In laboratory frame S, event A occurs at x_A = 0, t_A = 0 and event B at x_B = 330 m, t_B = 0. Frame S′ moves in the +x direction at speed 0.32c relative to S. Origins coincide at event A. Use c = 3.00 × 10⁸ m s⁻¹ and the Lorentz transformations x′ = γ(x − vt), t′ = γ(t − vx/c²), γ = 1/√(1 − v²/c²). Find Δx′ and Δt′ = t′_B − t′_A, identify which event occurs first in S′, and check the invariant interval (cΔt)² − (Δx)².

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Δx′ = 348.3 m; Δt′ = -3.715 × 10^-7 s. B occurs first in S′. c²Δt′² − Δx′² = −1.089 × 10^5 m² in both frames. The speed ratio is β = 0.32, so γ = 1/√(1 − β²) = 1.056. The two events have Δt = 0 and Δx = 330 m in the laboratory frame. Apply the transformation to the separation between events: Δx′ = γ(Δx − vΔt) = γΔx = 348.3 m. This is an event-coordinate separation, not a contracted rod length. The transformed time separation is Δt′ = γ(Δt − vΔx/c²) = −γβΔx/c = -3.715 × 10^-7 s. The negative value means B precedes A in S′. Evaluate c²Δt′² − Δx′² = γ²Δx²(β² − 1) = −Δx² = −1.089 × 10^5 m². This is identical to the laboratory value, checking both transformations with the stated sign convention. The events are spacelike separated, so changing their temporal order does not reverse a possible causal signal between them. Simultaneity is frame-dependent; a time-dilation formula alone cannot transform two spatially separated events.

Two simultaneous events in another frame · set 1

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