How gravity field shortened our metre

Abstract

We became interested in investigating the origin of the length of the metre as a basis of the metric system after reading Ken Alder’s 2002 book “The measure of all things”, where we learnt that the metre we use in all our length measurements is about 0.2 millimetres shorter than it was supposed to be. That does not sound like being too far off, but when one realizes that the metre was supposed to be one ten-millionth part of the Earth quadrant, the 0.2 mm translates to a 2 km error in the Earth quadrant. Even with the instruments available at the end of the eighteenth century, when these geodetic measurements were carried out by a couple of French geodesists, the error of 2 km in the Earth quadrant was a little too large to expect. Particularly so as both these gentlemen, Méchain and Delambre, had a reputation of being excellent and very careful observers. When we read in the two geodesists’ 1806 Opus Magnum, Base du système métrique décimal ou mesure de l’arc du méridian, that Legendre, when he compared their 1806 results with the results of their predecessors and contemporaries, said: “I’d suspect the irregularities of the local [gravity] attraction on the plumb-line [to be at play here]”, our interest was definitely aroused. Indeed, the analyses described in this paper confirm Legendre’s suspicion: while the error due to the neglect of gravity field is responsible for about 95% of the total error, the error in the meridian length measurements amounts to less than 2% of the total error and the error due to a wrong assumption of the Earth’s shape contributed about 3% of the total error. If the careful work of Méchain and Delambre was the only source of error, the metre nowadays would be too long only by less than 4 μm instead of being short by 197 μm. The bulk of the difference is made by not taking the effect of gravity field into account; that they could not have done because the needed gravity field features were not yet known.

Appendix: Derivations

The B in Eq. (3) is an “elliptical integral” multiplied by (1 − e²). Such an integral is solvable only through the development into a series of the sub-integral function $\left( {1 - e^{2 } \sin^{2} \varphi } \right)^{{\frac{ - 3}{2}}}$. The sub-integral function is of the type $\left( {1 + \alpha } \right)^{p}$ where $\alpha = - e^{2 } \sin^{2} \varphi$ is quite small and p =  − 3/2 (not an integer). Such functions can be developed into a series by the use of binomial theorem (Gullberg 1997) which reads

$$ \left( {1 + \alpha } \right)^{p} = \mathop \sum \limits_{n = 0}^{\infty } \left( {\begin{array}{*{20}c} p \\ n \\ \end{array} } \right)\alpha^{n} , $$

(A1)

where

$$ \left( {\begin{array}{*{20}c} p \\ n \\ \end{array} } \right) = \frac{{p\left( {p - 1} \right)\left( {p - 2} \right) \ldots \left( {p - n + 1} \right)}}{n!}. $$

(A2)

Thus, we can write, to a sufficient accuracy:

$$ \left( {1 + \alpha } \right)^{p} = 1 + \frac{3}{2} \alpha + \frac{15}{8} \alpha^{2} + \frac{105}{48 } \alpha^{3} + \cdots . $$

(A3)

Substituting this back into Eq. (3) we get

$$ \begin{aligned} B\left( {e;\varphi_{1} , \varphi_{2} } \right) & = \left( {1 - e^{2} } \right)\mathop \smallint \limits_{{\varphi_{1} }}^{{\varphi_{2} }} \left( {1 - e^{2 } \sin^{2} \varphi } \right)^{{\frac{ - 3}{2}}} {\text{d}}\varphi \\ & = \left( {1 - e^{2} } \right)\mathop \smallint \limits_{{\varphi_{1} }}^{{\varphi_{2} }} \left( {1 + \frac{3}{2}e^{2} \sin^{2} \varphi + \frac{15}{8}e^{4} \sin^{4} \varphi + \frac{105}{48}e^{6} \sin^{6} \varphi - \cdots } \right){\text{d}}\varphi . \\ \end{aligned} $$

(A4)

The individual integrals in this series can be now evaluated. We get (Spiegel 1968):

$$ \smallint \sin^{2} \varphi {\text{d}}\varphi = \frac{\varphi }{2} - \frac{1}{4}\sin 2\varphi , $$

(A5)

$$ \smallint \sin^{4} \varphi {\text{d}}\varphi = \frac{3}{8}\varphi - \frac{1}{4}\sin 2\varphi + \frac{1}{32}\sin 4\varphi , $$

(A6)

$$ \smallint \sin^{6} \varphi {\text{d}}\varphi = - \frac{1}{6}\sin^{5} \varphi \cos \varphi + \frac{15}{48}\varphi - \frac{5}{24}\sin 2\varphi + \frac{5}{192}\sin 4\varphi . $$

(A7)

After substitution back in Eq. (A4), these terms become quite complicated functions of φ₁ and φ₂ and we shall not show them here.

On the other hand, if we seek the length of the whole elliptical quadrant, and thus integrate within the limits of 0 and π/2, all terms containing sin2φ and sin 4φ will disappear. The term $\sin^{5} \varphi \cos \varphi$ will also disappear as it equals to 0 for the two limits. Substituting Eqs. (A5) to (A7) into Eq. (A4), we obtain

$$ \mathop \smallint \limits_{0}^{\pi /2} \left( {1 - e^{2 } \sin^{2} \varphi } \right)^{{\frac{ - 3}{2}}} {\text{d}}\varphi = \left( {\frac{\pi }{2} + \frac{3}{2}e^{2} \frac{1}{2}\pi /2 + \frac{15}{8}e^{4} \frac{3}{8}\pi /2 + \frac{105}{48}e^{6} \frac{15}{48}\pi /2 + \cdots } \right), $$

(A8)

which can be rewritten as

$$ \mathop \smallint \limits_{0}^{\pi /2} \left( {1 - e^{2 } \sin^{2} \varphi } \right)^{{\frac{ - 3}{2}}} {\text{d}}\varphi = \frac{\pi }{2}\left( {1 + \frac{3}{4}e^{2} + \frac{45}{64}e^{4} + \frac{1575}{2304} e^{6} + \cdots } \right). $$

(A9)

Taking care of the multiplication factor $\left( {1 - e^{2} } \right)$ in front of the integral, we get the full expression for B(e;0,π/2) = C(e):

$$ c\left( e \right) = \frac{\pi }{2}\left( {1 - e^{2} } \right)\left( {1 + \frac{3}{4}e^{2} + \frac{45}{64}e^{4} + \frac{1575}{2304}e^{6} + \cdots } \right) = \frac{\pi }{2}\left( {1 - \frac{1}{4}e^{2} - \frac{3}{64}e^{4} - \frac{5}{256}e^{6} + \cdots } \right). $$

(A10)

We note that the first (leading) term in this equation is nothing else but a quadrant of a unit circle!