Why do we talk about inertia tensor?

Question

When we talk about the inertia of a rigid body, in calculating the angular momentum as a function of the moment of inertia and angular velocity, the inertia tensor is introduced. But why is it a tensor?

The moment of inertia is defined as $$I=\int_V \! r^2 \, \mathrm{d}m.$$ where $r^2$ is the squared distance of the $dV$ from the axis of rotation and $m$ is the mass, why is $I$ a tensor?

According to "Manifolds, Tensor Analysis and Applications - Marsden, Ratiu and Abraham" a tensor over a Banach space $E$ is a multilinear map defined on the cartesian product of $r$ dual space $E*$ and $p$ space $E$ which takes values on $\mathbb{R}$. I can't see this definition on $I$.

$I$ seems to me a linear functional from $\mathbb{R}^3$ to $\mathbb{R}^+$ because it takes a function $r^2=x^2+y^2+z^2$ and gives out a scalar.

The question is:

Why we talk about a tensor of inertia and not a linear operator of inertia or a linear functional of inertia?

Answer 1

A rank 2 tensor is something that relates two vectors. In this case, the MMOI tensor relates the rotational velocity vector to the angular momentum vector.

Given a solid whose internal particles are designated with $\vec{r}$ in relation to the center of mass, you have the following volume integral to find the angular momentum of the body. $$ \vec{L} = \int \vec{r} \times (\vec{v}\, {\rm d}m) = \int \vec{r} \times (\vec{v} \rho \,{\rm d} V). $$ Here $\vec{v} = \vec{\omega} \times \vec{r}$ is the motion of each particle, so $$ \vec{L} = \int [\vec{r} \times ( \vec{\omega} \times \vec{r})] \rho\, {\rm d} V.$$ With the vector identity $\vec{a}\times ( \vec{b} \times \vec{c}) = \vec{b} ( \vec{a} \cdot \vec{c}) - \vec{c} ( \vec{a}\cdot \vec{b}) = ( \vec{a}\cdot \vec{c}) \vec{b} - (\vec{c} \odot \vec{a}) \vec{b} $, where $\cdot$ is the inner (dot) product, and $\odot$ is the outer product, $$ \vec{L}= \int \left( \vec{r} \cdot \vec{r} - \vec{r} \odot \vec{r} \right) \vec{\omega}\, \rho\,{\rm d} V.$$ And the integral is factored by the mass moment of inertia tensor $$ \vec{L} = \mathrm{I}\, \vec{\omega} $$ $$ \mathrm{I} \equiv \int \left( \vec{r} \cdot \vec{r} - \vec{r} \odot \vec{r} \right) \rho\, {\rm d} V.$$

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If $\vec{r} = \pmatrix{x \\y \\ z}$ then the integral is $$ \mathrm{I} \equiv \int \begin{bmatrix} y^2+z^2 & -x y & -x z \\ -x y & x^2+z^2 & - y z \\ -x z & - y z & x^2+y^2 \end{bmatrix} \rho\, {\rm d} V, $$ from which you are familiar with the 2D version $$ \mathrm{I}_{zz} \equiv \int (x^2+y^2) \rho\, {\rm d} V.$$

Answer 2

The moment of inertia you mentioned is only for a single, given axis of rotation. It's used to compute the angular momentum related to his axis:

$$L=I\omega$$

If you want to generalize the result in 3D space to compute the vector angular momentum as a function of the rotation vector, then:

$$\vec{L}=I\vec{\omega}$$

$I$ needs to become a $3\times 3$ matrix, which is called the inertia tensor.

If the rotation happens around a single axis $z$, you can choose a basis where one of the base vectors is alongisde this axis. Then the expression simplifies to:

$$L_z =\vec{L}.\vec{e}_z= \begin{pmatrix} a & b & c\\ e & f & g\\ i & j & k \end{pmatrix} \begin{pmatrix} 0\\ 0\\ \omega \end{pmatrix}.\vec{e}_z =k\omega$$

which is "single axis" result that you mentioned.

The basis in which $I$ is diagonal defines a specific set of axes that is sometimes useful is study the dynamics of the solid. In this basis, the (scalar) definition of $I$ that you gave can be used for each axis to compute three moments of inertie that are the eigenvalues of $I$.