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I am confused about a trivial concept. Let the rotation of a rigid body, say with one point fixed, be described by the equation $\vec{x}(t)=R(t)\vec{x}(0)$, with $R(0)=I$.

Then, at each instant there is only one real eigenvector of $R(t)$ with eigenvalue 1 that we may call $\vec{v}(t)$ and which we may take to be normalized. That vector $\vec{v}(t)$ is what geometrically we would call the (instantaneous) axis of the rotation.

Kinematically, however, the instantaneous axis of rotation is the line of points with vanishing instantaneous velocity $\dot{\vec{x}}(t)=\vec{\omega}(t)\times\vec{x}(t)=0$. That is the direction of $\vec{\omega}(t)$.

As is obvious (for example from the Rodrigues formula), in general $\vec{v}(t)$ and $\vec{\omega}(t)$ are not parallel. So, why are there two axes of rotation, and does $\vec{v}(t)$ play any role in the kinematics/dynamics of the motion?

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I will translate your post into the language of translations. Then I will answer this question about translations. Then I will answer your original question.

Translation Question

I am confused about a trivial concept. Let the displacement of a rigid body be described by the equation $\vec{x}(t)=\Delta \vec{x}(t) +\vec{x}(0)$, with $\Delta \vec{x}(0)=0$.

Then, at each instant there is only one unit vector in the direction of displacement that we may call $\hat{w}(t)$ and which we may take to be normalized. That vector $\hat{w}(t)$ is what geometrically we would call the (instantaneous) direction of velocity [Note: this sentence is wrong].

Kinematically, however, the instantaneous direction of velocity $\vec{v}$ is the derivative $\dot{\vec{x}}(t)=\dot{\Delta \vec{x}}(t)$. That is the direction of $\vec{v}(t)$.

As is obvious (for example an object not moving in a straight line), in general $\hat{w}(t)$ and $\vec{v}(t)$ are not parallel. So, why are there two directions for velocity, and does $\hat{w}(t)$ play any role in the kinematics/dynamics of the motion?

Answer to Translation Question

You are wrong that $\hat{w}$ is the direction of velocity. $\hat{w}$ was defined as the direction of $\Delta \vec{x}$, that is, the direction of the displacement. As Kevin said the total displacement is not needed because you only need to know a objects current position and velocity to get its motion in the future.

Answer to the Rotation Question

Now we are ready to answer the rotation question. We just translate the answer of the translation question to rotation language.

You are wrong that $\vec{v}$ is the direction of angular velocity. $\vec{v}$ was defined as the axis of rotation for $R$, that is, the direction of the rotation between the initial and final orientation. As Kevin said this rotation is not needed because you only need to know a objects current position and velocity (here we are talking about the velocity everywhere in the object, which for a rigid object can be summarized by a linear velocity and an angular velocity) to get its motion in the future.

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  • $\begingroup$ The definition of the axis of rotation as the eigenvector of the rotation matrix is given in Goldstein, 3rd ed., p. 156, third paragraph from the top. The definition of the axis of rotation as the direction of angular velocity is given by Landau & Lifshitz and others. I knew there had to be something wrong with Goldstein's definition. You and Kevin gave me the answer. $\endgroup$ – user32461 Nov 9 '13 at 19:43
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Cool question.

Let's try to formalize the general concepts presented in the other two responses thus far. Intuitively, the main idea is that the instantaneous axis of rotation is determined by the rotation that takes the object from its configuration at one instant to its configuration at the next instant as opposed to the rotation that takes it from its initial configuration to its configuration at some later time. Let's try to make this precise.

