velocity, givesSolving for ${\mathbf{a}}_{S^{\prime }}$ identifies the displayed terms: its left side is the acceleration relative to the rotating reference frame, the first right-hand term is the inertial acceleration, $-2\mathbf{\omega }\times {\mathbf{v}}_{S^{\prime }}$ is the Coriolis acceleration, and $-\mathbf{\omega }\times (\mathbf{\omega }\times \mathbf{r})$ is the outward centrifugal acceleration.

For a particle of mass $m$, replace ${\mathbf{a}}_S$ by the actual force divided by $m$ using Newton's second law. The remaining terms then appear as apparent forces in $S^{\prime }$. There is no Euler acceleration because $\mathbf{\omega }$ is constant and no translational acceleration term because the common origin is fixed in the inertial frame.

Work first in units $G=c=\hslash =k_B=1$. Black-hole thermodynamics starts with the laws of black-hole mechanics. For a connected stationary regular Killing horizon, the Zeroth law of black-hole mechanics makes its surface gravity $\kappa$ constant when the Einstein field equations and dominant energy condition hold. This parallels uniform equilibrium temperature. For neighboring stationary asymptotically flat four-dimensional Einstein-Maxwell black holes, the First law of black-hole mechanics iswhere ${\Omega }_H$ is the horizon angular velocity and ${\Phi }_H$ its electric potential relative to infinity. The last terms are rotational and electromagnetic work, analogous to the work terms in $\delta E=T\delta S+\text{work}$. The second law of black-hole mechanics, expressed by Hawking's area theorem, makes the total future horizon area nondecreasing for classical matter obeying the null energy condition and global assumptions such as strong asymptotic predictability. The third law of black-hole mechanics is an unattainability statement: subject to the usual regularity assumptions, the weak energy condition, and a bounded stress-energy tensor, no finite physical process reduces the surface gravity of a regular horizon to zero. It does not assert that extremal black holes have zero area or zero entropy.

Classical geometry alone fixes an analogy, not a nonzero physical temperature. Quantum field theory supplies Hawking temperature $T_H=\kappa /(2\pi )$. Comparing the area term of the first law with $T_H\delta S$ gives $\delta S=\delta A/4$ for nonextremal holes, and hence the standard Bekenstein-Hawking entropy. Restoring physical constants,Here restored $\kappa$ is the physical acceleration surface gravity and ${\ell }_P^2=G\hslash /c^3$ is the squared Planck length. The first-law comparison determines the area coefficient, leaving an additive entropy constant unspecified; the displayed formula is the usual convention. Its area scaling, rather than ordinary volume scaling, signals that the available thermodynamic degrees of freedom of a gravitating system are constrained by the horizon geometry.

Particle production explains the quantum input. For a real scalar field obeying the massless covariant wave equation, the Klein-Gordon inner product on solutions isThe wave equation makes its current divergence-free, so the product is independent of a Cauchy hypersurface when the boundary flux vanishes. Use suitably normalized wave packets or the appropriate continuum distributions. Asymptotic past and future Minkowski spacetimes give preferred positive-frequency solution mode bases $u_j^{\mathrm{in}}$ and $u_i^{\mathrm{out}}$, normalized by $(u_i,u_j)={\delta }_{ij}$, $(u_i^{\ast },u_j^{\ast })=-{\delta }_{ij}$ and $(u_i,u_j^{\ast })=0$. Between those regions, a nonstationary geometry generally has no preferred positive-frequency splitting.

The mode bases are related by a Bogoliubov transformation,The last identities follow from conservation of the Klein-Gordon inner product and are the canonical identities for a bosonic Bogoliubov transformation. Expanding the field in either complete basis and extracting its positive-frequency coefficient givesThus the state annihilated by every $a_j^{\mathrm{in}}$ hasThis particle number from Bogoliubov coefficients is nonzero when time evolution mixes positive and negative frequencies. It describes one state as vacuum in the early particle basis and populated in the late basis; it does not require an arbitrary choice of a vacuum at each intermediate time. For a finite implementable transformation the state remains a squeezed pure quantum state. In infinitely many modes, a common unitary bosonic Fock space implementation requires additional conditions, such as the beta map being a Hilbert-Schmidt operator; finite wave packet observables need not share every global divergence of an ideal continuum calculation.

