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Rotation and Minkowski diagrams.

  1. Nov 27, 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://img21.imageshack.us/img21/3051/drawing1lq.th.jpg [Broken]
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 27, 2009 #2
    Hello Grimble.

    Any point in the complete plane can be referred to a set of normal Cartesian coordinate axes. Quite often only the first quadrant is shown as in your diagram with the x and t axes. This is often called the first quadrant, and, moving clockwise around the origin we come to the second, third and fourth quadrants. All points in the first quadrant such as your defined point referred to the x,y axes have +ve x coordinates and +ve t coordinates. Referring to your x’ t’ axes the point lies in the second quadrant and such points have –ve x’ coordinates and +ve t’ coordinates . If you just ignore the rectangular axes, just read the coordinates off against the x’, t’ axes by drawing sets of lines parallel to each axis to form a non-rectangular grid pattern, not by dropping perpendiculars. That is what gives the difference between contravariant and covariant vector coordinates. It is at first a little confusing because with rectangular coordinates the two methods give the same result, contra and covariant are the same.But don't worry about that now.

  4. Nov 27, 2009 #3

    could you elaborate a little more on what gives the difference between contravariant and covariant in this context?

  5. Nov 27, 2009 #4
    Hello Mikkey W.

    I would be quite happy to. I am a little pushed for time this morning (GMT) but will try to get back to you. I am sure many here can help you and may well do so before I can reply.

  6. Nov 27, 2009 #5
    Hello MikeyW.

    This is by no means a rigorous or thorough explanation but its gives the idea, I hope.

    Take rectangular Cartesian coordinates in two dimensions. For simplicity take a point in the first quadrant. Join the point to the origin to make a position vector. Because the mathematics is not specific to spacetime diagrams, label the axes as standard x and y. We will just talk about the x coordinate as the same method applies to the y coordinate. Now if we drop a perpendicular (project the vector) from the point onto the x axis we get a value of the x coordinate of the vector. We could also draw a line parallel to the y axis and take its point of intersection with the x axis as the x coordinate of the vector. As you can see, in the case of rectangular (orthogonal) axes the result is the same.

    Now if you take oblique axes, for simplicity again take a vector in the first quadrant. Now if you drop a perpendicular (project) onto the x axis you get a different value for the x coordinate than if you drew a line through the point parallel to the y axis and took its intersection with the x axis as the x coordinate.

    Projecting onto the axes gives contravariant components and drawing parallels to the y axis and taking the intersection with the x axis, gives covariant components.

    This can of course be extended to three dimensions and higher but is more difficult to describe without the aid of diagrams for three dimensions and cannot be visualized at all for higher dimensions.

    Any corrections or additions from those more mathematically minded would be welcome.

  7. Nov 27, 2009 #6


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    I think this is the source of your confusion. A Lorentz boost is not a rotation. It's a distortion that takes a rectangle and sort of skews and distorts it into a parallelogram. http://www.lightandmatter.com/html_books/genrel/ch01/ch01.html#Section1.7 [Broken]

    There is a loose analogy between Lorentz boosts and rotations. They play similar logical roles, but they're not the same thing.

    It is also possible to use complex-valued coordinates to make the equations for Lorentz boosts look exactly like rotations. IMO this is basically a bad idea, and I think very few people these days teach it that way.
    Last edited by a moderator: May 4, 2017
  8. Nov 28, 2009 #7
    Thank you Matheinste, that is a lovely lucid explanation, I am struggling sometimes when terms are used that I am unfamiliar with and an explanation really saves me an awful lot of time looking them up, and finding over complex explanations!

  9. Nov 29, 2009 #8
    I'm sorry but that explanation leaves me a little confused, for there is a piece in the http://www.newworldencyclopedia.org...owski_formulation:_Introduction_of_spacetime" (not perhaps the most accurate source, I admit), that states:
    And anyway, in the last two diagrams in my original post you can see that the result of the rotation is that the moving frame is skewed into a parallelogram by its rotation, as I have envisaged it.

    p.s. I am sorry if this seems an odd way to pose questions but I find drawings much clearer than mathematical notations (in my 'minds eye' I can pick them up, turn them and manipulate them as if they were hologramatic displays, even quite complex ones). It is just the way my mind works.

    Last edited by a moderator: May 4, 2017
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