# Orthogonality of a Lorentz Boost Matrix in terms of an invariant

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?

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.
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.