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halcyon_zomboid
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- TL;DR
- My derivation produces a relativistic rocket equation with exhaust "efficiency" that disagrees with this one: https://en.wikipedia.org/wiki/Antimatter_rocket#Modified_relativistic_rocket_equation
Hi,
I'm looking at this relativistic rocket equation on Wikipedia. Something doesn't make sense here, and I can't find a derivation for this equation in the linked source, so I'm trying to derive it myself with limited success.
$$\frac{\mathrm{d}M}{M}=-\frac{\mathrm{d}v(1-v_\mathrm{ex}\frac{v}{c^2})}{(1-\frac{v^2}{c^2})(-v_\mathrm{ex}\frac{v^2}{c^2}+(1-a)v+av_\mathrm{ex})}
$$
This is purportedly the relativistic rocket equation when accounting for annihilation to usable exhaust. To quote Wikipedia, " ##a## is the fraction of the original (on board) propellant mass (non-relativistic) remaining after annihilation (i.e., ##a=0.22## for the charged pions)."
I can see that this reduces to the usual relativistic rocket equation when ##a=0##:
$$
\frac{\mathrm{d}M}{M}=-\frac{\mathrm{d}v}{v_\mathrm{ex}(1-\frac{v^2}{c^2})}
$$
However, I can't seem to derive the above equation. Here's my attempt (which succeeds for the normal relativistic rocket equation):
In the COM frame, the rocket's initial mass is ##M+\mathrm{d}M##. Its final mass is ##M## and its velocity in this frame is ##\mathrm{d}v'##. Let's say the exhaust has mass ##\mathrm{d}m## and velocity ##v_\mathrm{ex}##, so if all the exhaust were usable, energy conservation would give
$$(M+\mathrm{d}M)c^2-Mc^2=\frac{c^2\,\mathrm{d}m}{\sqrt{1-\frac{v_\mathrm{ex}^2}{c^2}}}
$$
(to first order in ##\mathrm{d}v'##). However, in reality, not all of the exhaust is usable. Let's say only a fraction ##\alpha## of the energy loss ##c^2\,\mathrm{d}M## is captured by beamed core exhaust pions:
$$
\alpha c^2\,\mathrm{d}M=\frac{c^2\,\mathrm{d}m}{\sqrt{1-\frac{v_\mathrm{ex}^2}{c^2}}}.
$$
Okay, now let's look at momentum. For simplicity's sake, we'll assume all the useless energy is radiated symmetrically, so there's no effect on momentum. The final momentum of the rocket is ##M\,\mathrm{d}v'## (to first order in ##\mathrm{d}v'##), and that of the exhaust is
$$
-\frac{v_\mathrm{ex}\,\mathrm{d}m}{\sqrt{1-\frac{v_\mathrm{ex}^2}{c^2}}}
=-\alpha v_\mathrm{ex}\,\mathrm{d}M
$$
But doesn't this then mean that
$$
\frac{\mathrm{d}M}{M}=-\frac{\mathrm{d}v'}{\alpha v_\mathrm{ex}}=-\frac{\mathrm{d}v}{\alpha v_\mathrm{ex}(1-\frac{v^2}{c^2})}
$$
when we go back to the lab frame? I get nothing resembling the equation at the start.
Is there something wrong with my derivation?
I'm looking at this relativistic rocket equation on Wikipedia. Something doesn't make sense here, and I can't find a derivation for this equation in the linked source, so I'm trying to derive it myself with limited success.
$$\frac{\mathrm{d}M}{M}=-\frac{\mathrm{d}v(1-v_\mathrm{ex}\frac{v}{c^2})}{(1-\frac{v^2}{c^2})(-v_\mathrm{ex}\frac{v^2}{c^2}+(1-a)v+av_\mathrm{ex})}
$$
This is purportedly the relativistic rocket equation when accounting for annihilation to usable exhaust. To quote Wikipedia, " ##a## is the fraction of the original (on board) propellant mass (non-relativistic) remaining after annihilation (i.e., ##a=0.22## for the charged pions)."
I can see that this reduces to the usual relativistic rocket equation when ##a=0##:
$$
\frac{\mathrm{d}M}{M}=-\frac{\mathrm{d}v}{v_\mathrm{ex}(1-\frac{v^2}{c^2})}
$$
However, I can't seem to derive the above equation. Here's my attempt (which succeeds for the normal relativistic rocket equation):
In the COM frame, the rocket's initial mass is ##M+\mathrm{d}M##. Its final mass is ##M## and its velocity in this frame is ##\mathrm{d}v'##. Let's say the exhaust has mass ##\mathrm{d}m## and velocity ##v_\mathrm{ex}##, so if all the exhaust were usable, energy conservation would give
$$(M+\mathrm{d}M)c^2-Mc^2=\frac{c^2\,\mathrm{d}m}{\sqrt{1-\frac{v_\mathrm{ex}^2}{c^2}}}
$$
(to first order in ##\mathrm{d}v'##). However, in reality, not all of the exhaust is usable. Let's say only a fraction ##\alpha## of the energy loss ##c^2\,\mathrm{d}M## is captured by beamed core exhaust pions:
$$
\alpha c^2\,\mathrm{d}M=\frac{c^2\,\mathrm{d}m}{\sqrt{1-\frac{v_\mathrm{ex}^2}{c^2}}}.
$$
Okay, now let's look at momentum. For simplicity's sake, we'll assume all the useless energy is radiated symmetrically, so there's no effect on momentum. The final momentum of the rocket is ##M\,\mathrm{d}v'## (to first order in ##\mathrm{d}v'##), and that of the exhaust is
$$
-\frac{v_\mathrm{ex}\,\mathrm{d}m}{\sqrt{1-\frac{v_\mathrm{ex}^2}{c^2}}}
=-\alpha v_\mathrm{ex}\,\mathrm{d}M
$$
But doesn't this then mean that
$$
\frac{\mathrm{d}M}{M}=-\frac{\mathrm{d}v'}{\alpha v_\mathrm{ex}}=-\frac{\mathrm{d}v}{\alpha v_\mathrm{ex}(1-\frac{v^2}{c^2})}
$$
when we go back to the lab frame? I get nothing resembling the equation at the start.
Is there something wrong with my derivation?