Motivation
I don't remember any reference, but I recall that many materials and papers related to robotics consider a rigid body; a rigid wheel, a rigid sphere but don't give the definition of it and the assumption of this comes from the simplification of the calculations. In this context, I was wondering about a formal definition of rigid sphere.
I am working on a personal project of robotics and in order to modeling the system the sphere used is considered rigid for a manageable level of complexity, but in real life it is a rubber sphere. Working on definition of rigid sphere, I initially first considered a rigid body transformation and thought it was enough, such that
$f:\mathbb{R}^3\rightarrow \mathbb{R}^3$ and $\forall p,q\in\mathbb{R}^3$ we have $\lVert f(p) – f(q)\rVert = \lVert p – q\rVert$. Besides, to preserve the orientation, consider that the cross product is also preserved, such that $g_* (\vec{u} \times \vec{v} ) = g_* (\vec{u}) \times g_* (\vec{v} ),\ \forall\ \vec{u},\vec{v} \in\mathbb{R}^3$. I could also state that the sphere is homogeneous, considering that the moment of inertia of the sphere $s$ around the axis $x,y,z$ are the same, i.e. $n_s = n_{s,x} = n_{s,y}= n_{s,z}$.
Trying to search about a way to define a rigid sphere, I found two concepts that I thought would help me: the special orthogonal group $SO(3)$ and the special Euclidean group $SE(3)$. Therefore, I considered the rotation matrices in $\mathbb{R}^{3\times 3}$, I thought that there would be something regarding a rigid sphere defining the space $SO(3) = \{R ∈ \mathbb{R}^{3\times 3}: RR^T = I, \det R = +1\}$, but couldn't relate to anything.
In this context, considering the special Euclidean group defined by
$$SE(3) = \{(R,p):R ∈ SO(3) \text{ and } p\in\mathbb{R}^3\} = SO(3) \times \mathbb{R}^3$$
such that, can't the configuration of the rigid sphere ($S$) be given by $S\in(R, p)$? Can I defined a rigid sphere that way? Or there is another approach for this? It is also important to note that I couldn't relate anything to the sphere's radius ($r_s$) and I am still not sure about that.
After that, I decided to think more about rigidity of a body. Thus, what I am looking for a rigid sphere is
rigid sphere: a solid body with no deformation
also, I am not considering slip and rolling constraints. And a idea that came next was about the isometry, basically trying to work correct with the explanation concerning $\lVert f(p) – f(q)\rVert = \lVert p – q\rVert$ as aforementioned. I couldn't complete it and relate to the sphere, but
What I have now is an isometry of three dimensions for $f:\mathbb{R}^3\rightarrow \mathbb{R}^3$ and $f: T_p \circ L $ such that $T_p$ is a translation $(p\in\mathbb{R}^3$ that $x \mapsto x + p$ and $L\in SO(3))$.
This is very incomplete, but perhaps a more correct path to choose?
I couldn't develop anything closer for the definition I think might be possible, but hope the edits are a bit better than before, but what I am looking for is precisely the definition a rigid sphere in space. I am sorry for not being able to give more information as I don't have much knowledge in the area.
Best Answer
Here is my best guess regarding your question. I will use some topological terminology in my answer. The meaning of all the words unknown to you that you encounter in my answer can be found by searching Wikipedia. (There is no way I can teach you all this staff within the confines of my answer.)
Fix a radius $r>0$ and the origin $o$ in the Euclidean 3-space $E^3$ (thus, making $E^3$ a vector space equipped with an inner product). Let $S(o,r)$ denote the round sphere of radius $r$ centered at $o$.
Then you consider a set $F$ consisting of homeomorphisms $E^3\to E^3$.
Examples: 1. $F$ consists of all diffeomorphisms of $E^3$, i.e. smooth maps which have smooth inverse.
Now, given a homeomorphism $f\in F$, let $f(S)$ denote the image of our sphere under the map $f$. Informally, one can think of $f$ as a nonrigid motion of $E^3$ that carries $S$ to $f(S)$.
Given our set $F$, one defines $\Sigma_F =\{f(S(o,r)): f\in F\}$, the set of all images of $S(o,r)$ under the maps $f$ from our family $F$. One can think of $\Sigma_F$ as a family of topological spheres in $E^3$. Typically, the shape of a topological sphere $S=f(S(o,r))$ is not one of the original round sphere. For instance, it can be a round sphere but of a different radius or an ellipsoid, or something even more complex. One usually equips $\Sigma_F$ with some topology; I will not do this here.
