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Rotation and Minkowski diagrams.

  1. Nov 26, 2009 #1
    Roger Penrose is reported to have said:
    And I understand that, in Minkowski space-time, the relative velocity between two inertial frames of reference is a rotation of the frames, with respect to one another, about a common origin.

    Now in the description of Minkowski diagrams given http://www.answers.com/topic/minkowski-diagram" it states:
    Yet I don't quite see how this can work if the coordinates observed are outside the axes for the primed frame.
    http://img8.imageshack.us/img8/1306/98894751.jpg [Broken]
    For instance, in the above diagram, how is the event at ct=5,x=1 to be read in the primed coordinates? Does it have a negative x' value?

    I have tried to determine how the rotation of the ct',x' plane is shewn in these diagrams and my first thought was that it was in the ct,x plane, as shewn by the angling of the ct' axis.
    But if this were so it would mean that the x' axis was rotated in the opposite direction! So that doesn't seem to work.

    Then I realised that it must be rotated out of the ct,x plane as shewn in the foolowing diagram.

    http://img16.imageshack.us/img16/5218/figure3g.jpg [Broken]
    With the narrowing of the angle, between the moving frame's axes, being a result of the primed frame transforming into a diamond shape, as it rotates, eventually approaching a straight line as the relative speed approaches 'c'.


    Drawing a more detailed diagram to shew two inertial frames with a relative velocity v = 0.6c seems to bear this line of reasoning out; An object(event?) with coordinates ct'=10,x'=6 dropped vertically on to the unprimed frame gives coordinates ct=8, x=4.8 as expected with γ = 1.25 and 1/γ= 0.8
    (x=γx', t=γt')

    http://img130.imageshack.us/img130/9692/drawing1z.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
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