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In Cartesian coordinates the magnetic field is $(bx,by,-2bz)$. Its divergence is $b+b-2b=0$, and all components of its curl vanish. Thus it satisfies the static vacuum magnetic equations $\nabla \cdot \mathbf{B}=0$ and $\nabla \times \mathbf{B}=0$, with no electric current in the region.

Fix clockwise orientation as viewed from above, looking down the positive $z$-axis. Its positive area normal is $-{\mathbf{e}}_z$. The loop's external magnetic flux is thereforeWith the opposite normal its flux has the opposite sign. On the circle $\mathbf{B}=ba{\mathbf{e}}_r-2bz{\mathbf{e}}_z$ and clockwise line element is $d\mathbf{\ell }=-a{\mathbf{e}}_{\phi }d\phi$. The Lorentz force element isThe radial terms cancel around the circle, soThis fixes the current/flux/force signs together.

Neglect self-inductance initially, so Faraday's law and Ohm's law give $RI=-K\dot{z}$. Including gravity then givesFor $R>0$ and $K\ne 0$, the terminal solution is $z=z_0-vt$ withThe general velocity relaxes exponentially to $-v$ on time scale $mR/K^2$. In the terminal state $I=Kv/R$, and the Joule-heating rate isThus all the lost gravitational potential energy becomes resistive heat when kinetic energy is constant.

Within the zero-inductance approximation $v\to 0$ as $R\to 0$; finite-speed motion would require an unbounded current. This limit is singular, so it is important to describe the physical ideal-conductor limit separately. With finite self-inductance $\mathcal{L}$, the circuit equation is $\mathcal{L}\dot{I}+RI=-K\dot{z}$. At $R=0$, total linked flux $\mathcal{L}I+Kz$ is conserved and there is no Joule heating. If $I=0$ initially at $z_0$, thenThis lossless limit of magnetic loop braking has undamped oscillations about $z_0-mg\mathcal{L}/K^2$, with magnetic energy storing and returning mechanical energy. It has no generic dissipative terminal descent. Thus the vanishing terminal speed is the formal resistive-model limit, not a justification for dropping inductance in a perfectly conducting moving loop.

Use signed velocities in the shock frame; the Mach number depends on their squares. For a genuine nonradiative perfect gas shock wave, the Rankine-Hugoniot conditions for a perfect gas give a common mass flux $j={\rho }_L{u}_L={\rho }_R{u}_R$, a common momentum flux $p+\rho u^2$, and a common specific stagnation enthalpy $\gamma p/[(\gamma -1)\rho ]+u^2/2$. Here $e$ in the given equation of state is specific internal energy, despite the source's informal “energy density” wording.

Put $r={\rho }_R/{\rho }_L$, $P=p_R/p_L$ and $M^2={\rho }_L{u}_L^2/(\gamma p_L)$. Mass and momentum give $u_R=u_L/r$ and $P=1+\gamma M^2(1-1/r)$. Substituting these into the energy relation and clearing denominators givesThe $r=1$ factor is the no-jump solution. On the nontrivial shock wave branch,In particular $r-1=2(M^2-1)/[(\gamma -1)M^2+2]$. With $\gamma >1$, compression is therefore equivalent to $M_L^2>1$. The other Mach number isPositive pressure makes its denominator positive, and $M_R^2-1$ has the opposite sign to $M_L^2-1$. Thus a nontrivial physical shock wave has a supersonic side and a subsonic side; the entropy-admissible flow goes from the former to the latter.

The shock compression ratio also yields $u_L{u}_R=(\gamma -1)u_L^2/(\gamma +1)+2\gamma p_L/[(\gamma +1){\rho }_L]$. Rearranging produces the first velocity identity in the question. The pressure relation gives the second identity in the equivalent formThe numerator here contains mass density, as the original PDF shows; the converted TeX incorrectly replaces that mass density by pressure.

For reflection, the once-shocked gas is the common left state in both applications. Before rebound, its velocity in the shock frame is $a=u_s-U_{+}<0$ and the gas ahead has velocity $-U_{+}$; after rebound, it is $b=u_s+U_{-}>0$ and the gas next to the wall has velocity $U_{-}$. In both cases the right velocity minus the left velocity equals $-u_s$. Applying the first identity therefore gives the same quadratic equation,with roots $a$ and $b$. Their opposite signs distinguish the two roots, so $\boxed{(u_s-U_{+})(u_s+U_{-})=-\gamma p_s/{\rho }_s}$.

