How to prove that rigid bodies have fixed areas during their movements?

Question

a rigid body is a pair of: a set $S$ of points with at least two distinct points on space $\mathbb R^3$ and a special continuous position function for each point of $S$ $f:S \times [t_1,t_2] \to \mathbb R^3$, such that $f(x,y,z,t)$ gives the position of point $(x,y,z)$ at the moment $t$ with the property that for any two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ and any instants $t_1$ and $t_2$ we have that: $$||f(x_1,y_1,z_1,t_1)-f(x_2,y_2,z_2,t_1)|| = ||f(x_1,y_1,z_1,t_2) - f(x_2,y_2,z_2,t_2)||$$

with this analytic approach, can we prove that the area of $S$ does not change over $t$? The set $S$ at the instant $t$ is given by $f(S,t)$.

I think this come to proving that the shape of $S$ doesn't change, but how to formalize this?

EDIT: is my definition of $f$ problematic?

Answer 1

I'm going to prove something more general, since the result doesn't depend on continuity of $f$ or $n=3$.

Lemma: Let $S\subseteq \mathbb{R}^n$, and let $f:S \to \mathbb{R}^n$ be such that $\|p-q\| = \|f(p) - f(q)\|$ whenever $p,q\in S$. Then $f$ is a composition of reflections, rotations and translations.

In particular $m(S) = m(f(S))$ whenever $S$ is measurable. This implies your version; fix $t_1$ and $t_2$ and consider that $f(\cdot,t_1)\mapsto f(\cdot,t_2)$ is a function satisfying the conditions of the lemma.

For more detail see Euclidean isometries. In this case the function is restricted in domain to $S$.

Proof:

Case 1: $S$ contains an affinely independent set of cardinality $n+1$.

Let $A = \{a_1,\ldots,a_{n+1}\} \subseteq S$ be an affinely independent set. We construct the composition of reflections, rotations and translations necessary to identify $A$ with $f(A)$:

1. Compose $f$ with a translation to identify $a_1$ with $f(a_1)$
2. Next, compose with a rotation about $a_1$ to identify $a_2$ with $f(a_2)$. This is possible since $\|a_1 - a_2\| = \|f(a_1) - f(a_2)\|$
3. Continue composing with rotations that fix all previously identified points, and identify the next point. Since affine rotations fix an $n-2$-dimensional affine subspace, and distances match, and the points are affinely independent, this is possible up to the second-to-last point.
4. Once $a_1,\ldots,a_n$ have all been identified, if $a_{n+1}$ and $f(a_{n+1})$ are already identified, great. Otherwise, consider the set $H$ of points $h$ such that $\|a_{n+1} - h\| = \|f(a_{n-1}) - h\|$ is a hyperplane which must contain $\{a_1,\ldots,a_n\}$. Therefore composing with a reflection across $H$ identifies $a_{n+1}$ with $f(a_{n+1})$ while leaving the rest fixed.

Let $\widehat{f}$ be the above composition of $f$ with translations, rotations and reflections. I claim $\widehat{f}$ is the identity on $S$. Suppose not, for a contradiction. Then there is $x\in S$ with $x\not = \widehat{f}(x)$. Note that $\widehat{f}$ is also an isometry, since it is a composition of isometries, so $x$ and $\widehat{f}(x)$ have the same distance to each point in $A$. Just as in step 4 above, this implies $A$ is contained in a hyperplane. But this is impossible, since $A$ is affinely independent of cardinality $n+1$. Therefore $\widehat{f}$ is the identity, which means $f$ is a combination of reflections, rotations and translations.

Case 2: $S$ contains no affinely independent set of cardinality $n+1$.

Given a maximal affinely independent set $A := \{a_1,\ldots,a_k\}$, we can identify $A$ with $f(A)$ as before (in fewer steps). Once again let $\widehat{f}$ be the composition of $f$ with these isometries, and suppose for a contradiction there is $x\in S$ with $x\not = \widehat{f}(x)$. As before $A$ must be contained in the hyperplane between $x$ and $\widehat{f}(x)$. As $x$ is not in this hyperplane, $x$ is affinely independent with $A$, contradicting the maximality of $A$, and we're done.