# Eigenvectors of Inertia tensor

1. Mar 12, 2015

### Derivator

Hi,

I've written a little fortran code that computes the three Eigenvectors $\vec{v}_1$, $\vec{v}_2$, $\vec{v}_3$ of the inertia tensor of a N-Particle system.
Now I observed something that I cannot explain analytically:
Assume the position vector $\vec{r}_i$ of each particle to be given with respect to the center of mass of the system.
Then define three new vectors $\vec{\omega}_j := (\vec{v}_j\times\vec{r}_1,\dots,\vec{v}_j\times\vec{r}_N)$ where j=1,...,3. These new vectors are of length 3*N.
Now, for a non-linear configuration of the $\vec{r}_i$ and N>=3, the $\vec{\omega}_j$ seem to be mutually orthogonal, that is $\vec{\omega}_j \cdot \vec{\omega}_i = 0$ for $i \neq j$ (At least, I obtain this numerically up to machine precision)

I have no analytical explanation for this...

The most promising ansatz I tried so far is:
$\vec{\omega}_i \cdot \vec{\omega}_j = \sum_{l=1}^N (\vec{v}_i\times \vec{r}_l)(\vec{v}_j \times \vec{r}_l) = \sum_{l=1}^N (\vec{v}_i\cdot\vec{v}_j)(\vec{r}_l\cdot\vec{r}_l)-(\vec{v}_i\cdot\vec{r}_l)(\vec{r}_l\cdot\vec{v}_j)= -\sum_{l=1}^N (\vec{v}_i\cdot\vec{r}_l)(\vec{r}_l\cdot\vec{v}_j)$

Where I have used the relation
$(\mathbf{a \times b})\mathbf {\cdot}(\mathbf{c}\times \mathbf{d}) = (\mathbf{a \cdot c})(\mathbf{b \cdot d}) - (\mathbf{a \cdot d})(\mathbf{b \cdot c})$

and the fact that the eigenvectors of the inertia tensor are mutually orthogonal:

$\vec{v}_i\cdot\vec{v}_j = 0$

Unfortunately, this is the point where I'm stuck. I dont see why $-\sum_{l=1}^N (\vec{v}_i\cdot\vec{r}_l)(\vec{r}_l\cdot\vec{v}_j)$ should vanish.

Best,
derivator

Last edited: Mar 12, 2015
2. Mar 12, 2015

### timthereaper

I'm a bit confused on how you're defining your $\vec{w}_i$. Can you explain? I'm not quite seeing how you're defining it (maybe it's that I don't understand the physical significance).

Also, the dot product won't vanish in your ansatz. The simplest case I can think of is considering a space in $\mathbb{R}^3$ when $N=1$. Dotting a random vector in the x-y plane with both the x-axis and y-axis. You'll get a vectors along the positive and/or negative z-axis. The dot product of those two vectors isn't zero.

Last edited: Mar 12, 2015
3. Mar 13, 2015

### Derivator

Hi,
the $\vec{\omega}_i$ are just the hypervectors build from the cross products of the eigenvectors of the inertia tensor and the particle positions (length 3*N). Frankly, I don't know, if they have any physical significance. They just happen to be an intermediate step in my calculation and I noticed their orthogonality. while playing around with my code.

Of corse, you are right, that for N=1 the dot product will not vanish. However, for N >=3 , as long as the particles are not in one line, it does.

Cheers,
derivator

4. Mar 14, 2015

### Haborix

Taking the eigenvectors to be normalized your last sum is a sum on the product of the i-th and j-th components (with the eigenvectors as the basis) of the position vectors of the particles. Since you have defined the position vectors relative to the center of mass you know the following holds for each component: $$\sum_l r_l^x=0.$$ I don't know the solution to the problem, but maybe this will help cast it in a new light.

5. Mar 25, 2015

### Derivator

Hi Haborix,

thanks a lot for your idea. Unfortunately, I also don't see how to use this relation...

Still hoping, someone might see why the sum is vanishing.

Cheers,
derivator

6. Mar 25, 2015

### Derivator

Maybe, I just had an idea:
Am I right, that the $-\sum_{l=1}^N (\vec{v}_i\cdot\vec{r}_l)(\vec{r}_l\cdot\vec{v}_j)$ are nothing else than the off-diagonal components of the inertia tensor when being expressed in the coordinate system defined by the eigenvectors of the inertia tensor? (for the components of the inertia tensor see: http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node64.html)
Now, if I did not overlook anything, then its trivial why the
$-\sum_{l=1}^N (\vec{v}_i\cdot\vec{r}_l)(\vec{r}_l\cdot\vec{v}_j)$ vanish: It's simply because the inertia tensor is (trivially) diagonal in its own `eigenframe'.

Let me know what you think.

Cheers,
derivator