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Worked relay arithmetic at $u = 2 c$ and $v = (9 / 10) c$ : arrival at $t = - 50 / (121 c)$ seconds

Example

Worked computation in the chart $x$ of an inertial frame of an affine Minkowski space (Oriented time-oriented affine Minkowski space (the spacetime primitives), Inertial frames and inertial charts of the affine Minkowski model) with $x^{0} = c t$ for a fixed constant $c > 0$ . Throughout this numerical example, $x^{a}$ and $L$ are numerical values in metres, $t$ and $Δ t$ are numerical values in seconds, and $c, u, v$ are numerical values in metres per second; units after a formula apply to its numerical value. Take $u : = 2 c$ , $v : = (9 / 10) c$ and $L : = 1$ (a length of $1$ m), so that $u > c$ , $0 < v < c$ and $v > 2 u / (1 + u^{2} / c^{2}) = (4 / 5) c$ . Let $A$ and $B$ have affine curves $a (t) = x^{- 1} (c t, 0, 0, 0)$ and $b (t) = x^{- 1} (c t, L + v t, 0, 0)$ for $t \in R$ . Their constant tangents $(c, 0, 0, 0)$ and $(c, v, 0, 0)$ are future timelike since $0 < v < c$ . Emit from $A$ at the event $A_{0} : = x^{- 1} (0, 0, 0, 0)$ a signal travelling at coordinate speed $u$ in the $+ x^{1}$ direction, and let $B$ re-emit immediately from the reception event $B_{1}$ a signal travelling at coordinate speed $u$ in the $- x^{1}$ direction of $B$ 's rest frame; this fixed-speed, immediate, lossless relay is the whole content of the hypothetical model, and no device, signal or measurement is asserted to exist.

(a) The reception time solves $u t = L + v t$ , so $t (B_{1}) = \frac{L}{u - v} = \frac{10}{11 c} s, x^{1} (B_{1}) = u t (B_{1}) = \frac{20}{11} m,$ and the displacement $B_{1} - A_{0} = (\frac{10}{11}, \frac{20}{11}, 0, 0)$ m has interval $η = - \frac{100}{121} + \frac{400}{121} = \frac{300}{121} > 0$ : it is spacelike.

(b) Write $β : = v / c = 9 / 10$ , $γ : = (1 - β^{2})^{- 1 / 2} = 10 / \sqrt{19}$ , and let $x^{'} = B (β) x$ be the chart of the inertial frame in which $B$ is at rest (The passive Lorentz boost with parameter beta, The passive boost is a proper orthochronous Lorentz map with inverse B(-beta)). In that chart the return displacement is $(c Δ t^{'}, - u Δ t^{'})$ with $Δ t^{'} > 0$ , and transforming back with the inverse boost $c Δ t = γ (c Δ t^{'} - β u Δ t^{'}), Δ x^{1} = γ (- u Δ t^{'} + β c Δ t^{'}),$ the condition that the arrival lies on $A$ 's worldline $x^{1} = 0$ gives $Δ x^{1} = - x^{1} (B_{1})$ , hence $Δ t^{'} = \frac{x^{1} (B_{1})}{γ (u - v)} > 0, Δ t = \frac{x^{1} (B_{1}) (1 - v u / c^{2})}{u - v} = \frac{20}{11} (- \frac{4}{5}) \frac{10}{11 c} = - \frac{160}{121 c} s .$

(c) The arrival event $A_{1}$ therefore satisfies $t (A_{1}) = t (B_{1}) + Δ t = \frac{10}{11 c} - \frac{160}{121 c} = - \frac{50}{121 c} s < 0,$ matching the closed formula $L (2 u - v - v u^{2} / c^{2}) / (u - v)^{2} = (- c / 2) / (121 c^{2} / 100) = - 50 / (121 c) s$ . The displacement $A_{1} - A_{0} = (- \frac{50}{121}, 0, 0, 0)$ m is past timelike, so the relayed signal is received strictly before it was emitted in the chart of the emitter; the return displacement in $x$ is $(- \frac{160}{121}, - \frac{220}{121}, 0, 0)$ m with $η = - (\frac{160}{121})^{2} + (\frac{220}{121})^{2} > 0$ , spacelike.

