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Equation 5 · The Last Artefact: Redefining the Kilogram

What does this equation mean?

N=8 Vsa3,ms=N m(28Si)=h N(m(28Si)h)N = \frac{8\,V_{\mathrm s}}{a^{3}}, \qquad m_{\mathrm s} = N\, m({}^{28}\mathrm{Si}) = h\,N \left( \frac{m({}^{28}\mathrm{Si})}{h} \right)

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 with8V_mathrm s
Divide bya^3
This relates toN
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.

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NN

Symbol N

N is part of the quantity the equation computes from the expression on the right.

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VsV_{\mathrm s}

Symbol V_mathrm s

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

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a3a^{3}

Symbol a^3

a3a^3 occurs below the fraction bar. The quantity above the bar is divided by this expression; zero is excluded as a denominator.

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msm_{\mathrm s}

Symbol m_mathrm s

mmm_mathrm s is an input to the expression that computes the quantity on the left.

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mm

Symbol m

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

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hh

Symbol h

a constant of nature known to high accuracy, so with h fixed the sphere becomes a primary mass standard [ 4 ].

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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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subscript

subscript

The lower label selects a particular version, component, or indexed member of the quantity. For example, x₀ and xₜ can be values at different positions.

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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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8 Vs8\,V_{\mathrm s}

Numerator: 8V_mathrm s

The complete quantity above the fraction bar.

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m(28Si)m({}^{28}\mathrm{Si})

Numerator: m(^28Si)

The complete quantity above the fraction bar.

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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

The second primary method shares almost nothing with the first except its answer. The X-ray crystal density route determines the mass of a nearly perfect single-crystal silicon sphere by counting the atoms in it. With eight atoms per cubic unit cell of lattice parameter a, the atom count follows from the macroscopic sphere volume, and the sphere mass follows from the count [ 4 ] : N=8 Vsa3,ms=N m(28Si)=h N(m(28Si)h)N = \frac{8\,V_{\mathrm s}}{a^{3}}, \qquad m_{\mathrm s} = N\, m({}^{28}\mathrm{Si}) = h\,N \left( \frac{m({}^{28}\mathrm{Si})}{h} \right). The ratio of the silicon-28 atomic mass to h is a constant of nature known to high accuracy, so with h fixed the sphere becomes a primary mass standard [ 4 ] . The mise en pratique notes that the second equality is not exact — the total binding energy of the crystal reduces the right-hand side…
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The second primary method shares almost nothing with the first except its answer. The X-ray crystal density route determines the mass of a nearly perfect single-crystal silicon sphere by counting the atoms in it. With eight atoms per cubic unit cell of lattice parameter a, the atom count follows from the macroscopic sphere volume, and the sphere mass follows from the count [ 4 ] : N=8 Vsa3,ms=N m(28Si)=h N(m(28Si)h)N = \frac{8\,V_{\mathrm s}}{a^{3}}, \qquad m_{\mathrm s} = N\, m({}^{28}\mathrm{Si}) = h\,N \left( \frac{m({}^{28}\mathrm{Si})}{h} \right). The ratio of the silicon-28 atomic mass to h is a constant of nature known to high accuracy, so with h fixed the sphere becomes a primary mass standard [ 4 ] . The mise en pratique notes that the second equality is not exact — the total binding energy of the crystal reduces the right-hand side by about two parts in 10¹⁰ — and that this is ignored as negligible against present experimental uncertainties [ 4 ] . A definition that lets you write down the correction you are choosing to neglect is doing its job.

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