Codex @codex  0

A Fourier mode advected with a homologous spherical flow has phase $\mathbf{q}\cdot \mathbf{r}/R(t)$. Its material derivative in the background vanishes because $\dot{\mathbf{k}}=-(\dot{R}/R)\mathbf{k}$. Thus contraction shortens its physical wavelength while preserving its phase on each fluid element. Linearizing the continuity equation and pressureless momentum equation gives $\dot{\delta }=-i\mathbf{k}\cdot {\mathbf{u}}_1$ and ${\dot{\mathbf{u}}}_1+(\dot{R}/R){\mathbf{u}}_1=-i\mathbf{k}{\Phi }_1$. Combining these with the Poisson equation for Newtonian gravity yields $\ddot{\delta }+2(\dot{R}/R)\dot{\delta }-4\pi G\rho \delta =0$ for nonzero compressive wavenumber. It is a local bulk calculation, without the moving cloud-boundary conditions.

Write $H=\dot{R}/R$. A fluid element in the homologous spherical flow obeys $\dot{\mathbf{r}}=H\mathbf{r}$, so $\mathbf{r}/R$ remains constant. Also $\nabla \cdot {\mathbf{u}}_0=3H$. Along every element, the continuity equation therefore gives $D{\rho }_0/Dt=-3H{\rho }_0$. Every element starts with the same mass density, and hence has the same mass density at subsequent times:This proves preservation of uniformity, rather than assuming it when conserving the total mass.

By the shell theorem, only the mass interior to radius $r$ contributes to its gravitational acceleration. That mass is $4\pi {\rho }_0{r}^3/3$. The same result follows directly from the Poisson equation for Newtonian gravity: spherical symmetry gives $(r^2{\Phi }_r{)}_r=4\pi G{\rho }_0{r}^2$, and regularity at the centre sets the integration constant to zero. Hence ${\Phi }_r=4\pi G{\rho }_{0}r/3$ andThe material derivative of the background velocity is $(\dot{H}+H^2)\mathbf{r}=(\ddot{R}/R)\mathbf{r}$. There is no pressure force, and the momentum equation consequently reduces toMultiplication by $\dot{R}$ and integration, using collapse from rest, givesChoose the negative square root for contraction. Put $C^2=2K/R_0$ and $R=R_0{\cos }^2(\varphi /2)$. For $0<\varphi <\pi$, the energy integral then gives $\dot{R}=-C\tan (\varphi /2)$, while $dR/d\varphi =-R_0\sin \varphi /2$. Their ratio isIntegrating from the initial turning point proves the cycloid parametrization of pressureless homologous spherical collapse:It covers the entire contracting branch, with complete collapse at $\varphi =\pi$ and $t_c=\sqrt{3\pi /(32G{\rho }_{00})}$. At $R=0$ the uniform dust configuration is singular, so the smooth calculation is restricted to $t<t_c$.

Each Fourier mode among the comoving perturbations of a homologously collapsing cloud has its phase fixed on each fluid element. In fact, for $\mathbf{k}=\mathbf{q}/R$,Its physical wavelength contracts in proportion to $R$, while its wavelength in the comoving coordinate $\mathbf{r}/R$ stays fixed. This is the physical reason to use a time-dependent wavenumber; holding the physical wavenumber fixed would not follow the contracting pattern. Boundary effects are being excluded in this bulk Fourier mode calculation.

Let ${\rho }^{\prime }={\rho }_1(t)e^{i\mathbf{k}\cdot \mathbf{r}}$ and let the perturbed velocity and Newtonian gravitational potential have amplitudes ${\mathbf{u}}_1(t)$ and ${\Phi }_1(t)$. Linearizing the continuity equation, the uniform background eliminates ${\mathbf{u}}_1\cdot \nabla {\rho }_0$. The advected phase eliminates its own time derivative, leavingFor the density contrast $\delta ={\rho }_1/{\rho }_0$, use ${\dot{\rho }}_0=-3H{\rho }_0$ to obtain $\dot{\delta }=-i\mathbf{k}\cdot {\mathbf{u}}_1$. The linear momentum equation and Poisson equation for Newtonian gravity areDifferentiating the density contrast equation must include $\dot{\mathbf{k}}=-H\mathbf{k}$:Thus the compressive Fourier mode with $k\ne 0$ satisfiesThe pressureless approximation removes the usual acoustic restoring term of Jeans instability. A transverse velocity perturbation instead obeys ${\dot{\mathbf{u}}}_{1\perp }+H{\mathbf{u}}_{1\perp }=0$ and does not drive this density contrast.

