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Curvature contractions. Use the index contraction stipulated in the question. In three dimensions,The same contraction in four dimensions givesThen the Ricci scalar is $R=4\Lambda$ and the Einstein tensor is $G_{\mu \nu }=-\Lambda g_{\mu \nu }$. Thus $R_{\mu \nu }=\Lambda g_{\mu \nu }$ is precisely the vacuum Einstein field equations $G_{\mu \nu }+\Lambda g_{\mu \nu }=0$.

The static solution. For $A=1+{\lambda }^2{r}^2$, $A^{\prime }=2{\lambda }^{2}r$ and $A^{\prime \prime }=2{\lambda }^2$. The given Ricci tensor components becomeEach equals $-3{\lambda }^2$ times the corresponding metric tensor component. ThereforeThis is Anti-de Sitter spacetime in static coordinates on anti-de Sitter spacetime, with curvature radius $1/\lambda$. The overall metric signature is opposite to the convention used in some descriptions of those coordinates; the geodesic constants below use the question's $(+,-,-,-)$ convention.

Conserved quantities and radial motion. In the equatorial plane, the geodesic Lagrangian is $L=A{\dot{t}}^2-{\dot{r}}^2/A-r^2{\dot{\varphi }}^2$. Its cyclic coordinates giveFor rest mass $m$, $mk=p_t$ is the conserved Killing energy and $mh=-p_{\varphi }$ is the axial angular momentum. Timelike normalization $L=1$ givesFor purely radial motion $h=0$, differentiating away from a turning point gives $\ddot{r}=-{\lambda }^{2}r$, and continuity extends that equation through the turning point. An outward launch from the origin at $\tau =0$ hasThis is the radial timelike geodesic in anti-de Sitter spacetime. Its first return to the origin isindependent of the outward launch speed for every nontrivial launch $k>1$. The coordinate $r$ is nonnegative, so this sine expression describes the excursion up to that return; subsequent passage through the origin requires changing the radial direction or using a signed diameter coordinate. The speed changes the amplitude, not the harmonic frequency.

Circular periods. The effective radial potential is $V_{\mathrm{eff}}=A(1+h^2/r^2)$. For a circle at $r_0$, $V_{\mathrm{eff}}^{\prime }=0$ gives $h^2={\lambda }^2{r}_0^4$, and normalization gives $k=A(r_0)$. Thus $\dot{t}=1$ and $\dot{\varphi }=\pm \lambda$. A stable circular timelike geodesic in anti-de Sitter spacetime therefore hasindependently of radius. This is the same isochronous behavior as the central harmonic radial equation. A static clock placed at the orbit's radius instead measures $\sqrt{1+{\lambda }^2{r}_0^2}\Delta t$, which does depend on radius; it is a different clock from the orbiting particle's proper time.

Local equivalence and tides. In Newtonian mechanics, inertial mass $m_{\mathrm{i}}$ measures the response to a force, whereas passive gravitational mass $m_{\mathrm{g}}$ couples to an imposed Newtonian gravitational field. A freely falling body obeysThe observed composition independence of free fall says that this ratio is universal. With a common choice of units it is one. This is the Newtonian content of the weak equivalence principle for test particles; it does not by itself specify all nongravitational physics.

The Einstein equivalence principle further says that local nongravitational experiments in a sufficiently small freely falling laboratory have the form predicted by special relativity, independently of the laboratory's location and motion. In such a local inertial frame, the metric tensor is locally the Minkowski metric and the Christoffel symbols vanish at the selected event. The strong equivalence principle extends the universality claim to local gravitational experiments and bodies with significant gravitational self-energy. None of these statements removes measurable tidal forces across a finite apparatus.

In an ideal small freely falling lift, a released ball initially floats relative to the cabin, because both undergo the same free fall. Two separated balls, however, need not retain their separation: they sample different parts of the gravitational field. For a point source, linearizing $\mathbf{g}=-GM\mathbf{r}/r^3$ across a small separation $\mathbf{\xi }$ gives the tidal tensor predictionA radial separation stretches with coefficient $2GM/r^3$, whereas a transverse separation is compressed with coefficient $-GM/r^3$. Initially parallel velocities therefore develop relative motion even though the individual bodies experience zero proper acceleration.

