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L2 transformation reduces to the L1 transformation

  1. Nov 16, 2003 #1

    I am having trouble answering the following question:

    "Show that the L2 transformation reduces to the L1 transformation when the two reference frames are in standard configuration."

    Am I wrong to assume that r = xi + yj + zk

    Any help would be beautiful!

    Thanx much
  2. jcsd
  3. Nov 16, 2003 #2


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    I assume that the source of this question defines what the L2 and L1 transformations are, as well as what two reference frames are being compared. Please elaborate.
  4. Nov 17, 2003 #3
    The L2 transformations are as follows:

    r' = r + γv^[(1 - 1/γ)(r.v^) - βct];

    ct' = γ(ct - r.β);

    where β = v/c & v^ is the unit vector in the direction of v.

    The L1 transformations are:

    x' = γ(x - βct);
    y' = y;
    z' = z;

    ct' = γ(ct - βx);

    where β = v/c.

    All are viewed in the S' frame.
  5. Nov 17, 2003 #4


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    We should also require that you define "standard configuration" but I'm going to assume that is with both reference frames moving in the direction of the x-axis :smile:.

    Yes, you are correct to right r= xi+ yj+ zk. Notice that the difference is that "xi+ yj+ zk" assumes some particular coordinate system ("standard configuration") while "r" does not.

    You may also assume ("standard configuration") that v= vi+ 0j+ 0k and that v^= i+ 0j+ 0k.
  6. Nov 17, 2003 #5
    Sorry. Yes standard configuration is when both reference frames move in the direction of the x-axis.

    Thanks I think I can solve it now.
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