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I have been doing questions recently involving Lorentz boosts. However I was wondering if the Lorentz boost matrix $Λ$ is orthogonal.

$$ \left[\begin{array}{cccc}\hat {ct} \\ \hat x\end{array}\right] = \left[\begin{array}{cccc}{\cosh \varphi} & {-\sinh \varphi} \\ {-\sinh \varphi} & {\cosh \varphi}\end{array}\right] \left[\begin{array}{cccc}{ct} \\ x\end{array}\right] =Λ(\varphi)\left[\begin{array}{cccc}{ct} \\ x\end{array}\right] $$

My understanding: For a matrix to be orthogonal $ΛΛ^T=Λ^TΛ=I$

That is that $Λ^T=Λ^{-1}$, however this is not the case with the given matrix here. So instead of using that definition could I prove it is orthogonal in terms of an invarient?

My attempt: If I denote $\eta $ to be a Minkowsi metric which is an invariant.

The matrix representing a Lorentz boost is orthogonal with respect to this Minkowski metric $$ \Lambda \eta \Lambda^T = \eta \text{ or } \Lambda^{-1} = \eta \Lambda^T\eta.$$

Is this a correct statement?

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2 Answers 2

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Yes your statement is correct. Rotations are isometries of 3D Euclidean space: they preserve the inner product defined using the Euclidean metric. Rotations + boosts are isometries of 4D Minkowski space: they preserve the inner product defined using the Minkowski metric (technically this isn't an inner product since it's not positive definite, it's a symmetric bilinear form).

Or, put differently: The Euclidean metric is left invariant under rotations and the Minkowski metric is left invariant under Lorentz transformations. For rotations this gives us $R^TR=1$, but this isn't the case for boosts.

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    $\begingroup$ Thanks so much for your help. $\endgroup$
    – user254780
    Commented Feb 27, 2020 at 17:47
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You are exactly right. The Lorentz group is not a subgroup of an orthogonal group, because those preserve Euclidean metrics; instead, it is part of the indefinite orthogonal group $O(3,1)$, which preserves the Minkowski metric. The condition for a matrix $\Lambda$ to be in this indefinite group is precisely that $g^{-1} \Lambda^T g = \Lambda^{-1}$. And since $g = g^{-1} = \eta$, this is precisely the condition you found.

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