The relativistic statement is geodesic deviation. With the Riemann curvature tensor conventiona connecting vector between neighboring freely falling worldlines satisfiesHere $D/d\tau$ is the covariant derivative along the reference geodesic, and $U^{\mu }$ its four-velocity. Curvature can be nonzero where the Christoffel symbols have been made zero. In the weak, static, slow-motion limit this equation reduces to ${\ddot{\xi }}^i=-{\partial }_i{\partial }_j\Phi {\xi }^j$, recovering the tidal tensor above. A finite falling lift can detect gravity through these relative accelerations; local equivalence does not imply a globally uniform field.

Weak-field connection and potential. Take $x^0=ct$ and the $+---$ metric signature. There is an index error in the printed formula for the Christoffel symbols: its first differentiated metric component must be $g_{\nu 0}$, not $g_{\mu 0}$. The correctly contracted expression isThe offending term vanishes under the stipulated time independence, so this correction does not change the intended calculation. Write $g_{\mu \nu }={\eta }_{\mu \nu }+h_{\mu \nu }$ and treat both $h$ and its derivatives as first order. To that order $g^{00}=1-h_{00}$, $g^{0i}=O(h)$, and $g^{ij}=-{\delta }^{ij}+O(h)$. A stationary metric tensor givesFor the temporal component the inverse-metric coefficient is already first order, soFor a spatial component the zeroth-order inverse metric contributes:The time component of the approximate geodesic equation says that $dt/d\tau$ is constant to the retained order. In its spatial component, convert the proper time derivatives to coordinate-time derivatives and cancel $(dt/d\tau {)}^2$:Comparison with the Newtonian equation $d^2\mathbf{x}/d{t}^2=-\nabla \Phi$ identifies $h_{00}=2\Phi /c^2+C$. For an isolated source, choose the gravitational potential to tend to zero at infinity and normalize the time coordinate there, giving $C=0$. ThusThe quantity $\Phi$ is the Newtonian gravitational potential, or gravitational potential energy per unit mass, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}={\mathrm{m}}^2{\mathrm{s}}^{-2}$. For a point source, $\Phi =-GM/r$. A static clock has $d\tau =\sqrt{g_{00}}dt\approx (1+\Phi /c^2)dt$, connecting the Newtonian limit with gravitational time dilation.

Frequency of the orbiting source. Treat the gas blob as a test particle on a timelike circular Schwarzschild geodesic, with pressure and other nongravitational forces negligible. Treat our very distant location as infinity, so the observer's own gravitational potential is negligible. These are the assumptions that make a unique answer in terms of $R_S/r$ possible.

Put $f=1-R_S/r$ and use a local static observer at the emission point as an intermediate measuring frame. The remote observer lies on the symmetry axis of the orbit. A ray arriving on that axis has zero conserved axial angular momentum, so its local direction is perpendicular to the emitter's azimuthal velocity. Bending within a meridional plane does not alter this orthogonality. The transverse relativistic Doppler effect therefore giveswhere $v$ is the orbital speed measured by that local static observer. The gravitational redshift between static Schwarzschild observers then givesA gravitational circular orbit has $v^2=GM/r+O((GM/r{)}^2/c^2)$ in the weak-field limit, so $v^2/c^2=R_S/(2r)+O((R_S/r{)}^2)$. Expanding both factors,Solving for the emitted frequency givesThe gravitational part is $R_S/(2r)$ and the transverse kinematic time dilation contributes another $R_S/(4r)$; the absence of line-of-sight motion does not remove this second effect.

As an exact check, the local speed on a timelike circular Schwarzschild geodesic satisfies $v^2/c^2=\mu /(r-2\mu )$, where $\mu =GM/c^2=R_S/2$. Therefore the face-on frequency shift of a circular Schwarzschild source iswhose expansion agrees with the boxed answer. If the gas is supported or accelerated by pressure or magnetic stresses rather than following a geodesic, its orbital speed is not fixed by the mass and radius alone; the appropriate first-order result is instead ${\nu }_{\mathrm{em}}\approx {\nu }_{\mathrm{ob}}[1+GM/(r{c}^2)+v^2/(2c^2)]$, with $v$ supplied independently.

Write $A(r)=1-2\mu /r$, with $\mu =GM/c^2$. On the equatorial plane, the geodesic Lagrangian isThe coordinates $t$ and $\varphi$ are cyclic, giving $\boxed{A\dot{t}=k,r^2\dot{\varphi }=h}$. The conserved physical energy per unit rest mass is $c^{2}k$, while $h$ is the signed angular momentum per unit rest mass. The radial Euler-Lagrange equation, with $A^{\prime }=2\mu /r^2$, isThese are the requested geodesic equations. The equatorial condition is consistent with the polar equation by the spherical symmetry of the Schwarzschild metric.

The normalization of the four-velocity is $c^2=c^{2}A{\dot{t}}^2-A^{-1}{\dot{r}}^2-r^2{\dot{\varphi }}^2$. Insert the two conserved quantities and multiply by $A$ to obtainThis is a one-dimensional radial energy equation. Half of its left side is radial kinetic energy per unit mass plus an effective potential: the Newtonian centrifugal and gravitational terms acquire the extra attractive contribution $-\mu h^2/r^3$.

A timelike circular Schwarzschild geodesic requires $\dot{r}=\ddot{r}=0$. Equivalently, set the derivative of $V(r)=h^2/r^2-2\mu h^2/r^3-2\mu c^2/r$ to zero. At $r=R$ this gives $h^2(R-3\mu )=\mu c^2{R}^2$, and substitution in the radial energy equation yieldsSuch timelike circular orbits require $R>3\mu$. Their coordinate angular velocity satisfies ${\Omega }^2=\mu c^2/R^3$, and their proper clock rate is $d{\tau }_s/dt=A/k=\sqrt{1-3\mu /R_s}$.

For the static North-Pole clock, $d{\tau }_E/dt=\sqrt{1-2\mu /R_E}$. Consecutive overhead passages occur one coordinate orbital period apart, and the repeated Earth-satellite geometry gives equal signal travel delays. The comparison of circular-orbit and static Schwarzschild clocks therefore givesThis includes both gravitational and orbital-motion effects; a separate Newtonian time-dilation correction must not be added again.

For an estimate, Kepler's third law gives $R_s=42100\mathrm{km}(3/24{)}^{2/3}=10525\mathrm{km}$. The same approximate geostationary radius and period give $\mu =4{\pi }^2{R}_{\mathrm{geo}}^3/(c^2{T}_{\mathrm{geo}}^2)\simeq 4.4\mathrm{mm}$. Expanding the ratio to first order,The satellite clock consequently gains about $0.7\mu \mathrm{s}$ over a three-hour orbit in this approximation. The surface observer is taken to be static, and the spacetime model is the spherical Schwarzschild metric specified in the problem.

For a particle on a circle, the tangential speed is $v=r\dot{\varphi }$, the moment of inertia is $I=m{r}^2$, and the angular momentum is $L=I\dot{\varphi }=mrv$. Hence its kinetic energy isThere is no potential energy variation along the circle in the stated model.

Applying the Hamiltonian operator to ${\psi }_n$ gives ${\psi }_n^{\prime \prime }=-n^2{\psi }_n$, and thereforeThe ground state $n=0$ has zero kinetic energy. Every positive energy level is of quantum degeneracy two, because $n$ and $-n$ correspond to opposite angular momenta with equal squared magnitude.

The confinement scale suggests a tangential momentum scale $\hslash /r$ and thus an excitation energy scale ${\hslash }^2/(m{r}^2)$, consistent with the Heisenberg uncertainty principle. More precisely the de Broglie wavelength must fit an integer number of times around the circumference: $2\pi r=|n|\lambda$, so $|p|=h/\lambda =|n|\hslash /r$. This explains both the $r^{-2}$ dependence and the $n^2$ sequence.

One must not infer a nonzero ground-state energy by applying $\Delta \varphi \Delta L\ge \hslash /2$ naively. Angle is periodic and is not a globally canonical position operator on the circle. The uniform $n=0$ wavefunction, with $\Delta L=0$, is a concrete counterexample to that naive inequality. The ordinary Cartesian Heisenberg uncertainty principle is not violated by the zero-angular momentum state.

A Stern-Gerlach experiment uses a nonuniform magnetic field to exert force $F_z\simeq {\mu }_z{\partial }_{z}B$ on a magnetic dipole. The two resolved beams reveal discrete projections of spin, rather than a continuum of classical dipole orientations. For the ground-state silver chemical atom, filled shells contribute no net angular momentum and the outer $5s$ electron has orbital angular momentum zero, leaving an effective spin-$1/2$ magnetic moment. This provides evidence for the electron's two-valued intrinsic spin.

Neutral silver chemical atoms can pass through the apparatus without the large charge-dependent Lorentz force that would deflect and broaden a free-electron beam. Neutrality isolates the small magnetic-gradient force far more cleanly. The later parts use an ideal spin filter model; a literal free-electron apparatus requires additional control of charge forces.

