Time — Question 1

Question 1

A count becomes a duration only with a calibrated period

The SI second is defined by fixing the numerical value of the specified unperturbed ground-state hyperfine transition frequency of caesium-133 to $$f_{\mathrm{Cs}}=9\,192\,631\,770\ \mathrm{Hz},\qquad 1\ \mathrm{Hz}=1\ \mathrm{s^{-1}}.$$ For an ideal periodic signal of constant frequency $f$, a count of $N$ complete periods spans $\Delta t=N/f$.

Two ideal signals each accumulate $N=22\,981\,579\,425$ periods. Signal A has the defining caesium frequency above. Signal B has an independently calibrated frequency $f_B=10\,000\,000\,000\ \mathrm{Hz}$. Treat both frequencies and counts as exact within this exercise.

1. Find the duration represented by each count. Explain why equal cycle counts do not imply equal elapsed times.

2. A counter is programmed to label every $9\,192\,631\,770$ input periods as “one second,” regardless of the input signal. What physical duration does one such label represent when B is connected? Does changing the input redefine the SI second?

3. A student says a stable pendulum could replace the atomic definition merely because it repeats. Identify the missing requirement for using its cycles to measure SI durations.

4. Does the defining frequency count radioactive decays? Explain what physical process the definition actually specifies, without needing a theory of atomic structure.

Question 1 - Solution

A count becomes a duration only with a calibrated period

(a) Use the frequency belonging to each signal.

$$\Delta t_A=\frac{22\,981\,579\,425}{9\,192\,631\,770}\ \mathrm{s} =\boxed{2.500000000\ \mathrm{s}},$$ $$\Delta t_B=\frac{22\,981\,579\,425}{10\,000\,000\,000}\ \mathrm{s} =\boxed{2.2981579425\ \mathrm{s}}.$$ B completes more cycles per second, so each of its periods is shorter. Counting the same number of shorter periods gives a shorter duration. The count alone has no time unit.

(b) The counter’s label does not determine the unit.

With B connected, one programmed “second” spans $$\frac{9\,192\,631\,770}{10\,000\,000\,000}\ \mathrm{s} =\boxed{0.919263177\ \mathrm{s}}.$$ The instrument would advance its displayed seconds too rapidly. It has the wrong conversion from input cycles to SI duration; the SI definition has not changed. It must use the calibrated B frequency, or an appropriate frequency ratio, to report SI seconds correctly.

(c) Repeatability alone supplies no SI calibration.

$$\Delta t=NP,\qquad P=\frac1f,\qquad\delta(\Delta t)=N\delta P\quad(N\text{ exact}).$$

A repeating pendulum can provide a useful time reference, but its period must be determined in SI seconds and its changes under use conditions controlled or bounded. A very stable but incorrectly assigned period gives repeatable, biased durations. The relevant requirement is a supported connection between its cycles and the SI second, not repetition alone.

(d) A transition frequency is not a decay count.

The definition refers to the frequency associated with a specified transition between atomic energy levels, under the stated unperturbed conditions. It does not define one second by a number of radioactive decays. The periods being counted are periods of a signal at that transition frequency, not a sequence of nuclei disappearing.