= Twin paradox with fixed reunion events
{title2=$\Delta\tau=\int_0^T\sqrt{1-v(t)^2/c^2}\,dt$}
Work in <Minkowski spacetime>, neglect <gravity>, and choose the stay-at-home clock's <inertial frame>. Let departure be $O=(t,x)=(0,0)$ and reunion be $R=(T,0)$. The home clock follows $x_H(t)=0$. A traveller moves outward with <velocity> $+v$ and returns with <velocity> $-v$, where $0<v<c$ and $c$ is the <speed of light>. The ideal <world line> has a sharp turnaround at $P=(T/2,vT/2)$:
$$
x_T(t)=
\begin{cases}
vt,&0\leq t\leq T/2,\\
v(T-t),&T/2\leq t\leq T.
\end{cases}
$$
Each straight leg lies in an <inertial frame>, but no single traveller <inertial frame> covers both legs.
For any differentiable <timelike curve> in this home <inertial frame>, its clock accumulates <proper time> through the <integral>
$$
\Delta\tau=\int_0^T\sqrt{1-\frac{v(t)^2}{c^2}}\,dt.
$$
Apply it separately to the traveller's two legs. With the <Lorentz factor> $\gamma=(1-v^2/c^2)^{-1/2}$,
$$
\Delta\tau_H=T,\qquad
\Delta\tau_T=\frac{T}{\gamma},\qquad
\Delta\tau_H-\Delta\tau_T=T\left(1-\frac1\gamma\right).
$$
For $v=0.6c$ and $T=10\,\mathrm{yr}$, the <Lorentz factor> is $\gamma=1.25$. The turnaround is $3$ light-years away at home time $5\,\mathrm{yr}$. The traveller records $4\,\mathrm{yr}$ outward and $4\,\mathrm{yr}$ returning: the reunited clocks read $10\,\mathrm{yr}$ and $8\,\mathrm{yr}$. These are illustrative model values, not experimental measurements.
\Image[/twin-paradox-reunion-spacetime.png]
{title=Fixed departure and reunion events: a 0.6c traveller records eight years while the inertial home clock records ten; turnaround simultaneity slices meet the home world line at 3.2 and 6.8 years}
The same <integral> gives a short geometric explanation: $\sqrt{1-v(t)^2/c^2}\leq1$, so every such competing <timelike curve> between these fixed <spacetime events> has $\Delta\tau\leq T$. Equality requires zero <velocity> throughout, apart from sets of zero duration. Thus the straight home <world line> maximizes <proper time> in this <Minkowski spacetime> problem. This statement concerns fixed endpoints and an inertial home clock; in a <gravitational field>, compare paths using the appropriate <metric tensor>.
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