According to the latest definition of the SI units in 2019, a second is defined as "$9,192,631,770$ periods of the transition between the two hyperfine levels of the caesium-133 atom $\Delta\nu_{{\rm Cs}}$," and $c$ is fixed as $299792458$ metres per second.

However I am puzzled by the fact that all uncertainties have disappeared since I graduated. Following wikipedia, the "realization" of the meter has a relative uncertainty of about $2×10^{-11}$, but I could not find information about the second and its uncertainty. From what I recall before the redefinition, $\Delta\nu_{{\rm Cs}}$ had a relative uncertainty of O($10^{-15}$), so is that how well do we know the duration of the 'realization' a second?

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    $\begingroup$ It may sound surprising but a second is a second long. $\endgroup$
    – user87745
    Commented Sep 29, 2021 at 22:34
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    $\begingroup$ Can you clarify what you're asking here? The duration of a second is not subject to uncertainty because it is defined exactly. $\endgroup$
    – J. Murray
    Commented Sep 29, 2021 at 22:42
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    $\begingroup$ Metrology is a strange subject. We know exactly what me mean by one ideal second, but we can never produce a measurement of exactly one second. $\endgroup$
    – Andrew
    Commented Sep 30, 2021 at 3:27
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    $\begingroup$ @DvijD.C. Every 1,000 milliseconds in Africa a second passes. $\endgroup$
    – MonkeyZeus
    Commented Oct 1, 2021 at 12:33
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    $\begingroup$ @MonkeyZeus does that account for time dilation? $\endgroup$
    – Sandejo
    Commented Oct 2, 2021 at 0:30

4 Answers 4


A second is a second long by definition, but if you measure any time in seconds, the number of seconds you infer will be subject to an error of at least $\mathcal O(10^{-15})$ because of the uncertainty of Caesium clocks as you correctly point out. This is true even if you make a higher precision measurement using new clocks like Quantum Clocks with uncertainties of $\mathcal O(10^{-18})$.

Edit: For completeness, I include this quote from here:

The second is defined by taking the fixed numerical value of the caesium frequency $\Delta$Cs, the unperturbed ground-state hyperfine transition frequency of the caesium 133 atom, to be 9,192,631,770 when expressed in the unit Hz, which is equal to s$^{–1}$.

and I note that one day this standard may change.

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    $\begingroup$ Can you elaborate on this more? That is very interesting. Is the 10^-15 a fundamental uncertainty in the Cs transition itself? E.g. line width due to uncertainty principle? I assume if that's the case they could eventually choose to redefine 1 s in terms of a process with less inherent uncertainty. $\endgroup$
    – RC_23
    Commented Sep 30, 2021 at 4:17
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    $\begingroup$ By "O(10^-15)," do you mean something like "within a couple of orders of magnitude of 10^-15"? I don't know if big O notation means something else in physics, but in mathematics, O(10^-15) means exactly the same thing as O(1), which means "bounded by a constant value." It would be completely accurate to say that the mass of the sun is O(10^-15) grams. $\endgroup$ Commented Sep 30, 2021 at 13:36
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    $\begingroup$ @TannerSwett In this usage, O(10^-15) means "error is less than or equal to 10^-15". It is an imprecise use of the big-O notation but is very common in physics. $\endgroup$ Commented Sep 30, 2021 at 16:31
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    $\begingroup$ @RossPresser I read it as "on the order of". I didn't even make the connection to big-oh until it was pointed out. $\endgroup$
    – JimmyJames
    Commented Sep 30, 2021 at 17:55
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    $\begingroup$ @AnoE Most jokes are funny to those who get them and confusing to those who don't $\endgroup$
    – TCooper
    Commented Sep 30, 2021 at 22:19

The second itself does have an uncertainty. When we're using it without uncertainty, we're basically using the following trick:

  • The time span of $n$ seconds is defined as $2\times n\times 9\,192\,631\,770$ transitions between the two Cs hyperfine levels.
  • Making $n$ big, we can still count exactly how many transitions have occured, so the error on the total time span is constant. A single second can then be computed as $\tfrac1n$ of that time, and the uncertainty becomes as small as we like. In other words, there is no uncertainty in the definition of the second, only in a concrete experiment we might perform to implement it.
  • $\begingroup$ I'm not sure I totally follow this. If we keep repeating this experiment multiple times, we could get different answers, no? The longer we run just increase the precision. $\endgroup$
    – JimmyJames
    Commented Sep 30, 2021 at 18:06
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    $\begingroup$ @Filippo Because the period of each actual transition deviates very slightly from the mean period. That's partly due to Doppler shift. $\endgroup$
    – PM 2Ring
    Commented Oct 1, 2021 at 7:57
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    $\begingroup$ @Filippo The deviation is pretty small, and improved technology has made the variation even smaller. In the definition, the Cs atoms are supposed to be at absolute zero. That's obviously not possible in practice, and although modern atomic fountains use laser cooled atoms, some warming is inevitable: you can't pump microwaves into the atoms without causing some heating, so there's thermal motion, and collisions. There's some discussion of changing the definition of the second to use a "quantum clock", with much higher frequency & better stability, but that will take years. $\endgroup$
    – PM 2Ring
    Commented Oct 1, 2021 at 8:24
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    $\begingroup$ @Filippo I think the periods should be identical at 0 K, but I'd like an expert to confirm that. $\endgroup$
    – PM 2Ring
    Commented Oct 1, 2021 at 8:44
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    $\begingroup$ @Filippo the transition is still not infinitely narrow at 0 K, every transition has what is called a natural linewidth due to the finite lifetime of the state and the time-energy uncertainty principle. This does not go away at 0 K and is a fundamental limit, which is why you want long-lived states for clock transitions. Of course you can still average down (and do more fancy tricks like Ramsey interferometry) to find the centre better, but the narrower transition you start with the less averaging you have to do $\endgroup$
    – llama
    Commented Oct 1, 2021 at 22:24

There's a shift from the old way of having standard examples of the units that everyone can compare against, to defining the units in terms of fundamental physics.

There was an uncertainty in measuring the speed of light. Then, the units were defined in terms of the speed of light, thus fixing the value exactly. Now there is still uncertainty in making the measurement when calibrating your clock or ruler, but the big advantage is that as such measurement technology improves you can immediately make use of it in a definitive way.

The big advantage of doing it this way is that you don't have to curate an artifact and have people go reference that specific object. This was a royal pain for the standard kilogram, which gained or lost weight for unexplained reasons.

When such a shift is made, it is done by fixing the value of the physical constant by definition, to the best-known value determined in the previous system.


Cited: "The second, symbol s, is the SI unit of time. It is defined by taking the fixed numerical value of the caesium frequency, ∆νCs, the unperturbed ground-state hyperfine transition frequency of the caesium 133 atom, to be 9 192 631 770 when expressed in the unit Hz, which is equal to s−1."

"The second, so defined, is the unit of proper time in the sense of the general theory of relativity."

Source: https://www.bipm.org/documents/20126/41483022/SI-Brochure-9-EN.pdf/2d2b50bf-f2b4-9661-f402-5f9d66e4b507

resp. https://en.wikipedia.org/wiki/SI_base_unit


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