Write $q_0=p_0/p_s$, $q_1=p_1/p_s$. The pressure identity applied to each shock wave givesSquaring the root-product relation and eliminating the velocities yieldsand hence the normal shock reflection at a rigid wall relationTaking $p_0/p_s\to 0$ gives $\boxed{p_1/p_s\to (3\gamma -1)/(\gamma -1)}$; for a monatomic gas this limiting amplification is six.

Set $a=2\pi G{\rho }_0$. The axisymmetric Poisson equation for Newtonian gravity integrates to $(R{\Phi }_R{)}^{\prime }=4\pi G{\rho }_{0}R$. Regularity on the axis fixes the integration constant, so ${\Phi }_R=aR$. Radial force balance in the rotating fluid givesWith the stated central pressure, $p_c={\rho }_0{a}^2/(4k^2)$, this is a perfect square:The free surface lies at the first zero of pressure. For the source's positive rotation parameter, $R_0=\sqrt{2\pi G{\rho }_0}/k$; allowing either rotation direction requires $|k|$ in the denominator. The inward effective gravity is ${\Phi }_R-R{\Omega }^2=R(a-k^2{R}^2)$, so $\boxed{g_{\mathrm{eff}}(R_0)=0}$. In particular $p_R(R_0)=0$. This is the critically rotating self-gravitating cylinder.

For a normal mode proportional to $e^{i\omega t+im\varphi }$, the background material derivative becomes $i\sigma$, where $\sigma =\omega +m\Omega (R)$. Linearizing the centrifugal term gives $-2\Omega u_{\varphi }$ in the radial equation. In the azimuthal equation the background velocity gradient and cylindrical geometric term together give $(2\Omega +R{\Omega }^{\prime })u_R=3\Omega u_R$. Since the interior mass density is constant, write $W=p^{\prime }/{\rho }_0+{\Phi }^{\prime }$. The two momentum equations and incompressible flow condition are consequentlyFor $m\ne 0$, the last two equations express the other amplitudes in terms of $u=u_R$:Differentiate the second expression using $R{\Omega }^{\prime }=\Omega$ and $R{\sigma }^{\prime }=m\Omega$, then substitute into the radial momentum equation. Before any division by $\sigma$, the result isAway from corotation this gives the required ordinary differential equationAt corotation the original coupled equations, or the undivided equation, must be used. A regular axisymmetric radial velocity would vanish by the incompressible flow condition; the requested dipolar calculation has $m=1$.

For $m=1$, use the amplitude normalization $u=A(\omega +kR)$, which remains useful when $\omega$ tends to zero. Substitution into the undivided equation gives zero identically. For nonzero $\omega$ this is the source's $1+kR/\omega$ after choosing $A=1/\omega$. The other amplitudes areThese formulas also satisfy the three original equations directly. By the regularity assumption in the question, this is the interior velocity family to which the free surface conditions must be applied.

The gravitational potential of a displaced cylindrical interface supplies an essential part of those conditions. For surface fluid displacement $\eta e^{im\varphi }$, the bulk mass density perturbation vanishes but the displaced mass density step contributes ${\rho }_0\eta \delta (R-R_0)$. The interior and exterior gravitational potential amplitudes solve Laplace's equation, are proportional to $R^m$ and $R^{-m}$ for $m\ge 1$, and are continuous at the surface. Integrating the Poisson equation for Newtonian gravity across the mass density step givesSince the two radial derivatives at the surface are $-m{\Phi }_s^{\prime }/R_0$ and $m{\Phi }_s^{\prime }/R_0$, respectively,It would be incorrect to set the entire potential perturbation to zero merely because the bulk mass density is unchanged.

Let ${\Omega }_s=k{R}_0$, so $a={\Omega }_s^2$. Zero fluid Lagrangian perturbation of surface pressure gives $p_s^{\prime }+\eta p_R(R_0)=0$, hence $p_s^{\prime }=0$. The surface momentum potential and kinematic condition for $m=1$ are thereforeIf $\omega +{\Omega }_s\ne 0$, the kinematic condition fixes $\eta =-iA$. The same value follows if the radial fluid Lagrangian displacement $u_R/[i(\omega +kR)]=-iA$ is required to extend continuously to the free surface, including a possible surface-corotation limit. Comparing the gravitational condition with the interior value of $W_s$ then givesfor a nontrivial perturbation. These dipolar perturbations of a critically rotating cylinder describe a displacement of the whole cylinder's axis. Translating an isolated self-gravitating system gives another equilibrium, so there is no external restoring force and the translation is neutral. A uniform drift of the centre also follows from invariance under a Galilean transformation, explaining the double zero of the frequency equation. The singular $1/\omega$ in the originally chosen velocity normalization does not obstruct the finite-amplitude limiting solution.

