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

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The determinant of the Lorentz transformation tensor, denoted as det|\Lambda\mu\upsilon|, is conclusively proven to be either +1 or -1, as established in the discussion surrounding MTW (Misner, Thorne, Wheeler) on page 89. This conclusion arises from the equation \(\Lambda^T \eta \Lambda = \eta\), leading to the relationship \((\det(\Lambda))^2 = 1\), which simplifies to \(\det(\Lambda) = \pm 1\). The discussion highlights the complexity involved in deriving this result through various combinations of equations derived from the properties of the Lorentz transformation.

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TerryW
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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
 
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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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