Past exam of the relativity course of the University of Cambridge / 2014 / Paper 1 / 3

Boost and velocity transformation. For the moving frame,The inverse has $v\mapsto -v$. Taking differentials of that inverse and dividing by $dt$ gives

Constant proper acceleration. For collinear motion, the instantaneous rest-frame acceleration is $a={\gamma }^{3}dv/dt$. Since $dt/d\tau =\gamma$,Write $v=c\tanh \eta$; then $d\eta /d\tau =a/c$. Starting from rest gives $\eta =a\tau /c$, $dt/d\tau =\cosh \eta$ and $dx/d\tau =c\sinh \eta$. integration with the stated initial origin yields

photon rocket mass ratio. In the instantaneous rest frame, ejecting backward photon energy $d{E}_{\gamma }$ gives, to first order, $c^{2}dM=-d{E}_{\gamma }$ and $Mcd\eta =d{E}_{\gamma }/c$. Hence $dM/M=-d\eta$, andHere $M$ includes all stored fuel and energy. This assumes complete conversion into a perfectly collimated photon exhaust without other losses.

photon-rocket fuel for acceleration and braking. With effectively instantaneous changes of speed, the cruise proper time is $D/(\gamma v)=10\mathrm{yr}$, givingBraking to rest costs another factor $e^{-\eta }$, so $M_f/M_i=e^{-2\eta }=(\sqrt{101}-10{)}^2=0.00248758$. If equal rest masses of matter and antimatter annihilate, the consumed antimatter is half the total consumed fuel:This is the fraction of initial total mass, not the antimatter fraction of the consumed matter–antimatter fuel, which is one half.

finite-acceleration correction to a relativistic cruise. Each constant-proper-acceleration leg has proper duration ${\tau }_a=c\eta /a$ and Earth-frame length $x_a=(c^2/a)(\gamma -1)$. ThusSubtracting the original $D/(\gamma v)$ givesFor $a=10\mathrm{m}{\mathrm{s}}^{-2}$, each accelerating leg takes about $2.85$ proper years and covers $8.60$ light years in the Earth frame, leaving positive cruise distance. The new total is about 13.98 proper years. Adding $2{\tau }_a$ to ten years would overcount the cruise segment replaced by acceleration and braking.