For gravitational collapse, the future is not globally Minkowski: it contains a black hole. The relevant late out-basis includes modes reaching future null infinity together with modes entering the future event horizon. Modes on infinity alone are not a complete Cauchy basis. Nevertheless the same mode-mixing calculation determines the outgoing particle flux. Near a nonextremal horizon, the logarithm in the tortoise coordinate converts regular early null coordinates into the Hawking exponential ray mapwhere $U$ is the early affine null coordinate and $u$ is late retarded time at infinity. Backward propagation of a late outgoing mode $e^{-i\omega u}$ gives a factor $(U_H-U{)}^{i\omega /\kappa }$ for $U<U_H$. Its positive- and negative-frequency Fourier integrals differ by analytic continuation around the logarithmic branch. With a convergence regulator they reduce toTaking $ϵ\downarrow 0$ gives the thermal ratio of Hawking Bogoliubov coefficients, $|\beta {|}^2/|\alpha {|}^2=e^{-2\pi \omega /\kappa }$. Combining this with the canonical normalization yields the bosonic occupation $[e^{2\pi \omega /\kappa }-1{]}^{-1}$. Late-time wave packets turn formal continuum coefficients into a finite number flux.

A Schwarzschild black hole has $\kappa =1/(4M)$, so this spectrum has $T_H=1/(8\pi M)$. The exterior curvature potential partly reflects the outgoing modes. Its transmission probabilities are the greybody factors, giving, for one massless scalar species in the stationary late-time approximation,Thus the horizon temperature is universal, while the spectrum received at infinity is not a perfect featureless blackbody radiation spectrum. The derivation assumes the near-horizon quantum state inherited from a regular collapse vacuum and ignores rapid backreaction over the timescale of the packets. Exponentially blueshifted precursor frequencies indicate the usual short-distance assumption in the semiclassical calculation, not an independently demonstrated quantum-gravity description of the endpoint.

The outgoing positive energy flux drives black-hole evaporation. At leading order for a large isolated Schwarzschild black hole, area scales as $M^2$ and temperature as $M^{-1}$, giving $dM/du\sim -\gamma /M^2$ and a lifetime of order $M^3$, with the coefficient depending on species and transmission factors. Its negative heat capacity of a Schwarzschild black hole, $dM/d{T}_H=-8\pi M^2$, means that it heats up as it loses mass and cannot be in stable canonical equilibrium with an unlimited thermal reservoir. The approximation fails when quantum-gravitational scales are reached and does not fix whether the endpoint is complete evaporation, a remnant, or something else.

Area loss does not contradict the classical area theorem, because the quantum stress-energy tensor need not obey the classical null energy condition; negative horizon energy accompanies the positive outgoing flux. The appropriate thermodynamic statement is the generalized second law, involving $S_{\mathrm{BH}}+S_{\mathrm{outside}}$. Finally, Hawking radiation raises the black hole information paradox: if a pure quantum state completely evaporates into an exactly thermal mixed quantum state and its partners disappear, the result conflicts with ordinary unitary time evolution. Early outgoing radiation can be mixed simply because of its entanglement with interior modes while the complete state remains a pure quantum state; that fact alone is not information loss. Resolving the fate of all correlations through the evaporation endpoint requires more than the leading semiclassical flux calculation.

Linearity of Stokes flow and rotational symmetry, including reflection in mathematics invariance of the spherical geometry, determine the possible rigid motions before any calculation. Translation is a polar vector; its linear dependence on a single vector $\mathbf{A}$ must be $c\mathbf{A}$. A traceless second-rank tensor $\mathbf{B}$ cannot produce a polar vector by an isotropic linear map: tensor contraction with the identity matrix vanishes and tensor contraction with the totally antisymmetric tensor vanishes because $\mathbf{B}$ is a symmetric second-rank tensor. Angular velocity is an axial vector. Neither $\mathbf{A}$ nor a symmetric $\mathbf{B}$ can produce one by an isotropic linear map. Products such as $\mathbf{A}\times \mathbf{BA}$ are excluded by Linearity of Stokes flow. Thus $\mathbf{\Omega }=0$.

