On the page 28 of the book "An Introduction to Quantum Field Theory" by Michael E. Peskin and Daniel V. Schroeder in the last paragraph it says: "When $(x−y)^2<0$ we can perform a Lorentz transformation taking $(x−y)→−(x−y)$. Note that if $(x−y)^2>0$ there is no continuous Lorentz transformation that takes $(x−y)→−(x−y)$."
Is the fact that there is no continuous transformation due to the necessity of having to cross over the null surface of the light cone (see fig. 2.4 in the book) to get from say, $t$ to $–t$, and therefore the transformation cannot be continuous? If so, if this could be explained I would appreciate it.
Answer
I know what figure you mean :-).
Yes, a Lorentz transformation cannot cross the null surface. Otherwise one could convert time-like into space-like distances. And since $\mathrm ds^2$ is Lorentz invariant, it cannot flip its sign either.
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