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When a Hamiltonian parameter varies slowly compared with the orbital period and no separatrix is crossed, the action changes only at higher adiabatic order.

The action is an adiabatic invariant under these assumptions; other slowly varying systems can have different adiabatic invariants.

In a slowly changing spherical gravitational potential, angular momentum is exactly conserved and the radial action is an adiabatic invariant. An initially circular orbit remains circular in the adiabatic limit, so $J_r=0$ and $L^2=GrM(r)$ give the displayed shell relation. If dark-matter shells do not cross, their enclosed dark mass is also conserved. A density profile alone does not specify this orbital model: eccentric-orbit populations require their full action distribution.

Use the usual spherical circular-shell adiabatic contraction approximation: initial dark-matter orbits are circular, slow accumulation keeps them circular, and dark shells do not cross. Angular momentum is conserved in a spherical potential, and the vanishing radial action stays zero as an adiabatic invariant. For a circular shell,Here $r_i$ is the initial shell radius and $r$ its final radius. The final flat circular speed $V=2v_{\mathrm{NFW},\mathrm{c},\max }$ gives $M_f(r)=V^{2}r/G$.

For an initial shell in the inner NFW profile, write $A=2\pi {\rho }_1{r}_1$, so $M_i(r_i)\simeq A{r}_i^2$. The invariant givesConservation of dark shell mass yieldsand differentiation givesTo express the requested ratios, set $K=F(x_{\max })\simeq 0.2162166$ and $B=8K\simeq 1.72973$, so $V^2/(GA)=B{r}_1$. Since the original cusp has ${\rho }_D(r)\simeq A/(2\pi r)$,The total density of the final spherical flat-rotation model is ${\rho }_{\mathrm{tot}}=V^2/(4\pi G{r}^2)$. Its dark fraction isSubtracting dark matter from the total gives the added-matter density, ${\rho }_B={\rho }_{\mathrm{tot}}(1-h)$, and thusThese expressions describe the leading inner approximation, not an extension to radii where the denominator becomes negative or the initial shell is outside the NFW cusp. As $r/r_1\to 0$, dark matter is enhanced relative to its original density while its fraction of the even steeper total density tends to zero. This is inner NFW contraction to a flat rotation curve.

The orbital assumptions matter. An initial density profile does not specify the phase-space distribution function. For eccentric orbits, slow evolution conserves their actions, but $rM(r)$ at an instantaneous radius is not generally their invariant. The above ratios are the intended circular-shell estimate; their numerical coefficients cannot be uniquely inferred from the density profile and slow accumulation alone.

Take rightward velocities as positive just before the right-wall collision. The incoming particle velocity is $v$ and the wall velocity is $-V$. Their total momentum divided by total mass gives the centre of mass velocityIn an elastic collision, relative velocity reverses, so the particle's signed outgoing velocity is $v_{\mathrm{out}}=2v_{\mathrm{cm}}-v$. For $M>m$ it is negative, and its speed isConsequently the exact change in speed isFor $m/M\to 0$ with $v,V$ fixed, this proves the requested leading increment $v^{\prime }-v\simeq 2V$. Reflection gives the same increment at the left wall.

This approximation is not uniform when $v/V$ is allowed to become arbitrarily large. To neglect recoil relative to the increment $2V$ requiresThe inequalities $m\ll M$ and $v\gg V$ alone do not imply it: for example, $m/M=10^{-3}$ and $v/V=10^3$ give an exact zero speed increment. The subsequent slowly moving-wall approximation therefore uses the consistent joint range $1\ll v/V\ll M/m$, or treats the walls as effectively prescribed massive reflectors. The assumed slow change of wall speed over many impacts embodies the same negligible-recoil requirement.

Within this approximation, the separation derivative is the right-wall velocity minus the left-wall velocity:For a time window containing many crossings but little separation change, $\ell /v\ll \Delta t\ll \ell /V$, the flight time between successive wall collisions is approximately $\ell /v$. Thus their number and the cumulative speed change arePassing to the averaged differential equation givesHence the particle-speed invariant between slowly approaching walls isIt is an adiabatic invariant for the rapid elastic bouncing motion while the separation and wall velocities vary slowly on the flight-time scale.

Between collisions the momentum is $p_0=\sqrt{2mE}$ or $-p_0$. In phase space the cycle runs right along the upper horizontal segment from $q=0$ to $q=b$, jumps vertically to $-p_0$ at the right wall, runs left along the lower segment, and jumps back at the left wall. The jumps have $dq=0$. ThereforeDifferentiation gives $dI/dE=2bm/\sqrt{2mE}=2b/v$, exactly the out-and-back period. Under slow wall motion the action $I$ is the adiabatic invariant; a convention using $I/(2\pi )$ is equivalent. HenceCompression raises the energy and expansion lowers it, provided the wall speed is small compared with the particle speed and the change per cycle is small.