Use the Unscaled Papkovich–Neuber representation, with harmonic functions $\mathbf{\Phi }$ and $\chi$:Indeed $\nabla \cdot \mathbf{u}=0$ and $\mu {\nabla }^2\mathbf{u}=\nabla p$ because each potential is a harmonic function. Put $r=|\mathbf{x}|$ and $S=\mathbf{x}\cdot \mathbf{Bx}$. The decaying harmonic functions with the necessary angular dependence are $\mathbf{A}\cdot \mathbf{x}/r^3$, $\mathbf{Bx}/r^3$ and $S/r^5$. The last is a harmonic function precisely because $\mathrm{tr}\mathbf{B}=0$. A $1/r$ Stokeslet is excluded by the force-free condition, and a rotlet by the torque-free condition. TryThe resulting Stokes flow isAt $r=a$, the coefficient of $\mathbf{n}(\mathbf{A}\cdot \mathbf{n})$ fixes $c=a^3/3$; the remaining constant vector gives $\mathbf{U}=-2\mathbf{A}/3$. In the $\mathbf{B}$ mode, matching $\mathbf{Bn}-(\mathbf{n}\cdot \mathbf{Bn})\mathbf{n}$ gives $e=a^4/2$ and $d=-a^2/2$. Therefore the complete two-mode tensorial squirmer flow isThe boundary condition is satisfied in the laboratory frame, so the Stokes flow tends to zero at infinity. The potential dipole in the $\mathbf{A}$ mode gives an irrotational flow and decays as $r^{-3}$; the leading $\mathbf{B}$ mode is a stresslet, decaying as $r^{-2}$. Neither carries a net force or torque. Direct surface integral averaging with the surface slip velocity formula gives the same translation, providing an independent check. Uniqueness of Stokes flow then identifies this decaying solution.

Let $\mathbf{e}=\mathbf{X}/R$, $s=\mathbf{e}\cdot \mathbf{Be}$, and $A_{\parallel }=\mathbf{A}\cdot \mathbf{e}$. Applied forces and couples below are forces and torques on the solid; by force balance their values are also the strengths exerted by that solid on the fluid. This fixes the signs in Faxén's first law and Faxén's rotational law. To hold the passive sphere fixed, set its translation and angular velocity to zero in those laws. The incident two-mode tensorial squirmer flow givesOnly the stresslet part has vorticity:Consequently, for fixed $\mathbf{A},\mathbf{B}$ as $a/R\to 0$,The leading force is the radial stresslet force. Its next correction, $O(\mu a^4|\mathbf{A}|/R^3)$, comes from the incident potential dipole. The next $\mathbf{B}$ correction, $O(\mu a^5\Vert \mathbf{B}\Vert /R^4)$, contains both the swimmer's $r^{-4}$ field and the Laplacian term in Faxén's first law. If the relevant coefficient vanishes, these next terms must be retained instead of calling the vanished term a nonzero leading approximation. Further exchanges in the method of reflections for Stokes flow are smaller still.

The passive sphere reflects primarily a Stokeslet with strength ${\mathbf{F}}_h$. At the swimmer, its velocity is $(\mathbf{I}+\mathbf{ee}){\mathbf{F}}_h/(8\pi \mu R)$. The unchanged surface slip velocity adds the free swimming velocity to the incident-flow contribution in Faxén's first law. ThusThe leading change is $O(a^3\Vert \mathbf{B}\Vert /R^3)$; its next correction is the displayed $O(a^4|\mathbf{A}|/R^4)$ term. Finite-radius Faxén's first law corrections, the passive sphere's reflected stresslet, and its rotlet contribute at order $a^5\Vert \mathbf{B}\Vert /R^5$ to translation.

