The Inertia Tensor

Alright… I think it’s finally time to face tensors.

I’ll be real — before I sat down and studied this properly, I Googled around, flipped through the tensor chapter in a math-methods-for-physics book, and… nope. Didn’t click.

The move, apparently, is: just grind through relativity, and at some point it hits you — “ohhh, that’s what a tensor is.”

Tensors show up everywhere in relativity too, so I think that’s just the deal. (sigh) Let’s just push through.

OK so. Physical quantities come in basically 3 flavors:

Scalar, vector, tensor. This is what the professor told us.

Hmm. Simply: a scalar is just the magnitude of a force.

A vector is the magnitude plus the direction, right?

And a tensor? Apparently it also tracks the plane the thing acts on… (yeah, I don’t really get it either lol)

But here’s the one thing I do get:

A tensor doesn’t change when the coordinate system changes.

A vector does — if you shift the origin from O to O’, the vector shifts with it, right?

Which makes writing equations annoying. Ugh.

A tensor? Doesn’t care. Coordinates change, tensor stays the same. It’s… kind of like a vector? But different. Definitely different.

The way I’m picturing it: take a sweet potato, spin it a certain way, and compute its moment of inertia tensor. Now take a different sweet potato over there, spin it the same way — or even flip it upside down —

As long as everything is rotating the exact same way, the tensor is the same. The same same.

And that tensor can be written as a matrix. That’s roughly the vibe.

So tensors can be computed ‘from any reference point’, which is why people call them ‘absolute’.

(Following me? If not, just keep reading — it’ll make sense by the end.)

Spoiler: I still don’t really know what a tensor is in full generality, but —

at least for the ‘moment of inertia tensor’, I think I finally get it after grinding through this!!!

OK. The moment of inertia tensor we’re going to work with is defined as this matrix:

$$\{I\} \quad = \quad \begin{pmatrix} I_{xx} & I_{xy} & I_{xz} \\ I_{yx} & I_{yy} & I_{yz} \\ I_{zx} & I_{zy} & I_{zz} \end{pmatrix}$$

The unit vector $\hat{n}$ can be written as a matrix too, right?

$$\hat{n} \quad = \quad \begin{pmatrix} \cos\alpha \\ \cos\beta \\ \cos\gamma \end{pmatrix}$$

You know — the $i, j, k$ unit vectors, direction cosines, all that stuff.

The (scalar) moment of inertia is a value tied to ‘some object’ and ‘some rotation axis’.

So $I$ changes depending on the axis, even for the exact same object. Worth keeping that in the back of your head.

The relationship between the moment of inertia tensor and the scalar moment of inertia is:

$$I \quad = \quad \tilde{n}\{I\}\hat{n} \\ \quad = \quad \begin{pmatrix} \cos\alpha & \cos\beta & \cos\gamma \end{pmatrix} \begin{pmatrix} I_{xx} & I_{xy} & I_{xz} \\ I_{yx} & I_{yy} & I_{yz} \\ I_{zx} & I_{zy} & I_{zz} \end{pmatrix} \begin{pmatrix} \cos\alpha \\ \cos\beta \\ \cos\gamma \end{pmatrix}$$

(If you want to see where this comes from, go back and look at how we derived the moment of inertia in the last post.)

Why are we bothering with the moment of inertia tensor? Because it’s convenient when you’re dealing with three-dimensional rigid bodies.

Think about why we beat vectors to death all the time — because they’re convenient. Same deal here: tensors (matrices, really) are convenient.

Let’s see why.

Today’s main goal — the finish line — is to write the angular momentum vector and the scalar rotational kinetic energy using the moment of inertia tensor!!

Angular Momentum Vector

So far we’ve been computing $I$ for rigid bodies moving in 3D, and picking up tensors along the way.

Now: how does angular momentum, rotational kinetic energy, etc., come out cleanly in tensor language?

And in a later post, the thing to really pay attention to is how insanely much cleaner everything gets once you use the principal axis (Principal Axis).

OK. Same as before — angular momentum is the sum over every particle. (Every little chunk of the sweet potato, one by one.)

$$\vec{L} \quad = \quad \sum \quad \vec{r}_{i} \times m_{i}\vec{v}_{i}$$

Wait — before, we wrote $v = r\omega$,

actually it was $\vec{v} = \vec{\omega} \times \vec{r}$, but for a flat planar rigid body the axis is perpendicular anyway, so we got away with just a scalar product!

Here I’d love to stay that simple, but $\vec{\omega}$ and $\vec{r}$ aren’t generally at 90 degrees anymore, so I can’t just collapse it down.

