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Euler's theorem of rotation states that any rigid body motion with one point fixed is equivalent to a rotation about some axis passing through that fixed point. Now it is often said that Euler's theorem of rotations implies the existence of an instantaneous axis of rotation. My question is, how can we prove that the instantaneous axis of rotation exists?

Consider a rigid body which is undergoing some motion with one point fixed. For any given times $t_1$ and $t_2$, let $\hat{u}(t_1,t_2)$ denote the unit vector parallel to the axis of the rotation that's equivalent to the motion of the the rigid body between time $t_1$ and time $t_2$. Then how can we prove that the limit of $\hat{u}(t_1,t_2)$ as $t_2$ goes to $t_1$ exists?

Also, having proved the limit does exist, how can we prove that unit vector we get points in the direction of the angular velocity vector of the rigid body?

This would all be much easier to prove if the angular velocity vector were the derivative of the angular displacement vector, but things don't work out so smoothly because rotations don't commute; see this journal paper for details.

EDIT: I just posted a follow-up question here.

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Keshav Srinivasan
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  • prove that it exists? Isn't that like proving that a coordinate system exists? And what is meant by instantaneous in terms of rotation axes? – docscience Oct 11 '16 at 21:52
  • @docscience No, it's not like proving that a coordinate system exists. It's about proving that the limit of limit of $\hat{u}(t_1,t_2)$ as $t_2$ goes to $t_1$ exists. As far as what is meant by an instantaneous axis of rotation, basically if you have a rigid body undergoing some motion with one point fixed, then at any given time $t$, there exists some axis passing through the fixed point such that at that exact time $t$, the rigid body will move as if it was rotating about that axis. – Keshav Srinivasan Oct 11 '16 at 22:33
  • @docscience To take a concrete example, if you watch a hula hoop rolling on its side, then at any given time the velocity of the part of the hula hoop touching the ground is 0, so you can think of the hula hoop's motion at that instant as a rotation about an axis passing through the point of contact, and pointed perpendicular to the plane of the hula hoop. – Keshav Srinivasan Oct 11 '16 at 22:34
  • It depends, for a rigid body do you still consider the transformation $$\vec{v}_A = \vec{v}_B + \vec{r} \times \vec{\omega}$$ because that implies an axis of rotation on it own. – John Alexiou Oct 11 '16 at 23:02
  • @docscience it is a good question, because in 4-dimensions NO rotation axis has to exist. For example, if $(x_1,x_2,x_3,x_4)$ is a point on a unit sphere, $(x_2,-x_1,x_4,-x_3)$ is a vector field of an instantaneous rotation that leaves no axis invariant. So it's very specific to three dimensions. (If you generalize it a bit, then it's specific to odd dimensions). –  Oct 12 '16 at 00:41
  • @NeuroFuzzy Yes I agree. I do prefer working with quaternions over Euler transformations. A good way to avoid singularities and lock-up. – docscience Oct 12 '16 at 14:47
  • @docscience You may be interested in my follow-up question: http://physics.stackexchange.com/q/285843/27396 – Keshav Srinivasan Oct 12 '16 at 15:26
  • @KeshavSrinivasan thanks. I left another question for you there. – docscience Oct 12 '16 at 16:36

1 Answers1

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Find the locus of points with no motion on a rigid body, if point A is fixed.

Without loss of generality place a coordinate system on A. Rigid body motion is defined if an arbitrary point maintains constant distance from A $$d = \sqrt{x^2+y^2+z^2}$$

This is found by setting $\frac{{\rm d}d}{{\rm d}t}=0$ which by the chain rule (with time derivative of $\frac{1}{2}d^2$) implies $$ x \frac{{\rm d}x}{{\rm d}t} + y \frac{{\rm d}y}{{\rm d}t}+ z \frac{{\rm d}z}{{\rm d}t} = 0$$

In vector form the above is $$\vec{r} \cdot \vec{v} =0$$ with $\vec{r}=(x,y,z)$ and $\vec{v} = \frac{{\rm d}\vec{r}}{{\rm d}t}$.

