Rigid bodies: proof of existence of internal forces that preserve the distances

Question

I am new to Physics and I have a pure Math background. I am currently studying mechanics and I have the following question regarding rigid bodies. I am posting here the 2D version of the question. If anyone is aware of the answer for the general 3D version please post that one.

Our rigid body is a collection of $N$ points $x_i(t)\in \mathbb{R}^2$, $i=1,\dots,N$, such that all distances $\|x_i(t)-x_j(t)\|$ are independent of the time $t$. The initial condition is that at time $t=0$, $x_i(0)=x_{i,0}$ and $x_i'(0)=0$. Furthermore, for all all times $t\ge 0$ we apply to each point $i$ an external force $F_i^{\text{ext}}(t)$. According to the standard rigid-body model, the points will exert forces on each other, which we denote by $F_{ij}(t)$ (force from $j$ to $i$ at time $t$) and which satisfy the following conditions:

1. They obey the strong form of Newton's third law, i.e., $F_{ij}(t)=-F_{ji}(t)$ and $F_{ij}(t)$ is parallel to $x_i(t)-x_j(t)$.
2. They preserve distances. Formally, let $$F_i(t)=F_i^{\text{ext}}(t)+\sum_{j\neq i}F_{ij}(t)$$ Then, the solutions of these $N$ (independent) ODEs: $$x_i''(t)=F_i(t), \ \ x_i'(0)=0,\ \ x_i(0)=x_{i,0}$$ (which can be solved via direct integration), satisfy $\|x_i(t)-x_j(t)\|= \|x_{i,0}-x_{j,0}\|$ for all $i,j,t$.

The question is, how do we know that such internal forces $F_{ij}(t)$ exist (mathematically speaking)?

The whole derivation of the laws of rigid-body motion from Newton's laws crucially relies on the existence of these forces. So, it is very important to know that they exist under natural assumptions. Some remarks:

• Some assumptions are surely needed. For example, consider $N=3$ points with equal mass $m$, initial positions $x_{1,0}=(0,0), x_{2,0}=(-1,0), \ x_{3,0}=(1,0)$, and external forces $F_1^{\text{ext}}(t)=(0,F)$ with $F$ independent of time, and zero external forces at the other two points. Here, the "natural" solution is that the 3 points go upwards with acceleration $\frac{F}{3m}$. However, this is impossible via central internal forces. This issue is easily avoided by assuming that the points are not collinear. Spivak in his "Physics for Mathematicians" book discusses this issue (page 177) but does not answer my question.
• Ideally, a proof should show that there exist internal forces that satisfy the conditions and $F_{ij}(t)$ is nonzero only if $i$ and $j$ are close.
• It is easy to show (I can say more on this if needed) that given the $F_i^{\text{ext}}(t)$, the existence of such internal forces is equivalent to the existence of a solution for a certain linear system of equations, which has much more unknowns than equations (for large $N$). However, this does not immediately imply the existence a solution.

Update: I realized that the solution actually exists in Spivak: Proposition 2 in page 181 proves it for equilibrium, and the general case can be reduced to equilibrium by applying the same subtraction trick that is used to derive d'Alembert's principle. One note: the proof of Spivak for Proposition 2 is fairly algebraic. A simple intuitive proof is possible and is given here (this proof is along the lines of the answer of @David Moews).

Answer 1

The existence of some internal forces is required in order to preserve the distances. Obviously they come in pairs action/reaction, so the obey the weak form of Newton's law. It is not required that they obey the strong form. For example, see the figure below:

In this triangle I choose all the posted forces to be equal in magnitude. They cancel each other at every point, so the net force and net torque on the $A, B, C$ points is zero. But these will not obey the strong form. Let's suppose now that you add some external forces, then compute the internal forces according to your original post. You can then add the internal forces that I have pasted in the figure, and it will not change the result (body will move the same way, no distances will change). But the internal forces will not be along the distances between the parts.

Answer 2

I'm going to answer this question in any number $n\ge 1$ of dimensions.

