Some Consequences of the Lorentz Equations — Question 5

Section 37-7 · Physics pages 6160–6161

Question 5 A hypothetical superluminal message reverses order in some frames As a hypothetical test, a signal leaves A at the origin and reaches B with ∆x = 2c∆t, where ∆t > 0. Use c∆t′ = γ(c∆t −β∆x), β = v/c, and γ = (1 −β2)−1/2. x ct hypothetical Spacetime sketch; the vertical coordinate is ct. Labels specify events and worldlines. (a) Find the boosts for which B precedes A. (b) At β = 0.80, calculate the transformed time difference and state the implication for treating this as an ordinary causal message.
Show solutionHide solution
Question 5 - Solution A hypothetical superluminal message reverses order in some frames x ct hypothetical (a) Method and calculation ∆t′ = γ∆t(1 −2β). 1/2 < β < 1 ⇒ ∆t′ < 0 . (b) Interpretation and further calculation β = 0.80 : ∆t′ = (5/3)(−0.60)∆t = −∆t . s2 = c2∆t2 −(2c∆t)2 = −3c2∆t2 < 0 . Physical check This is a counterfactual spacelike connection, not a permitted local signal in special relativity. Its order is not invariant.

Question 5

A hypothetical superluminal message reverses order in some frames

As a hypothetical test, a signal leaves A at the origin and reaches B with Δx=2cΔt\Delta x=2c\Delta t, where Δt>0\Delta t>0. Use cΔt′=γ(cΔt−βΔx)c\Delta t\prime=\gamma(c\Delta t-\beta\Delta x), β=v/c\beta=v/c, and γ=(1−β2)−1/2\gamma=(1-\beta^2)^{-1/2}.

Diagram shown in the original SVG worksheet.

  1. Find the boosts for which B precedes A.

  2. At β=0.80\beta=0.80, calculate the transformed time difference and state the implication for treating this as an ordinary causal message.

Question 5 - Solution

A hypothetical superluminal message reverses order in some frames

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Δt′=γΔt(1−2β).\Delta t\prime=\gamma\Delta t(1-2\beta).1/2<β<1⇒Δt′<0.\boxed{1/2<\beta<1\quad\Rightarrow\quad\Delta t\prime<0}.

(b) Interpretation and further calculation

β=0.80:Δt′=(5/3)(−0.60)Δt=−Δt.\beta=0.80:\quad\boxed{\Delta t\prime=(5/3)(-0.60)\Delta t=-\Delta t}.s2=c2Δt2−(2cΔt)2=−3c2Δt2<0.s^2=c^2\Delta t^2-(2c\Delta t)^2=\boxed{-3c^2\Delta t^2<0}.

Physical check

This is a counterfactual spacelike connection, not a permitted local signal in special relativity. Its order is not invariant.