Past exam of the mathematics course of the University of Cambridge / 2016 / ia / Paper 4 / 11B (Dynamics and Relativity)

In the Earth inertial frame, the return journey takes $2d/v$. Bob is at rest in this frame, while Alice's proper time is reduced by the Lorentz factor $\gamma =(1-v^2/c^2{)}^{-1/2}$. Neglecting the short turnaround interval, their increases in age are

The twin paradox is the apparent contradiction from applying reciprocal time dilation to claim that each twin must be younger than the other. Alice changes inertial frames at the turnaround, so a single comoving inertial frame cannot describe her whole journey; Bob uses one inertial frame throughout. The twins follow different world lines, and integrating proper time along those world lines gives the displayed, consistent ages. The change in Alice's notion of simultaneity at turnaround removes the apparent contradiction. Acceleration distinguishes the paths; its magnitude is not an additional local rule for aging.

The meetings occur where Bob's constant position $x=\alpha c{t}_0$ intersects Alice's positive-$x$ branch of the hyperbolic motion:Bob is at rest in this inertial frame, so his elapsed proper time is

Parametrize Alice's world line bySince $t$ increases with $\eta$, the proper time element isAt the meetings, $\cosh \eta =\alpha$, giving endpoints $\eta =\pm \mathrm{arcosh}\alpha$. Thus Alice's elapsed proper time isHere $\mathrm{arcosh}$ is the nonnegative real inverse hyperbolic cosine. As a check, if $\alpha =\cosh {\eta }_{\ast }$ with ${\eta }_{\ast }>0$, then Bob ages by $2t_0\sinh {\eta }_{\ast }$ and Alice by $2t_0{\eta }_{\ast }$; $\sinh {\eta }_{\ast }>{\eta }_{\ast }$ confirms that Bob again ages more.