Inertia tensor — principal axes, principal moments

1. WHAT is the inertia tensor — derived from scratch

We want $\vec{L}$ for a rigid body rotating about a fixed point with angular velocity $\vec{\omega}$ .

Start from one particle. Particle $a$ at position $\vec{r}_a$ , mass $m_a$ . In a rigid body its velocity is $\vec{v}_a = \vec{\omega}\times\vec{r}_a$ . Its angular momentum:

$\vec{L}_a = m_a\,\vec{r}_a\times\vec{v}_a = m_a\,\vec{r}_a\times(\vec{\omega}\times\vec{r}_a)$

Why this step? Angular momentum is $\vec{r}\times m\vec{v}$ by definition, and rigidity forces $\vec{v}=\vec\omega\times\vec r$ .

Use the BAC–CAB identity $\vec{A}\times(\vec{B}\times\vec{C}) = \vec{B}(\vec{A}\cdot\vec{C}) - \vec{C}(\vec{A}\cdot\vec{B})$ :

$\vec{L}_a = m_a\big[\vec{\omega}\,(r_a^2) - \vec{r}_a(\vec{r}_a\cdot\vec{\omega})\big]$

Why this step? This separates $\vec\omega$ out so we can read off a linear map acting on it.

Sum over particles (or integrate $\int dm$ ). Look at one component, say $L_x$ . With $\vec r=(x,y,z)$ and $\vec\omega=(\omega_x,\omega_y,\omega_z)$ :

$L_x = \sum_a m_a\big[\omega_x(x^2+y^2+z^2) - x_a(x_a\omega_x + y_a\omega_y + z_a\omega_z)\big]$ $= \omega_x\!\underbrace{\sum m_a(y^2+z^2)}_{I_{xx}} - \omega_y\!\underbrace{\sum m_a\,x y}_{-I_{xy}} - \omega_z\!\underbrace{\sum m_a\,x z}_{-I_{xz}}$

So $\vec{L} = \mathbf{I}\,\vec{\omega}$ where:

Formula The inertia tensor

$\mathbf{I}=\begin{pmatrix} I_{xx}&I_{xy}&I_{xz}\\ I_{yx}&I_{yy}&I_{yz}\\ I_{zx}&I_{zy}&I_{zz}\end{pmatrix}$ Moments of inertia (diagonal): $I_{xx}=\sum m_a(y_a^2+z_a^2),\quad I_{yy}=\sum m_a(x_a^2+z_a^2),\quad I_{zz}=\sum m_a(x_a^2+y_a^2)$ Products of inertia (off-diagonal): $I_{xy}=I_{yx}=-\sum m_a x_a y_a,\quad I_{xz}=-\sum m_a x_a z_a,\quad I_{yz}=-\sum m_a y_a z_a$ Continuous body: replace $\sum m_a \to \int dm$ .

2. WHY $\vec L$ isn't parallel to $\vec\omega$ — and the fix

Because $\mathbf I$ has off-diagonal terms, $\mathbf I\vec\omega$ generally points elsewhere. A symmetric real matrix can always be diagonalized by an orthogonal (rotation) matrix. That means there exists a special rotated coordinate frame where:

$\mathbf I' = \begin{pmatrix} I_1&0&0\\0&I_2&0\\0&0&I_3\end{pmatrix}$

WHY these are exactly the eigenvectors: " $\vec L$ parallel to $\vec\omega$ " means $\mathbf I\vec\omega = \lambda\vec\omega$ . That is the eigenvalue equation. So:

HOW to find them (recipe):

Write $\mathbf I$ in any convenient frame.
Solve the characteristic cubic $\det(\mathbf I - \lambda\mathbb 1)=0$ → three roots $I_1,I_2,I_3$ .
For each $\lambda=I_k$ , solve $(\mathbf I-I_k\mathbb1)\hat e_k=0$ → axis direction; normalize.
Shortcut: any symmetry axis of the body is automatically a principal axis.

Figure — Inertia tensor — principal axes, principal moments

3. Rotational KE & a key invariant

$T = \tfrac12\sum m_a v_a^2 = \tfrac12\,\vec\omega\cdot\vec L = \tfrac12\,\vec\omega^{\mathsf T}\mathbf I\,\vec\omega$

In principal axes this becomes the clean diagonal form:

$\boxed{\,T=\tfrac12(I_1\omega_1^2+I_2\omega_2^2+I_3\omega_3^2)\,}$

4. Worked examples

Worked example Example 1 — Two masses on the x-axis (build intuition)

Masses $m$ at $(d,0,0)$ and $(-d,0,0)$ .

$I_{xx}=\sum m(y^2+z^2)=0$ . Why? Both lie on the x-axis, zero distance from it.
$I_{yy}=I_{zz}=\sum m(x^2)= m d^2 + m d^2 = 2md^2$ .
Products: $I_{xy}=-\sum mxy=0$ (all $y=0$ ), similarly $I_{xz}=I_{yz}=0$ . $\mathbf I=\begin{pmatrix}0&0&0\\0&2md^2&0\\0&0&2md^2\end{pmatrix}$ Already diagonal → principal moments $0,\,2md^2,\,2md^2$ ; principal axes are $x,y,z$ . Why $I_1=0$ ? Point masses on a line have no extent off it — spinning about that line costs nothing.

Worked example Example 2 — Off-axis pair (see the tilt of

$\vec L$ ) Equal masses $m$ at $(a,a,0)$ and $(-a,-a,0)$ .

