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Determinant of LT matrix

  1. Jun 10, 2012 #1
    I have been asked to prove that the determinant of any matrix representing a Lorentz transformation is plus or minus 1. I can see that the determinant of the Lorentz transformation matrix is 1, but don't know how to prove +-1 in general. How to generalise the lorentz transformation? I've also read that rotations in the spatial planes also constitute L.T., that any transformation that keeps the metric invariant is an L.T.
  2. jcsd
  3. Jun 10, 2012 #2
    Since any arbitrary LT is either a boost between timelike and spacelike directions, a rotation between spacelike directions, or a combination of the two, I think all you need is to check that the determinant of a 3D rotation is [itex]\pm 1[/itex]. After that, if [itex]\underline L \underline R[/itex] both represent LT's, then what do you know about

    [tex]\det (\underline L \underline R) = ?[/tex]
  4. Jun 11, 2012 #3
    Reversal of the space or time axis produces a matrix of determinant -1. That proves that any LT matrix has determinant [itex]\pm 1[/itex]. det(LR)=det(L)det(R), so again, the resulting matrix has determinant [itex]\pm 1[/itex].
    Last edited: Jun 11, 2012
  5. Jun 11, 2012 #4
    How to prove that the matrix version of ηa'b'= Ama'Anb'ηmn is η= ATηA, where A is an LT matrix?
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