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On a steady phase-mixed orbit, line density on a Kepler orbit is proportional to $1/v$, so cross-sectional area is uniform in time. For an optically thin wire of blackbodies in radiative equilibrium,The proof uses specific angular momentum: $dt/r^2=df/h$, and $Ph=2\pi a^2\sqrt{1-e^2}$. If one instead imposes a line density proportional to $v$, area is weighted by $v^{2}dt$, yielding $f={\sigma }_{\mathrm{tot}}(1+e^2)/[4\pi a^2(1-e^2{)}^{3/2}]$. The two distributions agree only for a circular Kepler orbit.

The printed line-density statement is incorrect for a phase-mixed orbit. A steady line density on a Kepler orbit is inversely proportional to speed: for a cross-sectional-area current $J_{\sigma }$, the area per unit arc length is $d\sigma /ds=J_{\sigma }/v$. Equivalently, phase mixing gives $d\sigma =({\sigma }_{\mathrm{tot}}/P)dt$, uniform in mean anomaly, where $P=2\pi \sqrt{a^3/\mu }$ is the orbital period.

For an optically thin population of blackbodies in radiative equilibrium, a fragment absorbs $\sigma L_{\star }/(4\pi r^2)$ and reradiates the same luminosity. Thus the fractional luminosity of a phase-mixed eccentric wire isThe specific angular momentum gives $dt/r^2=d\theta /h$, so the integral is $2\pi /h$. Since $Ph=2\pi a^2\sqrt{1-e^2}$,This derives the intended result after explicitly correcting the density to $1/v$.

The error is consequential. If one instead imposes the literal density $d\sigma /ds\propto v$, its time weighting is $v^{2}dt$. Using $⟨v^2⟩=\mu /a$, $⟨r^{-2}⟩=[a^2\sqrt{1-e^2}{]}^{-1}$ and $⟨r^{-3}⟩=[a^3(1-e^2{)}^{3/2}{]}^{-1}$ givesAt $e=1/2$ this is $5/3$ times the printed result. The inverse-radius averages can also be obtained directly with eccentric anomaly $E$, $r=a(1-e\cos E)$ and $dt=(1-e\cos E)dE/n$.

For a phase-mixed orbit, probability in a short arc is $dt/T=ds/(Tv)$, so the line density on a Kepler orbit is inversely proportional to speed. At true anomaly $\theta$,Using the vis-viva equation, $v^2=(G{M}_{\star }/a)[1+2e\cos \theta ]+O(e^2)$ and hence $v=na[1+e\cos \theta ]+O(e^2)$. ThereforeFor one particle, the normalized line probability is $(2\pi a{)}^{-1}(1-e\cos \theta )+O(e^2)$. This is density per arc length; density per unit polar angle has a further factor $ds/d\theta \simeq a(1-e\cos \theta )$ and is proportional to $1-2e\cos \theta$.