[NOTES/ME-14006]-Tensor Nature of Moment of Inertia

The moment of inertia tensors relative to two sets of axes \(K, K{'}\) is given by \begin{eqnarray} I_{jk} &=&\sum_\alpha m_\alpha (\delta_{jk} |\vec{x}_\alpha|^2 - x_{\alpha j}x_{\alpha k}),\\ I_{jk}{'} &=&\sum_\alpha m_\alpha (\delta_{jk} |\vec{x}_\alpha{'}|^2 - x_{\alpha j}{'} x_{\alpha k}{'}). \end{eqnarray} We shall prove that if \(K,K{'}\) are related by a rotation, \begin{equation} I_{jk}{'} = R_{j\ell}R_{km} I_{\ell m} \end{equation} holds under a rotation of axes. This relation will show that the inertia tensor qualifies to be called a tensor of rank 2.

Proof of EQ02

In order to prove EQ02, we shall use he following results

• Under a rotation, the components of a vector transform as \begin{eqnarray} (x{'})_j &=& R_{j\ell} x_\ell \\ \therefore \qquad\qquad (x{'})_j(x{'})_k &=& R_{j\ell}\, R_{km}\, x_\ell\, x_m.  \end{eqnarray}
• The length of a vector does not change under rotation of axes. Therefore \begin{equation} (\vec{x}_\alpha{'})^2 = (\vec{x}_\alpha)^2.  \end{equation}
• The Kronecker delta is an invariant tensor of rank two. This means that \begin{equation} \delta_{jk} = R_{j\ell}\, R_{km}\, \delta_{\ell m}.  \end{equation}

Now begin with EQ01B and use EQ03-EQ05 to get \begin{eqnarray} I_{jk}{'} &=&\sum_\alpha m_\alpha (\delta_{jk} |\vec{x}{'}_\alpha|^2 - x{'}_{\alpha j} x{'}_{\alpha k}).\\ &=& \sum_\alpha m_\alpha (R_{j\ell} R_{km}\delta_{\ell m} |\vec{x}_\alpha|^2 - x{'}_{\alpha j} x{'}_{\alpha k}).\\ &=& \sum_\alpha m_\alpha (R_{j\ell} R_{km}\delta_{\ell m} |\vec{x}_\alpha|^2 - R_{j\ell}\,R_{km}\,x_{\alpha \ell} x_{\alpha m}),\\ &=& \sum_\alpha m_\alpha (R_{j\ell} R_{km}\big\{\delta_{\ell m} |\vec{x}_\alpha|^2 - x_{\alpha \ell} x_{\alpha m}\big\}),\\ &=& (R_{j\ell} R_{km} \sum_\alpha m_\alpha \big\{\delta_{\ell m} |\vec{x}_\alpha|^2 - x_{\alpha \ell} x_{\alpha m}\big\}),\\ &=& (R_{j\ell} R_{km}) I_{\ell m} . \end{eqnarray}

This completes the proof of EQ02, and hence the moment of inertia tensor transforms like a second rank tensor.