Inertia tensor — principal axes, principal moments: Visual walkthrough

Step 1 — One speck, going in a circle

WHAT. Picture a single tiny piece of the body: a dot of mass $m$ sitting at position $r$ . The word position vector $r$ just means "the arrow drawn from the pivot point to the dot." Its three numbers $(x, y, z)$ are how far along, sideways, and up the dot sits.

WHY. A rigid body is just millions of such dots glued together. If we understand the angular momentum of one dot, we add them all up at the end. Building from the smallest piece is the whole trick.

PICTURE. Below, the pivot is the black-dot origin. The pale-yellow arrow is $r$ . The body is turning, and the turning itself is described by another arrow, $ω$ (chalk blue), pointing along the axis of rotation. Its length is how fast we spin (radians per second); by the right-hand rule your curled fingers show the spin, your thumb is $ω$ .

Figure — Inertia tensor — principal axes, principal moments

Step 2 — How fast does the speck move? The cross product enters

WHAT. The velocity of the dot is $v = spin axis ω \times arm to the dot r .$ The symbol $\times$ is the cross product. For two arrows it outputs a third arrow that is perpendicular to both, with length $∣ ω ∣ ∣ r ∣ sin ϕ$ where $ϕ$ is the angle between them.

WHY this tool and not ordinary multiplication? We need the speed and direction a point sweeps when it goes around an axis. A point far from the axis moves fast; a point on the axis (arm parallel to $ω$ , so $sin ϕ = 0$ ) does not move at all. The cross product is precisely the operation whose output shrinks to zero when the arm lines up with the axis and grows with perpendicular distance — exactly the geometry of circular motion. No other simple product does that.

PICTURE. The blue axis $ω$ , the yellow arm $r$ , and the pink velocity $v$ form a right-handed corner. Notice $v$ is tangent to the circle the dot travels on.

Step 3 — Angular momentum of the speck

WHAT. Angular momentum of one dot is defined as $L_{a} = r \times (m v) = m r \times (ω \times r) .$ The quantity $m v$ (mass times velocity) is the ordinary momentum — "how much motion" the dot carries. Wrapping it in $r \times (\dots)$ gives the rotational version: momentum weighted by how far out and in which turning sense it acts.

WHY. $L = r \times (m v)$ is the definition of angular momentum — it is the rotational cousin of momentum, the thing that stays constant when no torque acts. We substituted $v = ω \times r$ from Step 2 because rigidity forces that relation, turning $L_{a}$ into something built purely from $ω$ and $r$ .

PICTURE. We now have a double cross product. On the board: $r$ crossed into the (blue) $v$ gives the (green) $L_{a}$ , which sticks out along the axis-ish direction — but watch, it will not always sit exactly on $ω$ .

Step 4 — Unfold the double cross product (BAC–CAB)

WHAT. A nested cross product $A \times (B \times C)$ can always be rewritten as $A \times (B \times C) = B (A \cdot C) - C (A \cdot B),$ the BAC–CAB identity. Here $A = r$ , $B = ω$ , $C = r$ , giving $L_{a} = m [pulls along the axis ω (r \cdot r) - pulls toward the dot r (r \cdot ω)] = m [ω r^{2} - r (r \cdot ω)] .$ The dot $\cdot$ is the dot product: $r \cdot ω = x ω_{x} + y ω_{y} + z ω_{z}$ , a single number measuring how much two arrows point the same way. And $r \cdot r = r^{2}$ is the arm's length squared.

WHY this tool? The whole goal is to pull $ω$ out into the open so we can see $L$ as "some machine acting on $ω$ ." BAC–CAB is the one identity that separates the nested product into a clean $(stuff) \cdot ω$ shape. The dot product enters because it is the natural "how-aligned" number that BAC–CAB spits out.

PICTURE. Two competing arrows: the first term $ω r^{2}$ (blue) tries to keep $L_{a}$ along the axis; the second term $- r (r \cdot ω)$ (pink) tugs it toward the dot. Their sum (green) is the true $L_{a}$ — tilted off the axis whenever the dot sits off to the side.

Step 5 — Read off one component: a matrix is hiding here

WHAT. Let us write only the $x$ -component of $L = \sum_{a} L_{a}$ (sum over all dots). Using $r^{2} = x^{2} + y^{2} + z^{2}$ and $r \cdot ω = x ω_{x} + y ω_{y} + z ω_{z}$ : $L_{x} = \sum_{a} m_{a} [ω_{x} (x^{2} + y^{2} + z^{2}) - x_{a} (x_{a} ω_{x} + y_{a} ω_{y} + z_{a} ω_{z})] .$ Group by which $ω$ each term multiplies: $L_{x} = ω_{x} I_{xx} \sum m (y^{2} + z^{2}) + ω_{y} I_{x y} (- \sum m x y) + ω_{z} I_{x z} (- \sum m x z) .$

WHY. $L_{x}$ came out as a sum of three coefficients, each times one component of $ω$ . That is exactly the top row of a matrix times a vector. Doing the same for $L_{y}$ and $L_{z}$ produces the other two rows. The coefficients — pure geometry of the mass, no $ω$ inside — are the entries of the inertia tensor.

