3
$\begingroup$

I have a problem with one of my study questions for an oral exam:

Euler’s equation of motion around the $z$ axis in two dimensions is $I_z\dot{\omega}_z = M_z$, whereas it in three dimensions is $I_z\dot{\omega}_z =-(I_y-I_x)\omega_x\omega_y+M_z$, assuming that the $xyz$ coordinate systems is aligned with the principal axis. Why does Euler’s equation of motion for axis $z$ contain the rotational velocities for axes $x$ and $y$?

How can one explain this physically? I mean I can derive Euler's equation of motion, but how can I illustrate that the angular velocities are changing in 3 dimensions?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
4
$\begingroup$

As explained on Wikipedia, the nice tensor form of the equations is $$ \mathbf{I} \cdot \dot{\boldsymbol\omega} + \boldsymbol\omega \times \left( \mathbf{I} \cdot \boldsymbol\omega \right) = \mathbf{M} $$ This reduces to your equations if one diagonalizes the tensor of the moment of inertia $I$ and labels the diagonal entries etc.

The three components are mixed with each other because quantities like $\vec\omega$ and $\vec M$ are really associated with rotations in space and rotations around the axis $x,y,z$ don't commute with each other – unlike translations. Translations commute with each other which is why the 3 components in $\vec F=m\vec a$ don't mix with each other.

For example, take the Earth, rotate it by 90 degrees around the $x$ axis, then 90 degrees around $y$ axis, then you rotate back by 90 degrees but first around $x$ axis again, so that you aren't undoing the $y$ rotation immediately, but then you undo the $y$ rotation, too. You don't get back where you have been: instead, you end up rotating the Earth around the $z$ axis. We say that rotations form the group $SO(3)$ which is non-abelian, $gh\neq hg$. The moment of force wants to rotate the rigid body around an axis but because it was already rotating around another axis given by $\vec \omega$ and the rotations don't commute with each other, the effect of the moment of force also influences the "remaining third" component.

A natural way to write the vectors $\vec \omega, \vec M$ is actually an "antisymmetric tensor" – they're "pseudovectors", not actual vectors. At any rate, when you correctly derive the equations, you should reproduce what Euler got.

Improve this answer
$\endgroup$
6
  • $\begingroup$ It might be good (although this is a very old question and the OP may already have done Nobel-earning work by now!) to point out that the $\boldsymbol\omega \times \left( \mathbf{I} \cdot \boldsymbol\omega \right)$ term is in fact the commutator $[\boldsymbol\omega ,\, \left( \mathbf{I} \cdot \boldsymbol\omega \right)]$ in the Adjoint representation, thus making the comparison with $\vec{F}=m\,\vec{a}$ precise. $\endgroup$
    – Selene Routley
    Commented Mar 11, 2017 at 10:14
  • $\begingroup$ What kind of a commutator? What is the algebra you are working with? Is your comment more than just the confusing renaming of the cross product as a "commutator"? $\endgroup$
    – Luboš Motl
    Commented Mar 11, 2017 at 16:13
  • $\begingroup$ Sorry to cause confusion. What I mean: if your using the Adjoint representation of $SU(2)$ to analyze rotations, then of course your rotation of $X\in\mathfrak{su}(2)$ is $X\mapsto \gamma\,X\,\gamma^{-1}$. So now let $X(t),\,Y(t) \in\mathfrak{su}(2)$ stand for spatial vectors in two relatively rotating frames linked by $\gamma(t)\in SU(2)$, then $Y=\gamma\,X\,\gamma^{-1}$ and $\dot{Y} = \gamma\,\left(\dot{X} + [\Omega,\,X]\right)\,\gamma^{-1}$, where $\Omega =\gamma^{-1}\,\gamma\in\mathfrak{su}(2)$ is the bivector representing the instantaneous angular velocity. $\dot{X} + [\Omega,\,X]$ ... $\endgroup$
    – Selene Routley
    Commented Mar 12, 2017 at 0:09
  • $\begingroup$ ... is of course the usual cross product formula $\mathrm{d}_t \mapsto \mathrm{d}_t + \omega\times$ for time derivatives in relatively rotating frames.Do the same thing for the Euclidean group's Adjoint representation, and the corresponding commutator term is nontrivial for rotations but vanishes for translations. Sorry to cause confusion: my supervisor Prof. John Stillwell was fond of calling $\mathfrak{su}(2)$ the cross product algebra for this reason and I'm not sure how widespread that is. More abstractly, although renaming the cross product a commutator would be confusing and .... $\endgroup$
    – Selene Routley
    Commented Mar 12, 2017 at 0:23
  • $\begingroup$ ...although that's not what I mean here, it wouldn't be too outlandish: the cross product is an abstract Lie bracket, in being billinear, skew-symmetric, and fulfilling the Jacobi identity. Construct a matrix representation thereof (always exists by these properties alone by Ado's theorem, but dead easy to write down from the cross product definition), exponentiate and the cross product thus defines $SO(3)$. $\endgroup$
    – Selene Routley
    Commented Mar 12, 2017 at 0:26
-1
$\begingroup$

If you are familiar aircraft flight dynamics, then please remember an aerobatic maneuver so called "Immelmann turn". Let us take x,y and z axis for longitudinal, lateral and vertical coordinate axes fixed to the aircraft. Then, if we pull control a stick and let the aircraft turn in vertical plane around the y axis and turn the stick to the right or left and make a rolling motion around the x axis at the same time. When the both two angular motion accomplished 180 degree turn, then the aircraft flies in the opposite direction before this maneuver. This means that the aircraft turned 180 degree around the z axis. This is my interpretation of the Euler equation of rigid body motion.

Improve this answer
$\endgroup$

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.