If there were an interstellar spacecraft being propelled by gamma rays from electron/positron annihilation, they say it would get harder to keep accelerating it the closer it got to the speed of light because of what used to be called relativistic mass increase. But wouldn't the onboard electrons and positrons get proportionally more massive as acceleration continues, leading to more energy released during annihilation, which would make up for the increasing mass of the spacecraft? I asked this somewhere one time a long time ago and somebody answered 'there is no such thing as a free lunch.' But I'm not talking about a free lunch: as the mass increases, including that of the onboard fuel, and mass turns to energy, then doesn't the increased energy released stay in step with the overall increase in mass of the craft?
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2$\begingroup$ Relativistic mass is a confusing and outdated concept. You don’t need it to understand anything, and without it you will have a much simpler understanding of relativity. The only mass that you need is the invariant mass. $\endgroup$– G. SmithSep 22, 2020 at 17:24
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$\begingroup$ Can you clarify what you mean by “make up for”. What is the additional energy supposed to compensate? $\endgroup$– DaleSep 22, 2020 at 20:01
2 Answers
If you're defining relativistic mass as $\gamma m$ (an old and misleading consideration), then that is the "mass" determined from the observer frame outside the ship and moving relative to the mass, but the mass isn't $\gamma m$ in the rest frame of the ship. The annihilation takes place in the ship frame where the mass of each particle is still only 511 keV/c$^2$.
What the outside observer will observe, moving with speed $\beta$ compared to the ship and its propulsion unit, is that the photons are Doppler shifted to different wavelengths and different momenta (both travelling at $c$ in opposite directions) so that the momentum of each electron/positron pair is conserved (now with $p=2\gamma m\beta c$ compared to zero in the ship.)
If you define $\gamma m$ as the relativistic mass, the total energy of a lump of fuel with rest mass $m$ is $E=\gamma mc^2$. After turning the mass into energy we have $E=pc$ where $p$ is the magnitude of the 3-momentum that can be transferred to the ship. But no matter how much the ship's momentum is increased, it will never reach the speed of light, since most of the momentum increase will only contribute to the ship's $\gamma$ factor.
It doesn't matter what method is used to accelerate the ship, it won't reach the speed of light even if you send in extra fuel or momentum-carrying particles from outside.
Side note: The full formula for the energy of a moving particle is $E=\sqrt{m^2c^4+p^2c^2}$. From this you can derive $E=mc^2$ for the special case of a stationary particle, and $E=pc$ for the special case of a massless particle. It's easy to verify that $\gamma mc^2$ is the same as $\sqrt{m^2c^4+p^2c^2}$, but the latter is much more helpful. It's better to always think of "mass" as the rest mass, since it is an invariant of motion.
