10
$\begingroup$

So apparently in May, all of the SI base units were redefined to be relative to the Planck Constant $h$, instead of relying on physical objects like the Kilogram Prototype in a Paris Basement. Planck's constant was defined as $$6.62607015\times10^{−34}\text{Js}$$

(It's not a measured value anymore. It is the value)

My question is: Why was this exact number chosen? In 2010, the measured value of $h$ was $$6.626\mathbf{\color{blue}{06957}}\times10^{−34}\text{Js}$$

So why did they choose this arbitrary value of $$6.626\mathbf{\color{blue}{07015}}\times10^{−34}$$

instead of something more exact like

$$6.626\mathbf{\color{blue}{07}}\times10^{-34}$$

That would have been within the margin of error so it could have worked just fine.

$\endgroup$
  • $\begingroup$ The number used depends on the system of measurement involved. The number of digits used indicates how precisely that number is known. $\endgroup$ – David White Nov 20 '19 at 22:36
  • $\begingroup$ @DavidWhite h is a fixed value now, not a measured constant. $\endgroup$ – SurpriseDog Nov 20 '19 at 22:37
  • $\begingroup$ So is the speed of light, but physicists didn't arbitrarily define its value as $3 x 10^8 m/s$. $\endgroup$ – David White Nov 20 '19 at 22:41
  • $\begingroup$ @DavidWhite I mean they could have done that with c instead of h and then everything would be relative to c. Right now c is defined by h which is defined as a number $\endgroup$ – SurpriseDog Nov 20 '19 at 22:43
  • 1
    $\begingroup$ Dig up the meeting minutes? If I had been king during that decision, I would have chosen the value of the Planck constant that would cause the least trouble. "Least trouble", to me, would be the number of instruments that would have to be recalibrated, and the precision to which the other fundamental constants would also have to be defined. $\endgroup$ – TimWescott Nov 20 '19 at 22:47
12
$\begingroup$

There are many, many instruments which are calibrated using the old definition of the kilogramme - the International Prototype Kilogramme (IPK) made of a platinum-iridium alloy.
So one kilogramme measured using the new definition had to be as close as possible to one kilogramme using old definition so as not to have to recalibrate all instruments which relied on the old definition of the kilogramme.

Using the old definition of the kilogramme (IPK) the numerical value of Planck’s constant was measured as accurately as possible using the Kibble (watt) balance and the X-ray crystal density method.

The two values that you quoted $6.62606957 \times 10^{−34}\, \rm kg\,m^2s^{-1}$ and later $ 6.62607015 \times 10^{−34}\, \rm kg\,m^2s^{-1}$ were the results of such measurements of Planck's constant.

As of 20 May 2019 the determination/definition was turned on its head with the value of Planck’s constant defined as $ 6.62607015 \times 10^{−34}\, \rm kg\,m^2s^{-1}$ and the IPK (made of a platinum-iridium alloy) having a measured value of one kilogramme to within one part in $10$ billion.

On page 131 of the BIPM brochure on the SI system of units it states:

The number chosen for the numerical value of the Planck constant in this definition is such that at the time of its adoption, the kilogram was equal to the mass of the international prototype, m(K) = 1 kg, with a relative standard uncertainty of $1 \times 10^{-8}$, which was the standard uncertainty of the combined best estimates of the value of the Planck constant at that time.

In future the new definition of one kilogramme via the defined exact value of Planck's constant will enable measurements to be made to see by how much the masses of the IPK and its daughters change with time.

—-

why did they choose this arbitrary value of 6.62607015×10−34.

This value was chosen to make one kilogramme using the old definition (IPK) as close as possible to one kilogramme using the new definition (via Planck's constant).

instead of something more exact like 6.62607x10-34.

This would have required the recalibration of many, many (accurate) instruments.

$\endgroup$
8
$\begingroup$

It is an updated experimental value that matches measurements by Kibble balances and by counting atoms in silicon spheres to determine Avogadro’s number. The 2010 measurement was presumably consistent with only one metrological approach. The new exact value is consistent with both.

See this NIST page, which says

The Kibble balance and Avogadro project measurements are not so much competing with as complementing each other to define the kilogram. Measurements from both experiments were used to determine the final value of $h$ for the redefined SI. That final value is $6.626070150\times 10^{-34}\text{ kg}\cdot\text{m}^2/\text{s}$.

$\endgroup$
1
$\begingroup$

Similarly you can ask about why speed of light was defined to be $299792458 ms^{-1}$ and not something else? Simple answer is that doing such would change the unit of length $m$. How? Simply if you make it (say) $1000000000ms^{-1}$ then the new meter would be around $3.33564$ times of the previous meter! This means that if you previously bought $1m$ of cloth in say $ 10 \$ $ then for the same price you can buy now $3.336m$ of cloth! No, in such case price of commodities would too rise. So you can see that most of the things may change (in one way or the other). But you may say "Hey! If most of the things are going to increase by the same factor then what's the problem? " . The reason is that even though professionals (like physicists, economists, etc.) would adapt to the situation in (say) I 5-7 days (though the recalibration of the instruments would take some time) but it would take months for a common man to adapt. Also it may have some drastic effects too (like people would start deceiving those who are adapted to previous units and unfamiliar / unknown to the new ones,etc.) . Otherwise physically it won't have any difference (just quantities would change their magnitudes). In a similar fashion you can think of Plank's constant.

So what scientists do is that measure the thing as accurately as possible (in previous units) and then set it as the new value so that much of the things don't change.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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