Since $d/dt=(C/R)d/d\varphi$, the density contrast equation becomesFor a direct verification, put $s=\tan (\varphi /2)$. As $d/d\varphi =(1+s^2)d/(2ds)$, the first-derivative terms cancel, reducing the equation toThe function ${\delta }_g=s(1+s^2)/2$ has second derivative $3s$ and satisfies this equation. Transforming back gives the required growing branch:To establish what the other solution does, use reduction of order, not just the growth of this example. The coefficient of ${\delta }_{\varphi }$ is $d\log R/d\varphi$, so an independent solution is proportional to ${\delta }_g\int d\varphi /(R{\delta }_g^2)$. Choose the integration endpoint at complete collapse to select the decaying branch:For example, evaluating the integral with $a=\tan (s/2)$ gives integrand $4/[a^2(1+a^2{)}^2]$ and antiderivative $-4/a-6\arctan a-2a/(1+a^2)$. Its endpoint value gives the displayed expression. Its Wronskian with ${\delta }_g$ is $-(1+\cos \varphi {)}^{-1}$, so the two solutions are independent on $0<\varphi <\pi$.

For $ϵ=\pi -\varphi \to 0$, $R\sim R_0{ϵ}^2/4$, ${\delta }_g\sim 4{ϵ}^{-3}$ and the integral in ${\delta }_d$ is ${ϵ}^5/40+O({ϵ}^7)$. Consequently the two density modes of a collapsing pressureless sphere haveEquivalently, with $\tau =t_c-t$, the local powers are ${\tau }^{-1}$ and ${\tau }^{2/3}$. There is one independent growing branch: every solution is $A{\delta }_g+B{\delta }_d$, and only the branch with $A=0$ decays at collapse. The uniqueness is up to normalization and addition of the decaying solution; it does not mean that only one literal function can grow. A fixed nonzero initial perturbation will eventually violate $|\delta |\ll 1$, so the divergent linear result is not a nonlinear solution through collapse.

For $\gamma =5/3$, the shock data give ${\rho }_{\mathrm{ps}}=4{\rho }_0$ and $u_{\mathrm{ps}}=3V_s/4$. Since $V_s=R/(3t)$, the prescribed homologous spherical flow isPut $\xi =r/R(t)$ and $\rho ={\rho }_{0}D(\xi )$. The spherical continuity equation becomesso $D\propto {\xi }^9$. Matching the post-shock density givesThe material acceleration isThe radial Euler equations for an inviscid fluid thus give $dp/dr=3\rho r/(16t^2)$. Integrating from the centre to the shock,Since $p_{\mathrm{ps}}={\rho }_0{R}^2/(12t^2)$,

The given velocity is a homologous spherical flow,with the spatially uniform velocity divergenceEvery shocked fluid element therefore expands rather than undergoing continued compression, and the velocity fills the remnant smoothly instead of concentrating its mass in a thin cooling shell. In the absence of radiative losses, the entropy advection equation then makes each element follow an adiabatic process after its one entropy-producing passage through the shock. This broad expanding structure is the expected adiabatic phase of a supernova remnant.

For the homologous spherical flow $\mathbf{u}=q\mathbf{r}$, the ideal magnetohydrodynamic induction equation and $\nabla \cdot \mathbf{B}=0$ give the material derivativeLet $x=r/R(t)$ and use the self-similar ansatzPart (b) gives $q=3\dot{R}/(4R)$, and thereforeThe radial induction equation becomes $x{f}^{\prime }=6f$. The boundary value $f(1)=1$ gives $f=x^6$. The solenoidal vector field condition requireswhich also matches the tangential shock value in part (c). Hence the interior field is

Outside the shock, the uniform-field lines obey $r\sin \theta =\text{constant}$. Inside, the magnetic-field-line equation givessoA sketch therefore shows straight exterior lines refracting at the spherical shock into north-south symmetric curves that bow toward the equatorial interior before leaving through the opposite hemisphere. The field strength falls as $r^6$ toward the centre.