How Does the Relativistic Rocket Equation Describe Velocity?

[shinobi20]

Homework Statement

Show the relativistic rocket equation, in contrast to the Newtonian case.

Relevant Equations

$E' = \frac{E}{\sqrt{1- v^2 / c^2}}$
$p' = \frac{p}{\sqrt{1- v^2 / c^2}}$

Show that, according to relativistic physics, the final velocity $v$ of a rocket accelerated by its rocket motor in empty space is given by

$\frac{M_i}{M} = \Big ( \frac{c+v}{c-v} \Big) ^ \frac{c}{2 v_{ex}}$

where $M_i$ is the initial mass of the rocket at launch (including the fuel), $M$ is the final mass at burnout, and $v_{ex}$ is the velocity, relative to the rocket, with which the exhaust gases emerge from the nozzle of the rocket motor. (Hint: If the rocket ejects a small mass $dm$ of exhaust, conservation of energy and momentum in the instantaneous rest frame $x'y'z't'$ of the rocket imply

$c^2 dM = \frac{-c^2 dm}{\sqrt{1- v_{ex}^2 / c^2}}, \quad \quad M dv' = \frac{v_{ex} dm}{\sqrt{1- v_{ex}^2 / c^2}}$.

Furthermore, the velocity addition formula implies

$v + dv = \frac{v + dv'}{1 + v dv' / c^2}\Bigg)$

Solution.

I am not sure why the problem went to the instantaneous rest frame of the rocket and from there applied the conservation of energy and momentum, but given that, I will deduce the energy and momentum relations given above.

At some instantaneous time $t$, the total mass is $M+dm$ and the velocity is zero. After the rocket spews out exhaust with mass $dm$ and velocity $v_{ex}$, the rocket with mass $M$ moves to the right with velocity $dv'$, all these with respect to the original instantaneous rest frame of the rocket.

$(M+dm)0 = M dv' - v_{ex} dm \quad \rightarrow \quad M dv' = v_{ex} dm$

for the relativistic case,

$M dv' = \frac{v_{ex} dm}{\sqrt{1- v_{ex}^2 / c^2}}$

The energy of the rocket initially is $(M+dm) c^2$, and so when the exhaust left the rocket the total energy of the system (rocket plus exhaust) must be the same, but their changes are related by,

$c^2 dM = - c^2 dm$

for the relativistic case,

$c^2 dM = \frac{- c^2 dm}{\sqrt{1- v_{ex}^2 / c^2}}$.

What I am confused about is why the problem stated the velocity addition in that way. I believe $v$ is the velocity of the rocket (the original instantaneous rest frame of the rocket) with respect to the laboratory frame so that when the rocket spewed out the exhaust, it gained a velocity $dv'$ with respect to the instantaneous rest frame of the rocket and so with respect to the lab frame, the velocity of the rocket after exhaust is $v+dv$ in which $dv$ is how the lab frame viewed $dv'$. I kind of understand and kind of not understanding this. Also, can anyone give me a hint on how to start deriving the original question of showing the mass ratio?

***I know there are other ways of deriving this relation, however, I want to stick with the given problem and probably gain the intuition it wants me to gain from how the problem was stated.

 

[shinobi20]

Actually, I have derived it already but I'm not so sure about everything.

From the velocity addition formula,

$v + dv = \frac{v + dv'}{1 + v dv' / c^2}$

$v + v^2 dv' / c^2 + dv = v + dv' \quad$ dropping the second order terms

we get,

$\frac{dv}{1 - v^2/c^2} = dv' \quad$ where $dv' = -v_{ex} \frac{dM}{M}$ (this was derived by dividing the energy to the momentum relation above).

$\frac{c^2}{v_{ex}} \frac{dv}{c^2 - v^2} = -\frac{dM}{M}$

integrating, we get

$\frac{c^2}{v_{ex}} \frac{1}{2c} \ln \mid \frac{v+c}{v-c}\mid = \ln \frac{M_i}{M}$

we can just interchange the $v - c$ to give $c - v$ in the denominator of the first natural log since its argument is enclosed by an absolute value giving us,

$\frac{c}{2v_{ex}} \ln \mid \frac{c+v}{c-v}\mid = \ln \frac{M_i}{M}$

$\frac{M_i}{M} = \Big( \frac{c+v}{c-v} \Big)^\frac{c}{2v_{ex}}$.

Can someone tell me if what I have done is correct? Also, is my argument for the derivation of the energy and momentum relations above correct? Why is going to the rocket frame and starting to argue from there much better and therefore leading to the velocity addition formula given above?

 

[PeroK]

Actually, I have derived it already but I'm not so sure about everything.

From the velocity addition formula,

$v + dv = \frac{v + dv'}{1 + v dv' / c^2}$

$v + v^2 dv' / c^2 + dv = v + dv' \quad$ dropping the second order terms

we get,

$\frac{dv}{1 - v^2/c^2} = dv' \quad$ where $dv' = -v_{ex} \frac{dM}{M}$ (this was derived by dividing the energy to the momentum relation above).

$\frac{c^2}{v_{ex}} \frac{dv}{c^2 - v^2} = -\frac{dM}{M}$

integrating, we get

$\frac{c^2}{v_{ex}} \frac{1}{2c} \ln \mid \frac{v+c}{v-c}\mid = \ln \frac{M_i}{M}$

we can just interchange the $v - c$ to give $c - v$ in the denominator of the first natural log since its argument is enclosed by an absolute value giving us,

$\frac{c}{2v_{ex}} \ln \mid \frac{c+v}{c-v}\mid = \ln \frac{M_i}{M}$

$\frac{M_i}{M} = \Big( \frac{c+v}{c-v} \Big)^\frac{c}{2v_{ex}}$.

Can someone tell me if what I have done is correct? Also, is my argument for the derivation of the energy and momentum relations above correct? Why is going to the rocket frame and starting to argue from there much better and therefore leading to the velocity addition formula given above?

It looks fine to me. You use the rocket frame because that is the frame in which the quantity $v_{ex}$ is measured. You have to start in this frame to use that data.