The twin paradox

The twin paradox is perhaps the most famous — and most discussed — example in special relativity. Catherine leaves Earth at $t=0$ and travels towards Alpha Centauri, at a distance $L=4$ light-years, at speed $v=0.9\,c$; on arrival, she reverses course and returns at the same speed. Her twin brother Simon waits for her at home, at rest on Earth. When Catherine returns, how many years have passed for her, and how many for him?

Catherine’s journey: outbound to Alpha Centauri and back, while Simon stays at rest on Earth.

A conceptual path to the result

We work in units where $c=1$ light-year/year. For Simon, who remains in Earth’s reference frame, Catherine travels with uniform motion: the outbound leg lasts $L/v = 4/0.9 \approx 4.44$ years, and the return the same, for a total of about $8.9$ years. For Catherine, instead, what matters is the proper time along her world line, which is obtained from the invariant applied to the outbound leg:

$(\Delta\tau)^2 = (\Delta t)^2 - (\Delta s)^2 = 4.44^2 - 4^2 \approx 3.7,$

giving $\Delta\tau_\text{outbound} \approx 1.9$ years; by symmetry the return takes the same, and the total for Catherine is about $3.9$ years. The full numerical calculation is carried out in a dedicated exercise; here the result alone suffices: on her return Catherine has aged by about $3.9$ years, Simon by about $8.9$. The difference is about 5 years. If they started out as identical twins, on her return Catherine is younger — and this is not an optical illusion: she has genuinely lived less time.

Why it is not really a paradox

Why it is not a "paradox"

At first sight it seems contradictory: couldn’t Catherine equally well claim that it is Simon who moved relative to her, and that therefore he should be the younger one? The answer is no, and the reason is that their roles are not symmetric. Catherine is the one who changed reference frame halfway through the journey, reversing course: she experienced an acceleration. Simon, instead, always remained in a single inertial frame. The invariant $(\Delta\tau)^2 = (\Delta t)^2 - (\Delta s)^2$ must be applied separately to the two legs, outbound and return, and the sum of Catherine’s two proper times turns out to be less than Simon’s total time. It is the asymmetry — only Catherine changes frame — that breaks the supposed paradox.

The explanation is a special case of the maximisation of proper time: Simon’s world line is “straight” (a single frame), Catherine’s is “broken” (two legs). Hence the golden rule: between two events connected by motion, proper time is maximal for a uniform straight-line trajectory and decreases if the traveller changes reference frame. In a phrase: “whoever stays put ages more”.

Links

Topics: Special relativity Concepts: Time dilation · Spacetime invariant Skills: Changing reference frame Methods: Minkowski diagram

Related exercises: Mr Rossi goes to the theatre · Problem — The reference frame of minimum time · The rabbit race