First at all, let me just say that I'm not a Physicist, I study mathematics. So, I have this question. If you have a spacelike four vector, is there any transformation that could change it to be a timelike four vector? I mean, I know that every Lorentz Transformation (LT) preserves this properties (timelike $\rightarrow$ timelike, spacelike $\rightarrow$ spacelike, etc.), but I was thinking in another frame $S'$, different from the former $S$, where a spacelike four-vector (in $S$) will be timelike (in $S'$). If it is possible to have this other frame then, the way to relate events between frames is not a LT? or I'm missing something?
$\begingroup$ Different frames at relate by a Poincaré transformation, that is a translation and a Lorentz transformation. So there is no physical way of transforming s space-like vector into a time-like one and vice versa. $\endgroup$– Phoenix87Feb 5, 2015 at 9:07
$\begingroup$ that was *are related $\endgroup$– Phoenix87Feb 5, 2015 at 10:47
$\begingroup$ Relevant? journals.aps.org/pr/abstract/10.1103/PhysRev.188.2287 $\endgroup$– Alfred CentauriFeb 5, 2015 at 11:51
The proper time, $\Delta\tau$, between two events is a conserved quantity in special relativity i.e. all observers will agree on its value. Since the definition of timelike is $(\Delta\tau)^2 \gt 0$, and the definition of spacelike is $(\Delta\tau)^2 \lt 0$, there can be no coordinate transformation that interconverts spacelike and timelike vectors.
$\begingroup$ Hi, John! I think a minor edit would make the answer perfect. Doesn't the definition of spacelike and timelike refer to $(\Delta\tau)^2$ being negative or positive? Or correspondingly, $\Delta\tau$ being imaginary or real. $\endgroup$– user87745Jun 19, 2016 at 19:47
Yes, there are transformations that take timelike vectors into spacelike vectors and viceversa. Consider $(t,x,y,z) \mapsto (x,t,y,z)$. You could event throw a Wick rotation, $(t,x,y,z) \mapsto (i\ t,x,y,z)$, as a transformation that takes timelike vectors and returns spacelike vectors.
Now, these transformation do not correspond to coordinate transformations between physical observers. This last fact is, as mentioned in a comment, a fundamental law of Physics.
2$\begingroup$ maybe add to your last sentence something like "and therefore this answer is wholly in keeping with the answers and comments that assert that proper time / proper length are conserved in SR." Maybe shorten my verbage a bit, though. Anyhow, I just mean make sure you avoid confusion for those who read both your correct answer and others which are equally correct. $\endgroup$ Feb 5, 2015 at 12:00
$\begingroup$ This was the most complet answer. Clarified many things, but generated other questions. You (physicist) actually solve a real(field) problem, through the complexification of the space? I don't know how this work for you. I mean, like you say, a frame in the complexification of the Minkowski space do not correspond to a "real observer". $\endgroup$– raulFeb 5, 2015 at 12:50
Let's think about this in terms of light cones. At any given point in space-time, we can consider the set of all possible light rays which pass through that particular point. If one now considers the set of all possible tangent vectors to those light rays (these are of course, null) then they form the light cone at that point.
If any one of those vectors is time-like then they will lie inside the cone, and space-like vectors will lie outside the cone. Null vectors lie on the cone.
I have not come across a transformation that would send one to the other, there may be some scope for considering a non-null foliation, where an integrable distribution of hypersurfaces can be either time-like or space-like but I concede that this isn't really answering your question.