Deriving the relativistic rocket equation

[timetraveller123]

Homework Statement

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

Homework Equations

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

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

 

[Asim Riaz]

$P_e = \gamma (P'_e + \beta \frac{E'_e}{c})\\P_e = \frac{1}{\sqrt{1 - \frac{(v+dv)^2}{c^2}}} (dm u + \beta \frac{dm c^2}{c})\\$where $\beta = \frac{v+dv}{c}$is that correct