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For a bounded conducting sphere with uniform magnetic diffusivity and an insulating exterior, an incompressible flow tangent to concentric spheres cannot sustain dynamo action. Neither the velocity nor the magnetic field needs to be axisymmetric. The radial magnetic scalar has a homogeneous advection-diffusion equation, so the poloidal field decays. Once it is absent, the toroidal magnetic potential also has a homogeneous scalar advection-diffusion equation. This is a theorem about spherical flow geometry, not about a general cylindrical flow with an axial component.

A torsional Alfvén wave couples azimuthal velocity to a toroidal magnetic field around a fixed poloidal magnetic field. Axisymmetric induction gives ${\partial }_t{B}_{\varphi }=R{\mathbf{B}}_p\cdot \nabla \Omega$ and magnetic tension gives $\rho R^2{\mu }_0{\partial }_t\Omega ={\mathbf{B}}_p\cdot \nabla (R{B}_{\varphi })$. Their combination is the displayed wave equation. Its local dispersion relation is ${\omega }^2=(\mathbf{k}\cdot {\mathbf{B}}_p{)}^2/({\mu }_0\rho )$.

For a small rigid sphere translating with velocity $\mathbf{V}(t)$, the integrated thickness source ${\rho }_0\int \mathbf{V}\cdot {\mathbf{n}}_{s}dS$ vanishes. The first spatial moment is instead ${\rho }_0\int y_j\mathbf{V}\cdot {\mathbf{n}}_{s}dS={\rho }_0\mathcal{V}{V}_j$, where $\mathcal{V}$ is the sphere volume. Expanding the retarded time across the compact source gives the displayed first nonzero thickness field. It is an acoustic dipole, at the same order as acoustic loading noise for a translating sphere. The added mass of a sphere supplies half this volume-mass coefficient through the loading term; their sum is $3{\rho }_0\mathcal{V}\mathbf{n}\cdot \ddot{\mathbf{V}}/(8\pi c_0^{3}r)$.

A two-sided line plume whose axial velocity and density anomaly both have triangular transverse profiles. For half-width $b$, ${\int }_{-b}^{b}f(x/b{)}^{n}dx=2b/(n+1)$. These profile moments set the different coefficients of stored mass, momentum, and their fluxes. It differs from a wall plume because both edges entrain ambient fluid.

Turbulent magnetic diffusivity is an effective transport coefficient produced by correlated small-scale velocity and magnetic field. It supplements microscopic magnetic diffusivity in a mean-field dynamo description. Its identification requires a specified mean-field geometry and averaging procedure; in a planar flux problem, the correlation $\overline{w{a}^{\prime }}=-{\eta }_t{B}_{0}f(z)$ describes transport relative to a prescribed uniform field.

Turbulent magnetic pumping is an effective transport of the mean magnetic field produced by correlations in inhomogeneous or anisotropic turbulence. Its contribution to the mean-field electromotive force is ${\mathcal{E}}_{\mathrm{pump}}=\mathbf{\gamma }\times \overline{\mathbf{B}}$. In the mean induction equation this enters as $\nabla \times (\mathbf{\gamma }\times \overline{\mathbf{B}})$, the same algebraic form as advection by a velocity $\mathbf{\gamma }$. Thus it supplements the mean material velocity without requiring an equal mean mass flow. Downward pumping can help retain or couple fields in a solar dynamo.

A circular turbulent round jet supplied with kinematic momentum flux $m_0$ but negligible buoyancy flux conserves $m=m_0$. A top-hat plume model and constant Batchelor entrainment give $dq/ds=2\alpha \sqrt{\pi m_0}$. From a point source with zero volume flux, $q=2\alpha \sqrt{\pi m_0}s$, radius $b=2\alpha s$, and axial velocity $V=m_0/q\propto s^{-1}$.