A cold, self-gravitating region in a molecular cloud can collapse when gravity overcomes thermal, turbulent and magnetic support. Increasing central density produces a protostar, while radiation removes part of the released gravitational energy. The cloud initially carries angular momentum; material that cannot fall radially to the centre settles into a rotating protoplanetary disk. Collisions and cooling reduce random vertical motion, while angular-momentum transport allows some material to accrete and some to move outwards. Bipolar outflows and magnetic stresses can remove additional angular momentum.

The schematic sequence is $\text{rotating cloud}\to \text{protostar plus disk}\to \text{accreting solids and gas}\to \text{planetary system plus cleared disk}$. In the cool disk, grains condense and collide; settling and concentration can produce planetesimals, whose gravitational growth leads to planetary embryos. Massive cores may capture gas before the disk disperses; sufficiently massive unstable disks can also form bound condensations. Migration and interactions modify the final orbits. Radiation, accretion and outflows eventually disperse the remaining gas. Thus planet formation uses the same rotating reservoir produced by ordinary star formation, although the efficiency and final architecture are not universal and individual growth stages remain active research topics.

For a central potential use $L=m|\dot{\mathbf{r}}{|}^2/2-V(|\mathbf{r}|)$. An infinitesimal rotation about any constant unit vector $\mathbf{n}$ has $\varphi =0$ and $\mathbf{\psi }=\mathbf{n}\times \mathbf{r}$. Noether's theorem gives $Q=m\dot{\mathbf{r}}\cdot (\mathbf{n}\times \mathbf{r})=\mathbf{n}\cdot \mathbf{L}$, where $\mathbf{L}=\mathbf{r}\times m\dot{\mathbf{r}}$. Since $\mathbf{n}$ is arbitrary, every component of angular momentum is conserved. A time-translation symmetry has $\varphi =1$, $\mathbf{\psi }=0$, giving $Q=-E$ andas conserved quantities.

To match the printed order $df/dt=\{H,f\}$, define the Poisson bracket with the sign conventionHamilton's equations imply $\dot{f}={\sum }_i(f_{q_i}{H}_{p_i}-f_{p_i}{H}_{q_i})=\{H,f{\}}_{\mathrm{paper}}$. With the other common Poisson bracket convention the correct statement is instead $\dot{f}=\{f,H{\}}_{\mathrm{standard}}$. For example $H=p^2/(2m)$, $f=q$ gives $\{H,q{\}}_{\mathrm{standard}}=-p/m$, contradicting the printed order unless its sign convention is adopted. Since $H_t=0$ and $\{H,H\}=0$, the Hamiltonian is conserved. Also ${\dot{p}}_{\varphi }=-H_{\varphi }=0$: this canonical momentum is the axial angular momentum.

A classical orbiting electron accelerates and therefore radiates according to the Larmor formula. Its mechanical energy becomes more negative, its radius decreases, and the orbit spirals inward rather than remaining a stable atom. The Bohr model replaces this prediction with postulated nonradiating stationary states and quantized orbital angular momentum, together with photon emission or absorption only between states, $h\nu =|E_n-E_m|$.

For a heavy nucleus, $m_{e}vr=n\hslash$. Since the de Broglie wavelength is $h/(m_{e}v)$, this impliesThe electronic wave fits an integer number of wavelengths around the orbit. This is a standing-wave consistency interpretation of the quantization rule, not a classical mechanism that prevents an accelerating charge from radiating. For finite nuclear mass, the two-body relative coordinate uses reduced mass $\mu =m_{e}M/(m_e+M)$; the next derivation retains that correction. In the limit $M\gg m_e$, it reduces to the fixed-nucleus electron description above.

Let $r$ be the electron-nucleus separation and use the reduced mass $\mu =m_{e}M/(m_e+M)$. Coulomb attraction supplies the relative centripetal force, and Bohr quantization fixes the relative angular momentum:Eliminating $v$ givesThe total energy is $E=\mu v^2/2-Z{e}^2/(4\pi {ϵ}_{0}r)=-Z{e}^2/(8\pi {ϵ}_{0}r)$, henceHere $a_0$ is the ordinary Bohr radius and $R_{\infty }$ the infinite-nuclear-mass Rydberg constant. The electron's radius about the centre of mass is $M{r}_n/(M+m_e)$; for a heavy nucleus this distinction and the reduced-mass correction are small, but the supplied nuclear mass permits them to be stated precisely.

