The Inertia Tensor and Principal Axes

One number is not enough

In Volume I the moment of inertia was a single number I for rotation about a fixed axis: L = I\omega and the energy was \tfrac{1}{2}I\omega^2. That works beautifully — as long as the axis is pinned in place. But a wrench flung across a room, or a satellite tumbling in orbit, has no pinned axis. The axis itself moves, and rotation about different directions meets different inertia. One number can no longer capture it.

Here is the central fact of this guide: in general the angular momentum \mathbf{L} and the angular velocity \boldsymbol{\omega} are not parallel. Flick an unbalanced body into a spin and \mathbf{L} points one way while \boldsymbol{\omega} points another; as the body turns, \boldsymbol{\omega} chases \mathbf{L} around, and the axis visibly wobbles. To relate two non-parallel vectors we need a machine that can rotate as well as rescale — a tensor.

Building the inertia tensor

Add up the angular momentum of every mass element, \mathbf{L}=\sum m\,\mathbf{r}\times\mathbf{v}, insert the rigid velocity field \mathbf{v}=\boldsymbol{\omega}\times\mathbf{r}, and expand the double cross product. Out drops a linear relation between \boldsymbol{\omega} and \mathbf{L}, with the coefficients gathered into a 3\times3 array — the inertia tensor. For a continuous body its components are integrals over the mass:

I_{ij} = \int \rho(\mathbf{r})\,\bigl(r^2\,\delta_{ij} - x_i x_j\bigr)\,dV

The components of the inertia tensor: the diagonal entries are the familiar moments of inertia, the off-diagonal entries are the products of inertia.

The diagonal entries are exactly the moments of inertia you know — for example I_{xx}=\int\rho\,(y^2+z^2)\,dV, the resistance to spinning about the x-axis. The off-diagonal entries, like I_{xy}=-\int\rho\,xy\,dV, are the products of inertia; they measure how lopsided the mass is, and they are what tilt \mathbf{L} away from \boldsymbol{\omega}. Because I_{ij}=I_{ji}, the tensor is symmetric: six independent numbers, not nine.

\mathbf{L} = \mathbf{I}\,\boldsymbol{\omega}

The inertia tensor is the machine that turns an angular velocity into an angular momentum — the rotational analogue of \mathbf{p}=m\mathbf{v}, but with a tensor in place of the scalar mass.

A tensor is a machine

The operational meaning of 'rank-2 tensor' is just this: feed it a vector, get a vector back, generally rotated and rescaled. The inertia tensor eats \boldsymbol{\omega} and hands you \mathbf{L}. Change your coordinate axes and its nine components change too, but in a fixed lawful way, \mathbf{I}' = R\,\mathbf{I}\,R^{\mathsf T}, so that the physical relation \mathbf{L}=\mathbf{I}\,\boldsymbol{\omega} holds in every frame. That transformation rule is what makes \mathbf{I} a genuine tensor rather than a random matrix.

Principal axes: making the tensor diagonal

Because \mathbf{I} is real and symmetric, a theorem of linear algebra guarantees it can be diagonalized: there always exist three mutually perpendicular principal axes for which every product of inertia vanishes. Spin the body about one of these and \mathbf{L} is parallel to \boldsymbol{\omega} again — no wobble. Finding them is an eigenvalue problem:

\mathbf{I}\,\hat{\mathbf{e}}_a = I_a\,\hat{\mathbf{e}}_a, \qquad a = 1,2,3

The principal axes are the eigenvectors of the inertia tensor; the principal moments I_1,I_2,I_3 are its eigenvalues.

You rarely have to grind through the algebra, because symmetry hands you the answer: any axis of rotational or mirror symmetry is automatically a principal axis. A uniform sphere has every axis principal (all three moments equal); a symmetric top like a cylinder or a discus has its symmetry axis plus any two perpendicular to it, with two of the three principal moments equal. In the principal frame the rotational energy loses all its cross terms:

T = \tfrac{1}{2}\bigl(I_1\omega_1^2 + I_2\omega_2^2 + I_3\omega_3^2\bigr)

In the principal-axis frame the rotational energy is beautifully diagonal — just three squared terms, one per axis.

Shifting the axis: the parallel-axis theorem

The tensor above is defined about the centre of mass, but often a body pivots about some other point — a door about its hinge, a pendulum about its support. The tensor version of the parallel-axis theorem shifts the reference point by a displacement \mathbf{d}:

I_{ij} = I_{ij}^{\text{cm}} + M\bigl(d^2\,\delta_{ij} - d_i d_j\bigr)

The inertia tensor about a point displaced by \mathbf{d} from the centre of mass — the tensor generalization of the familiar I = I_{\text{cm}} + M d^2.

Worked example: principal axes of a flat plate

Take a thin uniform rectangular plate of mass M, with sides a and b. Find its principal axes and principal moments.

Guess by symmetry. Put x and y along the plate's edges and z perpendicular to it, with the origin at the centre. Symmetry suggests these are the principal axes — now verify.
Check the products of inertia. For example I_{xy}=-\int\rho\,xy\,dA. The plate is symmetric under x\to-x, so the integrand is odd and the integral is zero. Because every point has z=0, I_{xz} and I_{yz} vanish too. All products of inertia are zero, so $x,y,z$ are the principal axes.
Compute the diagonal moments. Standard integrals give I_1=\tfrac{1}{12}Mb^2 (about x), I_2=\tfrac{1}{12}Ma^2 (about y), and I_3=\tfrac{1}{12}M(a^2+b^2) (about z).
Sanity check. Notice I_3 = I_1 + I_2. This perpendicular-axis relation holds for any flat lamina and is a quick way to catch arithmetic slips.
Read the physics. With a>b the smallest moment is I_1, so the plate is easiest to spin about the axis running along its longer edge and hardest about the perpendicular z-axis. All three moments differ, so this is an asymmetric top — the fully general case of the next guide.

The lesson generalizes: exploit symmetry first. Any plane of symmetry forces its normal to be a principal axis, sparing you the eigenvalue computation entirely.