$I_{xx}=\sum m\,y^2 = 2ma^2$ , $\;I_{yy}=\sum m\,x^2=2ma^2$ , $\;I_{zz}=\sum m(x^2+y^2)=4ma^2$ .
$I_{xy}=-\sum m\,xy = -(ma^2+ma^2)=-2ma^2$ . (Both terms: $a\cdot a$ and $(-a)(-a)$ , both $+$ .) $\mathbf I=2ma^2\begin{pmatrix}1&-1&0\\-1&1&0\\0&0&2\end{pmatrix}$ Spin about $x$ : $\vec\omega=(\omega,0,0)\Rightarrow \vec L=2ma^2(\omega,-\omega,0)$ — not parallel! Diagonalize the $2\times2$ block: eigenvalues $2ma^2(1\mp1)=0$ and $4ma^2$ , eigenvectors $\tfrac1{\sqrt2}(1,1,0)$ and $\tfrac1{\sqrt2}(1,-1,0)$ . Why these directions? The masses lie along the line $y=x$ ; that line ( $I=0$ ) and its perpendicular are the principal axes. Principal moments: $0,\,4ma^2,\,4ma^2$ .

5. Common mistakes (Steel-manned)

6. Flashcards

What does the inertia tensor relate?

It maps angular velocity to angular momentum:

\vec L=\mathbf I\vec\omega

Why is

\vec L

generally not parallel to

\vec\omega

Because

\mathbf I

has nonzero off-diagonal (product of inertia) terms, so it rotates/scales

\vec\omega

unequally.

Define principal axes.

Directions in which

\mathbf I

is diagonal; equivalently eigenvectors of

\mathbf I

where

\vec L\parallel\vec\omega

Define principal moments of inertia.

The eigenvalues

I_1,I_2,I_3

\mathbf I

— diagonal entries in the principal frame.

Formula for a diagonal moment, e.g.

I_{xx}

I_{xx}=\sum m_a(y_a^2+z_a^2)

(distance² from the x-axis).

Formula for a product of inertia

I_{xy}

I_{xy}=-\sum m_a x_a y_a

(note the minus sign).

Why is

\mathbf I

symmetric?

I_{xy}=-\sum mxy=I_{yx}

; products are symmetric in the two indices.

Equation that finds principal axes.

\mathbf I\hat e_k=I_k\hat e_k

with

\det(\mathbf I-I_k\mathbb1)=0

Rotational KE in principal axes.

T=\tfrac12(I_1\omega_1^2+I_2\omega_2^2+I_3\omega_3^2)

Frame-independent check on principal moments.

I_1+I_2+I_3=\operatorname{Tr}\mathbf I=2\sum m_a r_a^2

Shortcut for finding a principal axis.

Any axis of symmetry of the body is automatically a principal axis.

What is a spherical top?

A body with

I_1=I_2=I_3

(e.g. cube/sphere); every axis through CM is principal.

Recall Feynman: explain it to a 12-year-old

Imagine spinning a wonky lump of clay. If the lump is heavier on one side, when you twist it, it doesn't just spin straight — it wobbles and tries to flop sideways. The "wobble recipe" of the lump is a little grid of numbers (the inertia tensor). For most ways of spinning, the lump pushes its "spin-effort" sideways. But there are exactly three special straight lines through the lump where, if you spin around them, it spins clean and steady with no flop. Those three lines are the principal axes, and how hard it is to spin around each is the principal moment. Finding them is just asking: "which spin directions keep behaving themselves?"

Connections

Angular momentum — the quantity $\vec L=\mathbf I\vec\omega$ derived here.
Eigenvalues and eigenvectors — principal axes are eigenvectors of $\mathbf I$ .
Euler's equations of rigid body motion — written in the principal-axis frame.
Parallel axis theorem — shifts $\mathbf I$ to axes off the centre of mass.
Rigid body rotation — kinematics $\vec v=\vec\omega\times\vec r$ used in step 1.
Symmetric and orthogonal matrices — guarantee real eigenvalues & orthogonal axes.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, jab koi extended (phaila hua) body ghoomta hai, to har chhota tukda alag-alag tarike se rotation ko "resist" karta hai. Isi distribution ko capture karne ke liye hum inertia tensor $\mathbf{I}$ banate hain — ek $3\times3$ matrix. Single particle me $\vec L$ aur $\vec\omega$ parallel hote hain, par body me aksar nahi! Kyunki $\mathbf I$ ke andar off-diagonal (product of inertia) terms hote hain, jo $\vec\omega$ ko tilt kar dete hain. Yahi reason hai ki wonky object ghumate waqt wobble karta hai.

Ab kamaal ki baat: $\mathbf I$ ek symmetric matrix hai, isliye hamesha kuch special directions milti hain jahan ghoomne par $\vec L$ wapas $\vec\omega$ ke parallel ho jaata hai. In directions ko principal axes bolte hain, aur unke corresponding numbers ko principal moments $I_1, I_2, I_3$ . Mathematically ye bilkul eigenvalue problem hai: $\mathbf I\hat e_k = I_k\hat e_k$ . Jis direction me $\vec L\parallel\vec\omega$ , wahi eigenvector, aur scaling factor wahi eigenvalue.

Practical shortcut: agar body me koi symmetry axis hai, to woh apne aap principal axis ban jaata hai — alag se diagonalize karne ki zaroorat nahi. Aur ek check rakho: $I_1+I_2+I_3 =

Inertia tensor — principal axes, principal moments

1. WHAT is the inertia tensor — derived from scratch

2. WHY $\vec L$ isn't parallel to $\vec\omega$ — and the fix

3. Rotational KE & a key invariant

4. Worked examples

5. Common mistakes (Steel-manned)

6. Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Analytical Mechanics

Connections