The following is from the fifth Chapter Rigid Bodies of Spivak's Physics for Mathematicians. The post consists of a statement Spivak makes -with no proof- that I do not understand. For clarity, I've tried including the definitions and lemma that I believe may be relevant in regards to the result Spivak states.

Definition: let $b_1, \ldots ,b_K$ be a collection of points which we regard as a single object $b=(b_1,\ldots ,b_K)$ and let $F=(F_1,\ldots ,F_K)$ be a collection of forces where we regard $F_i$ as acting on $b_i$, then $b$ is in rigid equilibrium under the forces $F$ if there exist internal forces $F_{ij}=-F_{ji}$ which are multiples of $c_i-c_j$ so that $$F_i=-\sum_jF_{ij}.$$

Definition: a rigid motion $f$ of $b$ is a curve $t\to A(t)$ where $A(t)$ is an isometry of $\mathbb{R}^3$ such that

  1. The position function $c_i(t) = A(t)(b_i)$ of $b_i$ is smooth with $c_i(0)=b_i$.
  2. We have $$\langle c_i(t) - c_j(t), c_i(t) - c_j(t)\rangle \ \text{ stays constant} \tag{1}$$ so that the distance between the points $b_i$ and $b_j$ stays constant.

Definition: the configuration space $\mathcal{M}_b$ of $b$ is given by $$\big\{ \big(A(b_1),\ldots ,A(b_K)\big) : A \text{ is an orientation preserving isometry of }\mathbb{R}^3\big\}.$$

Lemma: we have that $\langle v_i(0) - v_j(0), b_i - b_j\rangle = 0$

Proof: Follows by differentiating $(1)$ and evaluating at zero. $$\square$$

Theorem: if we define the linear functions $\phi_{ij}$ on $(\mathbb{R}^3)^K$ by $$\phi_{ij}(v_1,\ldots ,v_K) = \langle v_i - v_j, b_i - b_j\rangle$$ then $$\mathcal{M}_b\subseteq \bigcap_{i,j}\ker \phi_{ij}.\tag{2}$$

It is this last Theorem that Spivak states without proof, and I'm confused on two grounds:

  1. In what sense are the functions $\phi_{ij}$ linear? If we substitute any of the arguments $v_i$ by $\mu v_i$ for some scalar $\mu$, the resulting output will not get scaled by $\mu$, so I do not understand how the $\phi_{ij}$ can be linear nor multilinear.

  2. How is $(2)$ derived? The elements of $\mathcal{M}_b$ are supposed to represent positions, not velocities. Thus, for an element $$m=\big( A(b_1), \ldots ,A(b_K)\big)\in \mathcal{M}_b$$ we get $$\phi_{ij}(m) = \langle A(b_i) - A(b_j), b_i - b_j\rangle$$ which need not be zero (compare with the Lemma).


1 Answer 1

  1. The $\phi_{ij}$ are linear in the usual sense. I don't see any issues with it. For any $v=(v_1,\dots, v_K), w=(w_1,\dots, w_K)\in (\Bbb{R}^n)^K$, and any scalar $\mu\in\Bbb{R}$, \begin{align} \phi_{ij}\left(\mu v+w\right)&:=\phi_{ij}\left((\mu v_1+w_1,\dots, \mu v_K+w_K)\right)\\ &:=\langle (\mu v_i+w_i)-(\mu v_j+w_j), b_i-b_j\rangle\\ &=\mu\langle v_i-v_j,b_i-b_j\rangle + \langle w_i-w_j,b_i-b_j\rangle \\ &=\mu\phi_{ij}(v) + \phi_{ij}(w), \end{align} where we have used bilinearity of the inner product $\langle\cdot,\cdot\rangle$.
  2. I just read the corresponding text in Spivak, and you must have mistyped what he wrote. He says that the configuration space of the point $b=(b_1,\dots, b_K)\in(\Bbb{R}^n)^K$ is denoted as $\mathcal{M}$. In the theorem that $\mathcal{M}_b\subset\bigcap_{ij}\ker\phi_{ij}$, the symbol $\mathcal{M}_b$ refers to the tangent space of $\mathcal{M}$ at the point $b$ (this is his notation for the tangent space to a manifold at a given point, and he also uses it in his other texts such as Calculus on Manifolds and his 5-Volume tome A Comprehensive Introduction to Differential Geometry). Just to be clear, $\mathcal{M}$ is an embedded submanifold of $(\Bbb{R}^n)^K$, so the tangent space at any given point, a-priori has an abstract definition, but because this is embedded in $(\Bbb{R}^n)^K$, we can use the identity chart to view this tangent space as an actual subspace of $(\Bbb{R}^n)^K$ (see this MSE answer of mine for more details) and hence it makes sense to say it is contained in the intersection of the kernels of the linear maps $\phi_{ij}$.

Of course, the theorem follows immediately from the preceding discussion in the text (particularly the lemma you quote).

  • $\begingroup$ I have to admit, when I first read your post, I was confused as well, because it seemed to be mixing up positions and velocities, something completely uncharacteristic for Spivak, and only then did I read the corresponding portion of the text and realized it's his notation which is tripping you up; $\mathcal{M}_b\equiv T_b\mathcal{M}$. $\endgroup$
    – peek-a-boo
    Commented Jul 16, 2022 at 15:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.