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Rotation of coordinate system in minkowsky spacetime

  • 27-06-2011 11:03AM
    #1
    bcvdbgfs
    Closed Accounts Posts: 8 ✭


    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??


Comments

  • #2
    Morbert
    Registered Users, Registered Users 2 Posts: 3,457 ✭✭✭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 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??

    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 )


  • #3
    antiselfdual
    Registered Users, Registered Users 2 Posts: 170 ✭✭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?


  • #4
    ZorbaTehZ
    Registered Users, Registered Users 2 Posts: 2,149 ✭✭✭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?


  • #5
    antiselfdual
    Registered Users, Registered Users 2 Posts: 170 ✭✭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...


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