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The present definition of the Coulomb, since May 20th 2019, is that one electron has a charge of exactly $1.602176634 \times 10^{-19} $C

Previously, the Coulomb was defined (through the Lorentz force and the Biot-Savart law) by the value of the vacuum magnetic permeability, that was defined as exactly $ 4\pi 10^{-7}$.

The accepted value of the charge of the electron, according to the best measurements at that time, was $1,6021766208 \times 10^{-19}$C with an error less than 98 on the two last digits, so between $1,6021766110 \times 10^{-19}$C and $1,6021766306 \times 10^{-19}$C

But the 28th General Conference on Weights and Measures decided to choose a value outside this range, the last three digits, at that precision, being 340, larger than 304.

The result of this choice is to transform the vacuum magnetic permeability into a quantity subject to measurement, rather than an arbitrary given value.

Per se, this is not unusual. For instance, instead of defining separately the second and the meter, and thus having the velocity of light a quantity to be measured, one has defined the second, fixed the exact value of the velocity of light and then the meter is obtained as an accessory quantity, with some inevitable experimental uncertainty.

But when that was done, the best value of the velocity was used and the old value of the meter was within the uncertainty.

But because the value taken for the charge of the electron was taken outside the uncertainty range of the best previous value (last digits 340 outside the range 110 to 306), the uncertainty range of the value of the vacuum magnetic permeability does not contain the old value.

Indeed if you look for it in

https://physics.nist.gov/cuu/Constants/Table/allascii.txt

you find

1.256 637 062 12 e-6 with an uncertainty 0.000 000 000 19 e-6

so the three last digits are between 193 and 231

But the old value was exactly $4\pi$, and even the lowest value of this range, divided by 4 gives 3.1415926548.., way above the value of $\pi$ 3.1415926535..

Can anyone explain to me the logic of this choice ?

To avoid any ambiguity let me state my point: I do not discuss the wisdom of turning the charge of the electron into the basic quantity, and thus turn the vacuum magnetic permeability as an accessory one to be measured. That is fine, per se.

If the chosen value for the charge of the electron had been such that $ 4\pi 10^{-7}$ was within the original range, and later, better measurements had reduced the uncertainty range and in that process left the old value $ 4\pi 10^{-7}$ outside the new range, I would have found that normal.

What I don't understand is why the old value was outside the uncertainty range from the start !


EDIT

Taking into account David Bayley's answer and various comments let me reformulate my question.

The value I got from the french Wiki page was obviously too old, and one can ignore it.

But I still see some "funny business" contrary to Jagerber's assertion.

If just before the change from $\mu_0$ being defined to $e$ becoming fundamental the most recent value had indeed been 1.6021766341(83) $ 10^{-19}$, then it was perfectly sensible to drop the last digit. But then the relative uncertainty on $e$ before the change would have been $52\times 10^{-10}$ doubled to about $ 10^{-8}$ for $\mu_0$ after the change.

But the figure given for the relative uncertainty is given by David Bayley as $2.3\times 10^{-10}$

In that case it would have been justified to keep one more digit for $e$ Indeed, they kept the digit 4 after the 3 even though the uncertainty on that digit was 8 (83 on the three last digits 341).

When was that figure $2.3\times 10^{-10}$ obtained ? What was the best value of $e$ at that time ?

Keeping the last digit at the time of the change would have reduced the value of $\mu_0$ by a relative amount of about $1.2\times 10^{-10}$. The inconsistency with the more recent value would have been somewhat reduced from 5 to 4 times the new precision of $1.5\times 10^{-10}$. Of course, they could not have known at the time but dropping the last digit would have been illogical if the uncertainty had already been as small as $2.3\times 10^{-10}$ at the time of the change.

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    $\begingroup$ The 2017 CODATA value of $e$ was $e=1.6021766341(83)\times10^{-19}\,\text{C}$ (iopscience.iop.org/article/10.1088/1681-7575/aa950a). From where did you obtain your accepted value for $e$? $\endgroup$
    – d_b
    Oct 7, 2022 at 2:53
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    $\begingroup$ Possible duplicate, with references. I haven’t tried to check your uncertainty analysis, because doing so for the linked answer required more concentration than I’ll have for the next few days. $\endgroup$
    – rob
    Oct 7, 2022 at 3:43
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    $\begingroup$ The SI systems is intellectual nonsense to rescue everyday engineering units like the meter and the second. While this is important for technical applications and space- and time-keeping (we don't want to arrive ten seconds late to an atmospheric Mars entry rendezvous), it is fairly irrelevant to physics how we are choosing these quantities. The important ones are those that we can't chose, like the fine structure constant. For metrological purposes we like to pick the most reliable phenomena, which change as technology improves. The errors we are chasing are therefor always yesterday's. $\endgroup$ Oct 7, 2022 at 4:43
  • $\begingroup$ FWIW, the SI second corresponds to the mean solar second from around 1820. See the end of physics.stackexchange.com/a/677946/123208 for details. $\endgroup$
    – PM 2Ring
    Oct 7, 2022 at 5:41
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    $\begingroup$ @rob My question is not a duplicate of the question you mention. There the question is whether $\mu_0$ is a defined quantity or not. I have perfectly understood that if the Coulomb is defined from the charge of the electron, $\mu_0$ is not a defined quantity, and is not anymore exactly $4\pi \times 10^{-7}$. My question is : why is that value more than one standard deviation from the acceped value already at the start ? $\endgroup$
    – Alfred
    Oct 7, 2022 at 6:10

1 Answer 1

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The defined value of $e$ is in almost perfect agreement with the best experimental evaluation at the time of the definition, but it is interesting (and mildly irritating) how the current value for $\mu_0$ disagrees with the old defined value.