Call an $N$-dimensional vector of forces, $F_1$ being exerted on point $1$, $F_2$ on $2$, $\dots$, $F_N$ on $N$, a force field $F=(F_1,\dots,F_N)$. If $F^{tot}$ is the force field of all forces on the points, then $F^{tot}=F^{ext}+F^{int},$ where the internal force field $F^{int}$ comes from summing the $F_{ij}$s you mention. Since $F_{ij}$ is along the direction of $x_i-x_j$ and satisfies Newton's Third Law, we can write $$ F^{int}_i=\sum_j q_{ij} (x_i-x_j), \qquad \hbox{where $ q_{ij}=q_{ji}$.} \qquad (1) $$ Given a force field $F$, we can define the resultant force $r(F)$, which is a linear function of $F$, by $$r(F):=\sum_i F_i.$$ Assuming that point $i$ has mass $m_i$ and defining the total mass $$M:=\sum_i m_i$$ and the center of mass of the points $$c:=\left.(\sum_i m_i x_i)\right/M,$$ we can define the resultant torque by $$\Gamma(F):=\sum_i (x_i-c) \otimes F_i - F_i \otimes (x_i-c).$$ The torque is an antisymmetric rank 2 tensor and is also a linear function of $F$. For the internal forces, since $q_{ij}$ is symmetric in $i$ and $j$ and $x_i-x_j$ is antisymmetric, $$ r(F^{int})=\sum_{i,j} q_{ij} (x_i-x_j) = 0 $$ and \begin{eqnarray*} \Gamma(F^{int}) &=& \sum_i (x_i-c) \otimes F^{int}_i - F^{int}_i \otimes (x_i-c)\\ &=& \sum_i x_i \otimes F^{int}_i - F^{int}_i \otimes x_i, \qquad \qquad \hbox{since $r(F^{int})=0$}\\ &=& \sum_{i,j} q_{ij} (x_i \otimes (x_i - x_j) - (x_i - x_j) \otimes x_i)\\ &=& \sum_{i,j} q_{ij} (x_j \otimes x_i - x_i \otimes x_j)\\ &=& 0, \\ &\ & \hbox{since $q_{ij}$ is symmetric in $i$ and $j$ and $x_j \otimes x_i - x_i \otimes x_j$ is antisymmetric.} \end{eqnarray*} Take a body-fixed coordinate system with the body's center of mass at the origin. Then, we can give point $i$ a position $y_i$ in this system. Since the body is rigid, $y_i$ does not change with time. We will have $$ x_i = R y_i + c, \qquad \qquad i=1, \dots, N, $$ for some time-dependent rotation matrix $R$ and center-of-mass position $c$. This implies that each distance $||x_i-x_j||$ is constant since then $$ ||x_i-x_j||=||( R y_i + c) - (R y_j + c)|| =||R(y_i-y_j)||=||y_i-y_j||. $$ Then, by Newton's Second Law, for each point $i$, $$ m_i \ddot x_i = m_i (\ddot R y_i + \ddot c) = F^{tot}_i = F^{ext}_i + F^{int}_i.\qquad \qquad (2) $$

First step

To solve for the motion, the first step is to use (2) to find $\ddot R$ and $\ddot c$, which we can compute from $r(F^{tot})=r(F^{ext})$ and $\Gamma(F^{tot})=\Gamma(F^{ext})$. In Physics For Mathematicians, Spivak discusses this (for 3 dimensions) on pp. 185-188.

