# What are the details that link an "inertia tensor" of a rigid body at a given point with the mathematical definition of a tensor?

I know how to operate with the matrix representation of an inertia tensor, understand how it changes with respect to basis transformations, but I'm having a hard time framing the concept of the inertia tensor of a rigid body at a given point in the larger framework of the mathematical definition of a tensor, at least with regard to this definition:

A $$k$$-tensor is a multilinear function from $$V×V×⋯×V$$ to the reals, where $$V$$ is a vector space and $$k$$ is the number of the $$V$$'s in the above Cartesian product. (Calculus on Manifolds, Michael Spivak, 1965, page 75).

In the case of an inertia tensor, which is of rank 2: given a rigid body $$B$$ and some point $$P$$, what is the inertia tensor $$T$$ of $$B$$ at $$P$$, as in what specific multilinear map is it? Which is the vector space $$V$$ such that $$V\times V$$ is the domain of this tensor? What is the significance of the real number $$T(v_1,v_2)$$, for any given vectors $$v_1$$ and $$v_2$$?

I‘ll first try to explain a bit more in depth than Spivak what tensors are, why they are cool and then tell you why the inertia tensor is one.

Let’s consider a vectorspace $$V$$ over let‘s say the real numbers $$\mathbb{R}$$, before we talk about tensors you need to be familiar with the dual space of $$V$$, denoted $$V^*$$. $$V^*$$ is the space of all linear functions that assign numbers to vectors so for $$f \in V$$, we have $$f: V \rightarrow \mathbb{R}$$. For a given basis of $$V$$ let‘s say $$\{e_1, \dots, e_n \}$$ we get a basis of $$V^*$$ often denoted $$\{e_1^*, \dots, e_n^* \}$$ the elements in that basis are uniquely determined by the property that: $$e_i^*(e_j)=\delta_{ij}$$.

Now we can consider the space of multilinear maps from some copies of $$V$$ and $$V^*$$ to the reals, i.e.:

$$\begin{equation} V^* \times \dots \times V^* \times V \times \dots \times V \rightarrow \mathbb{R} \end{equation}$$

But multilinear maps are lame, we understand linear maps much better. Instead of having a multilinear map that takes many different vectors and dual vectors as input we would like to have a linear map that takes one massive vector as input and this one vector should encompasses all the information of the other vectors combined. Where would this giant vector come from? From the tensor product space! In mathematics the tensor product space is even defined as the space that is so large that a single element in it can replace all the different vectors in the input of a multilinear map. This space is denoted:

$$\begin{equation} V^* \otimes \dots \otimes V^* \otimes V \otimes \dots \otimes V \end{equation}$$

And the space of multilinear maps could thus be written as:

$$\begin{equation} V^* \otimes \dots \otimes V^* \otimes V \otimes \dots \otimes V \rightarrow \mathbb{R} \end{equation}$$

But notice, we now have a map from a vector space (the tensor product space) to the real numbers, elements (these are the maps) of this vector space are elements of the dual space, so this is actually nothing but:

$$\begin{equation} (V^* \otimes \dots \otimes V^* \otimes V \otimes \dots \otimes V)^* \end{equation}$$

Now we can use that „the „$$^*$$“ is distributive“ and that $$V^{**} \cong V$$, so the space above can actually be written as:

$$\begin{equation} V \otimes \dots \otimes V \otimes V^* \otimes \dots \otimes V^* \end{equation}$$

Elements of this space are denoted tensors of rank (r,s) where r is the number of copies of $$V$$ and s the number of copies of $$V^*$$

Almost all objects in linear algebra can be described that way, for example linear maps are tensors of rank (1,1), i.e. elements in $$V \otimes V^*$$, why is that?

Well I can prove it to you we only need to define how an object in $$V \otimes V^*$$ acts on vectors, lets take an $$v \in V$$ and a $$x \otimes y^* \in V \otimes V^*$$ and we define $$x \otimes y^* (v):=y^*(w)x$$ why is this a linear map? Because for $$v,w \in V$$: $$x \otimes y^* (v+w):=y^*(w+v)x= (y^*(w)+y^*(v))x= y^*(w)x+y^*(v)x= x \otimes y^* (v)+ x \otimes y^* (w)$$, where we used that $$y^*$$ is a linear map.

Remember the basis $$\{e_1, \dots, e_n \}$$ and $$\{e_1^*, \dots, e_n^* \}$$? It turns out that $$\{e_1 \otimes e_1^*, e_1 \otimes e_2^*, \dots, e_2 \otimes e_1^*, \dots e_n\otimes e_n^* \}$$ is a basis of $$V \otimes V^*$$ general elements in $$V \otimes V^*$$ are linear combinations, i.e.:

$$\sum\limits_{i,j} A_{ij}e_i \otimes e_j^*$$

Where the $$A_{ij}$$ are the coefficients of the linear combination, i.e. just numbers. In physics people often only write the coefficients and ignore the basis vectors, putting the j as a subscript, indicating that it belongs to a dual vector and the i as a superscript, indicating that it belongs to a „normal“ vector.

$$\sum\limits_{i,j} A^{i}_{\ j}$$

And then they don‘t even write the sum, and call it Einstein sum convention:

$$A^{i}_{\ j}$$

and call the object above a tensor of rank (1,1).

So why is the inertia tensor a tensor? Because it is a linear map, i.e. an element of $$V \otimes V^*$$, i.e. a rank (1,1) tensor. It assigns to the vector that indicates the objects rotation $$\omega$$ the vector that indicates its angular momentum $$L$$. But it can also be thought used in a different way, as a (0,2) tensor, i.e. an element of $$V^* \otimes V^*$$, because if you use it to calculate an objects rotational energy you feed it two vectors, this operation is often written as $$\omega^T I \omega$$ (The two vectors just happen to be the same) and out comes a number, its Energy.

• You can also feed it a unit vector $\hat n$ and get the moment of inertia around the axis defined by $\hat n$. Oct 23, 2022 at 3:39