# Derivation of the inertia tensor

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:

http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node64.html

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.

• So how can $L = I\omega$ hold? Jan 2 '19 at 20:40
• 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. Jan 2 '19 at 20:42
• If the body rotate you transform the inertia with the rotation matrix $R$. $I\mapsto R^{T}\left( \overrightarrow {\varphi }\right) IR\left( \overrightarrow {\varphi }\right)$
– Eli
Jan 2 '19 at 22:04
• @DisplayName Okay I get it now. So the correct equation would be $L(t)=I(t)\omega(t)$. 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 $$b_j$$.