The 1972 paper Around-the-World Atomic Clocks: Observed Relativistic Time Gains by J. C. Hafele and others reports four cesium-beam atomic clocks flown around the world in October 1971, eastward and westward. Their readings were compared with the U.S. Naval Observatory reference time scale before and after travel. An electronic time-interval counter resolved differences to the nearest nanosecond; intercomparisons among the flying clocks continued during travel. The final ensemble results were $-59\pm 10\mathrm{ns}$ eastward and $+273\pm 7\mathrm{ns}$ westward; the quoted spreads are standard deviations. These are corrected clock offsets, not raw readings or aircraft travel times. Four clocks allowed identification of individual rate changes through patterns in their mutual differences. The authors also report magnetic shielding and laboratory orientation checks. Their analysis could not detect a disturbance common to every flying clock, and the small ensemble limits statistical conclusions. This is a concrete reunited-clock experiment with apparatus, readout and drift controls.^[ref]

The companion paper Around-the-World Atomic Clocks: Predicted Relativistic Time Gains by J. C. Hafele and others computes $-40\pm 23\mathrm{ns}$ eastward and $+275\pm 21\mathrm{ns}$ westward from the recorded flight paths. These are predictions, with flight-data uncertainty estimates. Both time dilation from motion and gravitational time dilation matter: aircraft clocks at altitude experience a different gravitational potential, while the rotating Earth means a ground clock is not the inertial home clock of our simple model. Eastward and westward motion change the aircraft's velocity relative to an approximately nonrotating inertial frame, not merely its speed relative to the ground. The measured agreement therefore tests combined special relativity and general relativity; the westward gain does not contradict the maximum-proper time result for fixed endpoints in Minkowski spacetime.^[ref]

A complementary microscopic test is Measurements of relativistic time dilatation for positive and negative muons in a circular orbit by J. Bailey and others. The authors measured muon decay lifetimes in a storage ring at Lorentz factor $\gamma =29.33$: $64.419\pm 0.058\mu \mathrm{s}$ for positive muons and $64.368\pm 0.029\mu \mathrm{s}$ for negative muons. The reported positive-muon lifetime agrees with the independently measured lifetime at rest and special relativity to a fractional error of $2\times 10^{-3}$ at 95% confidence. This is a decay-lifetime readout under continuous centripetal acceleration, rather than a reunion of portable clocks. It supports the clock hypothesis for this physical process and regime; the inspected abstract does not establish universal immunity of every clock mechanism to acceleration. ^[ref]

An instantaneous reversal is an idealization. A real traveller changes velocity over a finite time, follows a smooth world line, and has finite proper acceleration. Special relativity still calculates the elapsed proper time using the home inertial frame and the same integral; general relativity is not needed merely because the traveller accelerates in otherwise flat spacetime. The clock hypothesis identifies an ideal clock's reading with this geometric proper time.

One explicit smooth trip between the same $O$ and $R$ starts and ends at rest in the home inertial frame. ChooseIts velocity reverses continuously at $T/2$, and its proper acceleration is bounded because $v_{\max }<c$ and $dv/dt$ is finite. With $v_{\max }=0.6c$ and $T=10\mathrm{yr}$, numerical integration givesThis comparison fixes the reunion spacetime events and maximum speed, not the turnaround distance: the smooth trip reaches $v_{\max }T/\pi \simeq 1.910$ light-years, whereas the ideal trip reaches $3$ light-years. The smooth traveller spends less time at high speed and therefore accumulates more proper time than the ideal traveller, while still accumulating less than the home clock.

For a sufficiently short smooth reversal replacing the corner, the proper time approaches the piecewise inertial result. A reversal interval of home duration $\delta t$ contributes between zero and $\delta t$ to the clock reading; shrinking that interval cannot create a finite extra loss concentrated at the corner. The finite difference survives in the long coasting legs. Keeping the same reversal manoeuvre while lengthening those legs changes the age difference. Conversely, two travellers following mirrored paths with the same speed history accumulate equal proper times, even though both accelerate. Thus acceleration distinguishes paths and enables reunion here, but an amount of acceleration alone does not determine the age difference.

The changing simultaneity assignment also becomes continuous. At a traveller spacetime event $(t,x(t))$ with instantaneous velocity $v(t)$, the momentarily comoving inertial frame assigns home time $t_H=t-v(t)x(t)/c^2$. Differentiating givesDuring the reversal at positive $x$, the last term advances the assigned home time rapidly. It describes the traveller's changing relativity of simultaneity, not a locally measured alteration of the home clock's ticking.

