Some Consequences of the Lorentz Equations — Question 6

Section 37-7 · Physics pages 6162–6163

Question 6 Changing inertial frames at turnaround changes the distant simultaneity assignment A traveler turns around at Earth-frame event tT = 5.0 yr, xT = 4.0 light-years. The outgoing speed is +0.80c and the incoming speed is −0.80c. Idealize the turnaround as instantaneous. x ct turn Spacetime sketch; the vertical coordinate is ct. Labels specify events and worldlines. (a) Find the Earth times simultaneous with turnaround in the outgoing and incoming inertial frames. (b) Find the jump in assigned distant Earth time and explain whether Earth physically ages discontinuously.
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Question 6 - Solution Changing inertial frames at turnaround changes the distant simultaneity assignment x ct turn (a) Method and calculation Outgoing: tout E = tT −vxT /c2 = 5.0 −3.2 = 1.8 yr . Incoming: tin E = tT + vxT /c2 = 5.0 + 3.2 = 8.2 yr . (b) Interpretation and further calculation tin E −tout E = 2vxT /c2 = 6.4 yr . Earth clock worldline is continuous; the simultaneity slice changes. Physical check The jump is in a coordinate assignment to distant events, not in a local clock reading or a received light signal.

Question 6

Changing inertial frames at turnaround changes the distant simultaneity assignment

A traveler turns around at Earth-frame event tT=5.0yrt_T=5.0\,\mathrm{yr}, xT=4.0x_T=4.0 light-years. The outgoing speed is +0.80c+0.80c and the incoming speed is −0.80c-0.80c. Idealize the turnaround as instantaneous.

Diagram shown in the original SVG worksheet.

  1. Find the Earth times simultaneous with turnaround in the outgoing and incoming inertial frames.

  2. Find the jump in assigned distant Earth time and explain whether Earth physically ages discontinuously.

Question 6 - Solution

Changing inertial frames at turnaround changes the distant simultaneity assignment

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Outgoing: tEout=tT−vxT/c2=5.0−3.2=1.8yr.\text{Outgoing: }t_E^{\mathrm{out}}=t_T-vx_T/c^2=5.0-3.2=\boxed{1.8\,\mathrm{yr}}.Incoming: tEin=tT+vxT/c2=5.0+3.2=8.2yr.\text{Incoming: }t_E^{\mathrm{in}}=t_T+vx_T/c^2=5.0+3.2=\boxed{8.2\,\mathrm{yr}}.

(b) Interpretation and further calculation

tEin−tEout=2vxT/c2=6.4yr.\boxed{t_E^{\mathrm{in}}-t_E^{\mathrm{out}}=2vx_T/c^2=6.4\,\mathrm{yr}}.Earth clock worldline is continuous; the simultaneity slice changes.\text{Earth clock worldline is continuous; the simultaneity slice changes.}

Physical check

The jump is in a coordinate assignment to distant events, not in a local clock reading or a received light signal.