Rotation of coordinate system in minkowsky spacetime

Does performing a rotation of the usual coordinate system $ct,x$ in the minkowsky spacetime makes sense?

I guess it doesn't, but more than this i think that there is something that forbids it, since i could make coincident the 'lenght' axis of the non rotated coordinate system (observer A) with the 'time' axis of the rotated coordinate system (observer B), and that seems ridiculous to me (but you never know..)!

From this I suppose that the Lorentz trasformations has to have some particular costraint, so I checked the propriety of the Lorentz group (to which they all belong) but i couldn't see it!

Can you help me??
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 Recognitions: Gold Member Science Advisor In space, you can rotate the axes using a transformation $$x' = x \, \cos \theta - y \, \sin \theta$$$$y' = x \, \sin \theta + y \, \cos \theta$$ and this preserves the metric $$ds^2 = dx^2 + dy^2$$ You can't apply the above rotation to (ct,x) coordinates to get an inertial frame, but you can apply the transformation $$ct' = ct \, \cosh \phi - x \, \sinh \phi$$$$x' = -ct \, \sinh \phi + x \, \cosh \phi$$ We can call this a "hyperbolic rotation". Note this is actually nothing more or less than a Lorentz transformation with $v = c \, \tanh \phi$ and $\gamma = \cosh \phi$, and which preserves the metric $$ds^2 = c^2 \, dt^2 - dx^2$$
 The Lorentz Transformations are in fact rotations in 4-D Minkowsky space-time. The Lorentz transforms are constrained by the fact that v

Rotation of coordinate system in minkowsky spacetime

Thanks a lot pals!!!

So it's all becaouse it has always to be$$\eta_{\nu\mu}=\Lambda_{\nu}^{a}\Lambda_{\mu}^{b} \eta_{ab}$$ to preserve lenght and stuff!!! and my silly example obviously won't satisfy this!!!

But new question...all of the Lorents transformation are reconducible to hyperbolic rotations?
If so this means that both the lenght and time axis 'squeeze' by the same angle $\phi$ towards the world line of a particle which moves whit the speed of light! isn'it?
 The Lorentz transforms are the 6 rotations in the 4-D space time (3 rotations in space, 3 boosts) yes. However, there are 4 additional symmetries of Minkowski spacetime (3 translations, and 1 identity transform), and together these 10 symmetries make up the Poincare group. In all, there are 3 boosts, 3 rotations in space, 3 translations, and 1 identity. I don't know where are you are getting this "squeezing" from.
 Mentor There are four translations (three in space and one in time) and usually the identity isn't counted.
 Mentor The (homogeneous restricted) Lorentz group: 1) does not include translations; 2) has elements that are generated by boosts and spatial rotations. These elements are not necessarily rotations or boosts.
 Ah right, I miscounted. The identity is obtained by setting v=0 in a boost, or by rotating through angle 0, or by translating 0! 4 translations, 6 "rotations". George, what do you mean specifically that these elements are not necessarily rotations or boosts? You mean that the elements of the Lorentz group may not be pure rotations or boosts, and could be a combination? This property is guaranteed by the closure property of groups.