Is the moment of inertia matrix a tensor?

  • #1
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Homework Statement


Is the moment of inertia matrix a tensor? Hint: the dyadic product of two vectors transforms according to the rule for second order tensors.
  • [itex]I[/itex] is the inertia matrix
  • [itex]L[/itex] is the angular momentum
  • [itex]\omega[/itex] is the angular velocity

Homework Equations


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

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: [itex]L_i = I_{ij}w_j[/itex]. Now, is it ok to then write that [itex]L_i\frac{1}{w_j} = I_{ij}[/itex], so that [itex]I_{ij}[/itex] is a dyadic product of [itex]L[/itex] and the vector whose components are [itex]\frac{1}{w_i}[/itex]? 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!
 

Answers and Replies

  • #2
Geofleur
Science Advisor
Gold Member
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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.
 

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