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Use signed velocities in Skaro's inertial frame; the returning velocity $w$ must be negative. For the first two rockets, the Lorentz transformation of infinitesimal separations is $dx={\gamma }_u(d{x}^{\prime }+ud{t}^{\prime })$, $dt={\gamma }_u(d{t}^{\prime }+ud{x}^{\prime }/c^2)$. Dividing with $d{x}^{\prime }/d{t}^{\prime }=v^{\prime }$ gives the relativistic velocity-addition formulaFor the third rocket, its signed velocity in the second frame is $-w^{\prime \prime }$, so another application givesSince $0<v<c$ and $0<w^{\prime \prime }<c$, the denominator $1-v{w}^{\prime \prime }/c^2$ is positive. The third rocket returns from positive $x$ to Skaro if and only if $w<0$, henceEquality leaves it stationary in Skaro's frame, so it never returns from $x=2L$; a smaller $w^{\prime \prime }$ leaves it travelling outward.

Let $O$ denote departure, $A$ boarding the second rocket, $B$ boarding the third, and $C$ reunion with Skaro. In coordinates $(t,x)$, their positions on the worldline areIndeed, the first two coordinate durations are $L/u$ and $L/v$, and the returning leg has duration $-2L/w>0$. Equivalently, their $(ct,x)$ coordinates are obtained by multiplying each displayed time by $c$; these are the coordinates used for the spacetime diagram.

The diagram uses representative velocities satisfying the return condition; its event labels apply to the general formulas above. Skaro's elapsed time isOn a constant-velocity leg the traveller's proper time is $\Delta t/{\gamma }_V=\Delta t\sqrt{1-V^2/c^2}$, by time dilation. Neglecting the specified brief accelerations, Davros therefore ages byEvery contribution is positive because $w<0$. Each square-root factor is smaller than one, so $\Delta \tau <T$: Davros returns younger than a companion who remained on Skaro. The elapsed times compare the same departure and reunion events; they cannot be obtained by applying one time-dilation factor to the whole journey, whose worldline has three different velocities.

Write $u_i=⟨v_i⟩$ and use summation over repeated Cartesian indices. A mass-normalized galactic distribution function obeys the Collisionless Boltzmann equationAssume $f$ decays in velocity sufficiently fast that the necessary boundary terms vanish. Its zeroth velocity moment gives the continuity equationFor the first velocity moment, multiply by $v_j$ and integrate. In the gravitational term, integration by parts gives $\int v_j{\partial }_{v_i}f{d}^{3}v=-{\delta }_{ij}\rho$. ThereforeThe velocity-dispersion tensor is the centered second moment, so $⟨v_i{v}_j⟩=u_i{u}_j+{\sigma }_{ij}^2$. Expanding the streaming terms, the factor multiplying $u_j$ cancels by the continuity equation:Hence the Jeans equation isIts fluid analogue is the momentum equation with pressure tensor $P_{ij}=\rho {\sigma }_{ij}^2$. For isotropic random motion, $P_{ij}=P{\delta }_{ij}$, and this becomes the Euler equations for an inviscid fluid:The tensor form allows anisotropic stellar stresses. In a fluid an equation of state can provide a pressure closure; the Jeans equation alone does not determine the velocity-dispersion tensor of a collisionless stellar system.

Use relative energy $E=\psi -v^2/2$, with $\psi =-\varphi$. If the galactic distribution function depends on $E$ alone, rotations and sign reversals in velocity space leave it unchanged. Thus all mean velocities vanish, and the three component second moments are equal. The anisotropy term in the Spherical Jeans equation cancels. With constant radial velocity dispersion ${\sigma }_0$, it reduces toIntegration gives the isothermal collisionless tracer densityThe constant ${\rho }_0$ changes if a constant is added to the potential; this has no effect on the velocity distribution.

Apply the given untruncated Eddington inversion, retaining its lower limit $-\infty$. Define its integral before differentiation by $I(E)$. With $\psi =E-{\sigma }_0^2{s}^2$, the Gaussian integral givesDifferentiating and multiplying by $1/(\sqrt{8}{\pi }^2)$ yieldsIndeed, at a fixed position this iswhose integral over all velocities is $\rho$ and whose component second moments are ${\sigma }_0^2$. This checks both the normalization and the velocity dispersion.

