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In my special relativity book, there is a derivation of the velocity of a rocket. Let the rocket have rest mass $m$ (which decreases over time) and velocity $v$ in an inertial frame, and let the exhaust have speed $u$ in the accelerating frame of the rocket.

My book states that $$ \frac{d(mv\gamma)}{dv}=\left(\frac{v-u}{1-\frac{vu}{c^2}}\right)\frac{d(m\gamma)}{dv} $$ without any explanation, so I have to prove this myself. The left-hand side is easy. It is the change of the rocket's momentum. I can also understand that$\left(\frac{v-u}{1-\frac{vu}{c^2}}\right)$ is the speed of exhaust in inertial frame. The problem is the last one: $\frac{d(m\gamma)}{dv}$. It is the change in relativistic mass of the rocket. The basic idea of this equation is $$ \text{change in the momentum of rocket}=\text{change in the momentum of the exhausts} $$ so the RHS should be the momentum of the exhaust. However, I don't understand why $\frac{d(m\gamma)}{dv}$ is the mass of exhaust ejected; the rocket have lost $\frac{d(m\gamma)}{dv}$ kilograms of mass, but it doesn't mean that's the mass of the exhaust ejected, because exhaust have different velocity from the rocket, and mass depends on velocity. So I feel that it is not valid to assume $$ \text{mass loss of the rocket}=\text{mass of propellant ejected}. $$

So, is the equation in my book correct? Have I understood anything wrong?

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Let $$w=\frac{v-u}{1-\frac{uv}{c^2}}, \gamma_w=\frac{1}{(1-w^2/c^2)^{1/2}},\gamma=\frac{1}{(1-v^2/c^2)^{1/2}} $$ By conservation of momentum, we have $$ d(m\gamma v)+w\gamma_wdm=0 $$ By conservation of energy, we have $$ d(m\gamma c^2)+\gamma_wc^2dm=0, \gamma_wdm=-d(m\gamma ). $$ Substituing, we eliminate $\gamma_w$, and $$ d(m\gamma v)-wd(m\gamma)=0, $$ so the equation in the book is proven. In the question, the sign difference of the momentum is ignored.

Now we need to solve the equation. Keep in mind that $u$ is constant while $v$ is a function of $m$. We have $$ d(mv\gamma)=vd(m\gamma)+m\gamma dv=wd(m\gamma),\\ \Rightarrow m\gamma dv=(w-v)d(m \gamma ),\\ \frac{dv}{w-v}=\frac{d(m\gamma)}{m\gamma}. $$

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