 Twin paradox simultaneity change

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.