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Lorentz transformations

  1. Mar 29, 2006 #1
    Givne the Lorentz transformations (LTs)}, [tex] x'^{\mu} = L_{\nu}^{\mu} x^{\nu} [/tex], between the coordinates, [tex] x^{\mu} = (ct , \vec{r}) [/tex] of an event as seen by O, and coordinates, [tex] x'^{\mu} = (ct', \vec{r'}) [/tex] of the same event as seen by an inertial observer O', show that if we write the inverse transformation as [tex] x^\alpha = \tilde{L}_{\beta}^{\alpha} x'^{\beta},\mbox{then} \ L_{\omega}^{\alpha} \tilde{L}_{\beta}^{\omega} = \delta_{\beta}^{\alpha}[/tex]

    WELL from the inverse transformation we ca figure out that
    1... [tex] \frac{\partial x^{\alpha}}{\partial x'^{\omega}} = \tilde{L}_{\beta}^{\alpha} \frac{\partial x'^{\beta}}{\partial x'^{\omega}} [/tex]

    also
    2... [tex] \frac{\partial x'^{\alpha}}{\partial x^{\omega}} =L_{\beta}^{\alpha} \frac{\partial x^{\beta}}{\partial x^{\omega}} [/tex]

    there is a notation problem here that i am trying to resolve as well...
    do i simply rearrange for L and tilde L andmultiply out??
    mroe to come as i type it out
     
    Last edited: Mar 29, 2006
  2. jcsd
  3. Mar 29, 2006 #2

    Hurkyl

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    You didn't finish the statement of the problem!

    Anyways, why did you differentiate anything?
     
  4. Mar 29, 2006 #3
    p.s. corrected now
    there was wayyy too much typing involved with the statement of the problem! AHH
     
  5. Mar 30, 2006 #4

    dextercioby

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    Speaking of notation, since [itex] \Lambda [/itex] is a matrix, we use these conventions to denote its elements

    [tex]\Lambda^{\mu}{}_{\nu} ^[/tex]

    for the direct transformation matrix.

    and

    [tex] \Lambda_{\nu}{}^{\mu} [/tex]

    for its transposed.

    However, since both [itex] \Lambda [/itex] and its transposed belong to [itex]\mbox{O(1,3)} [/itex], then the inverse of [itex] \Lambda [/itex] is equal to its transposed.

    Daniel.
     
  6. Mar 30, 2006 #5

    dextercioby

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    There's something spooky with the latex compiler today....


    Daniel.
     
  7. Mar 30, 2006 #6

    nrqed

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    there is no need to get into partial derivatives. You only need to use [tex] x'^{\mu} = L_{\nu}^{\mu} x^{\nu} [/tex] and [tex] x^\alpha = \tilde{L}_{\beta}^{\alpha} x'^{\beta} [/tex]....Putting one into the other gives
    [tex] x^\alpha = \tilde{L}_{\beta}^{\alpha} L_{\nu}^{\beta} x^{\nu} [/tex]. But of course, when [tex] \alpha = \nu [/tex] the two sides are equal..Obviously, [tex] x^\alpha = \delta^\alpha_\nu x^\nu [/tex]. Therefore....
     
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