Redefinition of everything on May 20th, 2019 [closed]

A couple of issues:

1. So after May 20th, 2019, what exactly will be the defined value of $$\hbar$$?
2. What will be the defined number of elementary charges in a Coulomb?
3. Then $$\mu_0$$ and $$\epsilon_0$$ will not be defined values even though their product will be defined exactly as $$c^{-2}$$, right? What will be their values and the standard error (the two little digits in parenths that come after the last digit)?
4. Because $$G$$ is much sloppier, that will remain unchanged (including the standard error), right?

and

1. This could have been asked anytime since 1983, but given the definition of the second and the meter (which is unchanged on May 20), why don't they just define the meter as 9192631770/299792458 = 30.663318988498371 wavelengths "of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium-133 atom" at 0K in vacuo? How does a modern meter stick work? If one were to do a critical physical measurement and needed a perfect reference of length, would not they be comparing any measured length to this specific EM radiation in vacuo to have the most solid direct reference length?

So to have a tight reference of time, length, and mass, an experimenter would need an atomic clock with $${}^{133}$$Cs radiation, a vacuum chamber, to get both time and length, and a Kibble balance to get mass, right?

closed as too broad by David Z♦May 4 at 7:47

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To address some of the remaining nontrivial issues:

1. What will be the defined number of elementary charges in a Coulomb?

The final set of constants uses $$e = 1.602\,176\,634\times 10^{-19}$$ (exactly). This is not the inverse of an integer, so the Coulomb cannot be formed as an assembly of an integer number of elementary charges.

The closest you can get is $$\left\lfloor \frac{1\:\mathrm{C}}{e} \right\rfloor = 6\,241\,509\,074\,460\,762\,607,$$ but this number is essentially useless. (It is arguably of some use as a paedagogical construct in introductory courses, but to be frank, if you want to use it that way, you need to think very carefully about what misconceptions you might be seeding that will take substantial work to roll back further down the line.)

I think that the Coulomb should be simply defined as the charge contained in a collection of 6241509074460762608 elementary charges. Together 6241509074460762608 electrons posses exactly -1 Coulomb of charge

No, this is incorrect. Together, 6241509074460762608 get pretty close to 1 coulomb of charge, but the relationship is not exact.

That is simply how they should define it, if they're gonna fix $$e$$ to a constant.

It is obviously your prerogative to hold onto your opinions, even if they are wrong. In this instance, I would argue that an opinion emitted from a position that's informed only by the trappings of the old way of doing things, and without trying to find out why the professional metrologists are doing things the way they are, is essentially useless.

The constant $$1\:\mathrm{C}/e$$ (which you want to define as an exact integer) is used approximately none of the time. The constant $$e/1\:\mathrm{C}$$ (which you want to define as an infinite recurring decimal) is used all of the time. Making the used-none-of-the-time constant simpler to the enormous detriment of the used-all-the-time constant makes absolutely no sense.

1. Then $$\mu_0$$ and $$\epsilon_0$$ will not be defined values even though their product will be defined exactly as $$c^{-2}$$, right? What will be their values and the standard error (the two little digits in parenths that come after the last digit)?

The effect of the redefinition on all the other constants in the pantheon is given by this table in the Wikipedia page. The two particular constants you've asked about will take on the values of \begin{align} \epsilon_0 = \frac{e^{2}}{2hc\alpha } \ \text{ and } \ \mu_0 = \frac{2h\alpha }{ce^{2}}, \end{align} where $$e$$, $$h$$ and $$c$$ are exact, and the uncertainty comes in exclusively through $$\alpha$$, the fine-structure constant. The value of the latter is determined by CODATA, and it is measured independently of the realization of the units themselves.

It is probably instructive to expand a bit more on how $$\alpha$$ is measured, to help bring to the surface the ways in which this is not a circular definition; I'll follow this publication and this one. Basically, the idea is to invert the definition of the Rydberg constant, $$R_\infty = \alpha^2 \frac{m_e c}{h}$$ (OK, something close to the definition), to get the fine-structure constant as (the square root of) $$\alpha^2 = \frac{R_\infty \, h}{m_e c},$$ and then use a chain of other measurements to nail down this value to high precision, in the form $$\alpha^2 = \frac{R_\infty}{c}\frac{A_{\mathrm{r}}(e)}{A_{\mathrm{r}}(X)} \frac{h}{m(X)},$$ where:

