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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?

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    $\begingroup$ Don't have time at the moment to write up an answer but: Moment of inertia and Moment of Inertia Tensor $\endgroup$ Jun 19 at 18:01
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    $\begingroup$ Does the answer given in Understanding and Expressing the Definition of Inertia Tensor in the Language of Differential Geometry answer your question? $\endgroup$
    – peek-a-boo
    Jun 20 at 7:16
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    $\begingroup$ Note also that Abraham and Marsden have a massive text on classical mechanics, which you might be interested in glossing through (it's a little rough going). A much more readable text I found is Curtis and Miller's Differential Manifolds and Theoretical Physics. There is a chapter on rigid body motion, which you may be interested in. $\endgroup$
    – peek-a-boo
    Jun 20 at 7:28
  • $\begingroup$ @peek-a-boo The answer given in that post is way too difficult for me, I just understand that the Inertia tensor is a $(0,2)$ tensor, so a map which takes two vectors and gives a scalar, but can't understand what vectors it takes. $\endgroup$
    – Salmone
    Jun 20 at 21:56
  • $\begingroup$ You can imagine it just takes two vectors in $E=\Bbb{R}^3$ and produces a number according to a certain integral formula. $\endgroup$
    – peek-a-boo
    Jun 20 at 22:32

2 Answers 2

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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 $$ \vec{L} = \mathrm{I}\, \vec{\omega} $$ $$ \mathrm{I} \equiv \int \left( \vec{r} \cdot \vec{r} - \vec{r} \odot \vec{r} \right) \rho\, {\rm d} V.$$


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 $$ \mathrm{I}_{zz} \equiv \int (x^2+y^2) \rho\, {\rm d} V.$$

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  • $\begingroup$ @mr_e_man - yeah I fixed the outer product. Thank you. $\endgroup$ Jun 20 at 17:52
  • $\begingroup$ I don't understand so much your firt sentence, "A rank 2 tensor is something that relates two vectors", so why don't we talk for example of an operator of inertia to which a matrix can be associated anyway? $\endgroup$
    – Salmone
    Jun 20 at 20:08
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    $\begingroup$ I just stating the definition of a tensor. And for reference read answers on mathematics. A tensor maps vectors into vectors, or said otherwise, is a relationship (a 1:1 mapping in reality) between vectors. $\endgroup$ Jun 20 at 20:36
  • $\begingroup$ For example in the definition $\mathrm{I}_{\rm zz} \equiv \int (x^2+y^2) \rho {\rm d} V$, what are the two vectors the tensor "takes"? $\endgroup$
    – Salmone
    Jun 20 at 21:52
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    $\begingroup$ @Salmone - ${\rm I}_{\rm zz}$ is just a single component of the MMOI tensor. This scalar value can itself be considered a rank 0 tensor, relating rotational speed along the z-axis to angular momentum along the z-axis. $\endgroup$ Jun 21 at 12:38
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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 $I$.

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  • $\begingroup$ I understood the part about the 3D speech but I still can't understand why $I$ is a tensor. $I$ could be represented by a $3x3$ matrix but why is it a tensor? Couldn't it be just a linear functional? $\endgroup$
    – Salmone
    Jun 19 at 21:34
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    $\begingroup$ While there is a mathematical difference between a matrix and a tensor, in this context it matters very little. In physics, the difference is mostly in the behavior with relation to rotations: roughly speaking, the tensor must transform properly under the rotation to keep the same physical meaning before and after (see passive and active transformations in group theory). $\endgroup$
    – Miyase
    Jun 19 at 21:36
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    $\begingroup$ You'll probably want to open a new question for that. In a nutshell, mathematically it's a multilinear map. But in physics we add an additional constraint, in that it must behave in a certain way under rotations, i.e. be part of a representation of the group of rotations in $\mathbb{R}^3$. It's important so that we can write formulas that don't depend on the choice of coordinates. $\endgroup$
    – Miyase
    Jun 19 at 21:42
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    $\begingroup$ They can be non zero, they simply vanish from the equation because the rotation vector only has non-zero component on $z$, and because angular momentum is projected on $z$. See my answer, all tensor elements except $k$ vanish from the result without having to be zero. $\endgroup$
    – Miyase
    Jun 20 at 15:16
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    $\begingroup$ The very definition of the moment of inertia prevents it from being zero. It means that all three eigenvalues of the inertia tensors (the three moments of inertia) are strictly positive. $\endgroup$
    – Miyase
    Jun 20 at 15:32

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