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Derivator

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Hi,

I've written a little fortran code that computes the three Eigenvectors [itex]\vec{v}_1[/itex], [itex]\vec{v}_2[/itex], [itex]\vec{v}_3[/itex] of the inertia tensor of a N-Particle system.

Now I observed something that I cannot explain analytically:

Assume the position vector [itex]\vec{r}_i[/itex] of each particle to be given with respect to the center of mass of the system.

Then define three new vectors [itex]\vec{\omega}_j := (\vec{v}_j\times\vec{r}_1,\dots,\vec{v}_j\times\vec{r}_N)[/itex] where j=1,...,3. These new vectors are of length 3*N.

Now, for a non-linear configuration of the [itex]\vec{r}_i[/itex] and N>=3, the [itex]\vec{\omega}_j [/itex] seem to be mutually orthogonal, that is [itex]\vec{\omega}_j \cdot \vec{\omega}_i = 0 [/itex] for [itex]i \neq j[/itex] (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:

[itex]\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)[/itex]

Where I have used the relation

[itex](\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})[/itex]

(see: http://en.wikipedia.org/wiki/Quadruple_product)

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

[itex]\vec{v}_i\cdot\vec{v}_j = 0[/itex]

Unfortunately, this is the point where I'm stuck. I don't see why [itex]-\sum_{l=1}^N (\vec{v}_i\cdot\vec{r}_l)(\vec{r}_l\cdot\vec{v}_j)[/itex] should vanish.derivator

I've written a little fortran code that computes the three Eigenvectors [itex]\vec{v}_1[/itex], [itex]\vec{v}_2[/itex], [itex]\vec{v}_3[/itex] of the inertia tensor of a N-Particle system.

Now I observed something that I cannot explain analytically:

Assume the position vector [itex]\vec{r}_i[/itex] of each particle to be given with respect to the center of mass of the system.

Then define three new vectors [itex]\vec{\omega}_j := (\vec{v}_j\times\vec{r}_1,\dots,\vec{v}_j\times\vec{r}_N)[/itex] where j=1,...,3. These new vectors are of length 3*N.

Now, for a non-linear configuration of the [itex]\vec{r}_i[/itex] and N>=3, the [itex]\vec{\omega}_j [/itex] seem to be mutually orthogonal, that is [itex]\vec{\omega}_j \cdot \vec{\omega}_i = 0 [/itex] for [itex]i \neq j[/itex] (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:

[itex]\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)[/itex]

Where I have used the relation

[itex](\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})[/itex]

(see: http://en.wikipedia.org/wiki/Quadruple_product)

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

[itex]\vec{v}_i\cdot\vec{v}_j = 0[/itex]

Unfortunately, this is the point where I'm stuck. I don't see why [itex]-\sum_{l=1}^N (\vec{v}_i\cdot\vec{r}_l)(\vec{r}_l\cdot\vec{v}_j)[/itex] should vanish.derivator

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