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The International System of Units — Question 5

Section 1-2 · Physics pages 40–41

Question 5 An exact speed does not make an exact distance A pulse travels to a stationary mirror and returns along the same vacuum path. Use the model L = c∆t/2, with exact c = 299 792 458 m s−1. Pulse unit Mirror outgoing returning L Vacuum path, schematic. L is the one-way distance from the pulse unit’s reference plane to the mirror. After correcting all known delays, the round-trip time is reported as ∆t = (2.000 ± 0.004) × 10−8 s. For this question the ± gives a closed hard bound, not a standard deviation. Assume all residual error is represented by this time bound and that the stated model applies. (a) Calculate the central distance and the exact endpoint bounds implied by this model. Give a sensible shorter report that still encloses the allowed interval. (b) Find the relative half-width of the distance interval. Why is it not zero when c is exact? (c) A scientist later discovers an uncorrected positive electronic delay. Would the original distance estimate be too large or too small? Does the fixed value of c protect against that error?
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Question 5 - Solution An exact speed does not make an exact distance Pulse unit Mirror outgoing returning L (a) Transform the whole interval. The model is increasing in ∆t. Its central value is L0 = 299 792 458(2.000 × 10−8) 2 m = 2.99792458 m. The half-width is δL = c(0.004 × 10−8 s)/2 = 0.00599584916 m. Hence L ∈[2.99192873084, 3.00392042916] m . A convenient outward-rounded interval is [2.9919, 3.0040] m . The many digits in the unrounded endpoints express arithmetic on the supplied bounds, not the precision of the apparatus. (b) The measured time retains its uncertainty. δL L0 = δt ∆t0 = 0.004 2.000 = 0.002 = 0.2% . An exact conversion factor contributes no additional uncertainty but does not remove uncertainty from its input. (c) Diagnose a failure of the stated error model. Linferred −Ltrue = c 2 δtdelay > 0 (δtdelay > 0). An extra positive delay increases the reported round-trip time and therefore makes the inferred distance too large. If it was not included in the given bound, that bound must be reconsidered. Exact c neither identifies nor cancels an electronic delay; the realization still needs calibrated timing and a valid error model.

Question 5

An exact speed does not make an exact distance

A pulse travels to a stationary mirror and returns along the same vacuum path. Use the model L=cΔt/2L=c\Delta t/2, with exact c=299792458ms−1c=299\,792\,458\ \mathrm{m\,s^{-1}}.

Diagram shown in the original SVG worksheet.

After correcting all known delays, the round-trip time is reported as Δt=(2.000±0.004)×10−8s.\Delta t=(2.000\pm0.004)\times10^{-8}\ \mathrm{s}. For this question the ±\pm gives a closed hard bound, not a standard deviation. Assume all residual error is represented by this time bound and that the stated model applies.

  1. Calculate the central distance and the exact endpoint bounds implied by this model. Give a sensible shorter report that still encloses the allowed interval.

  2. Find the relative half-width of the distance interval. Why is it not zero when cc is exact?

  3. A scientist later discovers an uncorrected positive electronic delay. Would the original distance estimate be too large or too small? Does the fixed value of cc protect against that error?

Question 5 - Solution

An exact speed does not make an exact distance

Diagram shown in the original SVG worksheet.

(a) Transform the whole interval.

The model is increasing in Δt\Delta t. Its central value is L0=299792458(2.000×10−8)2m=2.99792458m.L_0=\frac{299\,792\,458(2.000\times10^{-8})}{2}\ \mathrm{m} =2.99792458\ \mathrm{m}. The half-width is δL=c(0.004×10−8s)/2=0.00599584916m\delta L=c(0.004\times10^{-8}\ \mathrm{s})/2 =0.00599584916\ \mathrm{m}. Hence L∈[2.99192873084,3.00392042916]m.\boxed{L\in[2.99192873084,\ 3.00392042916]\ \mathrm{m}}. A convenient outward-rounded interval is [2.9919,3.0040]m\boxed{[2.9919,\ 3.0040]\ \mathrm{m}}. The many digits in the unrounded endpoints express arithmetic on the supplied bounds, not the precision of the apparatus.

(b) The measured time retains its uncertainty.

δLL0=δtΔt0=0.0042.000=0.002=0.2%.\frac{\delta L}{L_0}=\frac{\delta t}{\Delta t_0}=\frac{0.004}{2.000} =0.002=\boxed{0.2\%}. An exact conversion factor contributes no additional uncertainty but does not remove uncertainty from its input.

(c) Diagnose a failure of the stated error model.

Linferred−Ltrue=c2δtdelay>0(δtdelay>0).L_{\mathrm{inferred}}-L_{\mathrm{true}}=\frac c2\,\delta t_{\mathrm{delay}}>0\quad(\delta t_{\mathrm{delay}}>0).

An extra positive delay increases the reported round-trip time and therefore makes the inferred distance too large. If it was not included in the given bound, that bound must be reconsidered. Exact cc neither identifies nor cancels an electronic delay; the realization still needs calibrated timing and a valid error model.

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