For a Stokeslet at the origin,The curl of the first term is $\mathbf{F}\times \mathbf{x}/r^3$; that of the second is also $\mathbf{F}\times \mathbf{x}/r^3$, before multiplying by $1/(8\pi \mu )$. Hence ${\mathbf{\omega }}_F=\mathbf{F}\times \mathbf{x}/(4\pi \mu r^3)$. At the swimmer use displacement $-\mathbf{X}$. The leading $\mathbf{B}$ force is parallel to $\mathbf{X}$, so its Stokeslet has zero vorticity there. This eliminates the putative $O(a^3\Vert \mathbf{B}\Vert /R^4)$ rotation. The potential dipole itself has no vorticity, but the holding force it induces is not radial. Applying Faxén's rotational law to its reflected Stokeslet yieldsThe passive sphere's rotlet and reflected stresslet can affect that last order, so they cannot restore the larger missing rotation. When $\mathbf{A}\times \mathbf{X}=0$, the boxed coefficient is zero and the higher-order terms determine any rotation.

A star of declination $\delta =\theta$ crosses the zenith. Its direction rotates around the celestial pole at ${\Omega }_{\mathrm{sid}}$, but its actual angular velocity on the celestial sphere is ${\Omega }_{\mathrm{sid}}\cos \theta$. Consequently the small-angle crossing-gap estimate becomesFor clarity, the factor $\cos \theta$ is the radius of the star's daily circle on a unit celestial sphere; it does not modify the sidereal hour angle rate.

The estimate assumes the required slew is nearly $180^{\circ }$. A more exact ideal symmetric reacquisition calculation is possible. Let the endpoints have hour angles $-H,+H$ and suppose $0<\theta <\pi /2$, $H$ small enough that both endpoints are above the astronomical horizon. The horizontal direction components are $n=\sin \theta \cos \theta (1-\cos H)$ and $w=\pm \cos \theta \sin H$. Their shortest azimuth separation isThe minimum ideal gap satisfies $2H{\Omega }_{\mathrm{az}}/{\Omega }_{\mathrm{sid}}=\Delta \eta (H)$; its angular separation is ${\varphi }_{\mathrm{full}}=2\arcsin (\cos \theta \sin H)$. Expanding for a fast drive gives the boxed expression. At an equatorial site $\Delta \eta =\pi$ exactly. At a geographic pole the direction with $\delta =\theta$ is stationary, so the crossing argument is inapplicable and the limit is zero. A full near-zenith rate-limited footprint still requires specifying the trajectory and drive model.

Let $q=\sigma /L$, with $L>0$, and let $Z=X^1+i{X}^2$. Direct differentiation gives $\ddot{Z}=Z^{\prime \prime }=-Z/L^2$, while the time coordinate also satisfies the wave equation. Using the Minkowski metric,Thus both Virasoro constraints hold, and the induced worldsheet metric is $g_{\mu \nu }={\cos }^{2}q\mathrm{diag}(-1,1)$. The configuration solves the Nambu–Goto action equations in its interior.

Place the fixed end at $\sigma =0$. The first free endpoint has $Z^{\prime }=e^{it/L}\cos q=0$ at $q=\pi /2$. For the usual single unfolded rotating Nambu–Goto string,This parameter interval need not equal the earlier interval $[0,\pi ]$, since that interval was a coordinate convention. The proper length is measured on a constant-$X^0$ slice; the local motion is perpendicular to the string, so there is no longitudinal Lorentz contraction.

All points have the same phase $t/L$, hence the segment rotates rigidly with angular velocity $\Omega =1/L$. At radius $r=L\sin q$, its speed is $v=r/L$. Therefore the free tip moves at the speed of light. This is consistent with its Neumann boundary condition: $X^{\prime }=0$ and the Virasoro constraints imply ${\dot{X}}^2=0$ at the free tip. The null endpoint is a boundary limit, not a nondegenerate interior point.

