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For $N$ equal weak shocks with individual pressure ratio $1+\delta$ and total ratio $P_s=(1+\delta {)}^N$, entropy production in a perfect-gas shock givesAt fixed $P_s>1$, this tends to zero as the compression is divided into increasingly many smaller weak shocks. A single finite normal shock wave instead gives the positive value $\log P_s-\gamma \log D(P_s)$. Gradual compression can consequently approach reversible isentropic flow while an abrupt compression generates finite entropy.

The local conservation laws. The first equation is the radial Euler equation: the material acceleration $Du/Dt={\partial }_{t}u+u{\partial }_{r}u$ is caused by the pressure gradient force per unit mass, $-{\partial }_{r}p/\rho$. It neglects gravity and viscosity. The second equation is mass conservation in spherical coordinates,The term $2\rho u/r$ appears because the area of a spherical surface grows as $r^2$.

To obtain the third equation, let $e=p/[(\gamma -1)\rho ]$ be the specific internal energy of an ideal gas with constant heat capacity ratio $\gamma$. The first law of thermodynamics for a smooth adiabatically moving element isDifferentiating $e$ and collecting terms givesand henceThus the entropy parameter $p/{\rho }^{\gamma }$ is conserved along each smooth pathline; for fixed composition, specific entropy differs from $c_v\ln (p/{\rho }^{\gamma })$ only by a constant. This is local isentropic flow behind the shock. A physical shock wave itself generates entropy and must be treated with jump conditions, rather than by applying this smooth equation across the discontinuity.

The front-radius scaling. In the self-similar blast wave approximation there is no scale set by ambient pressure, cooling, gravity, or the initial source radius. Write $R\propto E^a{\rho }_0^b{t}^d$. Using $[E]=\mathsf{M}{\mathsf{L}}^2{\mathsf{T}}^{-2}$ and $[{\rho }_0]=\mathsf{M}{\mathsf{L}}^{-3}$, dimensional analysis givesThus $a=1/5$, $b=-1/5$, $d=2/5$ andThe dimensionless coefficient ${\xi }_0$ depends on $\gamma$ and the resolved similarity profiles; dimensions alone do not fix it. Define the length scale $\ell (t)=(E{t}^2/{\rho }_0{)}^{1/5}$, so $\xi =r/\ell$ and the front is at the fixed coordinate $\xi ={\xi }_0$.

Kinetic energy. Substitute $\rho ={\rho }_0{f}_{\rho }(\xi )$ and $u=(r/t)f_u(\xi )$ into the volume integral. Since $r=\ell \xi$ and $dr=\ell d\xi$,But ${\rho }_0{\ell }^5/t^2=E$, so the first of the spherical blast-wave similarity energy integrals isThe conclusion requires a fixed, integrable similarity profile, as in the finite-energy blast solution. Expansion increases the swept-up mass while reducing the characteristic squared speed, leaving this product constant.

Pressure and internal energy. The derivatives of the similarity variable areWriting a prime for differentiation with respect to $\xi$, the material acceleration becomesThe radial momentum equation therefore requires a pressure gradient on the scale ${\rho }_0\ell /t^2$. Integrating over radius gives pressure similarity in a spherical blast wave:whereThe prefactor has units $\mathrm{kg}{\mathrm{m}}^{-1}{\mathrm{s}}^{-2}$, as required for pressure. Strictly, the momentum equation alone permits an additional pressure uniform in space and arbitrary in time. In the scale-free strong-blast solution, shock matching and the remaining thermodynamic equation fix that freedom with the same similarity scaling; any allowed constant of integration is included in $f_p$. An independently imposed external pressure would introduce another scale and invalidate the simple dimensional ansatz.

For constant $\gamma >1$, the internal energy is consequentlyIf $E$ is the total excess explosion energy, its normalization isThe two constant fractions need not be equal. No explicit solution for $f_p$ is needed to establish their time independence.

