Homework Help: Deriving the relativistic rocket equation

1. Jan 10, 2018

timetraveller123

1. The problem statement, all variables and given/known data
so i read morin's derivation of rocket equation propelled by photons now i want to try for relativistic mass ejection but i am having some problems

let subscript e denote quantities of ejected material and subscript of r denote quantities of rocket

2. Relevant equations
$P = \gamma (P' + \frac{v E'}{c^2})\\ P = \gamma m v\\ E = \gamma m c^2$

3. The attempt at a solution
syaing the rocket is currently moving at speed v then if ejects dm at exhaust speed u its mass becomes v + dv and mass of m-dm

then as viewed from the rocket is
conservation of energy becomes
$E'_e = mc^2 -(m-dm)c^2 = dm c^2$
as mass is not conserved other form of energy of the ejected mass maybe written as
$E'_e = \gamma _ e m' c^2$
where m' is the observed mass of ejected mass
so
$dm = \gamma _e m'$
thus the momentum observed from rocket is
$p_e' = \gamma _e m' u = dm u$
is that correct please correct me if its wrong
so as viewed from the ground frame is
$P_e = \gamma (P'_e + \beta \frac{E'_e}{c})$
now my question is if the gamma corresponds to speed v or v+dv similary for the beta
$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\\ \gamma = \frac{1}{\sqrt{1 - \frac{(v+dv)^2}{c^2}}}\\$
which one is the correct one morin used the first one but why that
because by the time rocket measured the momentum of the ejected mass its already moving v + dv with respect to ground so shouldn't it be the second gamma

Last edited: Jan 10, 2018
2. Jan 15, 2018

PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.