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Meaning of Lorentz Transformations purely mathematically ?

  1. Oct 31, 2009 #1
    Couple days ago, we get a lecture in relativity, I read quite a lot about it before so there was nothing new except one thing : our professor first started to conclude Lorentz transformation totally in a mathematical way by assuming gamma*(x-v*t) … (what I discovered that it is a known method but somehow I didn't read about before)

    Anyway during the formulation of Lorentz transformations, we proving that (1-gamma^2)/v^2=constant (which is actually 1/c^2) , this conclusion stopped me a lot, and I searched Internet for couple days to get an answer to may question in vain: from purely mathematical point of view , not speaking about Einstein relativity nor speed of light, what that means!? If I will have a question of calculating when some even accrue in another reference frame in some totally abstract space-time like Euclidean space, what this constant will mean? We can't say it is max allowed speed in this abstract 2D plane! Because there is not any, and saying put it simply to zero also not quite acceptable from pure math point of view, because we have to prove it! Actually it's very important question to me, and I can't accept mathematically just to put it zero to get Galilean transformation, hope that somebody has a clue how to understand this!

    Thanks in advance.
     
  2. jcsd
  3. Oct 31, 2009 #2

    haushofer

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    I must say I can't really follow your description of your question. But maybe it's a good idea to look at an analog which you do know: rotations in the plane.

    A rotation in the 2D plane {x,y} is described by a 2x2 matrix R which (and this is important!) leaves the Euclidean interval [itex]l^2 = x^2 + y^2[/itex] invariant. This puts a restriction on R; it has to be orthogonal, which you can immediately check.

    Now, look at 2D spacetime {t,x}. One axiom of SR is that the speed of light is constant for every inertial observer. This leads you to say that we have an invariant interval

    [tex]
    s^2 = c^2 t^2 -x^2
    [/tex]

    or infinitesimally

    [tex]
    ds^2= c^2dt^2 - dx^2
    [/tex]

    This describes "rotations between the t and x-axis" in the same way as the two dimensional spatial rotations are rotations between x and y. However, we can't describe these spacetime rotations with cosines and sines and angles which are between 0 and 2pi.

    You can check for yourself that we need hyperbolic functions with arguments which are in the real line, not just between 0 and 2pi. This is due to the minus sign. I think this is what it means to put c as a constant mathematically: you define an invariant line element which is preserved by transformations which are called "Lorentz transformations".

    Is this what you're looking for?
     
  4. Oct 31, 2009 #3

    haushofer

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    Now these Lorentz transformations can be looked upon in two different ways:

    *Passive: The event isn't changed, but a Lorentz transformation connects two different coordinate systems (observers) which look at the same event P in spacetime
    *Active: The event IS changed, but the coordinate system doesn't change

    The reason why this distinction is there, is because (from my understanding of it!) mathematically you're talking about diffeomorphisms on the spacetime manifold which connects different events (points) on the spacetime manifold. This formulation can be done coordinate-free. This is called "the active point of view". However, these "active transformations" can be used to induce the passive point of view. They seem to be interchangable, but I must confess this whole business is still not very clear to me in the general case.
     
  5. Oct 31, 2009 #4
    Re: Meajavascript:;ning of Lorentz Transformations purely mathematically ?

    Dear haushofer

    I must also confess that i thought that I understand SR very good, but this new way of obtaining LT blow my mind!

    Regrading your first post, as i said i want to speak about that purely mathematically! that means no speed of light! and this exactly what raising the question:

    if [tex]\grave{x}=\gamma (x-vt) [/tex]
    we can rpove mathematicly that
    [tex]\frac{v^{2}}{1-(\frac{dx}{dt})^{2}}=\frac{\grave{v}^{2}}{1-(\frac{\grave{dx}}{dt})^{2}}}=constant=c^{2}[/tex]

    (where c2 is written just to show it relation to speed of light in reality)
    so my question was what this mathematically means? what this constant means in a euclidean space? not in real space! how we can understand this constant mathematically! what it will be in simple 2D euclidean space? I can't see any sense of it because there can't be a limit for a speed in it , but although the problem is how to prove that there is really no limit mathematically! and more, how we even conclude that there is a limit for a speed when we didn't put such assumption when we started deriving LT, there is just something I can't get it!

    This exactly what i speaking about! when we derived LT, "t" doesn't mean time ! it can be abstractly any variable that some other variable depends on (for example we can define speed of f(x,y,z) relative to z by taking differential so there will be no time!), time in the basically has no meaning for math! so purely mathematically saying that LT is morphisms of space-time has no sense for me if i will think about it purely mathematically!
     
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