# Relativistic acceleration of a rocket?

I'm trying to solve a physics brain-teaser and getting nowhere fast. My knowledge of special relativity is very basic.

In the problem a rocket is flying linearly at relativistic speeds (say $v>0.9c$), in the absence of external force fields. The rocket loses mass $m$ which is converted to mechanical work $W$ done on the rocket, acc:

$$dW=-\alpha dm$$

Where $\alpha$ is an efficiency coefficient ($\mathrm{Jkg^{-1}}$).

In the absence external force fields all mechanical work is converted to kinetic energy, so:

$$dW=dK$$

Relativistic kinetic energy is given by:

$$K=\frac{mc^2}{\sqrt{1-v^2/c^2}}-mc^2$$

I tried to set up a differential equation as follows:

$$-\alpha dm=\frac{(m+dm)c^2}{\sqrt{1-(v+dv)^2/c^2}}-(m+dm)c^2-\frac{mc^2}{\sqrt{1-v^2/c^2}}+mc^2$$

Even with some reworking and elimination of higher order infinitesimals that's very unyielding and I'm not sure its even correct. In fact it doesn't take the $K$ of the mass $dm$ into account. Probably conservation of momentum is also needed to be take into consideration.

Any help would be much appreciated.

The way to do this is to note that the acceleration $a'$ in the inertial frame of the spectators is related to the proper acceleration in the rest frame of the rocket, $a$, by:
$$a' = \frac{a}{\gamma^3}$$
So just work out the acceleration felt by the observers on the rocket in their rest frame, which is regular Newtonian mechanics, then feed the expression for $a$ into the equation above and solve the (probably exceedingly messy) resulting equation of motion.