Take $u<0$ for inward radial velocity, and assume a constant specific-heat ratio $\gamma >1$. The isentropic flow has $p=K{\rho }^{\gamma }$ and adiabatic sound speed $c^2=\gamma K{\rho }^{\gamma -1}$. Steady spherical mass conservation and the radial Euler equation for fluid motion giveEliminating the mass density derivative gives $(u-c^2/u)u^{\prime }=2c^2/r-GM/r^2$, orDifferentiating $c^2\propto {\rho }^{\gamma -1}$ gives the corresponding sound speed equation:At a smooth sonic point, $u_s^2=c_s^2$. A finite derivative requires the numerator of the flow equation to vanish too. ThusThis regularity requirement concerns a smooth transonic branch, rather than an arbitrary singular solution reaching unit Mach number.

Integrating the momentum equation using the specific enthalpy $c^2/(\gamma -1)$ yields the Bernoulli equationFor a finite positive mass accretion rate, its large-radius solution hasThese expressions satisfy both the flow equations asymptotically and show $u\to 0$ and $\rho \to {\rho }_{\infty }$. In particular $u^{\prime }/u\sim -2/r$ agrees with the first boxed equation. Combining the Bernoulli equation with sonic regularity additionally gives $c_s^2=2c_{\infty }^2/(5-3\gamma )$, so a finite positive sonic point occurs in the range identified below.

For the small-radius inward free fall branch, neglecting enthalpy against $GM/r$ in the Bernoulli equation predicts $u^2\sim 2GM/r$. The continuity equation then fixes $\rho \propto r^{-3/2}$, and consequently $c^2\propto r^{-3(\gamma -1)/2}$. The neglected-to-leading ratio isIt tends to zero precisely whenThis makes the assumed supersonic flow and leading free fall self-consistent. At equality, the ratio is of constant order rather than vanishing; above it, the proposed asymptotic balance fails. The isothermal limit $\gamma =1$ also permits free fall, with logarithmic rather than power-law enthalpy.

For the stellar funnel, write the magnetic scalar potential as $\psi =\mu \cos \theta /r^2$, where $R=r\sin \theta$ and $z=r\cos \theta$. Its magnetic dipole field has $B_r=-2\mu \cos \theta /r^3$ and $B_{\theta }=-\mu \sin \theta /r^3$, so a magnetic-field-line equation isThe edge of the polar flux tube meets the stellar surface at ${\theta }_{\ast }\simeq a/R_{\ast }$. Thus, near the pole, ${\theta }^2\simeq {\theta }_{\ast }^{2}r/R_{\ast }$, and its transverse radius is $R\simeq r\theta$. This proves the dipolar flux-tube area lawEquivalently magnetic flux conservation gives $A|\mathbf{B}|=\mathrm{constant}$ and $|\mathbf{B}|\propto r^{-3}$. The stipulated small cross section keeps the tube close to the polar axis, where its direction is approximately radial; the approximation is not valid all the way to arbitrarily large $r$.

Outside the star the prescribed field is a gradient, hence $\nabla \times \mathbf{B}=0$ and $\mathbf{j}=0$. Its Lorentz force density is therefore $\mathbf{j}\times \mathbf{B}=0$. The magnetic-pressure and magnetic-tension parts cancel in this exact current-free dipole. Field-aligned velocity also has $\mathbf{u}\times \mathbf{B}=0$, consistent with a steady prescribed field. The following calculation uses the narrow polar funnel approximation.

Now steady mass conservation reads $\rho uA=\mathrm{constant}$. With $A\propto r^3$, replace the spherical factor 2 by 3:The Bernoulli equation is unchanged because the magnetic field supplies no work or force. On a gravity-dominated free fall branch, $u^2\sim 2GM/r$, but $\rho \propto r^{-5/2}$ and $c^2\propto r^{-5(\gamma -1)/2}$. HenceThe same bound follows from regular transonic accretion in a power-law tube: $u_s^2=c_s^2=GM/(3r_s)$ and the reservoir-matched Bernoulli equation gives $c_s^2=2c_{\infty }^2/(7-5\gamma )$. The stellar surface cuts off the mathematical small-$r$ limit; the criterion refers to a funnel that becomes gravity dominated and crosses its sonic point before reaching that surface.

