I Deriving the relativistic rocket equation with exhaust efficiency

AI Thread Summary
The discussion revolves around the relativistic rocket equation and its derivation, particularly when factoring in the annihilation of propellant to usable exhaust. The user is attempting to derive the equation from Wikipedia but is struggling to connect it with the standard form of the relativistic rocket equation. They successfully reduce the equation to its simpler form when the annihilation factor is zero but cannot reconcile their derivation with the more complex version that includes the fraction of usable exhaust. Questions arise regarding the definitions of exhaust velocity and the variables used in the equation, indicating a need for clarification on these terms to resolve the derivation issues. The conversation highlights the complexities of relativistic physics and the challenges in understanding advanced concepts.
halcyon_zomboid
Messages
2
Reaction score
0
TL;DR Summary
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?
 
Physics news on Phys.org
What are ##v_{ex}## and ##v## in your formula ? I don' t find the former in Wikipedia you referred.
 
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Thread 'Gauss' law seems to imply instantaneous electric field (version 2)'
This argument is another version of my previous post. Imagine that we have two long vertical wires connected either side of a charged sphere. We connect the two wires to the charged sphere simultaneously so that it is discharged by equal and opposite currents. Using the Lorenz gauge ##\nabla\cdot\mathbf{A}+(1/c^2)\partial \phi/\partial t=0##, Maxwell's equations have the following retarded wave solutions in the scalar and vector potentials. Starting from Gauss's law $$\nabla \cdot...
Back
Top