In Goldstein's Classical Mechanics book in the chapter about the dynamics of rigid bodies the equation

$$\dfrac{dL_i}{dt}+\epsilon_{ijk}\omega_jL_k = N_i$$

is presented. Now, in one exercise, we are asked to show that for general nonrigid motion, if the rotating axes coincide with the principal axes of inertia then the equation reduces to



$$I_i = \int dV \rho(\mathbf{r})\epsilon_{ijk}x_jv_k',$$

being $\mathbf{r}$ a point in the body and $\mathbf{v}'$ the velocity of the point $\mathbf{r}$ with respect to the body set of axes.

Now, by definition of the inertia tensor, one has $I(\omega)=\mathbf{L}$. In that case, if one picks the inertia tensor in the body basis, since the basis vectors point along the principal axes the matrix of $I$ is diagonal. In that case

$$L_r = I_r\omega_r,$$

being $\omega_r$ the components of $\omega$ on the body basis.

But something is wrong here. With this I can only get

$$\dfrac{d(I_i\omega_i)}{dt}+\epsilon_{ijk}\omega_j\omega_k I_k = N_i.$$

The term with the derivative is missing. Also, I have no idea how $I_i$ can be found to be given by that formula.

I think I may be confusing the components of things in different bases.

So, what am I doing wrong here? How can one derive that relation and show that $I_i$ can be written that way?

  • $\begingroup$ It's a typo. Your equation is correct. $\endgroup$ – user17116 Oct 28 '15 at 20:25
  • $\begingroup$ But then how does one show that $I_i$ can be written like that? The only thing I know is that if $\{\mathbf{u}_i\}$ are the basis vectors of the rotating system, then $I(\mathbf{u}_i)=I_i \mathbf{u}_i$, but this doesn't seem to get me very far. $\endgroup$ – user1620696 Oct 28 '15 at 22:03
  • $\begingroup$ Google Goldstein errata. You should find more details there. $\endgroup$ – user17116 Oct 28 '15 at 22:04
  • $\begingroup$ The second term seems redundant as when the angular velocity is aligned with a principal axis it is zero. $\endgroup$ – Borun Chowdhury Jan 24 '17 at 9:09

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