So I am currently reading Fowles and Cassidy and there is something I'm confused about in the section about geometric description of free rotation of a rigid body. I will present the stuff first that I am confused about and then I will ask my question.
The book says
..with zero torque the angular momentum of the body, as seen from the outside must remain constant in direction and magnitude according to the general principle of conservation of angular momentum. With respect to rotating axes fixed in the body , however the direction of the angular momentum may change, altough its magnitude must remain constant.
From those statements we derive two equations
$I_{1}^2w_{1}^2 + I_{2}^2w_{2}^2 + I_{3}^2w_{3}^2 = L^2 = constant$ (Equation 1)
$I_{1}w_{1}^2 + I_{2}w_{2}^2 + I_{3}w_{3}^2 = 2T_{rot} = constant$. (Equation 2)
These are equations of two ellipsoids whose principal axes coincide with the principal axes of the body. The first ellipsoid has principal diameters in the ratios $I_{1}^{-1} : I_{2}^{-1} : I_{3}^{-1}$.
The second ellipsoid has principal diameters in the ratios $I_{1}^{-1/2} : I_{2}^{-1/2} : I_{3}^{-1/2}$.
So
- I don't see how come it follows from conservation of angular momentum must remain constant ?
- I don't understand how come with respect to rotating axes fixed in the body it may experience angular momentum that will change in direction I don't understand and I don't fully visualize it, so it would be perfect if someone can explain this with some also clear visualization of why this happens.
Lastly, I don't see how those two ellipsoids principal axes coincide with the principal axes of the body, also I don't understand what do they mean by "principal diameters" ?
Best Answer
The statement is really about the transformation between inertial co-ordinates and co-ordinates fixed to the body. This is expressed by:
$$D_t = d_t + \omega(t)\times\tag{1}$$
where $D_t$ is the "total" derivative, i.e. the time derivative in the inertial frame, $d_t$ is the time derivative in the frame fixed to the body.
Since there there are no torques on the body, angular momentum is conserved in the inertial frame so that $D_t L=0$. Therefore we get:
$$d_t L = -\omega(t)\times L\tag{2}$$
So, in the frame fixed to the body, the time derivative of $L$ is given by a cross product with $L$, so the time derivative is always at right angles to $L$. Therefore, $|L|$ is constant, but $L$ can certainly change direction by (2): its head is constrained to a sphere.
So this reasoning yields $I_{1}^2w_{1}^2 + I_{2}^2w_{2}^2 + I_{3}^2w_{3}^2 = const.$
The other equation $I_{1}w_{1}^2 + I_{2}w_{2}^2 + I_{3}w_{3}^2 = 2T_{rot}$ does not follow from the paragraph you referenced, but it is correct nonetheless: it is instead a statement of conservation of rotational kinetic energy, not of angular momentum.
BTW (1) is derived from the Leibnitz rule applied to the matrix equation $X=\exp(H(t))\,Y(t)$ where $H$ is a $3\times 3$ skew-symmetric matrix; $Y$ are the co-ordinates in the frame fixed to the body, $X$ are those in the inertial frame and $\exp(H)$ is the total rotation operator: we think of the co-ordinates written as $3\times 1$ column vectors. At $t=0$ (when our co-ordinate frames are instantaneously aligned) this yields $\dot{X} = \Omega(t)\,Y + \dot{Y}$, where $\Omega(t)$ is derived by a complicated Lie-theoretical formula from the skew symmetric $H$, but $\Omega$ is nonetheless a skew-symmetrix $3\times 3$ matrix, and can thus be represented as a cross product $\omega\times Y$.