I am trying to understand the inertia tensor of rigid bodies but I don't quite understand how it is derived. This is what I tried:

Consider a rigid body consisting of $N$ point masses acted upon by forces such that the centre of mass isn't moving (the body is only rotating). Let $(b_j)_{j=1}^N \subseteq \mathbb{R^3}$ be the positions of the point masses at some time and $(m_j)_{j=1}^n \subseteq \mathbb{R^+}$ their masses. Then the position $x_j: \mathbb{R} \to \mathbb{R}^3$ of the $j$-th point mass is given by $x_j(t)=B(t)b_j$ for some $B: \mathbb{R} \ni t \mapsto B(t)\in SO(3)$. The angular momentum is $$L(t)=\sum_{j=1}^{n}m_j x_j(t) \times \dot{x_j}(t)=\sum_{j=1}^{n}m_j x_j(t) \times (\omega(t) \times x_j(t)) = \sum_{j=1}^{n} m_j (||b_j||^2\omega(t) - \langle \omega(t) \ , \ B(t)b_j\rangle B(t)b_j).$$ As I understand it the inertia tensor $I \in \mathbb{R}^{3 \times 3}$ satisfies $L(t)=I\omega(t)$. I can see that for fixed $t \in \mathbb{R}$ the map $\omega(t) \mapsto L(\omega(t))$ is linear, but why is $I$ independent of $B$? The expression for $L$ contains $B(t)$ still. The inertia tensor of a body should be the same no matter how it is rotating.


The inertia tensor of a body will change with rotation. The easiest way to rotate a rod is about its axis, and if I turn the rod on its side the same thing will be true along the new axis.

Here is a derivation of the inertia tensor:


Take a look at the integral for the component $I_{xy}$,

$$ I_{xy} = -\int xy \, \mathrm{dm} $$

If I rotate $x\rightarrow y$ and $y \rightarrow -x$, $I_{xy}$ is changed to $-I_{xy}$, indicating that the inertia tensor changes with rotation.

  • $\begingroup$ So how can $L = I\omega$ hold? $\endgroup$ Jan 2 '19 at 20:40
  • $\begingroup$ In general it only holds instantaneously, but one example otherwise is when $\omega$ happens to be an eigenvector of $I$, and that eigenvector isn't changing with rotation: then, it can hold with the same $I$ as the object rotates. Going back to the rod example, rotating the rod about its axis does not change its inertia tensor. $\endgroup$ Jan 2 '19 at 20:42
  • $\begingroup$ If the body rotate you transform the inertia with the rotation matrix $R$. $I\mapsto R^{T}\left( \overrightarrow {\varphi }\right) IR\left( \overrightarrow {\varphi }\right) $ $\endgroup$
    – Eli
    Jan 2 '19 at 22:04
  • $\begingroup$ @DisplayName Okay I get it now. So the correct equation would be $L(t)=I(t)\omega(t)$. $\endgroup$ Jan 3 '19 at 16:35

Your angular momentum $L(t)$ is still represented in the non-rotating coordinate frame. If you transform the coordinates of $L(t)$ to the rotating frame by using the linear transformation $B(t)^T$, that is, $$B(t)^TL(t)=L(t)_{rot}$$ you obtain a constant inertia tensor while having $\omega(t)$ transform to $B(t)^T \omega(t)=\omega(t)_{rot}$. You need to play with the inner product and the adjoint operator to have the inner product in terms of $\omega(t)_{rot}$ and $b_j$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.