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Spacelike-separated events: the spacelike coordinate inequality in every inertial chart, and an explicit family of inertial frames realizing every time-order sign

Statement

Let $(A, V, η, C^{+})$ be the oriented time-oriented affine Minkowski space (Oriented time-oriented affine Minkowski space (the spacetime primitives)) and let $p, q \in A$ with $p \neq q$ and $q - p$ spacelike (Timelike, null, spacelike and causal vectors of a Lorentz vector space). Then:

(a) In the chart $x$ of every inertial frame the coordinate differences $D : = x (q) - x (p)$ satisfy $∣ D^{0} ∣ < ∣ D_{s} ∣$ , where $∣ D_{s} ∣$ is the Euclidean norm of the spatial part $D_{s} : = (D^{1}, D^{2}, D^{3})$ of $D$ (The Euclidean spatial structure and cross product of an inertial frame); in particular $D_{s} \neq 0$ .

(b) There is an inertial frame whose chart $x_{0}$ satisfies $x_{0}^{0} (q) = x_{0}^{0} (p)$ : the two events are simultaneous in that frame.

(c) There are inertial frames whose charts $x_{+}$ and $x_{-}$ satisfy $x_{+}^{0} (q) > x_{+}^{0} (p)$ and $x_{-}^{0} (q) < x_{-}^{0} (p)$ .

(d) Explicit family. Fix an inertial frame with chart $x$ and let $e : = D_{s} / ∣ D_{s} ∣$ , so that $e$ is a unit spatial vector of that frame. For every real number $b$ with $∣ b ∣ < 1$ let $B (b e)$ be the passive boost of The passive Lorentz boost with parameter beta with parameter $b e$ , and let $x_{b}$ be the chart of the inertial frame whose ordered basis is $e_{a}^{'} : = B (b e)^{- 1} e_{a}$ and whose origin event is the same event $o$ . Then $x_{b} = B (b e) \circ x$ (the linear map acting on coordinate columns) and $x_{b}^{0} (q) - x_{b}^{0} (p) = γ_{b} (D^{0} - b ∣ D_{s} ∣), γ_{b} : = (1 - b^{2})^{- 1 / 2} .$ This number is positive for $b < D^{0} / ∣ D_{s} ∣$ , zero for $b = D^{0} / ∣ D_{s} ∣$ and negative for $b > D^{0} / ∣ D_{s} ∣$ ; since $- 1 < D^{0} / ∣ D_{s} ∣ < 1$ by (a), all three cases occur among values $∣ b ∣ < 1$ , and the family $b \mapsto x_{b}$ depends continuously on $b$ .

(e) Scope. The lemma concerns the inertial charts of the model; "simultaneous in the frame" means equality of the frame's time coordinate $x^{0}$ , and no synchronization procedure, observer convention or physical identification is asserted. No choice principle is used.

Facts & Assumptions

Given: The oriented time-oriented affine Minkowski space $(A, V, η, C^{+})$ , events $p, q \in A$ with $p \neq q$ and $q - p$ spacelike, a fixed inertial frame $(o; e_{0}, e_{1}, e_{2}, e_{3})$ with chart $x$ , the coordinate differences $D = x (q) - x (p)$ and $D_{s} = (D^{1}, D^{2}, D^{3})$ , the number $∣ D_{s} ∣$ and the unit vector $e = D_{s} / ∣ D_{s} ∣$ .

[L1]

An inertial frame is a tuple $(o; e_{0}, e_{1}, e_{2}, e_{3})$ with $o \in A$ and $(e_{0}, e_{1}, e_{2}, e_{3})$ an $η$ -orthonormal, future, positively oriented ordered basis of $V$ ; its chart is $x (r) = E^{- 1} (r - o)$ with $E (u) = \sum_{a} u^{a} e_{a}$ , and the components $x^{a} (r)$ are the inertial coordinates of $r$ (Inertial frames and inertial charts of the affine Minkowski model).

[L2]

In an $η$ -orthonormal basis with coordinates $v = v^{0} e_{0} + \dots + v^{3} e_{3}$ one has $η (v, v) = - (v^{0})^{2} + ∣ v_{s} ∣^{2}$ , where $∣ v_{s} ∣ = \sqrt{(v^{1})^{2} + (v^{2})^{2} + (v^{3})^{2}}$ is the Euclidean norm of the spatial part; for a fixed inertial frame the spatial part $S = span (e_{1}, e_{2}, e_{3})$ carries the positive-definite form $δ = η ∣_{S}$ and the norm $∣ \cdot ∣$ (The timelike set has exactly two convex components, described in orthonormal coordinates, The Euclidean spatial structure and cross product of an inertial frame).