In this conformal gauge, $P_m=T{\dot{X}}_m$, so the energy density per $\sigma$ is $-P_0=T$. Equivalently, the energy density per proper length is $T/\sqrt{1-r^2/L^2}$. Thusand the rotational kinetic energy isThe angular momentum about the fixed point isTherefore the Regge trajectory isThe endpoint conditions alone also admit folded extensions with parameter length $(n+\frac{1}{2})\pi L$, $n\ge 0$. Their proper length, energy and angular momentum are $(2n+1)$ times the values above, and $\beta =1/[(2n+1)\pi T]$. The displayed answer uses the implicit unfolded-segment convention.

Let $\{{\mathbf{e}}_i(t)\}$ be an orthonormal basis fixed in $S^{\prime }$ and write $\mathbf{a}=a_i{\mathbf{e}}_i$. A basis vector rotating with angular velocity $\mathbf{\omega }$ satisfiesDifferentiating the components and the basis vectors in $S$ therefore gives the rotating-frame derivative formula

Reflection in $y=0$ preserves the prescribed translational velocity and axial angular velocity. A force transforms as a polar vector, whereas a torque transforms as an axial vector; consequently $F_y=G_x=G_z=0$. Reflection in $x=0$ reverses both imposed motions but preserves the normal force component. The linearity of Stokes flow then gives $F_z=0$. Thus only $F_x$ and $G_y$ can be nonzero.

There is a sign inconsistency in the printed question. In right-handed coordinates, $\mathbf{\Omega }=\Omega {\mathbf{e}}_y$ gives a rotational velocity $-\Omega a{\mathbf{e}}_x$ at the sphere's lowest point. The printed shear and force formulas instead use the opposite rotational sense. Below, let $\omega$ be the right-handed component along $+{\mathbf{e}}_y$; the paper's displayed formulas are recovered by setting $\omega =-\Omega$.

In the translating frame, the parabolic gap and leading boundary velocities areThe no-slip boundary condition and lubrication theory giveThe gap is stationary in this frame, so ${\nabla }_{\parallel }\cdot \mathbf{q}=0$. The Reynolds lubrication equation is thereforeSince $h_x=x/a$ and ${\nabla }_{\parallel }^{2}h=2/a$,Differentiating the velocity profile gives the leading shear stressesThe first expression becomes the printed formula after $\omega =-\Omega$.

For the logarithmic lubrication resistance of a sphere near a wall, write $\ell =\log (1/ϵ)$. The logarithmic region is $a\sqrt{ϵ}\ll r\ll a$, where $h\sim r^2/(2a)$. ConsequentlyThese follow from $\int r/hdr=a\ell +O(a)$ and $\int r^3/h^{2}dr=2a^2\ell +O(a^2)$, with a fixed small outer cutoff. The general logarithmic radial numerator is $r^{2n-1}$ over $h^n$; the numerator in the printed integration hint appears to have a typographical error.

The force exerted on the fluid through the upper gap boundary includes pressure acting on its slope:The pressure cancels the $x^2/h^2$ term, givingPressure exerts no torque about the centre of a sphere because its traction is radial. To logarithmic order, the shear has lever arm $-a{\mathbf{e}}_z$, soThe other components vanish by parity: $G_x$ involves an integral of the $xy$ term, and $G_z$ has an integrand odd in $y$. The combined hydrodynamic resistance matrix isIts equal off-diagonal coefficients are required by the Lorentz reciprocal theorem for Stokes flow. It is a positive-definite matrix, as required by viscous dissipation.

In the rotational convention of the printed force formula, set $\Omega =-\omega$ and measure the scalar couple along $-{\mathbf{e}}_y$, so $G=-G_y$. ThenFor sedimentation under the buoyancy-adjusted weight $F$, a uniform sphere is torque-free. Thus $\omega a=U/4$, and the force balance givesThe rotational scalar in the paper's displayed-force convention is $\Omega =-U/(4a)$. The sphere rolls in the right-handed $+{\mathbf{e}}_y$ sense but still slips: its lowest point has laboratory velocity $U-a\omega =3U/4$. The signs and coefficients of the right-handed resistance agree with the rigid-wall terms of Bertin et al., equations (4.4)–(4.5), after reversing the forces and torques there from fluid-on-sphere to sphere-on-fluid.

Let $e_1,e_2,e_3$ be a principal body frame, let $I_1,I_2,I_3$ be the corresponding principal moments of inertia, and write the angular velocity as $\mathbf{\omega }={\sum }_i{\omega }_i{e}_i$. The body-frame components of the angular momentum areNo external torque acts, so $\mathbf{L}$ is constant in the fixed space frame. Since a body-fixed basis vector satisfies ${\dot{e}}_i=\mathbf{\omega }\times e_i$,Taking body-frame components gives the Euler equations for a torque-free rigid body:

In a frame with unit angular velocity, transforming the acceleration introduces the Coriolis acceleration and centrifugal acceleration. Radiation reduces the attraction of $M_1$ by the factor $1-{\beta }_1$, soWith the effective potentialits Cartesian components are exactlyThis is the photogravitational restricted three-body problem.

Let a change of orthonormal coordinates be represented by a rotation matrix $R$. Since both angular momentum and angular velocity are vectors,Using $L_i=I_{ij}{\omega }_j$ in both frames givesfor every vector $\mathbf{\omega }$. Multiplication by $R^{-1}=R^T$ yieldswhich is exactly the transformation law for a rank-two tensor. Thus $I_{ij}$ is a rank-two tensor.

Put $\mathbf{r}=\mathbf{x}-\mathbf{a}$. The vector triple product givesThereforeAt the centre of mass,Expanding the first formula, the terms linear in $x_i$ vanish because the centre of mass is the origin, while $\rho {\int }_{B}dV=M$. Hence the parallel axis theorem in tensor form is

The uniform rod is invariant under inversion through its centre. This centrosymmetry makes the translation-rotation blocks of its generalized hydrodynamic resistance matrix vanish: a polar translation velocity changes sign under inversion, whereas an axial angular velocity does not, so an invariant linear coupling between them is impossible.

Gravity acts through the centre of mass and supplies no torque. The remaining rotational resistance is positive definite, so zero applied torque implies zero angular velocity. Thus an initially inclined rod keeps its orientation while settling.

In the inertial frame, rigid corotation gives $\mathbf{u}=\mathbf{\Omega }\times \mathbf{r}$. The ideal magnetohydrodynamics condition is the motional electric fieldApply Gauss's law in Gaussian units and use the divergence and curl of a cross product:The exterior field is produced by currents inside the star, so $\nabla \times {\mathbf{B}}_0=0$ there, while rotation with constant angular velocity has $\nabla \times \mathbf{u}=2\mathbf{\Omega }$. Hence the required Goldreich-Julian charge density isHere $n_q$ denotes electric charge per unit volume, as in the question. If each carrier has charge $q$, its signed number density is $n_q/q$.

A fixed space frame is an inertial orthonormal frame whose axes remain fixed in space. A principal body frame is attached to the rigid body and aligned with the principal axes of its inertia tensor, so that the tensor is diagonal with entries $I_1,I_2,I_3$.

If $e_i(t)$ are the body axes and $\omega$ is the angular velocity, then the derivative of a body-fixed basis vector isIn the principal frame,For torque-free motion, the angular momentum has zero space derivative. ThereforeTaking body-frame components gives the Euler equations for a torque-free rigid body:

For an axisymmetric body, put $I_1=I_2=I$ and let $e_3$ be its symmetry axis. Decompose $\omega ={\omega }_1{e}_1+{\omega }_2{e}_2+{\omega }_3{e}_3$. ThenThus $L$ is a linear combination of $\omega$ and $e_3$, proving that the angular momentum, angular velocity, and symmetry axis are always coplanar. This is the coplanarity in a torque-free axisymmetric rigid body.