Past exam of the philosophy of space-time course of the University of Oxford / 2021 / 3 / Solution

The age comparison is a local comparison of clocks at two shared spacetime events: departure and reunion. No convention about the simultaneity of distant events is needed to read two adjacent clocks. The result is determined by the proper time along each complete world line. For an ideal clock following a timelike curve in the home observer's inertial frame,The home observer stays at $x=0$, whereas the traveler goes outward at velocity $v$ and returns at velocity $-v$. Ignoring the duration of a short turnaround gives home age $T$ and traveler age $T/\gamma$. The twin paradox with fixed reunion events derives this directly. There is no contradiction because the traveler does not remain in one inertial frame throughout the trip.

Relativity of simultaneity becomes essential to a different question: how can the traveler consistently describe the home clock as running slowly during both straight legs? The statement about a distant clock's rate refers to a sequence of remote events selected as simultaneous by a specified inertial frame. Changing that frame changes the selected sequence. Adding the two slow-rate comparisons while silently preserving one distant-time assignment is the error.

Let turnaround be $P=(t,x)=(T/2,vT/2)$ in the home frame. Constant outward-frame time means $t-vx/c^2$ is constant. Thus the home event simultaneous with $P$ according to the outward frame occurs atFor the return frame, constant time means $t+vx/c^2$ is constant, so its simultaneous home event occurs atThese are two different remote spacetime events, even though both assignments pass through the same turnaround event. The twin paradox simultaneity change isEach leg contributes $T/(2{\gamma }^2)$ to the home clock advance assigned within that leg's frame. Including the reassignment gives $T/{\gamma }^2+v^{2}T/c^2=T$, exactly the home reading at reunion. For the illustrative values $v=0.6c$ and $T=10$ years, the traveler's elapsed proper time is eight years; the two leg assignments give $3.2+3.2$ years of home advance, and the frame change supplies the missing $3.6$ years. The distant home clock does not physically jump.

Nor should simultaneity assignments be confused with received light signals. An observer sees emissions after a propagation delay and measures a Doppler shift. The received history remains consistent through a finite turnaround, but it is not the same sequence of events as a constant-coordinate-time slice. The Lorentz transformation relates the latter; signal travel times must be included separately when discussing what is seen.

An instantaneous turnaround is only an ideal limit. In the twin paradox with finite acceleration, the world line and the momentarily comoving frame vary continuously. The distant home-time assignment is $t_H(t)=t-v(t)x(t)/c^2$, soThe reversal at positive $x$ makes the last contribution positive. Its integrated effect replaces the ideal reassignment. By the clock hypothesis, the traveler's own elapsed proper time has no extra term depending directly on acceleration; physical disturbances of a particular clock are a separate experimental issue.

Thus relativity of simultaneity explains the consistency of the traveler's successive inertial descriptions. It is not a convention-dependent cause of the final age difference. The invariance of reunion ages under clock resynchronization makes that distinction explicit: remote time labels can change while the two reunited clock readings remain unchanged.