2
$\begingroup$

According to this 2018 article in Physics Today, after May 2019 the new SI system fixes, or defines, these seven quantities:

enter image description here

If the dimensionless fine structure constant $\frac{e^2}{4\pi \epsilon_0 hc}$ has a definite value, then $\epsilon_0$ has a fixed value.

But since $$c=\frac{1}{\sqrt{\epsilon_0\mu_0}}$$ doesn't this also fix $\mu_0$?

So is $\mu_0$ no longer defined to be exactly $4\pi \times 10^{-7}$ H/m?

$\endgroup$
5
  • $\begingroup$ I shd probably see this, but since I do not see epsilon_0 in that list: how do these 7 quantities fix mu_0? $\endgroup$
    – lalala
    Commented Jul 31, 2021 at 15:08
  • 1
    $\begingroup$ @lalala good point, the question has been edited to make that clear $\endgroup$ Commented Jul 31, 2021 at 17:36
  • $\begingroup$ Thanks. This poses the interesting question if the fine structure constant actually has (in the sense of S. I.) a definite value. $\endgroup$
    – lalala
    Commented Jul 31, 2021 at 18:33
  • 1
    $\begingroup$ @lalala In the pre-2019 SI, $\mu_0$ and $\epsilon_0$ were defined, but the definition of the ampere (whence $e$ and $\alpha$) was tied to the International Prototype Kilogram. In the new SI, electromagnetism has one experimentally-determined parameter, and it's fashionable to choose $\alpha$ rather than $\mu_0$ or $\epsilon_0$ (see e.g. here for a reason why). $\endgroup$
    – rob
    Commented Jul 31, 2021 at 18:59
  • $\begingroup$ @lalala, that seems right from Rob and another way to look at it is, if $\mu_0$ were exactly $4\pi...$ and $c$ is fixed, then from the speed of light formula $\epsilon_0$ would be fixed, then everything in the fine structure constant $\alpha$ would be fixed, leaving no room for measuring anything!? $\endgroup$ Commented Jul 31, 2021 at 19:10

3 Answers 3

5
$\begingroup$

The organization which defines the SI is the Bureau Internationale de Poids et Mesures (BIPM). The current version of the SI brochure contains the definitive answer to your question (page 132):

The previous definition of the ampere was based on the force between two current carrying conductors and had the effect of fixing the value of the vacuum magnetic permeability $μ_0$ (also known as the magnetic constant) to be exactly $\rm 4π × 10^{−7}\, H\, m^{−1} = 4π × 10^{−7}\, N\, A^{−2}$, where H and N denote the coherent derived units henry and newton, respectively. The new definition of the ampere fixes the value of $e$ instead of $μ_0$. As a result, $μ_0$ must be determined experimentally.

It also follows that since the vacuum electric permittivity $ε_0$ (also known as the electric constant), the characteristic impedance of vacuum $Z_0$, and the admittance of vacuum $Y_0$ are equal to $1/μ_0c^2$, $μ_0c$, and $1/μ_0c$, respectively, the values of $ε_0$, $Z_0$, and $Y_0$ must now also be determined experimentally, and are affected by the same relative standard uncertainty as $μ_0$ since $c$ is exactly known. The product $ε_0μ_0 = 1/c^2$ and quotient $Z_0/μ_0 = c$ remain exact. At the time of adopting the present definition of the ampere, $μ_0$ was equal to $\rm 4π × 10^{−7}\, H/m$ with a relative standard uncertainty of $2.3 × 10^{−10}$.

(Minor comment: somehow copy-pasting has converted all of the $\epsilon$ and $\mu$ into an upright typeface.)

Citing the 2018 CODATA recommendations, the NIST Constants Database gives for $\mu_0$,

$$ \begin{align} \mu_0 &= 1.256\,637\,062\,12(19)\times10^{-6} \rm \, N\, A^{-2} \\ \frac{\mu_0}{4\pi\times10^{-7}\rm\,N\,A^{-2}} &= 1 + (540\pm150)\times10^{-12} \tag{NIST} \end{align} $$

Note that the BIPM text basically says

$$ \frac{\mu_0}{4\pi\cdots} = 1 + (0 \pm 230)\times10^{-12} \tag{BIPM} $$

so the BIPM and NIST/CODATA results are really only about two standard deviations apart.