Geometric examples of spaces $\Sigma_F$ that I personally like are given, for instance, by the mean curvature flow which describes evolution of, say, a smooth convex topological sphere $S$ in $E^3$ to a surface which is asymptotically a round sphere.
I will assume from now on that $F$ contains the entire group $E(3)$ of rigid motions of $E^3$, i.e. of the maps $L$ of the form $$ L({\mathbf x})= A{\mathbf x} + {\mathbf b}, $$ where $A\in O(3)$ and ${\mathbf b}\in {\mathbb R}^3$.
Finally, I can define a rigid sphere. Fix some topological sphere $S\in \Sigma_F$; for instance, it could be the sphere $S(o,r)$ itself.
Now, a rigid sphere in $E^3$ (defined in the context of a specific family of homeomorphisms $F$ and a specific topological sphere $S$ as above) is simply an element $f(S)\in \Sigma_F$ which is congruent to $S$.
In other words, a sphere is rigid if there exists an orthogonal matrix $A$ and a vector ${\mathbf b}$ such that $$ f(S)=\{ A{\mathbf x} + {\mathbf b}: {\mathbf x}\in S\}. $$
Caveats. 1. In this definition, $f$ itself need not be a rigid motion.
Depending on the context, one may require the matrix $A$ to have determinant $1$, i.e. be orientation-preserving.
In general, the space $R_{F,S}$ of rigid spheres (equipped with a suitable topology that I will not spell out) is not homeomorphic to the group $E(3)$ of rigid Euclidean motions. This is because a topological sphere $f(S)$ (in general) does not uniquely the rigid motion $L=A{\mathbf x} + {\mathbf b}\in E(3)$. For instance, if $S$ was round, then we can rotate it around the center without changing the sphere itself. However, if a sphere $S$ is chosen generically, then indeed, no (nontrivial) rigid motion will preserve it and, thus, $L$ is uniquely determined and one obtains a homeomorphism $$ R_{F,S}\to L, f(S)\mapsto L, $$
where $L(S)=f(S)$.
If you are to choose $S=S(o,r)$ then the corresponding rigid spheres are just round spheres of the radius $r$ in $E^3$. Moreover, $R_{F,S}$ is homeomorphic to $E^3$, sending a rigid sphere $f(S)$ to its center.
How does one think (informally) of all this? Imagine a sphere $S$ evolving in the 3-space according to some rules, in such a way that all the evolved spheres are still homeomorphic to $S$. Some of the evolved spheres happen to be congruent to $S$ itself. These are the rigid spheres with respect to this evolution process and the choice of the initial sphere $S$.
This definition of a rigid sphere is consistent with the notion of rigidity coming from the theory of bar-and-link configurations ("mechanisms") $C$ in $E^3$ (or even in the plane) that people in robotics like to study. Such configurations "evolve" in $E^3$ according to some external (or internal) forces, preserving lengths of the rods and in such a way that different rods do not cross through each other. A configuration $C$ is called (globally) rigid if all of its evolved configurations are congruent to $C$.
See for instance
Rigidity and Symmetry (Fields Institute Communications Book 70), 2014th Edition. By Robert Connelly (Editor), Asia Ivić Weiss (Editor), Walter Whiteley (Editor).
A (smooth) topological sphere can be approximated by a bar-and-link configuration $C$. If $C$ is rigid in the sense of the bar-and-link rigidity theory, then one can say that its rigid images in $E^3$ approximate congruent copies of our original sphere.
Another example of rigidity comes from differential geometry where one studies smooth topological spheres $f(S)$ in $E^3$ such that $f: S\to f(S)$ is an intrinsic isometry, i.e. an isometry of the intrinsic Riemannian metrics. A remarkable rigidity theorem due to Alexandrov and Pogorelov says that if $S$ and $f(S)$ are convex, then $f(S)$ is congruent to $S$ (by a rigid motion of $E^3$). In this setting, one says that the sphere $S$ itself is rigid.
Remarks. 1. It is totally possible that the rigidity of spheres you are interested in is Alexandrov--Pogorelov-type, I just could not decipher it from your question and, hence, went for the most general notion of rigidity that I know.