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In this document "a transparent derivation of rocket equation", the author writes (on the last page):

These are the classical and relativistic rocket equations. For the relativistic case, there is a maximum exhaust velocity for the reaction mass that is given by: $$ w=c\sqrt{e(2-e)} $$ where $e$ is the fuel mass fraction converted into kinetic energy of the reaction mass.

I am not even sure what "mass fraction converted into kinetic energy" mean. Why is there a maximum of $w$? How to derive it?

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Ahhh, that's just bad notation. The author in that paper uses a non-standard terminology such as variant mass vs. rest mass, whereas we today use relativistic energy and momentum relations with an invariant mass. As far as I know these two approaches are equivalent in calculations, however there is no concept of mass changing with velocity in SR.

However, mass can converted into energy in SR, as I tried to exemplify below.

So, let's assume the exhaust has constant relative velocity $-w$ w.r.t. the rocket. Assume that at time $t$, the rocket has velocity $v$ w.r.t. the lab frame. Let's change our frame to a rest frame going with $v$. In this frame, the rocket is momentarily at rest.

Now, let $dt$ time pass. In this interval, let a mass $dm_{exhaust}$ be exhausted. Let's assume some part of the original fuel supply that was lost in this interval got converted into energy, $dm_{energy}$. In total, the fuel supply decreased by the sum of these two ($=dm_{reaction}$).

(Note: This process always occurs, be it fusion in stars or an ordinary chemical reaction. Although this change in mass is usually too small to ever be included in the calculations.)

From conservation of energy in this frame, $$m\,c^2=\frac{dm_{exhaust}\,c^2}{\sqrt{1-w^2/c^2}}+\frac{(m-dm_{energy}-dm_{exhaust})\,c^2}{\sqrt{1-dv'^2/c^2}}$$ where $dv'$ is the additional speed of the rocket in the frame going with $v$.

I believe the author defined $e$ as $e := dm_{energy}/dm_{reaction}$. If we divide the equation by $dm_{reaction}\,c^2$ after cancelling the $m\,c^2$s while ignoring $O(dx^2)$ terms: $$0 = \frac{1-e}{\sqrt{1-w^2/c^2}}-1 \implies \sqrt{1-w^2/c^2} = 1-e$$ $$\implies w = c\,\sqrt{1-(1-e)^2}=c\,\sqrt{e(2-e)}.\tag*{$\blacksquare$}$$

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