(d) The computation is an arithmetic specialization of the hypothetical fixed-speed model (signals at fixed coordinate speed $u > c$ in each emitter's rest frame, immediate lossless relaying, inertial motion at $v < c$ , ordinary Lorentz kinematics of displacements); it is a conditional identity about that model and neither asserts that such signals or devices exist nor reports any measurement.

Facts & Assumptions

Given: An inertial frame with chart $x$ , $x^{0} = c t$ , a constant $c > 0$ , the numerical data $u = 2 c$ , $v = (9 / 10) c$ , $L = 1$ (numerical metres), $β = v / c = 9 / 10$ , the affine curves $a (t) = x^{- 1} (c t, 0, 0, 0)$ of $A$ and $b (t) = x^{- 1} (c t, L + v t, 0, 0)$ of $B$ , $t \in R$ , the emission event $A_{0} = x^{- 1} (0, 0, 0, 0)$ , and the hypothetical relay: outgoing leg of coordinate speed $u$ in the $+ x^{1}$ direction from $A_{0}$ , immediate re-emission at the reception event $B_{1}$ of a return leg of coordinate speed $u$ in the $- x^{1}$ direction of $B$ 's rest frame.

[L1]

Coordinate criterion: for events $p, q$ with $D = x (q) - x (p)$ , $q \in J^{+} (p)$ iff $D^{0} \geq ∣ D_{s} ∣$ , and $q \in I^{+} (p)$ iff $D^{0} > ∣ D_{s} ∣$ ; $J^{-} (p) = {q : q \leq p}$ (The closed future cone in inertial coordinates: description, finite sums, and the coordinate criterion for causal reachability, Causal reachability, causal futures and pasts, causal diamonds, and reachability by future causal paths in affine Minkowski space).

[L2]

For $∣ β ∣ < 1$ the passive boost acts on coordinates by $x^{' 0} = γ (x^{0} - β x^{1})$ , $x^{' 1} = γ (x^{1} - β x^{0})$ in the $(0, 1)$ -plane with $γ = (1 - β^{2})^{- 1 / 2}$ , its inverse is $B (- β)$ with inverse formulas $x^{0} = γ (x^{' 0} + β x^{' 1})$ , $x^{1} = γ (x^{' 1} + β x^{' 0})$ , and $B (β)$ is a proper orthochronous Lorentz map (The passive Lorentz boost with parameter beta, The passive boost is a proper orthochronous Lorentz map with inverse B(-beta)).

[L3]

The ordered basis $e_{a}^{'} : = B (β)^{- 1} e_{a}$ is $η$ -orthonormal, future and positively oriented, so $(o; e_{0}^{'}, e_{1}^{'}, e_{2}^{'}, e_{3}^{'})$ is an inertial frame; its chart satisfies $x^{'} (r) = B (β) x (r)$ (The passive boost is a proper orthochronous Lorentz map with inverse B(-beta), Lorentz maps are isomorphisms of determinant ±1 that preserve causal character and permute the cones; the Lorentz sets are groups, Inertial frames and inertial charts of the affine Minkowski model, Linear and affine Lorentz maps, and the Lorentz group).

[L4]

A displacement is spacelike iff $η (D, D) > 0$ and timelike iff $η (D, D) < 0$ ; for a timelike displacement with negative time component the reversed displacement has positive time component dominating its spatial part, giving $A_{0} \in J^{+} (A_{1})$ by [L1] (Timelike, null, spacelike and causal vectors of a Lorentz vector space).

Verification

technique · direct

1.1givenL4algebra

The reception event $B_{1}$ has $u t = x^{1} = L + v t$ , so $t (B_{1}) = L / (u - v) = 1 / ((2 - \frac{9}{10}) c) = \frac{10}{11 c}$ s and $x^{1} (B_{1}) = u t (B_{1}) = \frac{20}{11}$ m; the displacement is $B_{1} - A_{0} = (c t (B_{1}), x^{1} (B_{1}), 0, 0) = (\frac{10}{11}, \frac{20}{11}, 0, 0)$ m, with $η (D, D) = - (\frac{10}{11})^{2} + (\frac{20}{11})^{2} = \frac{300}{121} > 0$ and $D^{0} > 0$ , so it is spacelike with positive time component in $x$ ; this is (a).

