 Solution

Take the kaon decay as the common origin. A pion traveling freely with constant momentum has lab decay time $t_i=(E_i/m_{\pi }{c}^2){\tau }_i$ and position ${\mathbf{r}}_i=({\mathbf{p}}_i/m_{\pi }){\tau }_i$. The supplied proper survival times therefore giveConsequently the separation from the first decay event to the second isThe requested distance is the separation of the two events, not the sum of the two individual pion path lengths.

Here $|\Delta \mathbf{r}|>c|\Delta t|$, so the separation is a spacelike separation. There is an inertial frame in which the events are simultaneous; in that frame their distance equals the invariant spacelike length:One such frame moves along $\Delta \mathbf{r}$ with $\mathbf{V}=c^2\Delta t\Delta \mathbf{r}/|\Delta \mathbf{r}{|}^2$, of speed $0.71658c$. The simultaneity condition is then $\Delta t^{\prime }={\gamma }_V(\Delta t-\mathbf{V}\cdot \Delta \mathbf{r}/c^2)=0$.

The spacelike distance between daughter decay events provides an independent invariant check: the daughter four-velocity contraction givesIt produces the same distance without using the particular kaon laboratory boost. A simultaneous frame exists only when this expression is positive, as it is for these two survival times.