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Proving that the det of the Lorentz tensor is +1 or -1 (MTW p 89)

  1. Aug 16, 2013 #1

    TerryW

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    Gold Member

    I have spent much effort trying to prove that det|[itex]\Lambda[/itex][itex]\mu[/itex][itex]\upsilon[/itex]| = +1 or -1 (following a successful effort to prove (3.50g) on p87 of MTW)

    From the result of [itex]\Lambda[/itex]T[itex]\eta[/itex][itex]\Lambda[/itex] = [itex]\eta[/itex] I've produced four equations like:

    [itex]\Lambda[/itex]00[itex]\Lambda[/itex]00 - [itex]\Lambda[/itex]01[itex]\Lambda[/itex]01 - [itex]\Lambda[/itex]02[itex]\Lambda[/itex]02 - [itex]\Lambda[/itex]03[itex]\Lambda[/itex]03 = 1

    and six like:

    - [itex]\Lambda[/itex]00[itex]\Lambda[/itex]10 - [itex]\Lambda[/itex]01[itex]\Lambda[/itex]11 - [itex]\Lambda[/itex]02[itex]\Lambda[/itex]12 - [itex]\Lambda[/itex]03[itex]\Lambda[/itex]13 = 0

    I was hoping that by various combinations of products of these equations I would be able to find all the terms needed to produce det|[itex]\Lambda[/itex][itex]\mu[/itex][itex]\upsilon[/itex]|2 but I end up with a large number of terms so of which look to be OK plus lots of terms which I need to eliminate and can't. It is also very messy!

    Is there an elegant way of doing this?


    TerryW
     
  2. jcsd
  3. Aug 16, 2013 #2

    WannabeNewton

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    ##\det(\Lambda^{T}\eta \Lambda) = \det(\eta)\det(\Lambda^{T})\det(\Lambda) = \det(\eta)\\ \Rightarrow (\det(\Lambda))^{2} = 1\Rightarrow \det(\Lambda) = \pm 1##
     
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