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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.

General relativity is not required for the ideal twin paradox in Minkowski spacetime. The relevant observable is the difference in proper times recorded by clocks that meet at departure and reunion. For a world line $x(t)$ in one inertial frame, the clock hypothesis givesThe home clock with $x=0$ records $T$. A traveler coasting outward at $v$ for $T/2$ and inward at $-v$ for $T/2$ records $T/\gamma$, in the ideal short-turnaround limit. For $T=10$ years and $v=0.6c$, these readings are ten and eight years. The complete paths, rather than conflicting applications of reciprocal time dilation to one pair of inertial frames, determine the invariant difference.

An exactly instantaneous reversal is an idealization, not an essential premise. Replace it by a smooth timelike turnaround. The proper time accumulated in a reversal interval of home duration $\delta t$ lies between zero and $\delta t$, so its contribution vanishes as the interval shrinks. The finite clock difference persists. Repeatedly lengthening the coasting legs while keeping the same reversal maneuver also changes the clock difference; the difference is not a universal function only of the peak proper acceleration.

The usual apparent symmetry fails because no single traveler inertial frame covers both legs. In the instantaneous rest frame at a traveler event $(t,x(t))$, the home event assigned as simultaneous has timeHence a finite turnaround changes the home simultaneity assignment continuously. In the sharp-turnaround limit at $x=vT/2$, changing $v$ to $-v$ changes that assigned home time by $v^{2}T/c^2$. This is a change in a distant relativity of simultaneity assignment, not a sudden physical jump in the home clock's reading. The full twin paradox simultaneity change reconciles the two inertial coasting descriptions.

An accelerated-coordinate calculation is also available within special relativity. Nonzero connection coefficients in that chart do not establish nonzero Riemann curvature tensor. Thus neither using accelerated coordinates nor making the traveler accelerate requires the Einstein field equations. The clock hypothesis is a physical assumption about a sufficiently ideal clock, not a guarantee that every real apparatus survives arbitrary acceleration without disturbances. Experiments such as the muon storage-ring time-dilation experiments test clock behavior in specified regimes.

Real flights around Earth also compare different gravitational potentials and require gravitational corrections, as in the twin paradox clock experiments. Those refinements involve general relativity. They do not make the flat-spacetime reunion calculation incomplete on its stated assumptions. Even two mirrored, accelerating travelers can record equal proper times; the decisive quantity is the proper time of each entire path between the common events.

During either inertial leg, the traveller can correctly say that the moving home clock runs slowly. The missing step is deciding which distant home-clock reading is simultaneous with turnaround. By the Lorentz transformation, the outward inertial frame's constant-time line through $P$ obeys $t-vx/c^2=\text{constant}$. At $x=0$ it meets the home world line atFor the return inertial frame, use $t+vx/c^2=\text{constant}$ through the same spacetime event $P$; its home-clock assignment isThe dashed lines in the spacetime diagram illustrate this relativity of simultaneity. For $v=0.6c$ and $T=10\mathrm{yr}$, they assign $3.2\mathrm{yr}$ and $6.8\mathrm{yr}$ to the home clock. The difference of $3.6\mathrm{yr}$ is a change in the traveller's distant-time assignment, not a sudden physical advance of the home clock.

On each leg, the traveller's $4\mathrm{yr}$ of proper time corresponds to $4/\gamma =3.2\mathrm{yr}$ of home-clock advance according to that leg's inertial frame. Including the twin paradox simultaneity change gives $3.2+3.6+3.2=10\mathrm{yr}$, exactly the reading at reunion. Actual received light signals remain continuous across a physical turnaround; their arrival times are distinct from the distant spacetime events assigned as simultaneous by Einstein synchronisation.