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In classical mechanics about rotation of rigid object, the general problem is to study the rotation on a given axis so we need to figure out the moment of inertia around some axes. In 3-dimensional case, we have a matrix (i.e. moment of inertia tensor)

$$ I = \left( \begin{matrix} I_{xx} & I_{xy} & I_{xz}\\ I_{xy} & I_{yy} & I_{yz}\\ I_{zx} & I_{zy} & I_{zz} \end{matrix} \right) $$

I am curious what's the physical significance of the matrix element. I guess the moment of inertia in element $ij$ is the moment of inertia when the object is rotating about the axis $ij$. For example, $I_{xy}$ is the moment of inertia when the object is rotating about $xy$ axis and $I_{yy}$ is the moment of inertia when the object is rotating about $y$ axis, is that correct?

When I read further into the text, it then introduce the a method to diagonalize the moment of inertia tensor such that the non-vanishing elements only appears on diagonal. In that case, the text call the diagonal elements as the principal moment of inertia, my question is what's the physical significance of the principal moment of inertia?

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What, exactly, do you consider the $xy$ axis to be??? In a body-fixed frame you can always choose coordinates to make the inertia tensor diagonal. Then the diagonal components mean what you say - the moment of inertia for rotations about one of the principle axes. In other frames the inertia tensor will be off diagonal. This just represents the fact that you are using coordinates which are rotated relative to the principle axes, which are adapted to the shape of the body. –  Michael Brown Apr 12 '13 at 4:30
I think it's reducing 9 rotation axes to 3 components by finding the eigenvectors, so finding the three (orthogonal, likely), uncorrelated axes into which all rotations in the entire space can be decomposed. Looks like it's directly analogous to principal components, about which there is some significant literature: en.wikipedia.org/wiki/Principal_components –  Paul Apr 12 '13 at 4:31
The special directions are "principal" not "principle" axes. –  DarenW Apr 12 '13 at 5:01
@DarenW Thank you, of course you're right. Derp on my part - I'm in a bad headspace today. :) Unfortunately I can't edit my comment. –  Michael Brown Apr 12 '13 at 5:06
I feel like there's still something missing from the responses so far. Consider the analogy with the stress tensor. Yes we can diagonalize it. But the off-diagonal components do have a physical interpretation, as they are shears (compared to the compression/tensile stresses appearing on the diagonal). They are not just artifacts of a poor choice of coordinates. –  Chris White Apr 12 '13 at 5:19

2 Answers 2

Your guess is wrong, because first of all, there is no "xy"-axis, there's only an "xy"-plane, the plane perpendicular to the z-direction.

As you note, your text describes how the matrix can be made diagonal. Not sure how well the text explains this, but the way to get the matrix to be diagonal is to go to a new coordinate system: You started out with $x$, $y$ and $z$ in "some" way. But your initial $x, y$ and $z$ might not be the "natural" axis of your object. If you go to new axes $x', y'$ and $z'$ such that the tensor is diagonal, you are now in a coordinate system where the object will rotate freely only around those new axes. That's why they are called the principal axes.

The nice thing about a diagonal matrix is that matrix-vector products in them are very easy to calculate. For example, if you rotate with angular velocity $\omega$ around an axis $\vec \omega = (\omega_x, \omega_y, \omega_y)$, then the rotational energy of that is $$\frac{1}{2} \vec{\omega}^T \cdot I \cdot \vec{\omega}$$ If $I$ is diagonal, this simply becomes $$\frac{1}{2} \omega_x^2 I_{xx} + \frac{1}{2} \omega_y^2 I_{yy} + \frac{1}{2} \omega_z^2 I_{zz}$$

It is, however, correct that $I_{yy}$ is the moment of inertia for rotation around the $y$ axis. The off-diagonal elements would come into play if you don't go to the coordinate system that makes $I$ diagonal and then look at rotations around axes different from the coordinate axes. For example $\vec \omega = (\omega, \omega, 0)$. There, the rotational energy would be $$\frac{\omega^2}{2}(I_{xx} + I_{yy} + 2 I_{xy})$$.

So if you want to compute the rotational energy in a coordinate system where $I$ is not diagonal, you get all those pesky off-diagonal matrix elements in there, cluttering up your expression, whereas you can get rid of all of them if you transform the coordinate system so that they are $0$.

This diagonalization, btw, is something that comes up very often in all fields that deal with matrices, just because it makes dealing with those matrices so much easier, and the remaining diagonal entries (the "eigenvalues" of that matrix) contain a lot of useful information about the nature of the matrix.

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Thanks for your reply. It is not hard to understand the significance of the principal axes and moment of inertia. But like what you said, if I use the off-diagonal matrix, when calculate the rotational kinetic energy, I have to count all those off-diagonal elements, so should it be any physical significance of the off-diagonal entries? Otherwise, how do you understand the cross terms in energy calculation? –  user1285419 Apr 12 '13 at 12:38
I believe $\frac{\omega}{2}$ should be $\frac{1}{2} \omega^2$. –  Psi Apr 12 '13 at 13:28

For a non-spherical object, there is a unique direction along which the object is "longest", that is, to have the smallest moment of inertia if rotated about an axis with that direction. The material of the object are as close to that axis as can be, compared to other directions.

There's another direction perpendicular to that about which the moment of inertia is maximum.

Then finally we have an intermediate amount of moment of inertia in a third direction perpendicular to the previous two. I lied; that "intermediate" moment of inertia may be the same as the minimum one or maximum one, in which case you have some freedom to pick an arbitrary angle for one axis, but never mind this detail for present purposes.

A spherical object, of course, has the same moment of inertia about any axis, so is boring. You have freedom to pick axes however you like, but never mind that special case either, since it's not interesting.

For the non-special case, we have the unique directions for minimum, maximum, and intermediate moments of inertia. We could name these directions, the 'principal axes', with letters like, oh maybe: 'X', 'Y', and 'Z' and thus have the tensor $$ I = \left( \begin{matrix} I_{xx} & 0 & 0\\ 0 & I_{yy} & 0\\ 0&0 & I_{zz} \end{matrix} \right) $$ These three numbers are physically meaningful, giving a general overall measure of size and mass distribution of the object.

But maybe the object is positioned at some crazy angle with respect to things we care about, like our nice level tabletop, our local notion of 'east' and 'north'. So we must rotate the object and its various physical vectors and tensors (and spinors if it's a fermion). An arbitrary rotation is described by three angles (e.g. Euler angles). The fully general $I$ tensor then has six independent quantities. We see nine components, but they count as six due to always being a symmetric tensor.

The physical significance of the off-diagonal components is that you're using a coordinate system not aligned with the principal directions of the object. They tell us nothing interesting about the object itself.

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