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The ideal slow clock transport synchronization procedure reproduces Einstein synchronisation in an inertial frame, as a limiting result about proper time. It does not determine distant simultaneity without assumptions about clocks and their dynamics. Begin with identical ideal clocks together at $x=0$, with matching rates and readings. Leave one there and transport the other to a laboratory position $L$ at small constant speed $u$, neglecting gravity. Use Einstein-synchronised laboratory coordinates to calculate the result, rather than assuming that the transported clock already displays coordinate time.

The journey takes laboratory time $T=L/u$. The transported clock's proper time is $\tau =T\sqrt{1-u^2/c^2}$, giving the lagConsequentlyThe small time dilation accumulates over a long journey, but its quadratic dependence on $u$ dominates the journey's inverse dependence on $u$. Saying that the journey is arbitrarily long is therefore not an objection to this limit. For a nonuniform journey with $|v|\le u$, $\delta =\int (1-\sqrt{1-v^2/c^2})dt$; a simple monotonic path of fixed length has the same vanishing bound. Acceleration stages can be arranged with vanishing displacement and proper time deficit, assuming the clock hypothesis.^[ref]

This procedure establishes synchronization operationally for stationary laboratory clocks, once ideal-clock behaviour has empirical support. It does not show that every actual device is insensitive to transport: thermal shifts, acceleration-dependent mechanisms or differing gravitational potentials can spoil the idealization. Nor does the same construction define global synchronization for rotating laboratories, where path dependence is significant.

The conventionality question is separate. Resynchronize the same laboratory with $t_{\kappa }=t+\kappa x/c$, where $|\kappa |<1$. At the arrival event the transported clock approaches the reading $t$, whereas a stationary clock displaying $t_{\kappa }$ reads $t+\kappa L/c$. Slow transport has not eliminated that difference. The transformed metric tensor has an off-diagonal term, and the proper time calculation still gives the same physical clock reading: changing labels cannot change its ticks. This is the invariance of reunion ages under clock resynchronization, applied to transport comparisons.

Slow clock transport synchronization is thus a useful physical realization of a preferred conventional coordinate choice given Lorentz-invariant dynamics and ideal clocks. It does not independently prove that the one-way speed of light must be isotropic in every admissible synchronization. It supplies a reproducible convention with a strong physical rationale, rather than a convention-free observation of an absolute present.

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.