In Wikipedia's article on moments of inertia, I read that the product of inertia $I_{xy}$ can be interpreted as the moment of inertia about the x-axis when the object rotates around the y-axis. I don't really understand this. Could somebody give me more details?


1 Answer 1


This is related to an object in physics called tensor. You are probably familiar with vectors, which are objects with both direction and magnitude. Look for example at the momentum vector in 2D

$$\vec{p}=\begin{pmatrix}p_x \\ p_y\end{pmatrix}$$

It has two components and each component is associated with a specific direction in space. Consider now the following situation: imagine a fluid which flows in 2D, and you are interested in the momentum flux at each point in space. But wait, what is momentum flux? Flux is the flow of some quantity through a surface. In this case, we are looking at the flow of momentum through lines in 2D. Lets consider a specific line, for instance $x=x_{0}$, and try to see what flux passes it. Imagine the fluid flows with constant velocity at angle $45^{\circ}$ to the $x$ axis. Then each fluid particle that passes this imiginary line has momentum both in the $x$ direction and in the $y$ direction. It follows that through lines of constant $x$ the momentum flux has two components. The same is true for lines with constant $y$. It means that in order to describe the momentum flux fully, we need four numbers, which we tend to write in the form of a matrix

$$T=\begin{pmatrix} T_{xx} & T_{xy} \\ T_{yx} & T_{yy}\end{pmatrix}$$

Here $T_{ij}$ is the flux of the $i$-th component of the momentum through surface of constant $j$. This quantity is what we call a tensor.

The same happens in your case. The response of a material to rotation is related to its moment of inertia. In first-year undergraduate mechanics one usually treats bodies that spins about one of their principal axes. Thus $I$ is a scalar and the angular momentum of the body is given by $L=I\omega$. However, for a general body that spins, you can get angular momentum that is not aligned with the angular velocity. It means, as you said, that rotation in one direction can induce angular momentum in another direction. Lets consider the following case more closely

enter image description here

In the gif taken from here you can see an inclined disk rotating about the $\hat{z}$ axis. Our first goal is to find this disc moment of inertia. It is pretty clear that in the principal axes coordinates this tensor takes the form


In order to find the new moment of inertia, we need to change the basis. This is done using a rotation matrix of angle $\theta$ about the $x$ axis (for example)


such that the new $\hat{I}^{\prime}$ is given by


where I assumed for simplicity that $\theta=45^{\circ}$. Now if you rotate this body around the $\hat{z}$ axis with $\vec{\omega}^{\prime}=\omega\hat{z}$, you get

$$\vec{L}^{\prime}=\hat{I}^{\prime}\omega^{\prime}=\begin{pmatrix}0\\ \frac{I_1-I_0}{2}\\ \frac{I_1+I_0}{2}\end{pmatrix}\omega$$

You can see that the off-diagonal elements are merely the result of choosing this particular coordinate system. This means that you actually rotate your body in two different axes. You could, in principle, use $\hat{I}$ and rotate $\omega$ to get $$\vec{L}=\begin{pmatrix}I_{0}&0&0\\0&I_{0}&0\\0&0&I_{1}\end{pmatrix}\begin{pmatrix}0\\ -\frac{\omega}{\sqrt{2}}\\ \frac{\omega}{\sqrt{2}}\end{pmatrix}=\begin{pmatrix}0\\ -\frac{I_0}{\sqrt{2}}\\ \frac{I_1}{\sqrt{2}}\end{pmatrix}\omega$$

which is the same as before, written in the unprimed coordinates.

You could have asked what the difference between this and mass for example, why mass is not a tensor in classical mechanics. It is because mass is an intrinsic property of a bunch of material, in contrary to the moment of inertia that depends on the material configuration in space. Thus the same body can have different moment of inertia about different axes, and that leads to the off-diagonal elements in some coordinate systems.

  • $\begingroup$ The question is asking for a physical interpretation. This answer is more mathematical than physical. $\endgroup$ Commented Nov 14, 2017 at 12:36
  • $\begingroup$ @eranreches so, can we say tensor of second rank such as moment of inertia are nothing but just liner transformation matrices which explain transformation of certain physical quantity? So, what about tensor of 3rd and 4th rank? Please help me to understand . $\endgroup$ Commented Nov 11, 2022 at 5:45

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