My question is: what does the quoted passage mean, and is it correct?
This is from The Meaning of Relativity by Albert Einstein. Many years ago I took it to a math professor at UT Austin. He went through a derivation and concluded that it is incorrect. But I cannot remember the details. Quite honestly, I'm not even sure what it means.
Between $n$ points of space there are $\frac{n\left(n-1\right)}{2}$ distances, $s_{\mu\nu}$; between these and the 3n co-ordinates we have the relations $$ s_{\mu\nu}^{2}=\left(x_{1\left(\mu\right)}-x_{1\left(\nu\right)}\right)^{2}+\left(x_{2\left(\mu\right)}-x_{2\left(\nu\right)}\right)^{2}+\dots $$ From these $\frac{n\left(n-1\right)}{2}$ equations the $3n$ coordinates may be eliminated, and from this elimination at least $\frac{n\left(n-1\right)}{2}-3n$ equations will result. [footnote: In reality there are $\frac{n\left(n-1\right)}{2}-3n+6$ equations.] Since the $s_{\mu\nu}$ are measurable quantities, and by definition are independent of each other, these relations between the $s_{\mu\nu}$ are not necessary a priori.
This is the beginning of my effort to understand what the quoted passage means.
First we define the following:
\begin{align*} N= & 3n\\ R= & \frac{n\left(n-1\right)}{2}\\ E= & R-N+6 \end{align*}
I call $N$ the number of degrees of freedom prior to introducing constraints. If the $s_{\mu\nu`}$ are specified, they serve as constraints on the system, of which there are $R$. $E$ is what appears in the book as the "number of equations" after "eliminating" the $3n$ coordinates.
We examine a few values of these formulae: \begin{align*} n=1\implies & N=3;R=0;E=3\\ n=2\implies & N=6;R=1;E=1\\ n=3\implies & N=9;R=3;E=0\\ n=4\implies & N=12;R=6;E=0\\ n=5\implies & N=15;R=10;E=1\\ n=6\implies & N=18;R=15;E=3\\ n=7\implies & N=21;R=21;E=6\\ n=8\implies & N=24;R=28;E=10\\ n=9\implies & N=27;R=36;E=15 \end{align*}
Going on the naive argument that each $s_{\mu\nu}$ serves as a constraint which can be used to eliminate one degree of freedom, we see that when $n=7$ the system is fully determined. But that doesn't really make sense, since we always need six values to fully specify the configuration of a rigid body. I'm guessing that is the origin of the 6 in the footnote.
In the case of three points, assuming they are not collinear, we determine a triangle. To locate it in space, we need three values to determine its position, and three to determine its orientation. But simply specifying the coordinates of two vertices will not suffice, since the third vertex is free to rotate about that axis.
This is a thought experiment I came up with to exercise the ideas implied by the math:
Thought Experiment
We purchase $R$ 1-dimensional rigid rods from Ikea in a kit for representing $n$ points in 3-space. Each rod can be attached at its ends to the ends of any number of other rods. The ends of each rod are labeled with index values. Our instructions say to connect all ends with matching index values. We can identify the rods using the indices on the $s_{\mu\nu}.$ If the rods are indexed correctly, none will have $\mu=\nu;$ no two rods will have $\left(\mu,\nu\right)_{1}=\left(\mu,\nu\right)_{2},$ nor will they have $\left(\mu,\nu\right)_{1}=\left(\nu,\mu\right)_{2}.$ The first of these restrictions is the source of the $-1$ in $R;$ the second is the source of the $1/2$ in $R.$
There is no need to include the index name when labeling the ends of the rods. All we require is that every pair of unique index values appears exactly once.