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Counterexample: simultaneity is relative to the observer vector

Statement refuted

False claim: the simultaneity relation $η (n, q - p) = 0$ between events of the affine model is independent of the future unit timelike vector $n$ (Simultaneity slices and rest distance relative to a future unit timelike vector).

Counterexample. In an inertial chart let $p : = x^{- 1} (0, 0, 0, 0)$ and $q : = x^{- 1} (0, 1, 0, 0)$ , so that $d : = q - p$ is spacelike with $η (d, d) = 1$ , and let $n : = e_{0} = (1, 0, 0, 0)$ and $n^{'} : = γ (e_{0} + \frac{3}{5} e_{1}) = \frac{5}{4} (1, \frac{3}{5}, 0, 0)$ , a boost of speed $3 c / 5$ along the first axis. Then $η (n^{'}, n^{'}) = \frac{25}{16} (- 1 + \frac{9}{25}) = - 1$ , so $n^{'}$ is also a future unit timelike vector; $η (n, d) = 0$ , so $p$ and $q$ are simultaneous relative to $n$ and lie in the slice $H (p, n)$ ; and $η (n^{'}, d) = \frac{5}{4} \cdot \frac{3}{5} = \frac{3}{4} \neq 0$ , so $p$ and $q$ are not simultaneous relative to $n^{'}$ and lie in different slices for $n^{'}$ . The two slices $H (p, n)$ and $H (p, n^{'})$ are distinct three-dimensional affine hyperplanes through $p$ . Hence simultaneity in the sense of Simultaneity slices and rest distance relative to a future unit timelike vector depends on the observer vector and is not an observer-independent relation.

Facts & Assumptions

Given: The affine Minkowski model with an inertial frame $(o; e_{0}, e_{1}, e_{2}, e_{3})$ and its chart $x$ (Inertial frames and inertial charts of the affine Minkowski model, Timelike, null, spacelike and causal vectors of a Lorentz vector space), the events $p = x^{- 1} (0, 0, 0, 0) = o$ and $q = x^{- 1} (0, 1, 0, 0) = o + e_{1}$ , the displacement $d = q - p = e_{1}$ , the future unit vectors $n = e_{0}$ and $n^{'} = γ (e_{0} + \frac{3}{5} e_{1})$ with $γ = (1 - \frac{9}{25})^{- 1 / 2} = \frac{5}{4}$ , and the slices $H (p, n) = p + n^{⊥}$ , $H (p, n^{'}) = p + n^{' ⊥}$ (Simultaneity slices and rest distance relative to a future unit timelike vector).

[L1]

$η (e_{a}, e_{b}) = diag (- 1, 1, 1, 1)_{a b}$ for the inertial frame's basis, so $η$ is evaluated on coordinate vectors by $η (u, v) = - u^{0} v^{0} + u^{1} v^{1} + u^{2} v^{2} + u^{3} v^{3}$ (Inertial frames and inertial charts of the affine Minkowski model, Oriented time-oriented affine Minkowski space (the spacetime primitives)).

[L2]

The future unit timelike vectors are the vectors $n$ with $η (n, n) = - 1$ and $n \in C^{+}$ ; the first basis vector $e_{0}$ of an inertial frame is future unit, and proper orthochronous Lorentz maps preserve the causal classification and the time orientation (Timelike, null, spacelike and causal vectors of a Lorentz vector space, 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).

[L3]

The passive boost with parameter $β$ , $∣ β ∣ < 1$ , is a proper orthochronous Lorentz map with inverse the boost with parameter $- β$ , and its matrix in the inertial basis has entries $B^{0}_{0} = B^{1}_{1} = γ$ and $B^{0}_{1} = B^{1}_{0} = - γ β$ on the first two coordinate directions, so $B (- β) e_{0} = γ (e_{0} + β e_{1})$ for the first basis vector, with $γ = (1 - β^{2})^{- 1 / 2}$ (The passive Lorentz boost with parameter beta, The passive boost is a proper orthochronous Lorentz map with inverse B(-beta)).

[L4]

For a future unit $n$ , $n^{⊥} = {v : η (n, v) = 0}$ and $H (p, n) = p + n^{⊥} = {z : η (n, z - p) = 0}$ ; the slice contains $p$ and is a three-dimensional affine hyperplane (Simultaneity slices and rest distance relative to a future unit timelike vector, Norms of four-velocity and four-acceleration, and the observer rest-space decomposition clause (c), The timelike set has exactly two convex components, described in orthonormal coordinates).