For isentropic flow of a perfect gas, $p=K{\rho }^{\gamma }$ with $\gamma >1$ and $c^2=dp/d\rho$. The mass and momentum equations implyAdding and subtracting the appropriate multiples therefore provesThese are the Riemann invariants for one-dimensional isentropic flow.

In the piston-driven rarefaction wave, the incoming minus invariant has its initial value zero. Thus $c=c_0+(\gamma -1)u/2$. At the piston $u_p=-ft$, giving $c_p=c_0-(\gamma -1)ft/2$. A plus characteristic emitted at time $\tau$ has constant $u=-f\tau$ and $c=c_0-(\gamma -1)f\tau /2$, so its straight trajectory isThe first characteristic is $x=c_{0}t$, ahead of which the gas is undisturbed. Later characteristics have decreasing propagation speed and spread into a rarefaction. More explicitly ${\partial }_{\tau }x=-c_0-(\gamma +1)ft/2+\gamma f\tau <0$ for $\tau \le t<t_v$, with equality only at the vacuum endpoint. They never intersect, so no compressive shock forms.

Since $p/p_0=(c/c_0{)}^{2\gamma /(\gamma -1)}$, the vacuum formation at an accelerating withdrawing piston givesAt $t_v$, density, sound speed and pressure at the gas edge vanish, and the piston velocity is $\boxed{\dot{X}(t_v)=-2c_0/(\gamma -1)}$. Its speed is the positive magnitude $2c_0/(\gamma -1)$. Afterwards the gas edge continues at this limiting velocity, while the piston keeps accelerating away; a vacuum gap of width $f(t-t_v{)}^2/2$ opens and the gas pressure on the piston stays zero.

For smooth isentropic flow with $p=K{\rho }^{\gamma }$, $\gamma >1$, the continuity and momentum equations givewhere $c^2=dp/d\rho$. Adding and subtracting these equations showsThese are the Riemann invariants on the two characteristic families.

Before any shock, the incoming $C_{-}$ characteristics originate in the initially uniform gas, so their invariant is zero. Thus $c=c_0+(\gamma -1)u/2$ throughout the disturbed simple wave. A $C_{+}$ characteristic leaving the piston at time $\tau$ has constant $u=\dot{X}(\tau )$ and speed $u+c=c_0+(\gamma +1)\dot{X}(\tau )/2$. ThereforeThis parametrization applies for $0\le \tau \le t$ in $X(t)\le x\le c_{0}t$ as long as the characteristics remain ordered. The gas for $x\ge c_{0}t$ is still at rest. The values $\tau =t$ and $\tau =0$ give the piston and leading disturbance respectively.

Characteristic crossing first occurs when the derivative of the displayed $x(\tau ,t)$ with respect to $\tau$ vanishes. With $U(\tau )=\dot{X}(\tau )$ this givesFor the printed $X=2c_0{t}^3/(3T^2)$, $U=2c_0{\tau }^2/T^2$, soIts minimum over $\tau >0$ occurs at ${\tau }_{\ast }=T/\sqrt{3\gamma +1}$ and satisfies $t_s>{\tau }_{\ast }$. Hence the first shock forms atThe characteristic formula is restricted to earlier times; crossing afterwards must be replaced by a shock satisfying the conservation laws.