Use the relative star–planet separation and specific angular momentum per unit reduced mass,For $r=l/(1+e\cos \theta )$, differentiation gives $\dot{r}=Je\sin \theta /l$, while the transverse speed is $r\dot{\theta }=J/r=J(1+e\cos \theta )/l$. Adding the squared orthogonal components yieldsThe Binet equation for the relative two-body inverse-square force is $u^{\prime \prime }+u=G(M+m)/J^2$, with $u=1/r$. The conic has $u^{\prime \prime }+u=1/l$, soThe printed division by $l$ is an error: $J^2$ has dimensions $\mathrm{lengt}{\mathrm{h}}^4/tim{\mathrm{e}}^2$, while $G(M+m)/l$ has dimensions of speed squared. A circular orbit independently checks the correction: $J^2=r^2{v}^2=G(M+m)r$, with $l=r$.

For a closed ellipse, $l=a(1-e^2)$ and $1+2e\cos \theta +e^2=2l/r-(1-e^2)$. Substitution now gives the vis-viva equation,This is relative orbital speed; either body's speed about the barycentre follows by the corresponding mass-ratio factor.

For an infinitesimal continuous symmetry $\delta q_i=\epsilon {\xi }_i(q,t)$, allow the Lagrangian to change by a total derivative, $\delta L=\epsilon dF/dt$. Direct variation givesOn a solution of the Euler-Lagrange equations, the first term vanishes. This proves Noether's theorem in the fixed-time form:Spatial translations yield total momentum, and rotations yield angular momentum. Time translations yield energy when the Lagrangian has no explicit time dependence; this follows either from the extended time-transformation form of the theorem or by differentiating $H={\sum }_i{p}_i{\dot{q}}_i-L$ to obtain $\dot{H}=-\partial L/\partial t$.

The central pair potential is invariant under simultaneous spatial translations, simultaneous rotations and time translations. The corresponding time-independent Noether charges are total momentum $P=p_1+p_2$, total angular momentum about a fixed origin, and total energy $H$.

In one dimension this supplies $P$ and $H$: two charges. In two dimensions it supplies $P_x,P_y,L_{xy},H$: four charges. In three dimensions it supplies three momentum components, three angular-momentum components and $H$: seven charges. For generic states these listed charges are independent in those dimensions. Galilean boosts can also be expressed by explicitly time-dependent conserved combinations, but are not part of the requested translation-rotation-energy count.

A classically accelerating Coulomb electron radiates and loses orbital energy, so a stable nonradiating orbit is impossible within that unmodified theory. Bohr instead postulates stationary nonradiating orbits, quantized angular momentum $\mu vr=n\hslash$ and photon emission/absorption with $h\nu =E_i-E_f$. Here $\mu =m_{e}M/(m_e+M)$ describes relative motion. The angular-momentum condition is the standing-wave relation $\boxed{2\pi r=n{\lambda }_{\mathrm{dB}}}$ for the relative orbit. These are additional quantum postulates, not consequences of classical radiation theory.

The Bohr model assumes Coulomb attraction and classical circular motion between transitions; selected stationary orbits do not radiate; their angular momentum is $mvr=n\hslash$; and a transition exchanges a photon of energy $h\nu =E_i-E_f$. The first two dynamical equations giveHere $n=1,2,\dots$, $m$ is the electron mass under the fixed-proton approximation, $e$ its charge magnitude, and ${ϵ}_0$ the vacuum permittivity. The stationary-orbit postulate deliberately suspends the classical radiation prediction; this is a historical model, not a consistent deduction from classical electrodynamics.

In two spatial dimensions there are only two kinetic matrices and one mass matrix. The three Pauli matrices square to identity and anticommute pairwise, sois a two-component Dirac Hamiltonian with $H_{2+1}^2=c^2(p_x^2+p_y^2)+m^2{c}^4$. It therefore reproduces the stated first-order equation and its Klein-Gordon square. Requiring a fourth anticommuting matrix in three spatial dimensions is precisely what prevented this two-component construction above.

Printed angular-momentum qualification. Define the dimensionless spin $s_i=-i{\epsilon }_{ijk}{\alpha }_j{\alpha }_k/4$ and the physical orbital angular momentum ${\ell }_i={\epsilon }_{ijk}{x}_j{p}_k$, with $[x_i,p_j]=i\hslash {\delta }_{ij}$. ThenThe source prints $L_i=i{\ell }_i$ instead. Taken literally, it is anti-Hermitian, has incompatible units with the dimensionless spin when they are added, and givesIt cannot yield the requested usual angular-momentum algebra or conservation law. For a concrete counterexample set $\hslash =1$ and take a wave with momentum along $z$. Sincethe printed $J_i=i{\ell }_i+s_i$ obeys $[H_D,J_x]=(1+i)c{\alpha }_y{p}_z\ne 0$.

