# Is the moment of inertia matrix a tensor?

Tags:
1. Oct 20, 2015

### krabbie

1. The problem statement, all variables and given/known data
Is the moment of inertia matrix a tensor? Hint: the dyadic product of two vectors transforms according to the rule for second order tensors.
• $I$ is the inertia matrix
• $L$ is the angular momentum
• $\omega$ is the angular velocity

2. Relevant equations
The transformation rule for a second order tensor is: $I'_{ij} = C_{ip}C_{jq}I_{pq}$. A dyadic product of two vectors $u$ and $v$ is a matrix of the form: $A_{ij} = u_iv_j$.

3. The attempt at a solution
In a previous homework, we proved that a dyadic product transforms according to the rule for second order tensors. I would like to show that the moment of inertia matrix is a dyadic product. However, we never defined it that way in class yet, so I am unsure of how exactly this would work. We have: $L_i = I_{ij}w_j$. Now, is it ok to then write that $L_i\frac{1}{w_j} = I_{ij}$, so that $I_{ij}$ is a dyadic product of $L$ and the vector whose components are $\frac{1}{w_i}$? That feels wrong to me, because dividing out by something with a dummy variable like that doesn't seem like it should be valid, but I'm not actually sure!

2. Oct 20, 2015

### Geofleur

You definitely can't divide by the $\omega_j$ that way, because it is being summed over. For example, $L_1 = I_{11}\omega_1 + I_{12}\omega_2 + I_{13}\omega_3$. This last equation could be divided by $\omega_1$, for example, but that would not eliminate the $\omega$'s from the right hand side.

I am not sure whether the moment of inertia tensor can be represented simply as a dyadic product. I would think it would be easier to use something like $I_{ij} = \sum_k m_k (r_k^2 \delta_{ij} - x_{k,i}x_{k,j})$ and then show that each term is a tensor under the relevant transformations.