Updating SI Base Units

Now all defined as a function of universal physical constants

https://www.bipm.org/en/measurement-units/si-promotion
If you are not familiar with the International System of Units (SI), I recommend you to check out this page of the Bureau International des Poids et Mesures (BIPM). You can also take a look at this fantastic collection of school posters from the NPL.
If you want more information, @fqsaja1 recommends that we take a look at David Newell’s excellent article published in Physics Today magazine.

The International System of Units (SI) underwent a revision in 2018, with the redefinition of four of its seven base units, namely the kilogram (kg), the ampere (A), the kelvin (K) and the mole (mol).

This update was made to define all the units as a function of universal physical constants, something that had already been achieved previously for the second (1967) and the metre (1983). The following table summarises the relationship between each unit and the universal constant on which it is based:

Unit (symbol) Universal constant (symbol)
Second (s) Transition frequency of the caesium 133 atom ($\Delta\nu_\mathrm{Cs}$)
Metre (m) Speed of light in vacuum ($c$)
Kilogram (kg) Planck constant ($h$)
Ampere (A) Elementary charge ($e$)
Kelvin (K) Boltzmann constant ($k$)
Mole (mol) Avogadro constant ($N_\mathrm A$)
Candela (cd) Luminous efficacy of radiation of frequency $540\times 10^{12}\thinspace\mathrm{Hz}$ ($K_\mathrm{cd}$)

Current Definitions of SI Base Units

Second (s)

$$ 1\thinspace \mathrm s = \frac{9192631770}{\Delta\nu_\mathrm{Cs}}, $$

where $\Delta\nu_\mathrm{Cs} = 9192631770\thinspace\mathrm{Hz}$ is the unperturbed ground state hyperfine transition frequency of the caesium 133 atom.

The second is therefore the duration of 9192631770 periods of radiation corresponding to the transition between the two hyperfine levels of the unperturbed ground state of the caesium 133 atom.

Metre (m)

$$ 1\thinspace\mathrm m = \frac{9192631770}{299792458}\frac{c}{\Delta\nu_\mathrm{Cs}}, $$

where $c = 299792458\thinspace\mathrm{m\cdot s^{-1}}$ is the speed of light in vacuum.

The metre is therefore the length of the path travelled by light in vacuum during a time interval of 1/299792458 of a second.

Kilogram (kg)

$$ 1\thinspace\mathrm{kg} = \frac{(299792458)^2}{(6.62607015\times 10^{-34})(9192631770)}\frac{h\Delta\nu_\mathrm{Cs}}{c^2}, $$

where $h = 6.62607015\times 10^{-34}\thinspace\mathrm{kg\cdot m^2\cdot s^{-1}}$ is the Planck constant.

The kilogram is therefore defined as a function of Planck’s constant value, $h$.

Ampere (A)

$$ 1\thinspace\mathrm{A} = \left(\frac{e}{1.602176634\times 10^{-19}}\right)\thinspace\mathrm{s^{-1}}, $$

where $e = 1.602176634\times 10^{-19}\thinspace\mathrm{A\cdot s}$ is the elementary charge.

The ampere is therefore the electrical current corresponding to the flux of $1/(1.602176634\times 10^{-19}) = 6.241509074\times 10^{18}$ elementary charges per second.

Kelvin (K)

$$ 1\thinspace\mathrm{K} = \frac{1.380649\times 10^{-23}}{(6.62607015\times 10^{-34})(9192631770)}\frac{h\Delta\nu_\mathrm{Cs}}{k}, $$

where $k = 1.380649\times 10^{-23}\thinspace\mathrm{kg\cdot m^2\cdot s^{-2}\cdot K^{-1}}$ is the Boltzmann constant.

The kelvin is therefore equal to the thermodynamic temperature variation that results in a thermal energy variation $kT$ of $1.380649\times 10^{-23}\thinspace\mathrm J$.

Mole (mol)

$$ 1\thinspace\mathrm{mol} = \frac{6.02214076\times 10^{23}}{N_\mathrm A}, $$

where $N_\mathrm A = 6.02214076\times 10^{23}\thinspace\mathrm{mol^{-1}}$ is the Avogadro constant.

The mole is therefore the amount of substance in a system that contains $6.02214076\times 10^{23}$ specified elementary entities.

Candela (cd)

$$ 1\thinspace\mathrm{cd} = \frac{1}{(6.62607015\times 10^{-34})(9192631770)^2 683}h(\Delta\nu_\mathrm{Cs})^2 K_\mathrm{cd}, $$

where $K_\mathrm{cd} = 683\thinspace\mathrm{cd\cdot sr\cdot kg^{-1}\cdot m^{-2}\cdot s^3}$ is the luminous efficacy of monochromatic radiation of frequency $540\times 10^{12}\thinspace\mathrm{Hz}$.

The candela is therefore the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency $540\times 10^{12}\thinspace\mathrm{Hz}$ and has a radiant intensity in that direction of $(1/683)\thinspace\mathrm{W/sr}$.
Rodrigo Alcaraz de la Osa
Rodrigo Alcaraz de la Osa
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I have a PhD in Physics and I teach Physics and Chemistry at IES PeΓ±acastillo in Cantabria (Spain).

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