Rotation of coordinate system in minkowsky spacetime

bcvdbgfs

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 , 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??

Morbert

bcvdbgfs wrote: »

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 , 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??

I'm not seeing any immediate problem. You can perform a Lorentz transformation on a 1+1 D system (ct,x).

t' = g (t - vx/c^2)
x' = g (x - v t )

antiselfdual

bcvdbgfs wrote: »

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??

The constraint on a Lorentz transformation is that it leaves the Minkowski "distance"

[latex]
ds^2 = - c^2 dt^2 + dx^2
[/latex]

invariant, i.e. if we have some Lorentz transformation to new coordinates [latex]t^\prime, x^\prime[/latex] (which are functions of the original [latex]t,x[/latex]) then we have to have

[latex]
ds^2 = - c^2 (dt^\prime)^2 + (dx^\prime) ^2 = - c^2 dt^2 + dx^2
[/latex]

You can check this is true for the Lorentz transformation Morbert posted for instance... and that it's not for a rotation of both time and space coordinates. Does that answer your question?

ZorbaTehZ

antiselfdual wrote: »

You can check this is true for the Lorentz transformation Morbert posted for instance... and that it's not for a rotation of both time and space coordinates. Does that answer your question?

Suppose you consider a rotation in the [latex]tx[/latex]-plane, then the transformation

[latex]x=x'\cosh \psi +ct' \sinh \psi[/latex]
[latex]ct=x' \sinh \psi +ct' \cosh \psi[/latex]

would preserve the metric. Granted, the idea of a rotation in tx-plane doesn't have physical meaning (at least to me.) Or am I totally missing something here?

antiselfdual

I was assuming the OP meant an ordinary (trigonometric) rotation of the time and space coordinates together, not a hyperbolic rotation (which I'm used to just calling a boost), otherwise I'm not sure what exactly the issue is...