I will here display the derivation for the relativistic rocket equation: \(v = ctanh[(\frac{v_{ex}}{c})ln(\frac{m_{i}}{m})]\) This does not employ the use of the ridiculous relativistic mass concept. Start with relativistic conservation of momentum and energy relating the initial and final states of the rocket and exhaust for a small element \(m_{fex}\) burned off. momentum \(\gamma mv = (m + dm)[\gamma v + d(\gamma v)] + m_{fex}\gamma _{fex}v_{fex}\) energy \(\gamma mc^{2} = (m + dm)(\gamma + d\gamma)c^{2} + m_{fex}\gamma _{fex}c^{2}\) Simplified \(0 = \gamma vdm + md(\gamma v) + m_{fex}\gamma _{fex}v_{fex}\) \(0 = \gamma dm + md\gamma + m_{fex}\gamma_{fex}\) Eliminate \(m_{fex}\gamma _{fex}\) by substitution from the second into the first \(0 = \gamma vdm + md(\gamma v) – (\gamma dm + md\gamma )v_{fex}\) Insert relativistic velocity addition \(0 = \gamma vdm + md(\gamma v) – (\gamma dm + md\gamma )[\frac{v – v_{ex}}{1 – \frac{vv_{ex}}{c^{2}}}]\) Simplify \(0 = [\gamma vdm + md(\gamma v)](1 – \frac{vv_{ex}}{c^{2}}) – (\gamma dm + md\gamma )(v – v_{ex})\) Switch variables to rapidity \(v = ctanh\theta \) \(0 = [sinh\theta dm + md(sinh\theta )][1 – tanh\theta (\frac{v_{ex}}{c})]c – [cosh\theta (dm) + md(cosh\theta )](tanh\theta – \frac{v_{ex}}{c})c\) After much simplification this reduces to \(d\theta = – (\frac{v_{ex}}{c})\frac{dm}{m}\) After integration \(\Delta \theta = (\frac{v_{ex}}{c})ln(\frac{m_{i}}{m})\) Choosing initial conditions of v = 0 when \(m = m_{i}\) and switching variables back from rapidity to velocity \(v = ctanh[(\frac{v_{ex}}{c})ln(\frac{m_{i}}{m})]\) QED

In all the excitement you forgot to specify a frame of reference. Of course, in Tril' Relativity maybe it doesn't make any difference where the observer is? Is your observer riding in the rocket or sitting in a comfy chair on Earth watching the rocket whizz by?

Hi Trilairian, Some of the tex seems to be broken - I see question marks! From the beginning: Why not set \(dm = -m_{fex}\) ? \(0 = \gamma vdm + md(\gamma v) - (\gamma dm + md\gamma )v_{fex}\) You seem to have used a peculiar minus character. It looks like an em-dash or something? Anyway, mimetex doesn't seem to like it. For the rest, it's interesting that the equation is in m rather than t. I haven't seen it this way before, only the constant acceleration version in terms of t. Checking that the two are equivalent in the case of constant acceleration should be about my level. I'll give it a shot when I have a little time.

Its working fine for my browser. I don't see any question marks. Because they are not equal. In relativistic rockets just like in decays, the mass of the initial conglomerate is not the sum of masses of the daughter particles, but is the sum of their relativistic energy's as reckoned by the center of mass frame. You don't need constant proper acceleration for that equation to be correct, just so long as the rocket never turns around to slow down ejecting fuel in the other direction. If you do want to look at the case of constant proper acceleration, in terms of this rocket it's not to hard. When the force is in the direction of motion you have that proper acceleration \(\alpha \) is equal to \(\gamma ^{3}a\)where \(a\) is the coordinate acceleration. \(\alpha = \gamma ^{3}a\) \(\alpha = \gamma ^{3}\frac{dv}{dt} = \gamma ^{3}\frac{dv}{d\theta }\frac{d\theta }{dt'}\frac{dt'}{dt}\) \(\alpha = \gamma ^{3}\frac{dv}{d\theta }\frac{d\theta }{dt'}\gamma ^{-1}\) \(v = ctanh\theta \) so \(\frac{dv}{d\theta } = \frac{c}{cosh^{2}\theta }\) resulting in \(\alpha = cosh^{3}\theta \frac{c}{cosh^{2}\theta }\frac{d\theta }{dt'}\frac{1}{cosh\theta }\) so \(\alpha = c\frac{d\theta }{dt'}\) So far that is correct whether or not the proper acceleration is constant which I think is a cool result. Now in the case that it is contant then \(\Delta \theta = \frac{\alpha \Delta t'}{c}\) and given our initial conditions this then can be input into \(v = ctanh\theta \) resulting in \(v = ctanh(\frac{\alpha t'}{c})\)

Odd. They work for me in IE, but not in Firefox. I wouldn't have thought it would be browser specific... must be the codepage or something. I see... that makes sense, after some cogitation. It would be easy to confuse oneself into thinking that the difference comes from the relativistic mass of the fuel or something... (that's the path I went down before figuring it out). It might help for amateurs like me to spell it out more by including the mass difference as a separate term, being energy spent in ejecting the fuel. Something like this: \(m\) is mass of rocket after ejecting the fuel element \(m_f\) is the fuel exhaust. \(m_e\) is the mass consumed in ejecting the exhaust. Momentum: Momentum before = momentum after \(\gamma mv + \gamma m_f v + \gamma m_e v = m[\gamma v + d(\gamma v)] + \gamma_f m_f v_f\) Similarly for energy: \(\gamma mc^2 + \gamma m_f c^2 + \gamma m_e c^2 = m(\gamma + d\gamma)c^2 + \gamma_f m_f c^2\) It certainly looks cool, but I need time to follow it through slowly. Please Register or Log in to view the hidden image! Thanks!

I just chose m to be the mass of the rocket before an element was burned off. m + dm to be the mass of the rocket after it was burned off and \(m_{fex}\) to be the mass of the element that came off. I just chose to represent it that way, but I suppose I should have explained all that in detail.

Believe me, there a LOT of things, such as specifying where the observer is when it is writing the math, that you should be explaining in detail.

No, that he is an inertial frame observer which is all that matters, not "where" he is, is obvious to anyone who would understand the math anyway. You don't so you don't matter.