• The Rydberg constant itself can be measured directly from precision spectroscopy of optical transitions in hydrogen and deuterium. (The Rydberg constant is basically already an inverse wavelength, and it can be tightly tied to the spectra. The use of two different isotopes then eliminates the dependence on the reduced mass.)
• The speed of light $$c$$, of course, has an exact value.
• The electron mass $$m_e$$ can be measured most precisely through the relative atomic mass of the electron, $$A_{\mathrm{r}}(e)$$, and this (as of CODATA-14) is measured by observing the dynamics of electrons and ions in ion traps, coupled with some relatively elaborate atomic physics (see this paper for more details), ideally in reference to a separate atom's relative atomic mass $$A_\mathrm{r}(X)$$.
• Finally, the quantity $$h/m(X)$$ (the Planck constant divided by the mass of the $$X$$ atom used for reference above), can be measured directly through recoil spectroscopy of the optical transitions in the relevant atom $$-$$ basically, by seeing how much the Doppler shift of the photon recoil changes the frequency of transitions where a net momentum is imparted, as compared to (say) crossed-beam two-photon schemes where no net photon is given to the atom.

Overall, the upshot is that we know $$\alpha$$, from independent measurements which are ultimately unit-independent, to a relative precision of about $$10^{-10}$$. This relative uncertainty then gets transferred to $$\epsilon_0$$, $$\mu_0$$, and the rest of the cast.

(This is probably a little bit too much detail, but I hope it's helpful.)

1. Because $$G$$ is much sloppier, that will remain unchanged (including the standard error), right?

Very much so. The relative uncertainty in $$G$$ is of the order of $$10^{-4}$$. The only changes from the shift happen to constants that are currently known to relative uncertainties of the order of $$10^{10}$$ or so.

1. This could have been asked anytime since 1983, but given the definition of the second and the meter (which is unchanged on May 20), why don't they just define the meter as 9192631770/299792458 = 30.663318988498371 wavelengths "of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium-133 atom" at 0K in vacuo?

You've written this as if the equation $$\frac{9192631770}{299792458} = 30.663318988498371$$ were exact. The left-hand side is the ratio of two integers, so the right-hand side is an infinite recurring decimal with a period of about 21 million digits. It does not terminate.

If you meant to put an approximate equality there, $$9192631770/299792458 \approx 30.663318988498371$$, so the left-hand side forms the real definition and the right-hand side is what it's approximately equal to, then you've basically produced a definition which is equivalent to the current one, but much harder to use.

The wavelength of light of a given frequency can be used as a length standard (though if you're going to do that, using a microwave transition instead of an optical one basically means that you've decided to throw away some six significant figures of precision in your measurement), and indeed one such standard was used between 1960 and 1983. The change to the current definition was forced by the fact that timing-based length metrology had left the interferometer-based wavelength standard far, far behind in precision (i.e.: the uncertainty in the distance measurements to the Apollo reflectors on the Moon was completely driven by the uncertainty in the implementation of the meter).

How does a modern meter stick work?

Through a variety of different methods; length metrology is obviously an extremely varied field, and the precise method of choice will depend on exactly what you're measuring and how precise of a measurement you need of how long of a distance. Some current methods do boil down to a comparison of the measured length to a chosen wavelength (generally not the caesium hyperfine-transition wavelength), and others don't. Keeping the definition general, neutral and simple makes it easier to work with all of the methods.

More generally, I'd suggest starting with the Wikipedia page and its references before attempting to throw stones at the existing definitions.

• "The constant 1C/e (which you want to define as an exact integer) is used approximately none of the time." i wonder if that will be the case in the future. even now, with some two slit experiments, they are detecting a single electron. it seems to me that if you can detect a single electron crossing a boundary and you wanted to calibrate a precision ammeter, you would want to, for a reference, calibrate it with a known amount of charge (which is quantized to $e$) passing the boundary, providing that you can count those discrete units of charge. – robert bristow-johnson May 3 at 22:52
• @robertbristow-johnson Ammeters based on single-electron tunnelling are already a thing and they achieve decent precision, though they are still not competitive with devices based on the Josephson and quantum Hall effects. They are essentially only useful for microscopic amounts of charge, and you will never get to charge measurements of the order of a Coulomb. (It's also worth pointing out that to measure 1C/$e$ to a precision where the difference matters, you'd need twice as many significant figures as the existing measurements.) – Emilio Pisanty May 4 at 0:11
• And, in any case, with single-electron devices the question is "I've seen $n$ electrons go through, what's the charge?" and not "I want to get to this artificial quantity charge that divides cleanly into the Coulomb, how many electrons need to pass"; the former benefits from a clean $e/$1C, and it's the latter (which we don't use) that benefits from a clean 1C$/e$. So: no. – Emilio Pisanty May 4 at 0:13
• "No, this is incorrect. Together, 6241509074460762608 get pretty close to 1 coulomb of charge, but the relationship is not exact." He said in the very text you quoted before this "I think that it should be so defined" - in other words, it was a statement of opinion, and thus not "incorrect" (or "correct", either). You realize this soon after, but even so, this bit is still suggesting that he's making a statement of fact about the definition, which is wrong. – The_Sympathizer May 4 at 2:25
• @The_Sympathizer OP's text has a clear equality there. If he wants to re-write the text so that there is no ambiguity, I can re-evaluate that passage. – Emilio Pisanty May 4 at 11:57
1. The new value of $$\hbar$$ will be consistent with the current CODATA value; it's at the BIPM website.