2.1step 1.1L2L3algebra

With $β = v / c = 9 / 10$ , the basis $e_{a}^{'} = B (β)^{- 1} e_{a}$ is the basis of an inertial frame by [L3], and its chart satisfies $x^{'} = B (β) x$ with inverse $x = B (- β) x^{'}$ , that is $x^{0} = γ (x^{' 0} + β x^{' 1})$ , $x^{1} = γ (x^{' 1} + β x^{' 0})$ ; moreover $γ = (1 - 81 / 100)^{- 1 / 2} = (19 / 100)^{- 1 / 2} = 10 / \sqrt{19}$ .

3.1step 1.1step 2.1L2L3algebra

In the chart $x^{'}$ the worldline of $B$ is constant in $x^{' 1}$ , since $x^{' 1} = γ (x^{1} - \frac{9}{10} x^{0}) = γ (L + v t - \frac{9}{10} c t) = γ L$ ; the return leg moves in the $- x^{' 1}$ direction at coordinate speed $u$ , so its displacement in $x^{'}$ is $(c Δ t^{'}, - u Δ t^{'})$ for a single $Δ t^{'}$ , and applying the inverse formula of step 2.1 to its $x^{1}$ -component gives $Δ x^{1} = γ (β c Δ t^{'} - u Δ t^{'}) = γ (v - u) Δ t^{'}$ .

4.1step 3.1L2algebra

The arrival event lies on the worldline $x^{1} = 0$ of $A$ , so $Δ x^{1} = - x^{1} (B_{1}) = - \frac{20}{11}$ m, and step 3.1 gives $Δ t^{'} = x^{1} (B_{1}) / (γ (u - v)) > 0$ ; substituting into the time component of the inverse formula gives $Δ t = \frac{Δ x^{0}}{c} = γ Δ t^{'} (1 - \frac{v u}{c^{2}}) = \frac{x^{1} (B_{1}) (1 - v u / c^{2})}{u - v}$ , and with $v u / c^{2} = \frac{9}{10} \cdot 2 = \frac{9}{5}$ this is $\frac{20}{11} (- \frac{4}{5}) \frac{10}{11 c} = - \frac{160}{121 c}$ s; this is (b).

5.1step 1.1step 4.1L1L4algebra

Adding the two legs gives $t (A_{1}) = t (B_{1}) + Δ t = \frac{10}{11 c} - \frac{160}{121 c} = \frac{110 - 160}{121 c} = - \frac{50}{121 c}$ s, and the closed formula $L (2 u - v - v u^{2} / c^{2}) / (u - v)^{2}$ evaluates as $(4 c - \frac{9}{10} c - \frac{9}{10} \cdot 4 c) / (\frac{11}{10} c)^{2} = (- \frac{c}{2}) / (\frac{121}{100} c^{2}) = - \frac{50}{121 c}$ s, the same number. The displacement $A_{1} - A_{0} = (- \frac{50}{121}, 0, 0, 0)$ m has $η = - (\frac{50}{121})^{2} < 0$ , so it is timelike, and the reversed displacement $A_{0} - A_{1} = (\frac{50}{121}, 0, 0, 0)$ has time component $\frac{50}{121} > 0 = ∣ D_{s} ∣$ , so $A_{0} \in I^{+} (A_{1})$ by [L1]: the arrival is in the strict chronological past of the emission. The return displacement in $x$ is $(- \frac{160}{121}, - \frac{220}{121}, 0, 0)$ m because its $x^{1}$ -component is $- \frac{20}{11} = - \frac{220}{121}$ m, and its interval is $η = - (\frac{160}{121})^{2} + (\frac{220}{121})^{2} > 0$ , so it is spacelike; this is (c).

6.1step 5.1given∎

Steps 1.1 to 5.1 are an arithmetic specialization of the hypothetical fixed-speed relay model with its stated inputs, and no step uses any property beyond the Lorentz kinematics of the boost and the causal classification of the two displacements; the reception of the relayed signal strictly before its emission in the chart of the emitter is therefore a conditional conclusion of that model, not a report of a measurement and not an assertion that superluminal signals or the devices realizing the relay exist; this is (d).

Worked relay arithmetic at $u = 2 c$ and $v = (9 / 10) c$ : arrival at $t = - 50 / (121 c)$ seconds

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

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Worked relay arithmetic at u=2cu=2c and v=(9/10)cv=(9/10)c: arrival at t=−50/(121c)t=−50/(121c) seconds

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources

Worked relay arithmetic at $u = 2 c$ and $v = (9 / 10) c$ : arrival at $t = - 50 / (121 c)$ seconds