Fine. Let me pull $m_i$ out front before I substitute anything.

Like so:

$$\vec{L} \quad = \quad \sum \quad m_{i}\vec{r}_{i} \times \vec{v}_{i}$$

$$\text{and substituting } \vec{v}_{i} = \vec{w} \times \vec{r}_{i}$$

$$= \sum \quad m_{i}\vec{r}_{i} \times (\vec{w} \times \vec{r}_{i})$$

$$\text{BAC-CAB rule time!!!}$$

$$= \quad \sum \quad \left\{ \vec{w} \cdot (m_{i}r_{i}^{2}) - \vec{r}_{i}(m_{i}\vec{r}_{i})\vec{w} \right\}$$

$$= \sum \quad (m_{i}r_{i}^{2}) \cdot \vec{w} \quad - \quad \sum \quad m_{i}\vec{r}_{i}(\vec{r}_{i}) \cdot \vec{w}$$

OK, we’ve gotten this far. Now $\vec{\omega}$ has nothing to do with the index $i$, so I want to pull it out front… can I?

(With the math I know right now — honestly, probably not?)

Whatever!!! Let’s just yank it out and see what happens!!!!!!!!!!!

$$\left[ \sum \quad (m_{i}r_{i}^{2}) \quad - \quad \sum \quad m_{i}\vec{r}_{i}(\vec{r}_{i}) \right] \cdot \vec{w}$$

In that second term — you absolutely cannot squish $\vec{r}_i \vec{r}_i$ into $r_i^2$!!

That kind of product — two vectors sitting next to each other with no dot, no cross — is called a dyad product. And it turns out it’s a kind of tensor.

Ohhh. So —

$$\sum \quad (m_{i}r_{i}^{2}) \cdot \vec{w} \quad - \quad \sum \quad m_{i}\vec{r}_{i}(\vec{r}_{i}) \cdot \vec{w}$$

the red piece is a tensor. Meaning: tensor times vector gives you a vector. Got it??

But what about the front term?

The front term is a scalar, so we need to multiply it by the unit tensor $\{1\}$ to get it into the same matrix shape!!

Which means we need to know a little about dyad products.

The unit tensor $\{1\}$, in dyad product form, is $ii + jj + kk$.

Quick detour — dyad products

$$\text{A product of the form } \vec{a}\vec{b} \text{ is different from the dot product } \vec{a} \cdot \vec{b}.$$

$$\vec{a}\vec{b} = \begin{pmatrix} a_x b_x & a_x b_y & a_x b_z \\ a_y b_x & a_y b_y & a_y b_z \\ a_z b_x & a_z b_y & a_z b_z \end{pmatrix}$$

$$\text{If a vector } \vec{C} \text{ hits it:}$$

$$\begin{pmatrix} a_x b_x & a_x b_y & a_x b_z \\ a_y b_x & a_y b_y & a_y b_z \\ a_z b_x & a_z b_y & a_z b_z \end{pmatrix} \begin{pmatrix} C_x \\ C_y \\ C_z \end{pmatrix} = \begin{pmatrix} a_x(b_x C_x + b_y C_y + b_z C_z) \\ a_y(b_x C_x + b_y C_y + b_z C_z) \\ a_z(b_x C_x + b_y C_y + b_z C_z) \end{pmatrix} \text{ — and the result is a vector}$$$$\text{See the dot product hiding in there????}$$

$$\begin{pmatrix} a_x(b_x C_x + b_y C_y + b_z C_z) \\ a_y(b_x C_x + b_y C_y + b_z C_z) \\ a_z(b_x C_x + b_y C_y + b_z C_z) \end{pmatrix} = \vec{a}(\vec{b} \cdot \vec{C})$$

$$\text{So so so so } (\vec{a}\vec{b}) \cdot \vec{C} = \vec{a}(\vec{b} \cdot \vec{C})$$$$\text{One more thing!!! Almost there!!!}$$

$$\vec{r_i}\vec{r_i} = \begin{pmatrix} r_{ix}r_{ix} & r_{ix}r_{iy} & r_{ix}r_{iz} \\ r_{iy}r_{ix} & r_{iy}r_{iy} & r_{iy}r_{iz} \\ r_{iz}r_{ix} & r_{iz}r_{iy} & r_{iz}r_{iz} \end{pmatrix}$$

