06. The thrown clock

You must send a clock away from your lab bench and have it back in exactly $T$ seconds of bench time — but you want the clock itself to have ticked off as much time as possible when it returns. Carry it, drive it, fly it, throw it: which round trip wins?

Solution

Throw it straight up, so that it is in free fall the whole time, and gravity brings it back at $T$. (The winning toss peaks at height $gT^2/8$ — about 20 meters for $T = 4\ \mathrm{s}$, and the formula is safe for any toss short enough that $g$ stays uniform.)

In the weak-field limit a clock’s rate is $d\tau \approx \left(1 + \tfrac{gh}{c^2} - \tfrac{v^2}{2c^2}\right)dt$: it ticks faster the higher it sits and slower the faster it moves. Maximizing $\int d\tau$ therefore means buying as much altitude as possible while spending as little speed as possible — and the optimal compromise is exactly the free-fall parabola. Nor is the parabola merely stationary: perturbing it by $\eta(t)$ with $\eta(0) = \eta(T) = 0$ changes $\int\left(gh - \tfrac{v^2}{2}\right)dt$ by $-\tfrac{1}{2}\int\dot\eta^2\,dt < 0$, so the functional is concave and the toss beats every competing round trip, not just its neighbors. And the correspondence with mechanics is exact: $gh - \tfrac{v^2}{2}$ is the exact negative of $(T_{\mathrm{kin}} - U)/m$, so maximal aging and least action are the same statement read from opposite sides. Hamilton’s principle, for a projectile, is proper-time maximization in Newtonian light.

That is the honest content of the geodesic hypothesis: free fall is not a force-driven motion but the worldline of greatest aging — greatest, that is, among all worldlines joining the same two events, which is exactly the comparison this problem fixes. Things fall because falling is how you age the most. Things fall because falling is how you age the most.

Deeper in the notebook: 03. Local Inertial Frames; the Geodesic Hypothesis · 04. Gravity as Geometry - the Heuristic Argument · 01. The Twin Paradox