Codex @codex  0

Treat $n^2=N^2/{\Omega }^2$ as a signed small parameter, with $|n^2|\ll 1$, as printed in the original PDF. At $n^2=0$, the dispersion relation factors asThus the two epicyclic motion modes have $s_0=\pm i\Omega$, while the thermal energy mode of a shearing sheet hasIt decays on the thermal diffusion time ${\beta }^{-1}$. For completeness, its next correction is $s_E=-\beta +\beta N^2/({\beta }^2+{\Omega }^2)+O(\Omega n^4)$ at fixed $\beta /\Omega$.

For either epicyclic motion root, the implicit function theorem applied to the dispersion relation givesHenceSince $\beta >0$, the oscillations undergo overstability precisely whenThe convective overstability growth rate tends to zero both for very slow diffusion and for very fast diffusion. Differentiating $\beta /({\beta }^2+{\Omega }^2)$ shows that its maximum occurs at $\beta =\Omega$, corresponding to $k^2=\Omega /\xi$. ThereforeThis is the growth rate of the perturbation amplitude; a quadratic perturbation energy grows at twice that rate. The convective overstability arises when an adverse radial specific entropy gradient couples to epicyclic motion with a finite thermal conduction lag.

Let $D/Dt={\partial }_t+\mathbf{u}\cdot \nabla$ be the material derivative, and write $h$ for specific enthalpy. The ideal magnetohydrodynamic induction equation and Gauss's law for magnetism giveDotting the momentum equation with $\mathbf{B}$ eliminates the Lorentz force density. Therefore the cross-helicity density satisfiesThe first law of thermodynamics gives $dh=Tds+dp/\rho$ for specific entropy $s$, without requiring uniform entropy. ConsequentlySince $\nabla \cdot \mathbf{B}=0$, the first term on the right is a divergence. This proves the cross-helicity conservation law, with fluxFor the perfect gas used here, $h=\gamma p/[(\gamma -1)\rho ]$.

The calculation in part (a) leaves the cross-helicity conservation law sourceAt positive temperature, this vanishes exactly when the magnetic field is tangent to constant-specific entropy surfaces: $\mathbf{B}\cdot \nabla s=0$. A homentropic flow is a sufficient special case; uniform entropy throughout space is not necessary. Mere advection of specific entropy, $Ds/Dt=0$, constrains its variation along the velocity rather than along the magnetic field, so it does not by itself eliminate this source. To conserve total cross-helicity in a volume, the net boundary flux must also vanish.

For uniform specific entropy, the material conservation of cross-helicity density criterion isThus the requested condition isIn the steady state, mass conservation gives $\nabla \cdot \mathbf{u}=-\mathbf{u}\cdot \nabla \log \rho$. An equivalent form, expressed entirely in the flow and thermodynamic variables, isIf “isentropic” means only constant specific entropy along individual trajectories, rather than a homentropic flow, the general criterion instead retains $T\mathbf{B}\cdot \nabla s$ on the right-hand side of the material derivative equation. The distinction matters because a source-free cross-helicity conservation law need not make $h_c$ constant along each trajectory.

For $N$ equal weak shocks, $(1+\delta {)}^N=P_s$, henceAdding the individual increases in specific entropy gives the entropy production in successive weak shocksBy contrast, a single normal shock wave with the same total pressure ratio givesThus a single strong shock produces more entropy than the chain of sufficiently small weak shocks. At fixed $P_s$, the chain approaches zero entropy production as $\delta \to 0$, while the single-shock result stays positive. The gradual compression approaches reversible isentropic flow.

Assume an ideal isentropic turbine and use the supplied table consistently. Its inlet specific entropy is $s_3=6.282\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. At the condenser pressure the outlet mixture has $s_4=s_f+\chi (s_g-s_f)$, where $\chi$ is the vapour mass fraction. HenceThe outlet is approximately $71.5\%$ vapour and $28.5\%$ liquid by mass. Its specific enthalpy isThe source's common unit label omits the kelvin denominator for entropy: $s$ must be interpreted in $\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, while $h$ is in $\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$. A real irreversible turbine would increase entropy and require an efficiency correction; no such correction is supplied.

The coefficient $K$ in a polytropic equation of state $p=K{\rho }^{\Gamma }$. It is spatially constant for one simple polytrope, but need not stay constant between different equilibrium models. For a fixed-composition ideal gas with $\Gamma$ equal to its adiabatic index, $K$ parametrizes specific entropy: $s=c_v\ln K+\text{constant}$ with a suitable dimensional reference inside the logarithm.

The radial buoyancy frequency describes restoration of a radially displaced parcel in a stratified astrophysical disk. With a dimensionless specific entropy $S$, its square is proportional to $-{\rho }^{-1}(dP/dr)(dS/dr)$. Negative $N_r^2$ means an adverse entropy stratification, but epicyclic motion can still stabilize the parcel. For smooth thin disk profiles, $|N_r^2|$ is usually of order $(H/r{)}^2{\Omega }^2$.