$$\text{If } \vec{w} \text{ hits it:}$$

$$\begin{pmatrix} r_{ix}r_{ix} & r_{ix}r_{iy} & r_{ix}r_{iz} \\ r_{iy}r_{ix} & r_{iy}r_{iy} & r_{iy}r_{iz} \\ r_{iz}r_{ix} & r_{iz}r_{iy} & r_{iz}r_{iz} \end{pmatrix} \begin{pmatrix} w_x \\ w_y \\ w_z \end{pmatrix} = \begin{pmatrix} r_{ix}(r_{ix}w_x + r_{iy}w_y + r_{iz}w_z) \\ r_{iy}(r_{ix}w_x + r_{iy}w_y + r_{iz}w_z) \\ r_{iz}(r_{ix}w_x + r_{iy}w_y + r_{iz}w_z) \end{pmatrix} = \vec{r_i}(\vec{r_i} \cdot \vec{w})$$

$$\text{Ohhhhhh — so that's why earlier when I recklessly yanked } \vec{w} \text{ to the back, nothing blew up~}$$

OK, back up to the top!!!

$$\vec{L} \quad = \quad \left[ \sum \quad (m_{i}r_{i}^{2}) \quad - \quad \sum \quad m_{i}\vec{r}_{i}(\vec{r}_{i}) \right] \cdot \vec{w}$$

$$\quad = \quad \{I\} \vec{w}$$

$$\text{Ohhhh — the angular momentum vector, written with the moment of inertia tensor!!!}$$

$$\text{So this relationship just... holds.}$$

$$\ll \text{It's literally the same } \vec{L} = I\vec{w} \text{ we already know, lol} \gg$$

Rotational Kinetic Energy

Now let’s do energy in tensor form. This one’s easy, heh.

$$\text{Rotational kinetic energy} \quad T \quad = \quad \frac{1}{2}\sum \quad m_i v_i^2 \quad = \frac{1}{2}\sum \quad m_i \vec{v}_i \cdot \vec{v}_i$$

$$\text{Substitute } \vec{v}_i = \vec{w} \times \vec{r}_i \text{ — here we go!!}$$

$$= \frac{1}{2}\sum \quad \vec{v}_i \cdot m_i \vec{v}_i \quad = \quad \frac{1}{2}\sum \quad (\vec{w} \times \vec{r}_i) \cdot (m_i \vec{v}_i)$$

$$\text{You know you can swap the dot and the cross in a scalar triple product, right? (This trick shows up all over the place — E\&M, wherever.)}$$

$$= \quad \frac{1}{2}\sum \quad \vec{w} \cdot (\vec{r}_i \times m_i \vec{v}_i) \quad = \frac{1}{2}\sum \quad \vec{w} \cdot \left\{ \vec{r}_i \times (m_i \vec{v}_i) \right\}$$

$$= \frac{1}{2}\vec{w} \cdot \sum \quad \left\{ \vec{r}_i \times (m_i \vec{v}_i) \right\} \quad \text{(that red bit — I've seen this before... position cross momentum is... ??)}$$

$$= \frac{1}{2}\vec{w} \cdot \vec{L} \quad \text{(angular momentum!!~)}$$

OK OK OK OK OK OK OK OK

Let me just copy-paste the result from above:

$$\vec{L} \quad = \quad \left[ \sum \quad (m_{i}r_{i}^{2}) \quad - \quad \sum \quad m_{i}\vec{r}_{i}(\vec{r}_{i}) \right] \cdot \vec{w}$$

$$\quad = \quad \{I\} \vec{w}$$

$$\text{So:}$$

$$\text{Rotational kinetic energy} \quad T \quad = \frac{1}{2}\vec{w} \cdot \vec{L} \quad = \quad \frac{1}{2}\vec{w}\{I\}\vec{w}$$

$$\therefore T \quad = \frac{1}{2}\vec{w}\{I\}\vec{w}$$

$$\ll \text{Looks basically like the } \frac{1}{2}I\omega^{2} \text{ we already know, right? heh heh heh} \gg$$$$\text{One more — written out:}$$

$$T \quad = \frac{1}{2}\tilde{\vec{w}}\{I\}\vec{w} \quad = \quad \frac{1}{2}\begin{pmatrix} w_x & w_y & w_z \end{pmatrix} \begin{pmatrix} I_{xx} & I_{xy} & I_{xz} \\ I_{yx} & I_{yy} & I_{yz} \\ I_{zx} & I_{zy} & I_{zz} \end{pmatrix} \begin{pmatrix} w_x \\ w_y \\ w_z \end{pmatrix}$$

$$= \quad \frac{1}{2}(I_{xx}w_x^2 + I_{yy}w_y^2 + I_{zz}w_z^2 + 2I_{xy}w_x w_y + 2I_{xz}w_x w_z + 2I_{yz}w_y w_z)$$

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Originally written in Korean on my Naver blog (2015-01). Translated to English for gdpark.blog.