Use $v=|u_r|>0$ for the radial speed and distinguish the central mass $M$ from the Mach number $\mathcal{M}=v/c$. In each smooth nonzero-flow region, mass conservation, the adiabatic equation of state, and the Bernoulli equation giveHere $K>0$ is constant on a smooth isentropic flow branch, $c$ is the adiabatic sound speed, and $\mathcal{E}$ is the Bernoulli function. Either sign of $u_r$ is possible. The positive-radius nondimensionalization in the question requires $\mathcal{E}>0$; flows with $\mathcal{E}\le 0$ do not have these positive $r_s,x,\lambda$ and require another scaling.

Indeed, the supplied adiabatic process equations give $D(p{\rho }^{-3/2})/Dt=0$, so $K$ is spatially constant along a steady nonzero radial fluid flow. The adiabatic sound speed is $c^2=(\partial p/\partial \rho {)}_s$, and the specific enthalpy is $h=c^2/(\gamma -1)=2c^2$. The radial Euler momentum equation, $vdv/dr=-{\rho }^{-1}dp/dr-GM/r^2$, integrates to $\mathcal{E}$ because $dh=dp/\rho$ on this isentropic flow branch.

Put $D=[j(3K/2{)}^2{]}^{2/5}$. Eliminating the mass density from the continuity equation givesWith $y={\mathcal{M}}^{2/5}$, the Bernoulli equation becomesTherefore the physical constants of the spherical polytropic flow with adiabatic exponent three halves are

For $t>0$, $F^{\prime }(t)=4(t^5-1)/t^2$, so $F$ decreases from infinity to its unique minimum $F(1)=5$, then increases to infinity. This completely determines the algebraic curves. For $0<\lambda <1$, there are two disconnected branches at every $x>0$, one subsonic flow with $y<1$ and one supersonic flow with $y>1$. For $\lambda =1$, the line $y=x$ and a decreasing branch meet at $(1,1)$. Their slopes are $+1$ and $-1$, since the leading expansion is $F(1+\eta )=5+10{\eta }^2+O({\eta }^3)$. For $\lambda >1$, solutions exist only where $F(x)\ge 5\lambda$: if $x_{-}<1<x_{+}$ solve $F(x_{\pm })=5\lambda$, there is a forbidden interval $x_{-}<x<x_{+}$. At either endpoint the two branches turn at $y=1$ and have an infinite slope.

A smooth sonic point needs $F^{\prime }(x)=F^{\prime }(y)=0$; hence a regular sonic transition requires $\lambda =1$ and $r=r_s$. The critical speed of a polytropic flow is then $c_s^2=GM/(2r_s)=2\mathcal{E}$, and the allowed mass flux is $j=(2\mathcal{E}{)}^{5/2}{r}_s^2/(3K/2{)}^2$. Here the subscript on $c_s$ denotes evaluation at the sonic point, rather than the isothermal sound speed convention used in the next solution.

The simple transonic branch is $y=x$. Direct substitution givesThus the radial velocity has constant magnitude, while the mass density and pressure scale as $r^{-2}$ and $r^{-3}$. For an outward fluid flow, this is a subsonic flow becoming a supersonic flow as its adiabatic sound speed decreases.

On the other transonic branch, $y$ decreases as $x$ increases. At large $r$, the terms $4/y$ and $x^4$ dominate, giving $y\sim 4x^{-4}$. At small $r$, the terms $y^4$ and $4/x$ dominate, giving $y\sim 4^{1/4}{x}^{-1/4}$. The resulting asymptotic expansions areFor inward fluid flow, these are the nearly static reservoir at infinity and the inner free fall region of Bondi accretion. The specific enthalpy is smaller than $GM/r$ in the inner limit, which explains free fall. Reversing the radial velocity gives a decelerating outflow with the same profiles. The formal small-$r$ limit describes the point-mass model: if the spherical gravitating body has a finite surface, the exterior solution stops there.

