Trilairian
Registered Senior Member
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
$$v = ctanh[(\frac{v_{ex}}{c})ln(\frac{m_{i}}{m})]$$
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
Last edited: