= Fractional luminosity of a phase-mixed eccentric wire
{title2=$f$}
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>,
$$
f=\frac{\sigma_{\rm tot}}{4\pi P}\int_0^P\frac{dt}{r^2}
=\frac{\sigma_{\rm tot}}{4\pi a^2\sqrt{1-e^2}}.
$$
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 $v^2dt$, yielding $f=\sigma_{\rm tot}(1+e^2)/[4\pi a^2(1-e^2)^{3/2}]$. The two distributions agree only for a <circular Kepler orbit>.
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