As you already wrote, for any time $t$ we have the following for the position $\vec x(t)$ of a point in the body at time $t$: \begin{align} \vec x(t) = R(t)\vec x(0) \tag{1} \end{align} where $R(t)$ is a rotation for each $t$. Now consider the configuration of the object at a neighboring time $t+\epsilon$. How is the position of a point in the body at this later point related to its position at the point $t$? Well, notice that \begin{align} \vec x(t+\epsilon) &= R(t+\epsilon)\vec x(0) \\ &= \Big(R(t) + \dot R(t)\epsilon\Big)\vec x(0) + O(\epsilon^2) \\ &= \Big(I + \epsilon\dot R(t)R(t)^t\Big)\vec x(t) + O(\epsilon^2) \end{align} For small $\epsilon$, the order $\epsilon^2$ terms become insignificant, so the transformation that takes $\vec x(t)$ to $\vec x(t+\epsilon)$ is the gadget in big parentheses. Let's give it a name: \begin{align} \rho(t, \epsilon) = I+\epsilon\dot R(t)R(r)^t \end{align}

But what exactly is this guy? Well, our intuition tells us that it should be a rotation with axis $\vec\omega(t)$; in particular, $\vec\omega(t)$ should be an eigenvector of this transformation. To see that this is the case,

I make the following:

Claim. $\rho(t,\epsilon)$ takes precisely the form of a "small" rotation about the axis $\vec\omega(t)$. In particular, $\vec \omega(t)$ is an eigenvector of $\rho(t,\epsilon)$.

Proof. First, recall that omega can be defined as the vector that generates the motion of the points in the body as follows: \begin{align} \dot{\vec x}(t) = \vec\omega(t)\times\vec x(t) \tag{2} \end{align} But notice, using eq. $(1)$ that \begin{align} \dot{\vec x}(t) = \dot R(t) \vec x(0) = \dot R(t)R(t)^t\vec x(t) \tag{3} \end{align} Now we notice that the matrix $\dot R(t) R(t)^t$ is antisymmetric since $R(t)$ is an orthogonal matrix for each $t$ (I'll leave this for you to prove. It follows that there exist functions $w_x(t), w_y(t), w_z(t)$ for which \begin{align} \dot R(t)R(t)^t = \begin{pmatrix} 0 & -w_z(t) & w_y(t) \\ w_z(t) & 0 & -w_x(t) \\ -w_y(t) & w_x(t) & 0 \\ \end{pmatrix} \end{align} Using this expression and comparing $(2)$ and $(3)$, it is not hard to show that $w_x,w_y,x_z$ are precisely the components of the angular velocity $\vec\omega$. Putting this all together, we see that $\rho(t,\epsilon)$ can be written as follows: \begin{align} \rho(t,\epsilon) = I + \epsilon\vec\omega(t)\times \end{align} where $\vec\omega(t)\times$ is shorthand for the transformation mapping any vector $\vec v$ to $\vec\omega(t)\times v$. Now, it is easy to show that $\vec\omega(t)$ is an eigenvector of $\rho(t,\epsilon)$; \begin{align} \rho(t,\epsilon)\vec\omega(t) = (I+\epsilon \vec\omega(t)\times)\omega(t) = \vec\omega(t) + \epsilon\vec\omega(t)\times\vec\omega(t) = \vec\omega(t). \end{align} It remains to show that $\rho(t,\epsilon)$ has the form of a "small" rotation about the axis $\vec\omega(t)$. Well, what does such a rotation look like? Well, a general rotation by an angle $\theta$ about an axis defined by a unit vector $\hat n$ can be written as follows (I'll again leave it to you to either show this, or find out why this is true): \begin{align} R(\hat n, \theta) = \exp(\theta\hat n\cdot \vec J) \end{align} Where $\vec J = (J_x, J_y, J_z)$ is a vector of matrices, the so-called rotation generators, whose components are defined by \begin{align} (J_i)_{jk}= -\epsilon_{ijk} \end{align} and $\exp$ is the matrix exponential. Notice that to first order in $\theta$, this shows that \begin{align} R(\hat n, \theta) = I + \theta\hat n\cdot J_i \end{align} But now notice that if we apply this to any vector $\vec v$, then we obtain \begin{align} R(\hat n, \theta)\vec v &= (I+\theta\hat n\cdot \vec J)\vec v \\ &= \vec v + \theta n_i (J_i)_{jk}\vec v_k \hat e_j \\ &= \vec v - \theta n_i\epsilon_{ijk} v_k \hat e_j \\ &= \vec v + \theta \hat n\times \vec v \\ &= (I + \theta\hat n\times )\vec v \end{align} We now see by inspection that $\rho(t,\epsilon)$ has precisely the form of the first order approximation to a rotation $R(\hat n, \theta)$ if we simply take $\hat n = \hat \omega(t)$ and $\theta = \epsilon|\vec\omega(t)|$. $\blacksquare$

