Let us consider the most general motion of a rigid body. Two arbitrary points of the body, $i$ and $j$ must not change their distance $d_{ij}$ during motion. Therefore,$$(\vec{r}_j - \vec{r}_i)^2 = d_{ij}^2 = \text{const.}$$ Differentiating, we have $$(\vec{v}_j - \vec{v}_i) \cdot (\vec{r}_j - \vec{r}_i)=0.$$From here, we can conclude that the relative velocity can be written in the form $$(\vec{v}_j - \vec{v}_i) = \vec{\omega}_{ij} \times (\vec{r}_j - \vec{r}_i),$$ for some vector $\vec{\omega}_{ij}$, which, in general, depends on the pair of points $(i,j)$ in consideration.

Is there an easy way to show that $\vec{\omega}_{ij} = \vec{\omega}$ is, in fact, the same for all pairs of particles? It seems to me that it should be possible to prove this just by using linear algebra, without any physical considerations.

UPDATE #1: My attempt - consider three particles $i,j,k$ and write $$(\vec{v}_j - \vec{v}_i) = \vec{\omega}_{ij} \times (\vec{r}_j - \vec{r}_i)\\(\vec{v}_k - \vec{v}_j) = \vec{\omega}_{jk} \times (\vec{r}_k - \vec{r}_j)\\(\vec{v}_i - \vec{v}_k) = \vec{\omega}_{ki} \times (\vec{r}_i - \vec{r}_k)$$Adding these three equations and rearranging we have $$(\vec{\omega}_{ij}-\vec{\omega}_{ki}) \times \vec{r}_i + (\vec{\omega}_{jk}-\vec{\omega}_{ij}) \times \vec{r}_j + (\vec{\omega}_{ki}-\vec{\omega}_{jk}) \times \vec{r}_k = \vec{0}.$$ Now, in general, my position vectors $\vec{r}$ are linearly independent. Does this imply that the brackets must vanish?

  • $\begingroup$ All the treatment you need you can find at this site $\endgroup$
    – Sofia
    Commented Jan 12, 2015 at 23:38
  • $\begingroup$ I am familiar with the standard approach to rotations. Here, I'm trying to derive the rotation, not start from it. $\endgroup$
    – user17116
    Commented Jan 13, 2015 at 0:07
  • $\begingroup$ very well. Though, I see a problem. Let me assume for simplicity that you chose the origin of the axes somewhere on the axis of rotation. Assume also that the two points $i$ and $j$ don't belong to a same plane perpendicular to the rotation axis, to the contrary, they are one below the other s.t. $\vec r_j - \vec r_i$ is parallel to the rotation axis. After all, a rigid body has a thickness. Then, $\omega _{i,j}$ can't be parallel to the rotation axis, but perpendicular to the rotation axis. Though, we know that the angular velocity vector is parallel to the rotation axis. $\endgroup$
    – Sofia
    Commented Jan 13, 2015 at 0:25
  • $\begingroup$ @Sofia No problem, then $\vec{v}_j - \vec{v}_i = \vec{0}$, as it should be. $\endgroup$
    – user17116
    Commented Jan 13, 2015 at 0:28
  • 1
    $\begingroup$ I'm sorry, has the obvious already been discussed: $A=B\times C$ uniquely determines $B$ only up to $B+kC$ for constants $k$. Edit: Beat me to it. $\endgroup$
    – user12029
    Commented Jan 13, 2015 at 0:56

3 Answers 3


First, $V_{rel}=\omega \times r_{rel}$ implies that $$\omega=\frac{r_{rel}\times V_{rel}}{r_{rel}\cdot r_{rel}}+r_{rel}\frac{r_{rel} \cdot \omega}{r_{rel}\cdot r_{rel}}$$ (use $A\times (B\times C)=B(A\cdot C)-C(A \cdot B)$). So the question is whether the sets "spanned" by:

$$\frac{r_{rel}\times V_{rel}}{r_{rel}\cdot r_{rel}}+r_{rel}k$$

for $k\in\mathbb{R}$, all intersect for all possible combinations of vectors.

