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Equation 1 · The Last Artefact: Redefining the Kilogram

What does this equation mean?

1 kg=(h6.626 070 15×10−34) m−2 s1\ \mathrm{kg} = \left(\frac{h}{6.626\,070\,15 \times 10^{-34}}\right)\ \mathrm{m^{-2}\,s}

Read the formula alongside the article passage below. Each part has a deeper page with its role in the equation, the supporting passage and nearby citations.

Start withh
Divide by6.62607015 × 10^-34
This relates to1 kg
How to read the two sides of this formula. Follow the article passage for the meaning of each quantity.

This equation states an equality: the expressions on both sides have the same value under the article’s assumptions. Read the equation part by part below; each part has a contextual explanation and a link to its mathematical background.

Read it piece by piece

hh

Symbol h

h occurs above the fraction bar. The numerator is divided by the entire denominator below it.

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=

=

The expressions on both sides represent the same quantity under the stated assumptions.

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fraction

fraction

Divide the expression above the line by the one below it.

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multiplication

multiplication

Multiply the quantities on either side.

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superscript

superscript

A raised number can be a power. When it is a label or bound, it selects a case or the upper limit of a sum; the formula’s structure distinguishes these uses.

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6.626 070 15×10−346.626\,070\,15 \times 10^{-34}

Denominator: 6.62607015 × 10^-34

The complete quantity below the fraction bar; it must be nonzero for this division.

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How to interpret it

With a fixed numerator, increasing a nonzero denominator reduces the fraction. Read it with the definitions, units, and assumptions supplied by the article.

What the article says around this equation

Inverting that gives the unit explicitly in terms of three fixed constants: 1 kg=(h6.626 070 15×10−34) m−2 s1\ \mathrm{kg} = \left(\frac{h}{6.626\,070\,15 \times 10^{-34}}\right)\ \mathrm{m^{-2}\,s}. which the SI Brochure also writes as approximately 1.475 5214 × 10⁴⁰ times the quantity h ΔνCs divided by c squared [ 1 ] . Notice what the definition does not contain: any apparatus, any material, any temperature, any place. That is the point. “The use of a constant to define a unit disconnects definition from realization”, which “offers the possibility that completely different or new and superior practical realizations can be developed, as technologies evolve, without the need to change the definition” [ 1 ] .

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Sources cited in the surrounding passage

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