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  • $\begingroup$ Well said. The exponential can be easily computed exactly for any angle (by summing the series of even and odd powers separately) and it leads to the Rodrigues formula. Once that is done, one can construct $\dot{R}R^T$ explicitly as a function of $\theta(t)$ and $\widehat{n}(t)$, then compute explicitly $\widehat{\omega}$ also as a function of $\theta(t)$ and $\widehat{n}(t)$, and verify that $\widehat{\omega}\neq \widehat{n}$. The question is then: why is $\widehat{n}$ not the axis of the rotation? The answer is the one given by you and the preceding answers. $\endgroup$ – user32461 Nov 9 '13 at 22:18
  • $\begingroup$ +1 all this work, and just one lousy up-vote? $\endgroup$ – Physiks lover Aug 20 '14 at 22:19
  • $\begingroup$ @Physikslover Well that's very kind. Much appreciated! Ultimately I care most about people's understanding improving due to any of my posts, so if that happened with this post for you, then it's worth a lot more than one upvote. $\endgroup$ – joshphysics Aug 20 '14 at 23:20
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R represents the accumulated rotation, while omega is the instantaneous rotation. If you choose a time T, then "glue" v(T) (the real eigenvector of R, not the translational velocity!) as an arrow on the body at t=0, than at time t=T it will be pointing in the same direction. w(t) is just the direction for which the points are not moving at that time t. v and w are not always the same since the rotation axis may change from time 0 to time t (even in the absence of external forces!).

v(t) is not needed for dynamics. Physics has no memory of where an object has been, it only cares about the positions, velocities, masses, etc of the situation now. Our brains only have memory because they actively store information.

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If the orientation of a rigid body is described by a single rotation by an angle $\theta$, lets say about an axis $\hat{z}$ (the eigen vector) then its angular velocity is by definition equal to

$$ \vec{\omega} = \hat{z} \dot\theta $$

In fact, look at this answer (Angular position vector?) to see how to fully develop the angular velocity vector for multiple rotations. To answer your question, the eigen vectors $\hat{z}_i$ play a role in

$$ \vec{\omega}_n = \hat{z}_1 \dot\theta_1 + R_1 \left(\hat{z}_2 \dot\theta_2 + R_2 \left( \ldots R_{n-1} \hat{z}_n \dot{\theta}_n\right)\right) $$

The Rodriguez equation is not sufficient to describe the motion, only to derscribe the orientation. Here is how do get the true motion axis (actually a line for rigid body. At any instant it prescribes a screw motion. This motion is described by the following properties:

  1. 3D Line the rotation is about. This is described with Pluecker coordinates, which can easily provide the direction of the line, as well as a point on the line closest to the origin.
  2. Pitch, (how much parallel translation for a unit rotation)
  3. Magnitude (speed) of rotation.

In fact, a rigid body with measured velocity at a point A of $\vec{v}_A$ and angular velocity $\vec\omega$ the motion properties are found by

  1. Magnitude = $|\vec{\omega}|$
  2. Direction = $\frac{\vec{\omega}}{|\vec{\omega}|}$
  3. Pitch = $\frac{\vec{v}_A \cdot \vec{\omega}}{|\vec\omega|^2}$
  4. Point on line of motion relative to A = $\frac{\vec{\omega} \times \vec{v}_A }{|\vec\omega|^2}$

Note: A pure translation is a degenerate motion screw (twist), with $\vec{\omega}=0$ the same way that a pure torque couple is a degenerate screw force (wrench), with $\vec{F}=0$.

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