There is an admissibility qualification to the frequency conclusion as printed. Regular interior fluid Eulerian perturbation of velocity alone does not justify cancelling the surface factor $\omega +{\Omega }_s$. Eliminating $\eta$ without that cancellation instead givesThe additional surface-corotation case $\omega =-{\Omega }_s$ formally satisfies every displayed velocity equation and the usual linearized surface conditions withHere $u_R(R_0)=p^{\prime }(R_0)=0$, and the velocity is finite on the axis. However, its interior radial fluid Lagrangian displacement tends to $-iA$, whereas the undisplaced surface has $\eta =0$. Thus a regular single-frequency fluid Lagrangian displacement continuous up to the surface excludes this degenerate branch. The intended ${\omega }^2=0$ conclusion holds with that regular-displacement condition, or with surface corotation excluded; the printed velocity regularity assumption by itself leaves this formal exception.

The Gaussian integral over the full vertical velocity line is $\sqrt{2\pi }{\sigma }_z$. Thus the mass density of the self-gravitating isothermal slab isNeglecting radial derivatives makes the Poisson equation for Newtonian gravity ${\Phi }_{zz}=4\pi G\rho (z)$. Substituting $\Phi ={\sigma }_z^2\varphi$ and $z=z_0\zeta$, with $z_0^2={\sigma }_z^2/(8\pi G{\rho }_0)$, givesMultiply by ${\varphi }^{\prime }$ and integrate. The central value and symmetry under $z\mapsto -z$ fix the constant of integration:On the positive-$\zeta$ side, Newtonian gravity makes ${\varphi }^{\prime }\ge 0$. Put $y=e^{\varphi /2}$; then $4(y^{\prime }{)}^2=y^2-1$ with $y(0)=1$. Integration gives $2\mathrm{arcosh}y=\zeta$, and even reflection supplies the other side. ThereforeDirect differentiation verifies both the equation and central conditions. Uniqueness of the smooth initial value problem rules out other branches meeting those conditions. The surface density is $4{\rho }_0{z}_0$, and the conventional slab scale height is $2z_0$; the factor of two is only a definition of the length scale.

Use the relative potential $\Psi ={\Phi }_{\ast }-\Phi$ and integrate the stated, untruncated Maxwell-Boltzmann velocity distribution over ${\mathbb{R}}^3$. The normalization cancels its three Gaussian integrals:The gravitational potential sign reverses under the relative-potential convention, so the spherical Poisson equation for Newtonian gravity isAlso ${\Psi }^{\prime }=-Gm(r)/r^2$, with $m^{\prime }=4\pi r^2\rho$. Differentiating the density relation yieldsAn ideal gas with constant temperature has $P=\rho k_{B}T/m_0$. Its hydrostatic equilibrium equation is precisely the same whenTogether with the same enclosed-mass equation, central density and boundary conditions, this proves the isothermal stellar and gas density equivalence. In dimensionless terms, $y=-[\Psi -\Psi (0)]/{\sigma }^2$ and $x=r\sqrt{4\pi G\rho (0)}/\sigma$ obey $x^{-2}(x^2{y}^{\prime }{)}^{\prime }=e^{-y}$ with $y(0)=y^{\prime }(0)=0$ in both systems.

This calculation uses the distribution as printed for all velocities. Imposing an additional cutoff at positive relative energy changes the integral to ${\rho }_1[e^W\mathrm{erf}\sqrt{W}-2\sqrt{W}/\sqrt{\pi }]$, where $W=\Psi /{\sigma }^2\ge 0$, and no longer gives an exactly isothermal gas profile. An untruncated self-gravitating isothermal sphere also needs an outer modification for finite total mass. The function $\mathrm{erf}$ here is the error function. Finally, the half-line Gaussian integral printed in the note lacks a factor: the correct value is ${\int }_0^{\infty }{e}^{-\alpha x^2}dx=\frac{1}{2}\sqrt{\pi /\alpha }$.

Both models obtain their density by integrating a galactic distribution function over velocity. The planar model integrates one vertical velocity, whereas the spherical model integrates all three components. In each case the resulting density must also generate its own gravitational potential through the Poisson equation for Newtonian gravity. The two printed numerical subparts are treated separately below.