[L3]

A vector $v \in V$ is spacelike exactly when $η (v, v) > 0$ ; the zero vector is not spacelike, and every nonzero vector has exactly one causal character (timelike, null or spacelike) (Timelike, null, spacelike and causal vectors of a Lorentz vector space).

[L4]

For $β \in R^{3}$ with $∣ β ∣ < 1$ the passive boost is the linear map $B (β) : V \to V$ whose action on coordinates is $x^{' 0} = γ (x^{0} - β \cdot x)$ , $x^{'} = x + \frac{γ - 1}{∣ β ∣^{2}} (β \cdot x) β - γ β x^{0}$ for $β \neq 0$ and $B (0) = i d$ , where $γ = (1 - ∣ β ∣^{2})^{- 1 / 2}$ and $β \cdot x$ is the Euclidean scalar product of $β$ with the spatial coordinates of $x$ (The passive Lorentz boost with parameter beta, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[L5]

For every $∣ β ∣ < 1$ the boost $B (β)$ is a proper orthochronous Lorentz map, $B (- β) B (β) = i d = B (β) B (- β)$ , and $B (β)$ is injective with $B (β)^{- 1} = B (- β)$ (The passive boost is a proper orthochronous Lorentz map with inverse B(-beta)).

[L6]

A linear Lorentz map $L$ is a linear isomorphism preserving $η$ and the causal character of every nonzero vector; if $L$ is proper it preserves the oriented volume form, so it maps a positively oriented $η$ -orthonormal ordered basis to a positively oriented $η$ -orthonormal ordered basis, and it maps $C^{+}$ onto $C^{+}$ when it is orthochronous (Lorentz maps are isomorphisms of determinant ±1 that preserve causal character and permute the cones; the Lorentz sets are groups, Linear and affine Lorentz maps, and the Lorentz group).

[L7]

Every vector has unique coordinates in an ordered basis, and a linear map is determined by the images of the basis vectors; every linear isomorphism is a bijection (A finite list $v : n \to V$ is an ordered basis if and only if every $x \in V$ equals $\sum_{i < n} λ_{i} v_{i}$ for exactly one $λ : n \to F$ ; those scalars are the coordinates of $x$ in that ordered basis, Invertible linear maps, linear isomorphisms, and inverse linear maps).

[L8]

For $D_{s} \neq 0$ and $e = D_{s} / ∣ D_{s} ∣$ the Euclidean scalar product satisfies $e \cdot D_{s} = ∣ D_{s} ∣$ and $e \cdot (λ e) = λ$ for every $λ \in R$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

Proof

technique · direct

1.1givenL1L2

Fix the inertial frame and chart $x$ of the given data; by [L1] one has $D = E^{- 1} (q - p)$ , and in the $η$ -orthonormal basis $(e_{0}, e_{1}, e_{2}, e_{3})$ the interval of the pair is $η (q - p, q - p) = - (D^{0})^{2} + ∣ D_{s} ∣^{2}$ in the notation of [L2].

2.1step 1.1L3algebra

Since $q - p$ is spacelike, [L3] gives $η (q - p, q - p) > 0$ , so $- (D^{0})^{2} + ∣ D_{s} ∣^{2} > 0$ , that is $(D^{0})^{2} < ∣ D_{s} ∣^{2}$ ; hence $D_{s} \neq 0$ and $e : = D_{s} / ∣ D_{s} ∣$ is a well-defined unit spatial vector, the quotient $c : = D^{0} / ∣ D_{s} ∣$ satisfies $∣ c ∣ < 1$ and is an admissible boost parameter, and $(c + 1) / 2$ and $(c - 1) / 2$ also lie in $(- 1, 1)$ .