The 2018 CODATA analysis, which is basically an enormous least-squares fit to the entire fundamental-constants literature, became available on 20 May 2019, which was also the effective date of the new SI. There is a long journal article describing the analysis leading to each set of CODATA recommended values, some of which are open-access. It looks like this 2017 analysis was the basis of the BIPM's SI brochure quoted above, but the 2018 values include literature through part or all of 2018. The 2014 paper (pdf link) contains a much more complete description of the method.

Another answer here (in its v1) quotes from Wikipedia a slightly different value $$ \frac{\mu_0}{4\pi\cdots}-1 \overset{?}{=} (820\pm200)\times10^{-12} \tag{2018} $$ This value is found by using a 2018 result for the fine-structure constant and the definition $\mu_0 = 4\pi\alpha\hbar/e^2c$ using the new SI values. This result was published before the CODATA-2018 cutoff, and is therefore already included, averaged with all other world data, in the NIST result quoted above. People who aren't in the fundamental-constants business should probably use the NIST result.

There is a 2020 measurement of the fine-structure constant which reduces the relative uncertainty on $\alpha$ (and therefore on $\mu_0$) from $150\times10^{-12}$ to $81\times10^{-12}$. However the new result "differs by more than 5 standard deviations from the best available result [2018] from caesium recoil measurements," giving

$$ \frac{\mu_0}{4\pi\cdots}-1 \overset{?}{=} (-345\pm81)\times10^{-12} \tag{2020} $$

In general, when there is this much tension between experimental results, a data-aggregating body like CODATA will end up with inflated uncertainties in the best-fit analysis. You can stay tuned for future measurements and CODATA-2022, but the action seems to be in measurements of $\alpha$ rather than measurements of $\mu_0$.

$\endgroup$
2
$\begingroup$

According to Wikipedia we now have $$ \mu_0= 4\pi × 1.00000000082(20)×10^{-7} {\rm H⋅m^1} $$

$\endgroup$
3
  • $\begingroup$ That's interesting, so within 4 standard deviations it's different to $4\pi$... $\endgroup$ Commented Jul 31, 2021 at 12:52
  • $\begingroup$ So that every book about electromagnetism before 2017 that gives $$ \mu_0 \equiv 4 \pi \times 10^{-7} \mbox{ H m}^{-1}$$ will need to be updated! $\endgroup$
    – jim
    Commented Jul 31, 2021 at 15:39
  • 1
    $\begingroup$ @JohnHunter Not really, because the "$4\pi$" had its own uncertainty. See my answer for a clarification about what this value actually means; the Wikipedia page misrepresented it a little. $\endgroup$
    – rob
    Commented Jul 31, 2021 at 17:23
1
$\begingroup$

There is a comment at the Physics Today web site asking the same question (from P. Nelson - don't know who that is). Here's the response from Byron (another unknown commenter):

c and mu_0 are defined (and thus are exact) but epsilon_0 is derived. Since it is computed from only exact quantities, epsilon_0 is also exact. Similarly, the Ampere is still exact but has a slightly different value. The Newton is derivable from the Ampere and mu_0 on the electromagnetic side and from the kg (i.e. h), the m, and the s on the mechanical side.

(Byron's comment doesn't use MathJax, so I didn't convert the symbols).

According to a 2017 article at NIST, however, the permeability of free space is no longer a defined quantity, but a measured value, and not equal to 1 in the cgs system.

Also, this result from NIST and CODATA 2018 which has a relative uncertainty of $1.5\times 10^{-10}$ for $\mu_0$, so it's not exact. Apparently, commenter Byron on the Physics Today site is incorrect.

$\mu_0$, indeed has become a non-exact quantity.

$\endgroup$
3
  • $\begingroup$ Thanks, although now there seem different opinions, this one says it's exact, mikes answer says it has an error $\endgroup$ Commented Jul 31, 2021 at 12:59
  • $\begingroup$ who are you gonna trust, Wikipedia or Physics Today? What does the NIST site say, or do they? $\endgroup$
    – Bill N
    Commented Jul 31, 2021 at 13:00
  • $\begingroup$ Thanks, this seems right and was upvoted. $\endgroup$ Commented Aug 1, 2021 at 13:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.