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I'm quoting a passage from my notes:

The development of clocks based on atomic oscillations allowed measures of timing with accuracy on the order of $1$ part in $10^{14}$, corresponding to errors of less than one microsecond (one millionth of a second) per year.

I do not understand what the accuracy of $1$ part in $10^{14}$ means. Does it mean that the atomic clocks can tell us the time accurate and ceratain to $10^{-14}s$? How should I understand this? Moreover, what is meant by the error of one microsecond per year? Is it a kind of uncertainty in measurement? How should I understand it? I googled this topic and found information about the atomic clocks and also reviewed the definitions of accuracy and error; however, I'm not able to make any sensisble connection between the concepts. Please help me, thank you.

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2 Answers 2

up vote 5 down vote accepted

It means that if the clock begins set to reading $\theta[ A_M ]$ at mark $A_M$ then the clock is not good for all other (subsequent) indications and corresponding readings, but it is wrong at least for some of those; i.e. there are at least two distinct indications $A_P$ and $A_Q$ with corresponding (different) readings $\theta[ A_P ]$ and $\theta[ A_Q ]$ such that $$ \frac{\theta[ A_P ] - \theta[ A_M ]}{\theta[ A_Q ] - \theta[ A_M ]} \neq \frac{\tau A[ _M, _P ]}{\tau A[ _M, _Q ]}, $$ where $\tau A[ _M, _P ]$ is the duration of the clock from its "mark" indication $A_M$ until its subsequent indication $A_P$, and similartly $\tau A[ _M, _Q ]$ is the duration of the clock from its mark indication $A_M$ until its subsequent indication $A_Q$.

In order to quantify by how much a given clock is wrong (if it is not good), for a given set $\mathcal A$ of its (subsequent) indications, beginning with "mark" indication $A_M$ up to some particular final (ultimate) indication $A_U$, physicists can use for instance the method of least squares and determine the corresponding "sum $S_{\mathcal A}$ of squared residuals" as $$ S_{ \! \mathcal A } := \sum_{A_{ \mathcal N } \, = A_M}^{A_U} \left( \frac{\tau A[ _M, _{\mathcal N} ]}{\tau A[ _M, _U ]} \right)^2 - \frac{ \left( \sum\limits_{A_{ \mathcal N } \, = A_M}^{A_U} \left( \theta[ A_{ \mathcal N } ] - \theta[ A_M ] \right) \frac{\tau A[ _M, _{\mathcal N} ]}{\tau A[ _M, _U ]} \right)^2 }{\sum\limits_{A_{ \mathcal N } \, = A_M}^{A_U} \left( \theta[ A_{ \mathcal N } ] - \theta[ A_M ] \right)^2 }.$$

For the clock under consideration then $$ \sqrt{S_{ \! \mathcal A } } \le 10^{-14},$$ or factoring out the common denominator $\tau A[ _M, _U ] := \tau_{ \mathcal A }$: $$ \frac{\delta \tau_{ \mathcal A }}{\tau_{ \mathcal A }} \le 10^{-14},$$ for any set of (subsequent) indications $\mathcal A$ of the clock;
where $$ \delta \tau_{ \mathcal A } := \sqrt{ \sum_{A_{ \mathcal N } \, = A_M}^{A_U} \left( \tau A[ _M, _{\mathcal N} ] \right)^2 - \frac{ \left( \sum\limits_{A_{ \mathcal N } \, = A_M}^{A_U} \left( \theta[ A_{ \mathcal N } ] - \theta[ A_M ] \right) \tau A[ _M, _{\mathcal N} ] \right)^2 }{\sum\limits_{A_{ \mathcal N } \, = A_M}^{A_U} \left( \theta[ A_{ \mathcal N } ] - \theta[ A_M ] \right)^2 } }.$$

This kind of expression of "fractional errors" is very common in many fields of quantitative science.

Now, to be concrete, a year is about $3.156 \times 10^7 \,\mathrm{s}$, so after one year the clock will be wrong by no more that $$(3.156 \times 10^7 \,\mathrm{s}) \cdot 10^{-14} = 3.156 \times 10^{-7} \,\mathrm{s} = 0.3156 \,\mathrm{\mu s} \,.$$

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The error would be in the order of 10^-14. This is mathematically similar to the sense of errors you have on your hand watch, caused by mechanical inaccuracy - probably in the range of 1 second per week, or 1 second per year if its a Rolex :)

One should note however, that such a very small inaccuracy in time measurement in atomic clocks is perhaps less than the error that would be caused by relativistic effects for a person who spends a lot of time driving. suppose that the average velocity between an observer and an atomic clock is 30 m/s, this would give a relativistic time dilation of about 1.5 microseconds per year, which is bigger than the inaccuracy of the atomic clock. For a pilot the time dilation would be about 15 microseconds per year :)

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In reality the best mechanical watches (including Rolex) are only accurate to 1-2 seconds per day. Most quartz watches have rated accuracies on the order of 15 seconds per month, better ones can achieve 2-5 seconds per month. The Bulova Precisionist watch is specified as no more than 10 seconds per year. I have a Citizen watch which is reset once a day by the NIST radio signal from Colorado so that its long term error is zero and its short term error is determined by its internal quartz oscillator which in practice has a negligible error between resets. –  Barry Feb 9 '14 at 4:01
I was joking about the Rolex, however, for the price you pay to get them they should be more accurate than atomic clocks :) –  user40229 Feb 13 '14 at 4:06

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