Counterexample

technique · direct

1.1L1L2L3

The displacement is $d = q - p = e_{1}$ by [L1], and $η (d, d) = η (e_{1}, e_{1}) = 1$ , so $d$ is spacelike; the vector $n = e_{0}$ is future unit by [L2]; the vector $n^{'} = \frac{5}{4} (e_{0} + \frac{3}{5} e_{1})$ is the image $B (- 3 / 5) e_{0}$ of $e_{0}$ under the boost with parameter $β = 3 / 5$ by [L3], hence $n^{'}$ is future timelike by [L2], and its norm is $η (n^{'}, n^{'}) = \frac{25}{16} (η (e_{0}, e_{0}) + \frac{6}{5} η (e_{0}, e_{1}) + \frac{9}{25} η (e_{1}, e_{1})) = \frac{25}{16} (- 1 + \frac{9}{25}) = - 1$ by [L1], so $n^{'}$ is future unit; in components $n^{'} = \frac{5}{4} e_{0} + \frac{3}{4} e_{1}$ .

2.1L1L4step 1.1

Simultaneity relative to $n$ : $η (n, d) = η (e_{0}, e_{1}) = 0$ by [L1], so $q - p \in n^{⊥}$ and $q \in H (p, n)$ by [L4]; the events $p, q$ are simultaneous relative to $n$ .

2.2L1L4step 1.1

Non-simultaneity relative to $n^{'}$ : $η (n^{'}, d) = \frac{5}{4} η (e_{0}, e_{1}) + \frac{3}{4} η (e_{1}, e_{1}) = 0 + \frac{3}{4} = \frac{3}{4} \neq 0$ by [L1] and step 1.1, so $q - p \notin n^{' ⊥}$ and $q \notin H (p, n^{'})$ by [L4]; the events $p, q$ are not simultaneous relative to $n^{'}$ .

2.3L1L4step 1.1

The two slices are distinct: the vector $e_{1}$ satisfies $η (n, e_{1}) = 0$ , so $e_{1} \in n^{⊥}$ , while $η (n^{'}, e_{1}) = \frac{3}{4} \neq 0$ , so $e_{1} \notin n^{' ⊥}$ ; hence $n^{⊥} \neq n^{' ⊥}$ as subspaces of $V$ . Both $H (p, n) = p + n^{⊥}$ and $H (p, n^{'}) = p + n^{' ⊥}$ are three-dimensional affine hyperplanes through $p$ by [L4], and since their direction subspaces differ they are distinct sets.

3.1step 2.1step 2.2step 2.3∎

Conclusion: the single pair of events $p = x^{- 1} (0, 0, 0, 0)$ , $q = x^{- 1} (0, 1, 0, 0)$ is simultaneous relative to the future unit vector $n = e_{0}$ by step 2.1 and not simultaneous relative to the future unit vector $n^{'} = \frac{5}{4} (e_{0} + \frac{3}{5} e_{1})$ by step 2.2, and the corresponding slices through $p$ are distinct by step 2.3. Therefore the simultaneity relation defined by $η (n, q - p) = 0$ is not independent of the observer vector $n$ , and the claim of the Statement refuted section is false.

Remarks

Exact rational data. The boost speed $3 c / 5$ was chosen so that $γ = 5 / 4$ and the two pairings $η (n, d) = 0$ , $η (n^{'}, d) = 3 / 4$ are exact rational numbers; no numerical approximation is involved.
The witness for distinctness of the direction subspaces is $e_{1}$ . The vector $e_{1}$ lies in $n^{⊥}$ and not in $n^{' ⊥}$ ; the two rest spaces of distinct future unit vectors are in fact always distinct, indeed, if $n^{⊥} = n^{' ⊥}$ , then $h_{n} (n^{'}) \in n^{⊥}$ is orthogonal to every vector of $n^{⊥}$ (because both $n$ and $n^{'}$ are); positive definiteness forces $h_{n} (n^{'}) = 0$ . The decomposition theorem then makes $n^{'}$ a multiple of $n$ , and future unit normalization forces that multiple to be $1$ .

Counterexample: simultaneity is relative to the observer vector

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

Dependency tree · 2 levels

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