The physically consistent operator is ${\mathbf{J}}_{\mathrm{phys}}=\mathbf{\ell }+\hslash \mathbf{s}$. It satisfiesThe cancellation follows directly from the two mixed commutators above. In a spherical basis the free spinor can be labelled by energy, $j=1/2,3/2,\dots$, and $m_j=-j,\dots ,j$, together with a parity or equivalent Dirac angular label distinguishing the two orbital couplings $\ell =j\pm 1/2$. Alternatively one can choose a momentum and helicity basis. One cannot simultaneously assign all components of momentum and all angular labels as independent commuting observables.

For a neutral condensate with order parameter $\psi =\sqrt{n_s}{e}^{i\varphi }$, the superfluid velocity away from singularities is ${\mathbf{v}}_s=\hslash \nabla \varphi /m$. Single-valuedness around a closed contour gives $\oint \nabla \varphi \cdot d\mathbf{l}=2\pi n$, hence quantization of circulation:The mass is that of the condensed object: an atom for a bosonic atomic superfluid, or a pair for a paired-fermion condensate. A nonzero winding cannot be continuously removed while the order parameter remains nonzero throughout a spanning region. A vortex allows its phase to wind while its amplitude vanishes in a core.

For a charged condensate, minimal coupling gives $m{\mathbf{v}}_s+q\mathbf{A}=\hslash \nabla \varphi$. ThusUsing ${\mathbf{j}}_s=n_{s}q{\mathbf{v}}_s$ and ${\lambda }_L^2=m/({\mu }_0{n}_s{q}^2)$ gives the fluxoid and flux quantization in a superconductor relation $\Phi +{\mu }_0{\lambda }_L^2\oint {\mathbf{j}}_s\cdot d\mathbf{l}=nh/q$. The signs of oriented flux and winding depend on the signed pair charge; the magnitude of the elementary superconducting flux quantum is ${\Phi }_0=h/(2e)$. Flux alone is quantized when the current term is negligible on a contour deep in the bulk. A thin ring or a contour near a core can instead require the full fluxoid.

In a neutral superfluid, vortex flow carries kinetic energy and angular momentum. Rotation can favor vortices because equilibrium minimizes the rotating-frame quantity $E-\mathbf{\Omega }\cdot \mathbf{L}$. The core size is set by a coherence or healing scale. In a type-II superconductor, an Abrikosov vortex carries quantized magnetic flux and circulating supercurrent; its core scale is the coherence length and its magnetic screening scale is the penetration depth. Between the lower and upper critical fields, a vortex lattice can be favored. Flux pinning affects how that structure responds to current. Type-I bulk superconductors instead favor macroscopic normal/superconducting separation rather than the usual stable dilute Abrikosov-vortex regime.

Let the fixed direction of motion be the unit vector $\mathbf{n}$ and decompose $\mathbf{s}=s_{\parallel }\mathbf{n}+{\mathbf{s}}_{\perp }$. In ii the two parallel contributions cancel, whereas the transverse equation isFor the stated speed law, $1-v^2/c^2=e^{-2\Gamma \tau }$, hence $\gamma =e^{\Gamma \tau }$. Since the particle is initially at rest, $\gamma (0)=1$, andThe corresponding spin four-vector has components $s^0=\gamma \beta s_{\parallel }(0)$, $s_{\parallel }^{\mathrm{spatial}}=\gamma s_{\parallel }(0)$, and ${\mathbf{s}}_{\perp }^{\mathrm{spatial}}={\mathbf{s}}_{\perp }(0)$. Its squared norm is therefore $s_{\parallel }(0{)}^2+|{\mathbf{s}}_{\perp }(0){|}^2$, independent of time. The transverse decay of the source's laboratory three-vector is a boost-normalization effect, not a loss of proper angular momentum or a torque.

For a real increasing speed take $\Gamma >0$ and $\tau \ge 0$. The prescribed profile has a singular initial acceleration: $v\sim c\sqrt{2\Gamma \tau }$ near zero. The differential transport calculation applies for $\tau >0$ and the solution has the continuous initial limit displayed above. This trajectory should not be silently described as motion at constant proper acceleration.

The coordinate $\theta$ is cyclic. Its conjugate momentum is therefore a second conserved quantity,It differs from mechanical angular momentum $m{r}^2\dot{\theta }$. The magnetic contribution is essential to the rotational Noether charge in this axisymmetric gauge.