# Lorentz transformation for timelike and spacelike points [duplicate]

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 $(x-y)\rightarrow -(x-y)$. Note that if $(x-y)^2>0$ there is no continuous Lorentz transformation that takes $(x-y)\rightarrow -(x-y)$"

Why is this happening?

Note that the metric signature is (+ - - -), as given in the preface/introduction of the book. Now let us take an example, which we can later extend to all frames related by Lorentz transformations.

Take $x$ and $y$ to be events lying on the time axis in some suitable frame. Note that

$$(x-y)^2 = (x^{\mu} - y^{\mu})(x_{\mu} - y_{\mu}),$$

therefore in this frame $(x-y)^2 > 0.$ Now there is certainly no Lorentz transformation taking $t \to -t$, where $t$ is along time axis. We can now generalize this to all frames related by Lorentz symmetry, to generalize this for all $(x-y)^2 > 0,$ i.e. for all timelike events.

Similarly we can show the same for spacelike separated points too, lets say on either sides of the origin on the x-axis. The spacelike separated points have $(x-y)^2 < 0.$ We can show easily for a Lorentz frame moving with +v will have an opposite sign for $(x-y)$ in his frame than for a frame moving with -v, and both can be related by a continuous Lorentz transformation. Again, we can now generalize this to all frames related by Lorentz symmetry, to generalize this for all $(x-y)^2 < 0,$ i.e. for all spacelike events.

• Bruce (or anyone), does the fact that "when (x−y)2>0 there is no continuous Lorentz transformation that takes (x−y)→−(x−y)" have anything to do with the notion that one would have to cross over the null surface of the light cone (see fig. 2.4 in the book) to get from say t to -t (in Bruce's answer above). If so, if this could be explained I would appreciate it. Thx.
– user131743
Oct 1, 2016 at 4:01