For a regular periodic energy orbit, differentiate the action integral with respect to its energy. On a branch $H(p,q)=E$, implicit differentiation gives $\partial p/\partial E=1/H_p=1/\dot{q}$. ThusThe integral follows the entire orbit, with its orientation; at turning points this formula is interpreted by the limiting integral from each branch. It is valid for a smooth family of nonseparatrix periodic orbits.

An adiabatic invariant means that the action variable remains constant to leading order when the external parameter changes on a timescale much longer than the orbital period. Denote that parameter by $\lambda$. With $H(0,\lambda )=0$, the displayed identity integrates toIf the parameter change decreases $T(J,\lambda )$ for every intervening action $J$, this integrand increases pointwise. At the adiabatically fixed positive $I$, the energy increases. This uses the phrase “with any $I$” as a statement about the family of orbits, not merely a period decrease at a single selected action; that weaker premise alone would not determine the integral.

The central equation must be read as $\ddot{\mathbf{r}}=-\mu \mathbf{r}/r^3$; its printed left-hand expression lacks $=0$. Initially $\mu$ is constant. Dotting this equation with the velocity givesTaking its cross product with $\mathbf{r}$ gives $d(\mathbf{r}\times \mathbf{v})/dt=0$. In the fixed orbital plane the magnitude is $h=r^2\dot{\theta }$. These are conservation of specific orbital energy and specific angular momentum.

Let $f=\theta -\varpi$. Differentiating the polar equation of the Kepler orbit, with its elements initially constant, givesSubstitution into the specific orbital energy yields

For isotropic stellar mass loss with no recoil or drag, the force is still central, so $\mathbf{h}$ remains exactly constant, although $C$ does not. NowThe same algebraic relation between $C,h,e$ holds for the instantaneous osculating orbital elements. Differentiating it and using $h^2/(\mu r)=1+e\cos f$ givesand, for $e>0$,A particularly useful nonsingular description uses the eccentricity vectorThe vector equation remains meaningful at a circular orbit, where a longitude of pericentre is undefined. Its components along and perpendicular to $\mathbf{e}$ give the scalar formula above and

The osculating pericentre is $q=h^2/[\mu (1+e)]$. ThereforeFor mass loss, $\dot{\mu }<0$, this is nonnegative: pericentre does not decrease. It is positive except at pericentre itself, where $\cos f=1$ and $\dot{q}=0$. Thus a strictly opposite sign at every instant is not literally true; a pericentre passage with nonzero mass-loss rate is a counterexample to the strict wording. The intended monotonicity follows exactly, without requiring slow mass loss.

For adiabatic orbital expansion under isotropic mass loss, average over the unperturbed Kepler orbit and hold $\dot{\mu }/\mu$ constant to leading order during that orbit. If $E$ is the eccentric anomaly and $n=\sqrt{\mu /a^3}$, thenIt follows immediately thatAveraging with uniform true anomaly instead of uniform time would give the wrong result. Since $\mu a(1-e^2)=h^2$, conserved $h$ and constant secular orbital eccentricity giveThe pericentre and apocentre expand in the same secular proportion. These are leading adiabatic invariants, rather than exact invariants for arbitrary time-dependent $\mu$.

For the instantaneous precession fraction write $\theta =f+\varpi$ and $\dot{\theta }=h/r^2$. The adiabatic apsidal condition is controlled byIt is a signed fraction: apsidal advance contributes positively on the outward leg during mass loss, and negatively on the inward leg. Let ${\tau }_{\mu }=|\mu /\dot{\mu }|$. Ordinary orbit averaging requires ${\tau }_{\mu }\gg P$, together with negligible variation of the mass-loss law over one orbit. If the pericentre direction is also to vary negligibly relative to the instantaneous azimuthal motion, the maximum absolute fraction must be small. A precise condition for $0<e<1$ isThe maximizing cosine is $c=(1-\sqrt{1+8e^2})/(2e)$: differentiating the last factor gives $c+2e-e{c}^2=0$. Thus one can evaluate $F(e)$ explicitly by putting $\sqrt{1-c^2}/(1+ec{)}^2$ into the formula. Its useful limits areFor orbital eccentricity of order unity, mass loss much slower than an orbital period suffices; near a circular orbit, keeping the pericentre direction slowly varying additionally requires $n{\tau }_{\mu }e\gg 1$. The latter divergence is a singularity of the pericentre coordinate, not a divergence of the eccentricity vector dynamics. Indeed, at frozen $\dot{\mu }/\mu$, integrating the scalar evolution to first order gives $\delta e=-(\dot{\mu }/\mu )(1-e^2)\sin E/n$ up to an integration constant. Its absolute amplitude stays small for slow mass loss even when the scalar longitude of pericentre becomes unsuitable.

A constant-energy phase-space trajectory of the harmonic oscillator is the ellipseIts area is $\pi \sqrt{2E/(m{\omega }^2)}\sqrt{2mE}=2\pi E/\omega$. Thus the action and slow energy change areThe action variable is an adiabatic invariant when $|\dot{\omega }|\ll {\omega }^2$, with $\omega$ staying positive. Averaging over many short oscillations therefore keeps $I$ approximately fixed; an increase in frequency increases the energy in the same proportion.