About parallel and intersecting timelike worldlines Suppose that two straight timelike worldlines are (not) parallel with respect to some frame $S$. Will these worldlines remain (not) parallel with respect to any other system $S^{\prime}$ related to $S$ by some Lorentz transformation? Whatever the answer, how can this be formally demonstrated?
 A: Two parallel lines must always remain parallel. Let us say two objects are stationary in some reference frame at $x_{0}$ and $x_{1}$, then performing a Lorentz transformation to a primed frame with velocity $\beta$ yields $x'_{0} = \gamma x_{0} - \gamma \beta t$ and $x'_{1} = \gamma x_{1} - \gamma \beta t$.
Further, since $t’ = \gamma t- \gamma \beta x_{0}$, we have equivalently $t = t’/\gamma + \beta x_{0}$, and hence $x'_{0} = \gamma x_{0} - \gamma \beta^2 x_{0} - \beta t’$ and $x'_{1} = \gamma x_{1} - \gamma \beta^2 x_{1} - \beta t’$ which are clearly two parallel lines in the $t‘-x'$ plane, with angle $\arctan(-\beta)$ relative to the $x'$ axis, with the domain restricted to $[0, \pi]$.
From here it is easy to generalize to the case where they are not stationary to begin, but are parallel with some angle $\theta$ with respect to the $x$ axis. Let us say your transformation is $\Lambda_{f}$. Use the approach in the previous paragraph (working backward) to go back to the stationary frame (with a transformation $\Lambda_{inv}$, say), and then applying $\Lambda_{f} (\Lambda_{inv})^{-1}$, we get our result. Note that the fact that $(\Lambda_{inv})^{-1}$ is also a Lorentz transform and that the product of two Lorentz transforms is still another transform is guaranteed by the group structure.
A remark: you can think that if two lines were parallel in one frame but not another, then the act of changing the reference frame has somehow introduced one to accelerate with respect to the other, which is grossly against the spirit of relativity.