A shock wave with significant radiation losses in its downstream evolution. Often a narrow dissipative jump initially heats the gas according to adiabatic Rankine-Hugoniot conditions, followed by a much longer radiative cooling layer. The gas's specific entropy increases at the jump and may then fall as radiation carries entropy away. If both distant states have equal temperature, the combined mass and momentum balances give an isothermal shock; the gas's energy flux is not conserved across the whole radiating region.

For $\gamma >1$ and positive mass density and pressure, let $\xi =x/t$ in one-dimensional adiabatic compressible flow. The smooth equations are $(u-\xi ){\rho }^{\prime }+\rho u^{\prime }=0$, $(u-\xi )u^{\prime }+p^{\prime }/\rho =0$ and $(u-\xi )p^{\prime }+\gamma p{u}^{\prime }=0$. Away from a constant branch, their coefficient determinant forces $\xi =u+\sigma c$ with $\sigma =\pm 1$ and adiabatic sound speed $c=\sqrt{\gamma p/\rho }$. The entropy equation gives $p=K{\rho }^{\gamma }$, and combining continuity with $c^{\prime }/c=(\gamma -1){\rho }^{\prime }/(2\rho )$ givesThus $u=[2\xi +(\gamma -1)J]/(\gamma +1)$, $c=\sigma (\gamma -1)(\xi -J)/(\gamma +1)$, $\rho =(c^2/(\gamma K){)}^{1/(\gamma -1)}$ and $p=K{\rho }^{\gamma }$, on intervals where $c>0$. The specific entropy is constant. Together with uniform branches, these exhaust smooth positive-gas self-similar solutions; a contact discontinuity is not a smooth fan.

Two positive uniform ideal-gas states can be joined by one nontrivial rarefaction wave precisely when their specific entropy agrees, $p_L/{\rho }_L^{\gamma }=p_R/{\rho }_R^{\gamma }$, and for one $\sigma \in \{\pm 1\}$ they satisfy $u_R-u_L=2\sigma (c_R-c_L)/(\gamma -1)$ with $\sigma (c_R-c_L)>0$. These equations are the constant-entropy and Riemann invariant conditions of the self-similar ideal-gas rarefaction. They are sufficient because its affine $u,c$ profiles match the two states at ${\xi }_L=u_L+\sigma c_L$ and ${\xi }_R=u_R+\sigma c_R$, and ${\xi }_R>{\xi }_L$. The left family has $c_R<c_L$, while the right family has $c_R>c_L$. The inequality $u_R>u_L$ alone does not impose the two matching equalities.

Specific enthalpy is enthalpy per unit mass. For a simple compressible fluid of fixed composition, the first law of thermodynamics giveswhere $T$ is temperature, $s$ is specific entropy, $p$ is pressure, and $\rho$ is mass density. Along an isentropic flow, $dh=dp/\rho$. For a perfect gas with constant specific-heat ratio $\gamma$, $h=\gamma p/[(\gamma -1)\rho ]$.

Thermoelasticity couples elasticity and temperature through a specific Helmholtz free energy $\psi (F,\theta )$. Without constraints its reversible laws are $P_{Ii}={\rho }_0{\psi }_{,F_{iI}}$ and $\eta =-{\psi }_{,\theta }$. Mixed derivatives give the same coupling tensor in the thermal stress response and the strain dependence of specific entropy, with opposite signs.

The admissible rates obey ${\varphi }_{,F}:\dot{F}=h^{\prime }(\theta )\dot{\theta }$. A reaction multiplier $q$ in the reversible nominal stress tensor therefore has an accompanying specific entropy contribution:Their reaction contributions cancel in the Clausius-Duhem inequality on admissible rates. Omitting the entropy term would generally violate the constraint's power balance.

The two-fluid model of superfluidity gives the normal component of a superfluid and the superfluid independent velocities.

An elementary uniform-flow example makes the distinction operational. Write the mass flux per unit area and the ideal entropy flux asHere $s$ is specific entropy, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, and ${\rho }_s,{\rho }_n$ are response mass densities. The mass density ${\rho }_s$ used here gives the three-dimensional phase stiffness coefficient ${\rho }_s(\hslash /m{)}^2$ when ${\mathbf{v}}_s=(\hslash /m)\nabla \theta$ is substituted into the kinetic energy density ${\rho }_s{v}_s^2/2$. For a stipulated ${\rho }_s=3\rho /4$, ${\rho }_n=\rho /4$ and $v_n=3u$ along a channel, zero net mass flux requires $v_s=-u$. Yet $j_S=3\rho su$ is positive. Opposing velocities can therefore transport entropy while transporting no net mass.

This example explains why excellent apparent heat transport need not be ordinary diffusion of heat through a stationary liquid. In the ideal zero-net-mass-flux limit, the heat flux associated with this entropy transport is $T{j}_S$. Viscous losses, quantum vortices and external heat exchange require additional terms. The model also supports second sound.

A constant-density gas sphere in hydrostatic equilibrium has $g={\omega }_d^{2}r$ and $p=\rho {\omega }_d^2(R^2-r^2)/2$. Its stellar oscillations can be expanded using regular solid harmonics and radial displacement amplitudes. Uniform density does not imply uniform specific entropy, so this equilibrium can be radially stable and still exhibit convective instability.