PICTURE. The rows-and-columns grid below shows each coefficient landing in its slot. Diagonal boxes $I_{xx}, I_{y y}, I_{z z}$ (yellow) are the moments of inertia; off-diagonal boxes (pink) are the products of inertia, and they each carry a minus sign from the $- r (r \cdot ω)$ term of Step 4.

Step 6 — The tilt, made undeniable (worked degenerate + generic case)

WHAT. Take the smallest possible off-axis body: two equal masses $m$ at $(a, a, 0)$ and $(- a, - a, 0)$ . Plugging into Step 5's formulas, $I = 2 m a^{2} 1 - 1 0 - 1 10 002 .$ Now spin it about the $x$ -axis, $ω = (ω, 0, 0)$ : $L = I ω = 2 m a^{2} (ω, - ω, 0) .$ $ω$ points along $+ x$ , but $L$ points into the fourth quadrant — not parallel.

WHY show this? Everything before was symbols; this is the payoff you can literally see. The off-diagonal $- 1$ (a product of inertia) is what bent $L$ away from $ω$ . Kill that entry and the arrows would agree.

PICTURE. Blue $ω$ along $x$ ; green $L$ dragged down toward the mass line $y = x$ . The masses (pink dots) sit on that diagonal — $L$ leans toward where the mass actually is.

Step 7 — The fix: rotate until the tilt vanishes (principal axes)

WHAT. " $L$ parallel to $ω$ " is written mathematically as $I ω = λ ω,$ i.e. the matrix only stretches $ω$ by a number $λ$ , never turns it. This is the eigenvalue equation (see Eigenvalues and eigenvectors). A special $ω$ that satisfies it is an eigenvector; its stretch factor $λ$ is an eigenvalue. Because $I$ is real and symmetric, three mutually perpendicular such directions always exist — the principal axes — and their $λ$ 's are the principal moments $I_{1}, I_{2}, I_{3}$ .

WHY these directions? Rotating our coordinate frame to point along the eigenvectors zeroes out every off-diagonal box. In Step 6's example the masses lie along $y = x$ ; so $\frac{1}{2} (1, 1, 0)$ and its perpendicular $\frac{1}{2} (1, - 1, 0)$ are the principal axes, with $I = 2 m a^{2} (1 - 1) = 0 along the mass line, I = 2 m a^{2} (1 + 1) = 4 m a^{2} across it .$ Along either of these, spinning gives $L ∥ ω$ — the tilt is gone.

PICTURE. Same body, but now the blue $ω$ is placed along the diagonal principal axis; the green $L$ lies exactly on top of it. Rotating the frame diagonalised $I$ .

Step 8 — Every case, on one board (edge cases)

WHAT. Three shapes exhaust the possibilities, sorted by how many principal moments are equal.

Body	Principal moments	What $L$ does
all different $I_{1} \neq = I_{2} \neq = I_{3}$ (asymmetric top)	3 distinct axes	$L ∥ ω$ only on the 3 axes
two equal $I_{1} = I_{2} \neq = I_{3}$ (symmetric top)	axis $\overset{e}{^}_{3}$ + any pair $⊥$	any axis in the equal plane also principal
all equal $I_{1} = I_{2} = I_{3}$ (spherical top, e.g. cube)	every axis is principal	$L ∥ ω$ always

WHY it matters. The degenerate case $I_{1} = I_{2} = I_{3}$ means $I = I 1$ (identity times a number), so $I ω = I ω$ for every $ω$ — a cube behaves like a sphere. And the zero case (Step 6's mass line, $I = 0$ ): masses strung on a line offer no resistance to spinning about that line, since none of them is off it.

PICTURE. Three mini-sketches — asymmetric brick, symmetric cylinder, cube — with their principal axes drawn.

The one-picture summary

One dot → its velocity $ω \times r$ → its angular momentum $r \times m v$ → BAC–CAB splits it → summing over dots packs the geometry into the symmetric matrix $I$ → off-diagonals tilt $L$ → rotating to eigenvectors (principal axes) removes the tilt and restores $L = I_{k} ω$ .

Recall Feynman retelling (say it out loud)

"I grab one tiny bit of the spinning thing. Because the whole body turns together, that bit whips around at $v = ω \times r$ — fast if it's far from the axis, still if it's on it. Its spin-momentum is $r$ crossed with $m v$ . When I untangle that double cross with BAC–CAB, I get two tugs: one keeping the momentum along the axis, one dragging it toward wherever the mass sits. That second tug is the troublemaker: it tilts $L$ away from $ω$ . When I add up every bit, all the geometry collapses into one $3 \times 3$ table $I$ — big diagonal numbers say 'hard to spin about this axis,' the off-diagonal minus-numbers say 'and it'll lean sideways.' To make the leaning stop, I just twist my coordinate axes until those off-diagonal numbers hit zero. The directions where that happens are the principal axes — the eigenvectors — and there, beautifully, $L$ points the same way as $ω$ again, exactly like a single point mass."

Recall

Why is $L$ generally not parallel to $ω$ ? ::: The $- r (r \cdot ω)$ term (the off-diagonal products of inertia) drags $L$ toward the mass, off the axis. What operation turns " $L ∥ ω$ " into an equation? ::: The eigenvalue equation $I ω = λ ω$ . For the cube, why is every axis principal? ::: $I_{1} = I_{2} = I_{3} \Rightarrow I = I 1$ , so $I ω = I ω$ for all $ω$ .