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.


2 Answers 2


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? $\endgroup$ Commented Jan 2, 2019 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. $\endgroup$ Commented Jan 2, 2019 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
    Commented Jan 2, 2019 at 22:04
  • $\begingroup$ @DisplayName Okay I get it now. So the correct equation would be $L(t)=I(t)\omega(t)$. $\endgroup$ Commented Jan 3, 2019 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 $b_j$.


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