I'm working with data of biological cells, and we want to measure how the shape of a cell aligns with the normal vector of its surrounding tissue (where the cell is at the boundary).
One idea was to compare the normal vector $n$ with the inertia tensor $I = \int_\Omega \rho(x) (\Vert \mathbf{x}\Vert^2 \mathrm{Id_3} - \mathbf{x} \otimes \mathbf{x}) \, \mathrm{d}{\mathbf{x}}$.
If the inertia matrix has eigenvalues $\lambda_1 \geq \lambda_2 \geq \lambda_3$ with corresponding (ONB) of eigenvectors $v_1, v_2, v_3$, then the naive idea was to just compute the angle $$ \text{alignment} = \angle( v_1, n). $$
Now, that doesn't work well in cases where the two largest eigenvalues are close to each other. In that situation, the normal seems pretty much in alignment with the cell's body, but the naive alignment measure doesn't show it.
Question: Is there a geometric structure that we could make use of to compare the inertia matrix with a vector?
Ideally the comparison should be such that:
- it is invariant under scaling of the cell (e.g. invariant under space scaling $\mathbf{x} \mapsto \mu \mathbf{x}$)
- optimal: it should be easy to interpret, e.g. for example what is the maximal value and minimal value of the metric.
Intuitively I would like to compute $I \wedge n$ but I don't really know if that is justified.
Ideas:
- We could compute $n^T I n$ which would kind of tell us the kinetic energy of rotation around the normal axis. If the cell is aligned with that axis, then this value should be small.
- maybe something like $$\text{alignment} = \frac{\sum_{i=1}^3 \lambda_i \angle(v_i,n)}{\sum_{i=1}^3 \lambda_i}?$$