3.1step 2.1givenL1L4L5L6L7

Fix a real number $b$ with $∣ b ∣ < 1$ , put $β : = b e$ , $B_{b} : = B (β)$ and $e_{a}^{'} : = B_{b}^{- 1} e_{a} = B (- β) e_{a}$ for $a = 0, 1, 2, 3$ . By [L5] and [L6] the map $B_{b}^{- 1} = B (- β)$ is a proper orthochronous Lorentz map, so $(e_{0}^{'}, e_{1}^{'}, e_{2}^{'}, e_{3}^{'})$ is an $η$ -orthonormal, future, positively oriented ordered basis of $V$ , hence $(o; e_{0}^{'}, e_{1}^{'}, e_{2}^{'}, e_{3}^{'})$ is an inertial frame of the model by [L1]. Writing $x_{b}$ for its chart, $M_{b}$ for the matrix of $B_{b}$ in the ordered basis $(e_{0}, e_{1}, e_{2}, e_{3})$ and $E_{b} (u) : = \sum_{a} u^{a} e_{a}^{'}$ , one has $E_{b} (u) = B_{b}^{- 1} (E (u))$ for every $u \in R^{4}$ by linearity of $B_{b}^{- 1}$ , hence $E_{b}^{- 1} = E^{- 1} B_{b}$ and $x_{b} (r) = E_{b}^{- 1} (r - o) = E^{- 1} (B_{b} (r - o)) = E^{- 1} (B_{b} (E (x (r)))) = M_{b} x (r)$ for every $r \in A$ , because $r - o = E (x (r))$ and $E^{- 1} B_{b} E$ is the matrix map $u \mapsto M_{b} u$ .

4.1step 2.1step 3.1L4L8algebra

By linearity of $B_{b}$ , subtracting the chart identity of step 3.1 at $p$ from the identity at $q$ gives $x_{b} (q) - x_{b} (p) = M_{b} D$ , and its time component is $(M_{b} D)^{0} = γ_{b} (D^{0} - b ∣ D_{s} ∣)$ , where $γ_{b} : = (1 - b^{2})^{- 1 / 2} > 0$ : the first row of $M_{b}$ gives $(M_{b} D)^{0} = γ_{b} (D^{0} - β \cdot D)$ , and $β \cdot D = b (e \cdot D_{s}) = b ∣ D_{s} ∣$ by [L8]. Since $D^{0} = c ∣ D_{s} ∣$ , this reads $x_{b}^{0} (q) - x_{b}^{0} (p) = γ_{b} ∣ D_{s} ∣ (c - b)$ , which is positive for $b < c$ , zero for $b = c$ and negative for $b > c$ ; moreover each matrix entry of $M_{b}$ , namely $γ_{b}$ , $- γ_{b} b e_{i}$ and $δ_{j}^{i} + (γ_{b} - 1) e_{i} e_{j}$ , is a continuous function of $b$ on $(- 1, 1)$ , because $b \mapsto γ_{b}$ is continuous and positive there with $γ_{b} \to 1$ as $b \to 0$ , so every coordinate $x_{b}^{a} (r)$ is continuous in $b$ .

5.1step 2.1step 3.1step 4.1choose

The frame of step 3.1 with parameter $b : = c$ , admissible by step 2.1, has chart $x_{0} : = x_{c}$ satisfying $x_{0}^{0} (q) - x_{0}^{0} (p) = 0$ by step 4.1; this is assertion (b).

5.2step 2.1step 3.1step 4.1choosealgebra

The frames of step 3.1 with parameters $b_{+} : = (c - 1) / 2$ and $b_{-} : = (c + 1) / 2$ , both admissible by step 2.1, have $x_{b_{+}}^{0} (q) - x_{b_{+}}^{0} (p) = γ_{b_{+}} ∣ D_{s} ∣ (c - b_{+}) > 0$ and $x_{b_{-}}^{0} (q) - x_{b_{-}}^{0} (p) = γ_{b_{-}} ∣ D_{s} ∣ (c - b_{-}) < 0$ by step 4.1, since $c - b_{+} = (c + 1) / 2 > 0 > c - b_{-} = (c - 1) / 2$ ; taking $x_{+} : = x_{b_{+}}$ and $x_{-} : = x_{b_{-}}$ this is assertion (c).

6.1step 2.1step 3.1step 4.1step 5.1step 5.2∎

Assertion (a) is step 2.1 read in an arbitrary inertial chart; assertion (d) is the construction of steps 3.1 and 4.1, comprising the chart identity $x_{b} = B (b e) \circ x$ , the time-difference formula, the sign trichotomy relative to $c = D^{0} / ∣ D_{s} ∣$ and the continuous dependence on $b$ ; assertions (b) and (c) are steps 5.1 and 5.2. Only inertial charts of the model and one explicitly constructed basis per parameter are used, "simultaneous" is the equality of the frame's $x^{0}$ coordinate as required in (e), and no choice principle is used since each parameter is a single explicit real number.

Spacelike-separated events: the spacelike coordinate inequality in every inertial chart, and an explicit family of inertial frames realizing every time-order sign

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources