# How are these definitions of the inertia tensor the same?

I'm looking for some help in understanding the inertia tensor (not the physics, just the math).

I'm trying to figure out how to convert between the wedge product and tensor product definitions. Here's the wedge product definition $$\mathcal I(B) = \int d^3 \mathbf x \rho(\mathbf x)\ \mathbf x \wedge (\mathbf x \cdot B)$$ where $B$ is a bivector. Thus the inertia tensor is just a linear mapping between bivectors. Because I can draw of picture of this, I understand this definition much better.

There's also the tensor product definition: $$\mathcal I = \int d^3 \mathbf x \rho(\mathbf x)\ [(\mathbf x \cdot \mathbf x)E -\mathbf x \otimes \mathbf x)]$$ where $E=\mathbf e_1 \otimes \mathbf e_1 + \mathbf e_2 \otimes \mathbf e_2 + \mathbf e_3 \otimes \mathbf e_3$. I will admit I don't understand tensors well. From what I gather $\mathbf u \otimes \mathbf v = \mathbf u \mathbf v^T$.

How do these two definitions jive with one another?

Well, first you should recognize that we need some way of representing the $B$ you have in your first definition in the second definition. This is because the first definition is the definition of a linear mapping with the thing it maps (a bivector) and the second definition is just a linear mapping without the thing that it maps. In fact the second definition can just be thought of as a matrix (using the formula: $u \otimes v = uv^T$) which acts on column vectors. Why do we represent a bivector as a column vector, you ask? That's no problem, we'll just represent the bivector $B=b_1 e_2 \wedge e_3 + b_2 e_3 \wedge e_1 + b_3 e_1 \wedge e_2$ as the column matrix $b=\begin{bmatrix} b_1 \\ b_2 \\ b_3\end{bmatrix}$ (Note that because vectors and bivectors transform differently under inversions, the following should be taken as an indication of them being the same -- not a rigorous proof. To be a little more rigorous I'd have to introduce the Hodge (or algebraic) dual.).

Because constant vectors can be brought inside the integral, we really just need to then prove that $x \wedge (x\cdot B) = [(x\cdot x)E-x\otimes x]b$ (equality here being in the only sense that a bivector can equal a coordinate vector -- component-wise).

Considering dyads to just be matrices we see that $$[(x\cdot x)E-x\otimes x]b = \begin{bmatrix} x_2^2 + x_3^2 & -x_1x_2 & -x_1x_3 \\ -x_1x_2 & x_1^2 + x_3^2 & -x_2x_3 \\ -x_1x_3 & -x_2x_3 & x_1^2 + x_2 ^2\end{bmatrix}\begin{bmatrix} b_1 \\ b_2 \\ b_3\end{bmatrix} = \begin{bmatrix} \color{red}{b_1(x_2^2 + x_3^2) -b_2x_1x_2 -b_3x_1x_3} \\ \color{purple}{-b_1x_1x_2 +b_2(x_1^2 + x_3^2) -b_3x_2x_3} \\ \color{blue}{-b_1x_1x_3 -b_2x_2x_3 + b_3(x_1^2 + x_2^2)}\end{bmatrix}$$

Now consider the bivector $x\wedge (x\cdot B)$. This is just $$x\wedge (x\cdot B) = (x_1e_1 + x_2e_2 + x_3e_3) \wedge [(x_3b_2-x_2b_3)e_1 + (x_1b_3-x_3b_1)e_2+(x_2b_1-x_1b_2)e_3] \\ = [x_2(x_2b_1-x_1b_2)-x_3(x_1b_3-x_3b_1)]e_2\wedge e_3 + [x_3(x_3b_2-x_2b_3)-x_1(x_2b_1-x_1b_2)]e_3\wedge e_1 + [x_1(x_1b_3-x_3b_1)-x_2(x_3b_2-x_2b_3)]e_1 \wedge e_2 \\ = [\color{red}{b_1(x_2^2 + x_3^2) -b_2x_1x_2 -b_3x_1x_3}]e_2\wedge e_3 +[\color{purple}{-b_1x_1x_2 +b_2(x_1^2 + x_3^2) -b_3x_2x_3}]e_3\wedge e_1 + [\color{blue}{-b_1x_1x_3 -b_2x_2x_3 + b_3(x_1^2 + x_2^2)}]e_1\wedge e_2$$

Can you see now that both definitions are the same?

I think the key here is to understand what is meant by the tensor product. For any vectors $a, b, u, v$, the tensor product $a \otimes b$ means

$$(a \otimes b)(u,v) = (a \cdot u)(b \cdot v)$$

Or, perhaps, it might mean this instead:

$$(a \otimes b)(v) = a (b \cdot v)$$

The two notions are equivalent to each other, so mathematicians freely use one or the other depending on context. Physicists often tend to prefer the latter.

With this in mind, note that $E$ here is the identity, so in the tensor definition, we can write

$$I(v) = \int d^3x \, \rho(x) [v (x \cdot x) - x (x \cdot v)]$$

Use the BAC-CAB rule to transform this into a wedge product:

$$v (x \cdot x) - x (x \cdot v) = x \cdot (x \wedge v)$$

You should be able to multiply through by the pseudoscalar $i$ at this point to transform $v$ from vector to bivector, and we'll get the same expression as given in GA.

• Much simpler than the brute force way I did it. (+1)
– user137731
Apr 20 '15 at 18:00