So the velocity field has to be perpendicular to the location vector. An obvious solution to the above is

$$ \vec{v} = \vec{\omega} \times \vec{r}$$

$$ \vec{r} \cdot (\vec{\omega} \times \vec{r}) \equiv 0 $$

So we have established that vector a velocity field of the form $\vec{v} = \vec{\omega} \times \vec{r}$ describes rigid body motion. This also imposes the limitation that $\vec{\omega}$ is constant throughout the body because otherwise the above expression wouldn't be zero. Try it.

Now let's look at the points parallel to $\vec{\omega}$ with $\vec{r} = \lambda \,\vec{\omega}$. The velocities are

$$\vec{v}_\parallel = \vec{\omega} \times (\vec{r}) = \vec{\omega} \times \lambda \,\vec{\omega} \equiv 0$$

Now let's look at points perpendicular to the direction of $\vec{\omega}$ with a distance $d$

$$\vec{r} = \lambda \vec{\omega} + d \vec{n}$$ with the properties $\| \vec{n} \|=1$ and $\vec{n}\cdot\vec{\omega}=0$

$$\vec{v} = \vec{\omega} \times (\vec{r}) = \vec{\omega} \times ( \lambda \,\vec{\omega} + d \,\vec{n}) = d\,(\vec{\omega} \times \vec{n})$$

which is a vector perpendicular to both $\vec{\omega}$ and $\vec{n}$. This implies a tangential (hoop) direction for the velocity with the magnitude increasing linearly with distance.

We call this motion rotation.

Here is an interesting fact. If at some arbitrary point located at $\vec{r}$ the velocity vector is $\vec{v}$ and the body is rotating with $\vec{\omega}$ then the instant axis of rotation is located at

$$\vec{q} = \vec{r} + \frac{\vec{\omega} \times \vec{v}}{\| \vec{\omega} \|^2}$$

The proof is that $\vec{v} = \vec{\omega} \times (\vec{r}-\vec{q}) \rightarrow$

$$\require{cancel} \begin{align} \vec{v} & = \vec{\omega} \times \left(\vec{r}-\left(\vec{r} + \frac{\vec{\omega} \times \vec{v}}{\| \vec{\omega} \|^2}\right) \right) \\ & = -\frac{\vec{\omega} \times(\vec{\omega} \times \vec{v})}{\| \vec{\omega} \|^2} = -\frac{ \vec{\omega}( \vec{\omega} \cdot \vec{v} ) - \vec{v} (\vec{\omega} \cdot \vec{\omega})}{\| \vec{\omega} \|^2} = \frac{ \vec{v} \| \vec{\omega}\|^2}{\| \vec{\omega} \|^2} - \frac{\vec{\omega}(\vec{\omega}\cdot\vec{v})}{\| \vec{\omega} \|^2} \\ \vec{v} &= \vec{v} -\frac{\vec{\omega}(\cancel{\vec{\omega}\cdot\vec{v}})}{\| \vec{\omega} \|^2} & \vec{v} \equiv \vec{v} \end{align}$$

John Alexiou
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    Yeah, I agree that this shows that the rigid body is undergoing what we would call a rotation. But I'd like to connect this to Euler's theorem of rotations; I'd like to show that the limit of $\hat{u}(t_1,t_2)$ as $t_2$ goes to $t_1$ is parallel to the angular velocity vector, and first of all that this limit even exists. – Keshav Srinivasan Oct 12 '16 at 04:16
  • By the way, I just posted a follow-up question here: http://physics.stackexchange.com/q/285843/27396 – Keshav Srinivasan Oct 12 '16 at 14:16
  • Hey @John Alexion , could you explain what you meant here "Obvious solution to the above is" – tryst with freedom Jan 23 '21 at 21:28
  • The cross product produces a vector that is perpendicular to both arguments. Thus it is the obvious choice for finding the vector $\vec{v}$ that has $\vec{v} \cdot \vec{r} = 0$. – John Alexiou Jan 23 '21 at 22:01