The value of $e=1.602\,176\,6208(98)\times10^{−19}\,\textrm{C}$ you mention is an older 2014 CODATA value. (Note: Parentheses uncertainty notation is used for high precision results since it is more concise than $e=1.602\,176\,6208\times10^{−19} \pm 0.000\,000\,0098\,\textrm{C} \times10^{−19}$.) As @d_b notes in their comment, the last CODATA (2017) fitted value based on experimental measurements was $e=1.602\,176\,6341(83)\times10^{−19}\,\textrm{C}$, so the defined value of $1.602\,176\,634\times10^{−19}\,\textrm{C}$ just dropped the last insignificant digit.

At the time, this did have the effect of slightly shifting $\mu_0$ from its old defined value: $$\frac{\mu_0}{4 \pi \times 10^{-7} \textrm{N/A}^2} = 1 + 2.0(2.3) \times 10^{-10}\qquad \textrm{(CODATA 2017)}$$

The first subsequent CODATA (2018) adjustment of the constants, however, created a more significant shift of half a part per billion: $$\frac{\mu_0}{4 \pi \times 10^{-7} \textrm{N/A}^2} = 1 + 5.5(1.5) \times 10^{-10}\qquad\textrm{(CODATA 2018)}$$

This is just a consequence of improved determinations of the fine-structure constant $\alpha$. Since $\mu_0$ is now calculated from $\mu_0 = {4\pi \alpha \hbar}/{e^2 c} $, and $\hbar$, $e$, and $c$ are all now defined, the value $\mu_0$ changes with $\alpha$. The dimensionless fine-structure constant parameterizes the current strength of electromagnetism in our universe and its value can never be defined, only measured.

Notes: Comparing uncertainties before and after the 2019 SI redefinition can be confusing.

  • Before 2019, $h$ and $e$ were experimentally determined and $\epsilon_0$ was defined. The CODATA 2017 uncertainty in $e=\sqrt{2 h \alpha \epsilon_0 c}$ was dominated by the uncertainty in the Planck constant $h=6.626070150(69)\times10^{-34}\,\textrm{J s}$, not the uncertainty in the inverse fine-structure constant $1/\alpha=137.035999139(31)$ that was unchanged from the CODATA 2014 value. The fractional uncertainties in $h$ and $\alpha$ were respectively $104\times10^{-10}$ and $2.3\times10^{-10}$. Once $h$ and $e$ were defined with zero uncertainty in 2019, $\mu_0$ then had the same fractional uncertainty ($2.3\times10^{-10}$) as $\alpha$.

  • Because of new measurements, the CODATA 2018 value for the fine-structure constant is $1/\alpha=137.035999084(21)$, which has a fractional uncertainty of $1.5\times10^{-10}$, which also becomes the current fractional uncertainty of $\mu_0$.

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    $\begingroup$ Excellent answer, thanks. For readers who, like I do, find it easier to remember words than numbers, the 2018 value is larger than the previous definition by "half a part per billion." $\endgroup$
    – rob
    Oct 7, 2022 at 17:11
  • $\begingroup$ So the summary I take away is that there is no funny business and everything is as anyone would have expected. The only issue is that the OP had an incorrect value for $e$ prior to the change. From the comments it sounds like they got this value from the French Wikipedia page for elementary charge. Indeed, as of Oct 8, 2022, the French Wikipedia page reports the 2014 CODATA value as the last measured value of $e$, missing the 2017 update on which the redefinition was based. Someone who speaks French should clean up that Wikipedia page. $\endgroup$
    – Jagerber48
    Oct 8, 2022 at 17:53
  • $\begingroup$ Well, this is quite interesting. There is a point that is still quite unclear : if the relative uncertainty on the electron charge was, as you write it, 1.6021766341(83) just before they shifted to fixing the charge and letting $\mu_0$ be the secondary quantity, then the relative uncertainty on $\mu_0$ should have be the double of the relative uncertainty on $e$ so twice $5\times 10^{-9}$ or $10^{-8}$ and not $2.3\times 10^{-10}$. The shift itself would have been 80 times less, more or less compatible with $2\times 10^{-10}$ but the uncertainty much higher. $\endgroup$
    – Alfred
    Oct 8, 2022 at 22:06
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    $\begingroup$ And then let $e$ depend on the measurement of $\alpha$ ?The remaining uncertainty was entirely related to that of the measurement of $\alpha$, it did not make any difference at that point which was basic and which was accessory. I see what happened, but I still think the last step, switching from $\mu_0$ to $e$ was absurd, while on the contrary defining the mass unit from $h$ made sense rather than relying on a piece of platinum ! $\endgroup$
    – Alfred
    Oct 11, 2022 at 22:48
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    $\begingroup$ I am considering, next time I walk by the Pavillon de Breteuil, to put a wreath in memory of the late "étalon de masse en platine iridié"... RIP, son of the French Revolution. $\endgroup$
    – Alfred
    Oct 11, 2022 at 23:05

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