A stationary, nonradiative normal shock wave conserves the mass flux and the sum $v^2/2+h$, while the Newtonian gravitational potential is continuous across its thin layer. Hence $\mathcal{E}$ and $r_s$ are unchanged. The entropy production in a perfect-gas shock increases $K=p/{\rho }^{3/2}$, soMore explicitly, the Rankine-Hugoniot condition for $\gamma =3/2$ gives, with upstream $m={\mathcal{M}}_1^2>1$,Thus the shock wave jumps vertically, at the same $x$, from the supersonic flow to a subsonic flow on a larger-$\lambda$ curve. The figure illustrates an outward flow on $y=x$ with a shock wave at $x=1.4$; the downstream subsonic flow has $\lambda \simeq 1.224$ and continues to larger $x$. The shock wave radius is an additional boundary-data choice, rather than being determined by the smooth equations alone.

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.

The steady Bernoulli equation isAt the regular sonic point, part a givesDefineThe isentropic flow relation $c_s^2\propto {\rho }^{\gamma -1}$ and mass conservation between the stellar surface and the sonic point giveEquating the surface and sonic values of the Bernoulli function therefore yields the required relationAn outflow reaching infinity with positive terminal kinetic energy requires $E>0$, henceAt $\gamma =5/3$ the terminal state is marginal with $E=0$; larger $\gamma$ cannot support the stipulated wind to infinity.

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 $\gamma$, $h=\gamma p/[(\gamma -1)\rho ]$.

For steady spherical isentropic flow with $p=K{\rho }^{3/2}$ around a point mass, put $j=r^2\rho v>0$ and $\mathcal{E}=v^2/2+2c^2-GM/r>0$. Eliminating the mass density gives $c^2=[j(3K/2{)}^2{]}^{2/5}{r}^{-4/5}{\mathcal{M}}^{-2/5}$. With $r_s=GM/(4\mathcal{E})$, $x=(r/r_s{)}^{1/5}$, $y={\mathcal{M}}^{2/5}$ and $\lambda =[j(3K/2{)}^2{]}^{2/5}/(2\mathcal{E}{r}_s^{4/5})$, the Bernoulli equation reduces to $\lambda F(y)=F(x)$, where $F(t)=t^4+4/t$. Since $F$ has its unique minimum at $t=1$, a regular sonic point requires $\lambda =1$ and $r=r_s$. One transonic branch is $y=x$, which has $v=\sqrt{2\mathcal{E}}$, $\rho \propto r^{-2}$ and $p\propto r^{-3}$. The decreasing branch has inner free fall and an outer nearly static reservoir, giving Bondi accretion for inward fluid flow. A nonradiative stationary shock wave keeps $j,\mathcal{E},r_s$ fixed but raises $K$, so $\lambda$ increases by the factor $(K_2/K_1{)}^{4/5}$. The inward transonic solution is the spherical $n=2$, $\gamma =3/2$ specialization of transonic accretion in a power-law tube.

Consider steady isentropic flow toward a Newtonian gravitational potential $-GM/r$ through a tube of cross-sectional area $A(r)=C{r}^n$. Write $v>0$ for the inward speed and use a polytropic equation of state with specific-heat ratio $\gamma >1$. Mass conservation and the Euler equations for an inviscid fluid givewhere $c_s$ is the adiabatic sound speed. A regular sonic point therefore has $v_s=c_s$ and $r_s=GM/(n{c}_s^2)$. The Bernoulli equation, matched to a nearly stationary reservoir with sound speed $c_0$, givesFor $n>1/2$, a finite positive sonic point requires $1<\gamma <(2n+1)/(2n-1)$. Differentiating the flow equation at that point, with $x=r_s{v}_s^{\prime }/c_s$, givesIts discriminant is $4n[(2n+1)-(2n-1)\gamma ]$, so the same bound permits real regular slopes. The branch on which the Mach number rises inward selects the negative sign inFor a spherical tube $n=2$, this recovers the $5/3$ threshold of Bondi accretion; for a dipolar flux-tube area, $n=3$ gives $7/5$. The tube approximation must remain valid between the reservoir matching region and the accretor.