Some Consequences of the Lorentz Equations — Question 7

Section 37-7 · Physics pages 6164–6165

Question 7 The straight inertial path maximizes proper time between fixed events Two travelers share endpoints separated by T = 10.0 s and ∆x = 6.0c s in one laboratory. One travels at constant velocity; the other spends 5.0 s at 0.40c and 5.0 s at 0.80c. Transitions are negligible. x ct reunion Spacetime sketch; the vertical coordinate is ct. Labels specify events and worldlines. (a) Find both proper durations and check that the endpoints agree. (b) Use concavity of f(β) = √ 1 −β2 to explain the general comparison.
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Question 7 - Solution The straight inertial path maximizes proper time between fixed events x ct reunion (a) Method and calculation ¯β = ∆x/(cT) = 0.60, τstraight = 10 √ 1 −.602 = 8.00 s . τbroken = 5 √ .84 + 5 √ .36 = 7.58258 s . (b) Interpretation and further calculation ∆x/c = 5(.40) + 5(.80) = 6.0 s, f′′(β) = −(1 −β2)−3/2 < 0. ∫T 0 f(β(t)) dt ≤Tf(⟨β⟩) . Physical check The endpoint comparison differs from finding the smallest coordinate-time interval by changing frames. Here different worldlines connect fixed events.

Question 7

The straight inertial path maximizes proper time between fixed events

Two travelers share endpoints separated by T=10.0sT=10.0\,\mathrm s and Δx=6.0cs\Delta x=6.0c\,\mathrm s in one laboratory. One travels at constant velocity; the other spends 5.0 s at 0.40c0.40c and 5.0 s at 0.80c0.80c. Transitions are negligible.

Diagram shown in the original SVG worksheet.

  1. Find both proper durations and check that the endpoints agree.

  2. Use concavity of f(β)=1−β2f(\beta)=\sqrt{1-\beta^2} to explain the general comparison.

Question 7 - Solution

The straight inertial path maximizes proper time between fixed events

Diagram shown in the original SVG worksheet.

(a) Method and calculation

β‾=Δx/(cT)=0.60,τstraight=101−.602=8.00s.\bar\beta=\Delta x/(cT)=0.60,\quad\boxed{\tau_{\mathrm{straight}}=10\sqrt{1-.60^2}=8.00\,\mathrm s}.τbroken=5.84+5.36=7.58258s.\tau_{\mathrm{broken}}=5\sqrt{.84}+5\sqrt{.36}=\boxed{7.58258\,\mathrm s}.

(b) Interpretation and further calculation

Δx/c=5(.40)+5(.80)=6.0s,f′′(β)=−(1−β2)−3/2<0.\Delta x/c=5(.40)+5(.80)=6.0\,\mathrm s,\quad f\prime\prime(\beta)=-(1-\beta^2)^{-3/2}<0.∫0Tf(β(t))dt≤Tf(⟨β⟩).\boxed{\int_0^T f(\beta(t))\,dt\le T f(\langle\beta\rangle)}.

Physical check

The endpoint comparison differs from finding the smallest coordinate-time interval by changing frames. Here different worldlines connect fixed events.