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Counterexample: a spatial rotation is a proper orthochronous Lorentz map that is not a boost

Statement refuted

The statement "every proper orthochronous Lorentz map is a passive boost $B (β)$ for some $β$ with $∣ β ∣ < 1$ " is false.

Facts & Assumptions

Given: An inertial frame $(o; e_{0}, e_{1}, e_{2}, e_{3})$ with $η (e_{a}, e_{b}) = diag (- 1, 1, 1, 1)_{a b}$ , and the linear maps defined below by their action on the frame basis.

[L1]

A linear Lorentz map is a linear $L : V \to V$ with $η (L v, L w) = η (v, w)$ for all $v, w$ ; it is proper when $\det L = + 1$ and orthochronous when $L (C^{+}) = C^{+}$ (Linear and affine Lorentz maps, and the Lorentz group).

[L2]

A spatial rotation of the frame is a linear $R : V \to V$ with $R (e_{0}) = e_{0}$ , $R (S) \subseteq S$ , $δ (R w, R z) = δ (w, z)$ on the spatial three-space $S$ , and $\det (R ∣_{S}) = + 1$ in $(e_{1}, e_{2}, e_{3})$ (Spatial rotations of an inertial frame).

[L3]

The extension $\hat{R} = diag (1, R)$ of a determinant-one $δ$ -isometry $R : S \to S$ is a proper orthochronous Lorentz map (Unique boost-rotation factorization of proper orthochronous Lorentz maps).

[L4]

For every $∣ β ∣ < 1$ the passive boost satisfies $B (β) e_{0} = γ (e_{0} - β_{1} e_{1} - β_{2} e_{2} - β_{3} e_{3})$ and $B (0) = {id}_{V}$ ; the map $β \mapsto B (β) e_{0}$ is injective (The passive boost is a proper orthochronous Lorentz map with inverse B(-beta)).

[L5]

The determinant of a square matrix is computed by Laplace cofactor expansion, $\det (A) = \sum_{k} a_{k j} C_{k j} (A)$ along the column $j$ (Laplace expansion computes the determinant along every row and every column over a commutative ring, Deleted-row-and-column minors, cofactors, the cofactor matrix and the adjugate over a commutative ring); in particular $\det (\begin{array}{cc} a & b \\ c & d \end{array}) = a d - b c$ by expanding along the first column, since the $(0, 0)$ -cofactor is $d$ and the $(1, 0)$ -cofactor is $- b$ .

Counterexample

1.1L2L5algebra

Define $\hat{R} : V \to V$ by $\hat{R} e_{0} = e_{0}$ , $\hat{R} e_{1} = e_{2}$ , $\hat{R} e_{2} = - e_{1}$ , $\hat{R} e_{3} = e_{3}$ . In the ordered basis $(e_{0}, e_{1}, e_{2}, e_{3})$ its matrix is $(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & - 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array})$ , and its restriction to $S = span (e_{1}, e_{2}, e_{3})$ has the matrix $A = (\begin{array}{ccc} 0 & - 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array})$ in $(e_{1}, e_{2}, e_{3})$ . The columns of $A$ are orthonormal in the standard inner product on $R^{3}$ (they are $(0, 1, 0)$ , $(- 1, 0, 0)$ , $(0, 0, 1)$ ), so $A^{T} A = I_{3}$ and $A$ preserves the form $δ$ on $S$ ; and Laplace expansion along the third column gives $\det A = 1 \cdot \det (\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array}) = 1 \cdot (0 \cdot 0 - (- 1) \cdot 1) = 1$ by [L5]. Hence $\hat{R} (e_{0}) = e_{0}$ , $\hat{R} (S) = S$ , $δ (\hat{R} w, \hat{R} z) = δ (w, z)$ for all $w, z \in S$ , and $\det (\hat{R} ∣_{S}) = + 1$ : all four defining conditions of [L2] hold, so $\hat{R}$ is a spatial rotation of the frame.

2.1L1L3step 1.1

By step 1.1 and [L3], the extension $\hat{R} = diag (1, \hat{R} ∣_{S})$ is a proper orthochronous Lorentz map. In particular $\hat{R}$ is a member of $S O^{+} (V, η)$ in the sense of [L1], and it is not the identity because $\hat{R} e_{1} = e_{2} \neq e_{1}$ .

3.1L4step 1.1step 2.1algebra

Suppose, for contradiction, that $\hat{R} = B (β)$ for some $β$ with $∣ β ∣ < 1$ . Applying both sides to $e_{0}$ and using [L4] gives $γ (e_{0} - β_{1} e_{1} - β_{2} e_{2} - β_{3} e_{3}) = e_{0}$ ; comparing the $e_{0}$ coordinates gives $γ = 1$ and then comparing the spatial coordinates gives $γ β_{i} = 0$ for $i = 1, 2, 3$ , hence $β = 0$ and $B (β) = B (0) = {id}_{V}$ by [L4]. But $\hat{R} e_{1} = e_{2} \neq e_{1} = {id}_{V} (e_{1})$ , a contradiction. Hence $\hat{R}$ is a proper orthochronous Lorentz map that is not a passive boost, and the refuted statement is false.

4.1L3step 2.1algebra∎

Consequently the set of passive boosts is a proper subset of $S O^{+} (V, η)$ : the strict inclusion is witnessed by $\hat{R}$ . The factorization theorem $L = B (- β) {\hat{R}}_{β}$ of Unique boost-rotation factorization of proper orthochronous Lorentz maps therefore cannot drop its rotation factor in general: for $L = \hat{R}$ the theorem produces the factorization $\hat{R} = B (0) \hat{R}$ with rotation factor $\hat{R} ∣_{S} \neq {id}_{S}$ , and by clause (v) of that theorem $L$ is a passive boost if and only if that factor is the identity.

Counterexample: a spatial rotation is a proper orthochronous Lorentz map that is not a boost

Statement refuted

Facts & Assumptions

Counterexample

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