= Solution
{c}
Orient the $x$ axis along the displacement from the earlier event to the later one, so $\Delta x=d=1000\,\mathrm m$ and $\Delta t=10^{-6}\,\mathrm s$. For a parallel <Lorentz transformation>, <simultaneity> in the second <inertial frame> requires
$$
\Delta t'=\gamma\left(\Delta t-\frac{v\Delta x}{c^2}\right)=0,
\qquad
\boxed{v=\frac{c^2\Delta t}{d}=8.99\times10^7\,\mathrm{m\,s^{-1}}\simeq0.300c.}
$$
The second <inertial frame> moves towards the later event's location. The positive value $d^2-c^2\Delta t^2$ establishes <spacelike separation>, which permits this change of <simultaneity>. For a general <velocity> $\mathbf V$, the condition is $\mathbf V\cdot\Delta\mathbf r=c^2\Delta t$: only its component along the separation is fixed at the displayed value, and transverse components are allowed provided $|\mathbf V|<c$. Thus the parallel <velocity> has the smallest possible <speed>.
A physical <inertial frame> must have $|v|<c$, giving
$$
\boxed{|\Delta t|<\frac dc=3.336\,\mu\mathrm s.}
$$
\b[There is no attained maximum:] $d/c$ is the <supremum>. For the concrete boundary case $\Delta t=1000\,\mathrm m/c$, the events have <null separation> and the formula would require $v=c$; no <inertial frame> can make them simultaneous.
Back to article page