Since $r(F^{int})=0$, (2) implies that $$\sum_i m_i (\ddot R y_i + \ddot c) = r(F^{ext})\qquad\qquad(3)$$ and since by definition $$ Mc=\sum_i m_i x_i=R(\sum_i m_i y_i)+Mc$$ we must have $\sum_i m_i y_i=0$, so solving (3) for $\ddot c$ gives $$\ddot c=r(F^{ext})/M.$$ Since also $\Gamma(F^{int})=0$, (2) also implies that $$ \sum_i m_i ((R y_i) \otimes (\ddot R y_i + \ddot c) - (\ddot R y_i + \ddot c) \otimes (Ry_i)) = \Gamma(F^{ext}) $$ so since $\sum_i m_i y_i=0$, $$ \sum_i m_i ((R y_i) \otimes (\ddot R y_i) - (\ddot R y_i ) \otimes (Ry_i)) = \Gamma(F^{ext}).\qquad\qquad(4) $$ Define the body-fixed inertia symmetric rank 2 tensor by $$I:=\sum_i m_i y_i \otimes y_i. $$ It is constant. We're going to think of rank 2 tensors as matrices and we're taking vectors to be column vectors, so we can set $a\otimes b:=a b^T$. Then we can rewrite (4) as $$ R I \ddot R^T - \ddot R I R^T = \Gamma(F^{ext}).\qquad\qquad(5) $$ Considering only the rotational motion of point $i$, in an infinitesimal time $dt$ it will move from $R y_i$ to $R y_i + \dot R y_i dt$. Expressing the change in position in terms of the original position gives $$ \dot R y_i dt = (\dot R R^{-1}) (R y_i) dt.$$ Define the angular velocity rank 2 tensor as the transformation matrix for this expression, $$U:=\dot R R^{-1}.$$ Since $R$ is a rotation matrix, we have $R^{-1} = R^T$ so $$1= R R^T.$$ Differentiating this with respect to time gives $$0=\dot R R^T + R \dot R^T=U + U^T$$ so $U$ is antisymmetric. Conversely, if we start time with $R=1$ and integrate the equation $\dot R = U R$ for any time-varying antisymmetric matrix $U$, we will get a solution $R$ with $R R^T=1$, so $R$ will be a rotation matrix.

We have \begin{eqnarray*} \ddot R&=&(\dot R)\dot\ \\ &=& (U R)\dot\ \\ &=& \dot U R + U \dot R \\ &=& \dot U R + U^2 R.\qquad\qquad(6) \end{eqnarray*} If we define the body-fixed angular velocity tensor $V$ so as to make the same transformation as $U$, but in the body-fixed coordinate system, then $$ V:=R^T U R $$ so that $V$ is also antisymmetric and \begin{eqnarray*} \dot V&=& \dot R^T U R + R^T \dot U R + R^T U \dot R\\ &=& R^T U^T U R + R^T \dot U R + R^T U U R\\ &=& R^T \dot U R,\\ &\ & \hbox{since $U$ is antisymmetric.} \end{eqnarray*} Therefore, from (6), $$\ddot R=R \dot V + R V^2.$$ Plugging this into (5) and writing $$\Delta(M):=M-M^T$$ for the antisymmetrization of $M$, we get $$ \Delta(R I \dot V^T R^T + R I (V^T)^2 R^T)= \Gamma(F^{ext}). $$ Conjugate this by $R^T$ and use the antisymmetry of $V$ to get $$ \Delta(I V^2 - I \dot V)= R^T \Gamma(F^{ext}) R. $$ This is a form of Euler's rotation equations. To solve them, we can move $\Delta(I V^2)$ over to the right-hand side of the equation. We then want to solve $$ -\dot V I - I \dot V=A\qquad\qquad(7) $$ for an antisymmetric tensor $\dot V$, where $A$ is also an antisymmetric tensor. Now, if $s$ is any nonzero vector then $$ s^T I s = \sum_i m_i (s \cdot y_i)^2. $$ Assuming that the points don't lie in any hyperplane, there must be some $i$ with $s\cdot y_i\ne 0$, so this will be positive. Therefore $I$ is positive definite, and when we diagonalize it, which we can do by changing the coordinate system by some orthogonal matrix, we will get all diagonal entries positive. Working in the diagonalized coordinate system and writing $$A=(A_{ab}), \dot V=(\dot V_{ab}), I={\rm diag}(I_1,\dots,I_n), \qquad I_1, \dots, I_n > 0,$$ (7) becomes $$ -(I_a + I_b) \dot V_{ab} = A_{ab} $$ which has the antisymmetric solution $$ \dot V_{ab} = -A_{ab} / (I_a + I_b). $$ This gives us $\ddot R$.