Work in Minkowski spacetime, neglect gravity, and choose the stay-at-home clock's inertial frame. Let departure be $O=(t,x)=(0,0)$ and reunion be $R=(T,0)$. The home clock follows $x_H(t)=0$. A traveller moves outward with velocity $+v$ and returns with velocity $-v$, where $0<v<c$ and $c$ is the speed of light. The ideal world line has a sharp turnaround at $P=(T/2,vT/2)$:Each straight leg lies in an inertial frame, but no single traveller inertial frame covers both legs.

For any differentiable timelike curve in this home inertial frame, its clock accumulates proper time through the integralApply it separately to the traveller's two legs. With the Lorentz factor $\gamma =(1-v^2/c^2{)}^{-1/2}$,For $v=0.6c$ and $T=10\mathrm{yr}$, the Lorentz factor is $\gamma =1.25$. The turnaround is $3$ light-years away at home time $5\mathrm{yr}$. The traveller records $4\mathrm{yr}$ outward and $4\mathrm{yr}$ returning: the reunited clocks read $10\mathrm{yr}$ and $8\mathrm{yr}$. These are illustrative model values, not experimental measurements.

The same integral gives a short geometric explanation: $\sqrt{1-v(t{)}^2/c^2}\le 1$, so every such competing timelike curve between these fixed spacetime events has $\Delta \tau \le T$. Equality requires zero velocity throughout, apart from sets of zero duration. Thus the straight home world line maximizes proper time in this Minkowski spacetime problem. This statement concerns fixed endpoints and an inertial home clock; in a gravitational field, compare paths using the appropriate metric tensor.

For a nonrotating uniform-stratification fluid under the Boussinesq approximation driven at $0<\omega <N$ by a localized vertical-wall velocity, positive vertical wavenumbers radiate down and right and negative ones up and right. The real velocity separates into functions of $Z_{+}$ and $Z_{-}$. Its in-phase component is half the imposed boundary profile on each beam; the quadrature component is its Hilbert transform partner. Thus compact localization at maximum boundary speed does not imply compact localization of all components at every phase. A finite-duration source contains several frequencies and spreads energy over their different ray angles and group speeds.

In two-dimensional vortex dynamics, solve the Poisson equation for the streamfunction with the prescribed boundary conditions. On the plane its fundamental solution is $(2\pi {)}^{-1}\log |\mathbf{x}|$, so differentiating the displayed inverse gives the planar Biot-Savart kernel. The harmonic function ${\psi }_h$ represents boundary and global-flow data. An additive constant changes no velocity; circulation around holes and a uniform flow are additional degrees of freedom if their boundary or far-field values are not prescribed. The same inverse acts independently on each level in nonrotating layerwise-two-dimensional vortex dynamics.

For inviscid planar incompressible flow, write velocity as $(-{\psi }_y,{\psi }_x)$ and vorticity as $\zeta ={\nabla }^2\psi$. Taking curl of the Euler equations for an inviscid fluid removes pressure and gives the displayed transport equation, with $J(\psi ,\zeta )={\psi }_x{\zeta }_y-{\psi }_y{\zeta }_x$. The Poisson equation then recovers the streamfunction at each instant. A velocity tangent to the plane cannot stretch its perpendicular vorticity vector.

Let the lower fluid have thickness $h$ and viscosity $\mu$, and the upper fluid thickness $\alpha h$ and viscosity $\beta \mu$, with common density $\rho$. For steady flow down a plane inclined at angle $\theta$, set $H=(1+\alpha )h$, $A=\rho g\sin \theta /\mu$. The common hydrostatic pressure is $p_{\mathrm{atm}}+\rho g\cos \theta (H-y)$. No slip at the wall, continuous velocity and shear stress at $y=h$, and zero upper surface shear give $u_1=A(Hy-y^2/2)$ and $u_2(h+z)=A{h}^2(\alpha +1/2)+(A/\beta )(\alpha hz-z^2/2)$. These follow by integrating ${\mu }_i{u}_i^{\prime \prime }=-\rho g\sin \theta$ in each layer. The stress applied to the lower fluid equals the downslope weight of the upper layer, independent of its viscosity; the upper viscosity affects its shear rate but not that transmitted stress.