Let $w$ be the line of sight component of velocity. Integrate this Maxwell-Boltzmann velocity distribution over the two transverse components:At projected radius $R$, integrating along the line of sight simply replaces $\rho$ by the projected surface density $\Sigma (R)$. Dividing by $\Sigma (R)$ gives the spectroscopist's normalized Gaussian line-of-sight velocity distribution:It is a zero-mean normal distribution with width ${\sigma }_0$, independent of projected radius. This is the untruncated model specified by the lower limit in the inversion formula: no escape-energy cutoff has been imposed. A population restricted to positive binding energy would have truncated velocity tails and would not obey this exact Gaussian density law in a finite escape potential.

Let ${\mathbf{x}}^{\prime }=\mathbf{x}-\mathbf{v}t$, $t^{\prime }=t$ for a constant relative velocity $\mathbf{v}$. The Galilean transformation gives ${\nabla }^{\prime }=\nabla$ and ${\partial }_{t^{\prime }}={\partial }_t+\mathbf{v}\cdot \nabla$. In the magnetic Galilean limit of the pre-Maxwell equations, takeThe Ampère's law without displacement current and the solenoidal constraint are immediately unchanged. For Faraday's law, the constant-velocity vector identity is $\nabla \times (\mathbf{v}\times \mathbf{B})=\mathbf{v}\nabla \cdot \mathbf{B}-(\mathbf{v}\cdot \nabla )\mathbf{B}$. ThusThese are transformations of the magnetic nonrelativistic approximation. The exact charge-current transformation also contains the convective term $-{\rho }_e\mathbf{v}$ in ${\mathbf{J}}^{\prime }$; ignoring it requires the quasineutral magnetic limit. They are not exact transformations of the full Maxwell equations.

Since ${\mathbf{u}}^{\prime }=\mathbf{u}-\mathbf{v}$, the combination ${\mathbf{E}}^{\prime }+{\mathbf{u}}^{\prime }\times {\mathbf{B}}^{\prime }$ equals $\mathbf{E}+\mathbf{u}\times \mathbf{B}$. For an ideally conducting fluid the rest-frame electric field vanishes, so the ideal moving-conductor Ohm law is $\mathbf{E}+\mathbf{u}\times \mathbf{B}=0$. Substitution in Faraday's law givesThe ideal magnetohydrodynamic induction equation requires nonrelativistic fluid motion, negligible displacement current, and negligible nonideal terms in the electric-field relation. In particular, magnetic diffusion must be slow compared with advection: the magnetic Reynolds number $\mathrm{Rm}=UL/{\eta }_m\gg 1$, where ${\eta }_m=({\mu }_0\sigma {)}^{-1}$ is the magnetic diffusivity. The single-fluid moving-conductor Ohm law must apply on the scales considered, with corrections to that relation negligible; viscosity is not itself excluded by this induction equation.

For a smooth ideal evolution, magnetic flux freezing states that the magnetic flux through every material surface is conserved and that magnetic field lines are carried with the fluid. Their connectivity is preserved: ideal smooth advection does not reconnect them. These implications are stated here without proof, as requested.

For the isothermal equation of state, $p=c_s^2\rho$, with constant isothermal sound speed $c_s$. A nonzero steady plane-parallel flow has constant mass flux $\rho u=j$. Choose the positive flow direction for convenience. The continuity equation then gives ${\rho }^{\prime }/\rho =-u^{\prime }/u$, and the steady Euler momentum equation becomesIntegrating and absorbing the reference velocity into the constant yields, for the Mach number $\mathcal{M}=u/c_s$,For the opposite flow direction use $\mathcal{M}=|u|/c_s$; the same relation holds.