Second step

After doing all this, using (2) again to solve for $F^{int}$ will give us some value for $F^{int}$. Since we successfully solved (3) and (4), we will have $r(F^{int})=0$ and $\Gamma(F^{int})=0$. The second step is to find the $q_{ij}$s given $F^{int}$. To do this, as you point out, you need to make some assumptions. In Physics For Mathematicians, Spivak discusses these assumptions (for 3 dimensions) on p. 177 and proves that the $q_{ij}$s can be found in Proposition 2 (p. 181.)

I'm going to assume that:

1. Some pairs of points $i$ and $j$ are connected by imaginary massless rigid rods. These are the only points for which we're going to allow $q_{ij}$ to be nonzero. You can take close pairs of points $i$ and $j$ if you like.

2. For $i=1$, $\dots$, $\min(n,N)$, point $i$ is connected to all points 1, $\dots$, $i-1$, and these $i$ points don't lie in any affine $(i-2)$-dimensional subspace.

3. If $N>n$, for $i=n+1$, $\dots$, $N$, then there is no hyperplane containing the points $j<i$ connected to point $i$ together with point $i$. This implies that point $i$ is connected to at least $n$ points $j<i$.

Assumption 3 implies that, if $N>n$, points 1, $\dots$, $n+1$ don't lie in any hyperplane. Also, given assumption 3, we can delete rods so that each $i\ge n+1$ is connected to exactly $n$ points $j<i$ while keeping assumption 3 true.

The total set of force fields $F$ is a vector space of dimension $nN$. The requirement $r(F)=0$ obviously decreases this dimension to $n(N-1)$. In this case, we can simplify the definition of $\Gamma$ to $$\Gamma(F)=\sum_i x_i \otimes F_i - F_i \otimes x_i.$$

If $N=1$, since $r(F^{int})=0$, $F^{int}=0$ and there is nothing to prove. Otherwise, let $m:=\min(n+1, N)$. For points $i=2$, $\dots$, $m$ and $k=1$, $\dots$, $n$, we can let $F$ be a couple of forces, $\hat e_k$ on point 1 and $-\hat e_k$ on point $i$. This will give \begin{eqnarray*} \Gamma(F)&=&x_1 \otimes \hat e_k - \hat e_k \otimes x_1 - (x_i \otimes \hat e_k-\hat e_k \otimes x_i)\\ &=& (x_1 - x_i) \otimes \hat e_k - \hat e_k \otimes (x_1 - x_i). \end{eqnarray*} Assumption 2 (if $N\le n$) or 3 (if $N\ge n+1$) implies that the vectors $x_1-x_i$ are linearly independent for $i=2$, $\dots$, $m$. This implies that the range of $\Gamma(F)$ has dimension at least $(n-1)+(n-2)+\cdots+(n-(m-1))$. This means that the dimension of the vector space of force fields $F$ that we wish to find $q_{ij}$s for has dimension no more than $$ n(N-1) - ((n-1) + \cdots + (n-(m-1)). $$ The dimension of the vector space of possible $q_{ij}$s is $$ 1 + 2 + \cdots + (m-1) + (N-m)n, $$ which is the same. Therefore, since the map (1) from the $q_{ij}$s to $F^{int}$ is linear, it will do to prove that it has zero kernel. So, given some $q_{ij}$s that are not all zero, look at the largest $i$ where $q_{ij}\ne 0$ for some $j<i$. Then $$F^{int}_i=\sum_{j<i, q_{ij}\ne 0} q_{ij} (x_i-x_j).$$ Now, because of assumption 2 (if $i\le n$) or 3 (if $i\ge n+1$), the vectors $x_i-x_j$ which appear here are linearly independent, so $F^{int}_i$ must be nonzero. This completes the proof.