Challenge: the twin paradox

Problem

Challenge: the twin paradox. Anna leaves for a star and comes back. Beatrice stays on Earth. On her return Anna is younger. Beatrice objects that in her SR it was Beatrice moving away: why isn’t the paradox symmetric? Draw the world lines of the two twins on a $(t,s)$ diagram.

Solution

The apparent paradox. Motion is relative: if Anna sees Beatrice recede and then approach again, by symmetry each should see the other’s clock run slow. How, then, can Anna really return younger, without contradiction?

The symmetry breaking. The situation is not symmetric. Beatrice always stays in the same inertial frame: her world line is a vertical straight line. Anna, on the other hand, to return, must reverse course: she slows down, stops, and sets off in the opposite direction. At that point she changes inertial frame (she undergoes an acceleration). Her world line is bent, with a vertex at the turnaround point.

$\ev{\text{Anna changes reference frame (accelerates), Beatrice does not}}$

The geometric explanation. In spacetime, proper time is the “length” of the world line, measured with the invariant $\Delta\tau^2 = \Delta t^2 - \Delta s^2/c^2$. Unlike Euclidean geometry, here Anna’s bent path corresponds to a smaller proper time than Beatrice’s straight path. Anna accumulates less time: she returns younger.

Why it is not symmetric. Only Beatrice remains inertial for the whole trip, so only she can naively apply time dilation along the entire path. Anna, who changes frame, cannot: during the turnaround her notion of simultaneity “jumps” forward, and it is precisely that jump which explains why, on reunion, she is the younger one. The acceleration is the objective mark that distinguishes the two twins.

Links

Topics: Special relativity Concepts: Time dilation · Spacetime invariant Skills: Constructing Minkowski diagrams