A smooth sonic point has $u=c_s$, so its momentum equation requires ${\Phi }^{\prime }=0$. Moreover $f(\mathcal{M})={\mathcal{M}}^2/2-\ln \mathcal{M}$ satisfies $f^{\prime }(1)=0$, $f^{\prime \prime }(1)=2$ and has its minimum $f(1)=1/2$. Thus, if the flow reaches $\mathcal{M}=1$ at $x_{\ast }$,A sonic crossing is therefore possible at a maximum of the potential, withAt a nondegenerate maximum, the expansions give $(\mathcal{M}-1{)}^2=-{\Phi }^{\prime \prime }(x_{\ast })(x-x_{\ast }{)}^2/(2c_s^2)+\cdots$. Choosing either smooth sign gives a crossing slope ${\mathcal{M}}^{\prime }(x_{\ast })=\pm \sqrt{-{\Phi }^{\prime \prime }(x_{\ast })/(2c_s^2)}$, joining a subsonic flow to a supersonic flow, or the reverse. This establishes the plane-parallel isothermal sonic transition at a potential maximum. For a globally defined crossing, the same maximum must also be high enough for the Bernoulli relation to remain real over the intended domain.

A single rarefaction wave enforces two matching equalities: equal specific entropy and one common Riemann invariant. The inequality $u_R>u_L$ does not enforce either. If the entropies differ, one smooth isentropic fan cannot connect them; even if they agree, arbitrary end velocities and pressures need not satisfy either acoustic-family invariant.

The general positive-gas Riemann problem has a left acoustic wave, a middle contact discontinuity, and a right acoustic wave. Across the contact the two intermediate states have the same velocity $u_{\ast }$ and pressure $p_{\ast }$, but can have different mass densities and entropies. Each acoustic wave is a rarefaction wave or a shock wave, depending on the required intermediate pressure relative to its adjacent initial state. Shocks satisfy the conservative jump conditions and increase entropy. A sufficiently large pressure contrast can produce a shock on one side even when the initial velocities obey $u_R>u_L$.

For sufficiently rapid separation, the two rarefactions instead exhaust the gas and create vacuum between them. Their vacuum-edge velocities are $u_L+2c_L/(\gamma -1)$ and $u_R-2c_R/(\gamma -1)$, so a finite vacuum gap occurs whenEquality is the limiting case in which the vacuum edges meet. Thus the additional features can be two acoustic waves, a contact carrying an entropy jump, shocks, or a vacuum region, rather than one fan joining the two original uniform states.

First suppose $W[\mathbf{\xi }]>0$ for every nonzero admissible displacement. The conserved energy from part (b) then bounds both $T$ and $W$ by $E$. In particular,where $\Vert \mathbf{\xi }{\Vert }_{\rho }^2=⟨\mathbf{\xi },\mathbf{\xi }⟩$. The positive quadratic form $\Vert \dot{\mathbf{\xi }}{\Vert }_{\rho }^2+2W[\mathbf{\xi }]$ defines the perturbation energy norm and is conserved. Small perturbations in this norm remain small, proving energetic linear stability. Also $\Vert \mathbf{\xi }(t){\Vert }_{\rho }\le \Vert \mathbf{\xi }(0){\Vert }_{\rho }+\sqrt{2E}t$, excluding exponential displacement growth. For any admissible normal mode $\mathbf{\xi }=\mathbf{\zeta }{e}^{-i\omega t}$, multiplication of $-{\omega }^2\mathbf{\zeta }=F\mathbf{\zeta }$ by ${\mathbf{\zeta }}^{\ast }$ gives directlyso all such frequencies are real. No variational principle has been assumed.

To prove instability, suppose some admissible ${\mathbf{\xi }}_0$ has $W_0=W[{\mathbf{\xi }}_0]<0$, and put $N_0=\Vert {\mathbf{\xi }}_0{\Vert }_{\rho }^2>0$. Choose the initial velocityIts kinetic energy exactly cancels $W_0$, so $E=0$ for the entire subsequent motion. Write $N(t)=\Vert \mathbf{\xi }(t){\Vert }_{\rho }^2$. Since $F$ is self-adjoint,Zero energy gives $⟨\mathbf{\xi },F\mathbf{\xi }⟩=\Vert \dot{\mathbf{\xi }}{\Vert }_{\rho }^2$, and hence $N^{\prime \prime }=4\Vert \dot{\mathbf{\xi }}{\Vert }_{\rho }^2\ge 0$. Initially $N^{\prime }(0)=2\alpha N_0>0$, so $N$ stays positive and its logarithm is defined. The Cauchy-Schwarz inequality now givesTherefore $(\ln N{)}^{\prime }(t)\ge (\ln N{)}^{\prime }(0)=2\alpha$, and integration provesThis is an exponentially growing perturbation. Scaling the initial displacement and velocity down leaves $\alpha$ unchanged, so arbitrarily small initial perturbations display the same linear growth mechanism. It proves that a negative-energy direction gives exponential growth in a self-adjoint wave equation, without assuming a discrete spectrum or quoting a trial-function variational theorem.

