24.6: Definition of a Tensor

We have a definite rule for how vector components transform under a change of basis: $x_{i}^{\prime}=R_{i j} x_{j}$. What about the components of the inertia tensor $I_{i k}=\sum_{n} m_{n}\left(x_{n l}^{2} \delta_{i k}-x_{n i} x_{n k}\right)$?

We’ll do it in two parts, and one particle at a time. First, take that second term for one particle, it has the form $-m x_{i} x_{k}$. But we already know how vector components transform, so this must go to

$$\begin{equation} -m x_{i}^{\prime} x_{k}^{\prime}=R_{i l} R_{j m}\left(-m x_{l} x_{m}\right) \end{equation}$$

The same rotation matrix $R_{i j}$ is applied to all the particles, so we can add over n.

In fact, the inertia tensor is made up of elements exactly of this form in all nine places, plus diagonal terms $m r_{i}^{2}$, obviously invariant under rotation. To make this clear, we write the inertia tensor:

$$\begin{equation} \left[\begin{array}{ccc} \sum m\left(y^{2}+z^{2}\right) & -\sum m x y & -\sum m x z \\ -\sum m x y & \sum m\left(z^{2}+x^{2}\right) & -\sum m y z \\ -\sum m x z & -\sum m y z & \sum m\left(x^{2}+y^{2}\right) \end{array}\right]=\sum m\left(x^{2}+y^{2}+z^{2}\right) \mathbf{1}-\left[\begin{array}{c} \sum m x^{2} \sum m x y & \sum m x z \\ \sum m x y & \sum m y^{2} & \sum m y z \\ \sum m x z & \sum m y z & \sum m z^{2} \end{array}\right] \end{equation}$$

where 1 is the 3×3 identity matrix. (Not to be confused with $I!$)

Exercise: convince yourself that this is the same as $\begin{equation} \mathbf{I}=\sum m\left[\left(\mathbf{x}^{\mathbf{T}} \mathbf{x}\right) \mathbf{1}-\mathbf{x} \mathbf{x}^{\mathbf{T}}\right] \end{equation}$

This transformation property is the definition of a two-suffix Cartesian three-dimensional tensor: just as a vector in this space can be defined as an array of three components that are transformed under a change of basis by applying the rotation matrix, $x_{i}^{\prime}=R_{i j} x_{j}$, a tensor with two suffixes in the same space is a two-dimensional array of nine numbers that transform as

$T_{i j}^{\prime}=R_{i l} R_{j m} T_{l m}$

Writing this in matrix notation, and keeping an eye on the indices, we see that with the standard definition of a matrix product, $(\mathbf{A B})_{i j}=\mathbf{A}_{i k} \mathbf{B}_{k j}$

$$\begin{equation} \mathbf{T}^{\prime}=\mathbf{R} \mathbf{T} \mathbf{R}^{\mathbf{T}}=\mathbf{R} \mathbf{T} \mathbf{R}^{-\mathbf{1}} \end{equation}$$

(The transformation property for our tensor followed immediately from that for a vector, since our tensor is constructed from vectors, but by definition the same rule applies to all Cartesian tensors, which are not always expressible in terms of vector components.)