Time — Question 2

Question 2

Resetting a clock does not repair its rate

An electronic clock is compared with a reference whose elapsed SI time is $t$. Assume the electronic reading obeys the exact model $$C(t)=b+(1+\epsilon)t,$$ where $b$ is an offset in seconds and $\epsilon$ is a dimensionless constant rate error. The calibration points are $(t,C)=(0\ \mathrm{s},3.200\ \mathrm{s})$ and $(3600\ \mathrm{s},3603.380\ \mathrm{s})$. Use these data as exact within the constant-rate model.

1. Determine $b$ and $\epsilon$. Express the rate error in parts per million, where $1\ \mathrm{ppm}=10^{-6}$. Does the clock gain or lose time?

2. The clock reports a reading difference of $120.006\ \mathrm{s}$ between two events. Find the true elapsed time. Explain why subtracting $b$ from this difference would be incorrect.

3. At a comparison instant, an operator resets the displayed reading to match the reference but makes no rate adjustment. How much error accumulates during the next $24$ hours under the model? Use $1\ \mathrm{h}=3600\ \mathrm{s}$.

4. What evidence, beyond the two calibration points, would be useful before assuming that the same correction applies over a changing-temperature experiment?

Question 2 - Solution

Resetting a clock does not repair its rate

(a) Use the intercept for offset and differences for rate.

The first point gives $\boxed{b=3.200\ \mathrm{s}}$. The clock advances $3603.380-3.200=3600.180$ indicated seconds during $3600$ SI seconds, so $$1+\epsilon=\frac{3600.180}{3600}=1.000050, \qquad\boxed{\epsilon=5.0\times10^{-5}=50\ \mathrm{ppm}}.$$ Its positive rate error means that it gains time relative to the reference.

(b) A common offset cancels between timestamps.

$$\Delta C=[b+(1+\epsilon)t_2]-[b+(1+\epsilon)t_1]=(1+\epsilon)\Delta t.$$

For two events, $C(t_2)-C(t_1)=(1+\epsilon)(t_2-t_1)$. Hence $$\Delta t=\frac{120.006\ \mathrm{s}}{1.000050} =\boxed{120.000\ \mathrm{s}}.$$ The offset $b$ is present in both readings and has already canceled. Subtracting it again would introduce an error. The multiplicative rate error, in contrast, remains in their difference.

(c) Setting the phase leaves the rate unchanged.

After the reset the initial offset is zero, but the clock still gains $\epsilon$ seconds per SI second. Over $24(3600)=86\,400\ \mathrm{s}$, the accumulated error is $$(5.0\times10^{-5})(86\,400\ \mathrm{s}) =\boxed{4.320\ \mathrm{s}}\quad\text{fast}.$$

(d) Check whether a constant-rate model remains valid.

Compare against the reference at intermediate times and across the experiment’s temperature range, allowing the clock to reach relevant thermal conditions. Examine whether a single slope describes the readings within defensible bounds. Two points establish the parameters of an assumed straight-line model, but do not prove that temperature changes, transients, or aging leave that model valid.