2. Same for $$e$$.

3. From here:

the vacuum magnetic permeability $$\mu_0$$ is equal to $$4\pi × 10^{–7}\rm\, H\, m^{–1}$$ within a relative standard uncertainty equal to that of the recommended value of the fine-structure constant $$\alpha$$ at the time this Resolution was adopted, namely $$2.3 × 10^{–10}$$ and that in the future its value will be determined experimentally,

That also determines $$\epsilon_0$$, since $$\mu_0\epsilon_0 = 1/c^2$$ and $$\alpha\hbar c = e^2/4\pi\epsilon_0.$$

4. Since gravity is so challenging experimentally, it would be an error to try to fix $$G$$. But even if gravitation weren't a mess, $$G$$ is a coupling constant, like $$\alpha$$, rather than (as we spent the twentieth very century verifying) a unit-conversion constant like $$c$$.

Your fifth question is complex enough that you should probably ask it separately.

• rob, why did an admin delete all these comments following this answer? one of the comments by me had important content that was evidently missed by @DavidHammen evidenced by his comment above. – robert bristow-johnson May 3 at 19:19
• @robertbristow-johnson The deleted comments were about the error in v1 of the answer. I think David's remark is still useful on the corrected answer, even if the definition $\alpha\hbar c = e^2/4\pi\epsilon_0$ doesn't appear in the discussion. Or did you mean something different? – rob May 3 at 19:31
• but i had explicit expressions for both $\epsilon_0$ and $\mu_0$ in terms of the measured $\alpha$ and the defined $\hbar$, $c$, and $e$. i think that comment should have survived. and i am not so sure it's particularly considerate to use moderator tools to delete a comment of someone else's unless that comment was abusive or really bad in some other manner. – robert bristow-johnson May 3 at 19:39
• @robertbristow-johnson Comments are supposed to critique and improve answers, and are not intended to be permanent in any case. It is, in fact, one of the essential duties of moderators to clean up comments that are obsolete (or were never needed in the first place). In this case, you pointed out an error in the answer, then the error was fixed. After that, your comment became obsolete. Please do not take the deletion of a comment of yours as any sort of personal attack - it is generally not a judgement of the value of the comment at the time it was written. – ACuriousMind May 3 at 20:32
• @robertbristow-johnson See e.g. this meta question and links therein. If you'd like to continue this discussion about how commenting does and/or should work, I suggest you make a new post on Physics Meta so that the community can be involved. – rob May 3 at 21:59

I just wanna thank rob for pointing me to a root source. One additional click in gets us to:

... effective from 20 May 2019, the International System of Units, the SI, is the system of units in which:

• the unperturbed ground state hyperfine transition frequency of the caesium 133 atom $$\Delta \nu_\mathrm{Cs}$$ is 9 192 631 770 Hz,

• the speed of light in vacuum $$c$$ is 299 792 458 m/s,

• the Planck constant $$h$$ is 6.626 070 15 × 10-34 J s,

• the elementary charge $$e$$ is 1.602 176 634 × 10-19 C,

• the Boltzmann constant $$k$$ is 1.380 649 × 10-23 J/K,

• the Avogadro constant $$N_\mathrm{A}$$ is 6.022 140 76 × 1023 mol-1

• the luminous efficacy of monochromatic radiation of frequency 540 × 1012 Hz, $$K_\mathrm{cd}$$, is 683 lm/W,

That takes care of questions 1 and 2 (sorta, we need to reciprocate and express $$\frac{1}{e}$$ as a fixed and defined integer) and maybe 4. Question 3 is still not satisfied and Question 5 should be bumped to another physics.se post, I am told.

It seems to me that the answer to Q2 is 6 241 509 074 460 763 000, but I am dubious of the last 3 zeros and I think the integer that, when multiplied by 1602176634, most closely approximates 1028 is the answer to Q2. I don't have longer than double in MATLAB, so I gotta think a little how I will come up with that integer.

UPDATE, so from the kind assistance of @DavidHammen, I think that the Coulomb should be simply defined as the charge contained in a collection of 6241509074460762608 elementary charges. Together 6241509074460762608 electrons posses exactly -1 Coulomb of charge (keeping with the sign convention). In one second, the period required for radiation from the unperturbed ground state hyperfine transition of the caesium 133 atom to oscillate 9192631770 times, 6241509074460762608 electrons pass a boundary in a wire conducting exactly 1 Ampere of current. That is simply how they should define it, if they're gonna fix $$e$$ to a constant.