Time — Question 6

Question 6

Clock synchronization depends on the travel-time model

Clocks A and B have equal, correct rates. Let A supply the reference coordinate time $t$, and let B read $t+\theta$, with unknown constant offset $\theta$. A sends a request, B receives it and sends a reply, and A receives that reply. The four recorded timestamps are shown.

Let $d_f\ge0$ be the request’s travel time and $d_r\ge0$ the reply’s travel time. Assume all timestamps are exact and all processing between receipt and reply occurs at B. Relativistic effects and changes in clock rate are outside this model.

1. Find B’s processing duration and the sum $d_f+d_r$. Explain why the unknown offset cancels from the processing duration.

2. If $d_f=d_r$, find the inferred $\theta$ and state which clock is ahead.

3. Show that $d_f=0.005\ \mathrm{s}$ and $d_r=0.003\ \mathrm{s}$ also fit the timestamps with a different offset. Determine that offset.

4. Without equal-delay information, find the full range of $\theta$ consistent with nonnegative travel times. What does this imply about treating half the round-trip delay as an independently measured one-way delay?

Question 6 - Solution

Clock synchronization depends on the travel-time model

(a) Use differences on each clock.

B processes the message for $t_3-t_2=0.002\ \mathrm{s}$. Its constant offset appears in both timestamps and cancels. A’s round-trip interval is $0.010\ \mathrm{s}$, so $$\boxed{d_f+d_r=0.010\ \mathrm{s}-0.002\ \mathrm{s}=0.008\ \mathrm{s}}.$$

(b) Apply the extra symmetry assumption explicitly.

Equal paths give $d_f=d_r=0.004\ \mathrm{s}$. Since $t_2=t_1+d_f+\theta$, $$\boxed{\theta=(0.007-0.000-0.004)\ \mathrm{s}=0.003\ \mathrm{s}}.$$ Under that assumption B reads 3 ms ahead of A.

(c) Check an asymmetric alternative.

$$t_{\mathrm{reply,ref}}=0.009-0.002=0.007\ \mathrm s,\qquad t_{\mathrm{return}}=0.007+0.003=0.010\ \mathrm s.$$

With $d_f=0.005\ \mathrm{s}$, the same equation gives $\theta=0.002\ \mathrm{s}$. B’s reply timestamp 0.009 s then corresponds to reference time 0.007 s; adding the return delay 0.003 s gives A’s observed 0.010 s. Thus the alternative fits all four timestamps.

(d) Bound what the data identify without symmetry.

The general delay expressions are $$d_f=0.007\ \mathrm{s}-\theta,\qquad d_r=0.001\ \mathrm{s}+\theta.$$ Nonnegative delays require $\boxed{-0.001\ \mathrm{s}\le\theta\le0.007\ \mathrm{s}}$. The exchange identifies the sum of travel times, not their split. Half that sum is a one-way estimate only with an additional symmetry assumption or equivalent path information; precise timestamps alone do not remove this ambiguity.