Positive $W$ on nonzero displacements gives energetic stability; a negative direction gives instability. For mathematical precision, a uniform bound on the mass-weighted displacement norm follows if $W\ge \frac{1}{2}{\omega }_{\min }^2\Vert \mathbf{\xi }{\Vert }_{\rho }^2$ for some ${\omega }_{\min }>0$, giving $\Vert \mathbf{\xi }{\Vert }_{\rho }^2\le 2E/{\omega }_{\min }^2$. Strict positivity alone in an infinite-dimensional space need not supply such a uniform spectral gap; the standard magnetohydrodynamic energy principle asserts energetic stability and absence of growing modes. Cases with zero directions are marginal and require separate consideration.

Let the mean streamwise velocity be $U(y)$ and set $S=dU/dy>0$. The relevant covariance is $\overline{u^{\prime }{v}^{\prime }}<0$ on this half of the channel. The work of the Reynolds stress against the mean shear rate givesThe eddy viscosity relation is $-\overline{u^{\prime }{v}^{\prime }}={\nu }_{T}S$, soThese are the kinematic-stress conventions, with the covariance measured per unit mass; inserting a stress measured in force per area would require division by mass density.

Define the friction velocity by $u_{\ast }=(|{\tau }_w|/\rho {)}^{1/2}$, where ${\tau }_w$ is the total tangential wall stress. It is a stress-derived velocity scale, not the local mean velocity. In the overlap region, $y$ is small relative to the channel half-width, so the total stress is approximately its wall value. Molecular stress is small there, giving $-\overline{u^{\prime }{v}^{\prime }}\simeq u_{\ast }^2$.

Combining this Reynolds stress with the logarithmic shear gives $\mathcal{P}\simeq u_{\ast }^3/(\kappa y)$. For locally equilibrated wall turbulence, energy transfer to small scales occurs over the local eddy turnover time and is balanced by molecular dissipation; energy storage and the net energy-transport divergence are small in the leading overlap-layer balance. ThusThis is the local turbulent kinetic energy balance. It is a modeling approximation rather than a general identity for all wall positions. The constant-energy solution derived next is consistent with it: its modeled $k$ diffusion term is exactly zero, while the $ϵ$ diffusion term remains essential.

Equating the two expressions for eddy viscosity gives directlyFrom the preceding part, $u_{\ast }^2=\sqrt{C_{\mu }}k$. Replace one factor of $k$ using that identity, and then replace $u_{\ast }$ itself by $C_{\mu }^{1/4}{k}^{1/2}$. This proves all three equivalent dissipation expressions:Each is the same turbulent kinetic energy dissipation rate, with dimensions of velocity cubed divided by length. Its equivalent friction-velocity form is $u_{\ast }^3/(\kappa y)$.

The Minimum-dissipation theorem for Stokes flow compares fields with fixed geometry and boundary velocities. Moving the sphere changes the geometry, while sedimentation prescribes its effective weight rather than its velocity. It therefore supplies no rule for migrating toward a position of least dissipation. Indeed, at fixed force $W$, the power is $WV$, and a lower drag resistance gives a larger settling speed and larger power, rather than a smaller power. This is a fixed-force comparison of minimum viscous dissipation issue.

The proposed migration is itself false in ideal Stokes flow. Reflect the uniform vertical tube in a horizontal plane through the sphere centre. The sphere's lateral position is unchanged, but the applied axial force reverses. Reflection leaves a transverse translational velocity unchanged. By Linearity of Stokes flow, however, reversing the force reverses that transverse velocity. It must therefore be zero. This proves no lateral migration of a single sphere in a uniform tube in Stokes flow:Rotation and a wall-dependent vertical speed are allowed. Inertial lift, particle deformation or tube-end effects belong to different problems and do not follow from the dissipation theorem.