However, one can imagine situations where $V_{rel}$ varies in a way not consistent with rotations but consistent with $V_{rel}\cdot r_{rel}=0$. So clearly $\frac{d}{dt}(\vec{r}_j - \vec{r}_i)^2=0$ doesn't directly determine $V_{rel}$. Furthermore, there are a lot of linkages which preserve distances but don't give rotations. So, for example, if we have the vertices of a square, we can't expect to prove the theorem using only $r_{12}$, $r_{23}$, $r_{34}$, $r_{41}$. There must be a reason that our proof won't work considering that, but will work considering $r_{12}$, $r_{23}$, $r_{31}$. My point with this is just that I wouldn't expect an easy linear algebra solution given only conservation of $r_{ij}$. (Note: Why do there exist easy linear algebra solutions to rotation problems? Those usually depend on conservation of all possible inner products, which is much easier/nicer to work with)

I really don't want to work through the algebra, but I would do this, for a brute-force "dumb" method that I'd bet guarantees a proof:

  1. Start with a triangle $r_{12}$, $r_{23}$, $r_{31}$.
  2. Solve for $k$ by, say, writing $\omega_{21}=\omega_{31}$. There must be a trick in here that makes use of $r_{23}$ (because conservation of $r_{21}$ and $r_{31}$ alone does NOT imply a well defined angular velocity- it's a hinge not a rigid body then).
  3. To show that it's well defined for all other particles $r_{ij}$, you might write $r_{ij}$ in terms of the basis set $r_{21}$, $r_{31}$ and their cross product. Once you do that, pray for cancellations! I think they'd certainly have to cancel, but I don't see how and I haven't done the algebra.

Step 2 might be simplified using the conservation of inner products, which would state that $\frac{d}{dt}r_{ij}\cdot r_{mn}=0$ meaning $V_{ij}\cdot r_{mn}+r_{ij}\cdot V_{mn}=0$.


In my opinion, the most elegant way to show this is by noting that (this can be proven) the motion of any rigid body (system of particles in which pairwise distances between particles are constant) with one point fixed is generated by a time-dependent rotation: \begin{align} \mathbf r_i(t) = R(t) \mathbf r_i(0). \end{align} From here, one can show that there exists a time-dependent vector $\boldsymbol\omega(t)$ such that \begin{align} \dot{\mathbf r}_i(t) = \boldsymbol\omega(t)\times \mathbf r_i(t) \end{align} for all particles $i$ in the system. The fact that a single such $\boldsymbol\omega$ exists which works for all of the particles is the result you're looking for.

Perhaps this is begging the question since I haven't addressed how one proves the claim that the motion of all particles is generated by a single, time-dependent rotation, but it's true (and even goes by a name I can't recall), and I can try to outline the proof if you wish.

Related and potentially useful for further understanding: https://physics.stackexchange.com/a/85989/19976


The simplest solution is to show the trivial fact, that in a rigid body that rotates, the angular velocity is the same for any point,

$$\dot \theta _j = \dot \theta_i \ = \dot \theta \ .$$ As the solid rotates, two arbitrary points $i$ and $j$ describe circles. Their coordinates with respect to a system of rotating axes, are, respectively $x'_i, y'_i, z'_i$, and $x'_j, y'_j, z_j$, while with respect to the fixed axes they are $x_i, y_i, z_i$, respectively $x_j, y_j, z_j$. The axis $z$ is parallel to the rotation axis, s.t. the $z$ coordinate is the same in both the rotating and the stationary system.

The relation between the two systems of coordinates is,

(1) $x_i = x'_i cos(\theta_i) + y' sin(\theta_i), \ \ y_i = -x'_i sin(\theta_i) + y'_i cos(\theta_i), \ \ z_i = z'_i,$

and analogously for the point $j$. Since the distance between the two pints has to remain constant,

(2) $\frac {d}{dt}[(x_j - x_i)^2 + (y_j - y_i)^2 + (z_j - z_i)^2] = 0$.

One can see in (1) that only $\theta_i$ and $\theta_j$ depend on time, as the primed coordinates are fixed. For shorting the calculi let's notice that

$dx_i/dt = \dot \theta_i y_i$ $ \ \ $ and $ \ \ $ $dy_i/dt = -\dot \theta_i x_i$, $

and analogously for $j$. Then, performing the derivative (2),

$(\dot \theta_j y_j - \dot \theta_i y_i)(x_j - x_i) - (\dot \theta_j x_j - \dot \theta_i x_i)(y_j - y_i) = 0$.

Reducing all the similar terms one gets

(3) $(\dot \theta _j -\dot \theta _i)(x_iy_j - x_jy_i) = 0$.

In general the content of the 2nd pair of parentheses doesn't vanish, s.t. the angular velocities are equal.


Your Answer

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