So say we can describe some angular velocity $\boldsymbol{\omega}$ in canonical basis, $\{\mathbf{e}_i\}$, and a rotated basis, $\{\mathbf{\tilde{e}}_i\}$, like


and its coordinates in each basis and their basis are related by the rotational matrix $\mathbf{R}$ in the following way: $$\begin{aligned} \begin{pmatrix} \tilde{\omega}_1\\ \tilde{\omega}_1\\ \tilde{\omega}_1 \end{pmatrix}=\mathbf{R} \begin{pmatrix} \omega_1\\ \omega_2\\ \omega_3 \end{pmatrix} \end{aligned}\ \text{and}\ \begin{aligned} \begin{pmatrix} \mathbf{\tilde{e}}_1\\ \mathbf{\tilde{e}}_2\\ \mathbf{\tilde{e}}_3 \end{pmatrix}=\mathbf{(R^{-1})} \begin{pmatrix} \mathbf{e}_1\\ \mathbf{e}_2\\ \mathbf{e}_3 \end{pmatrix}. \end{aligned}$$

Then, what I want to ask is whether when calculating the angular momentum in canonical basis, $$\mathbf{L_e}=\mathcal{I_e}\boldsymbol{\omega_{e}}$$ and in rotated basis $$\mathbf{L_{\tilde{e}}}=\mathcal{I_{\tilde{e}}}\boldsymbol{\omega_{\tilde{e}}},$$

would $\mathcal{I_e}=\mathcal{I_{\tilde{e}}}$?

Because I thought that maybe if $$(\mathcal{I_{ij}})_e=\iiint_V \rho(r^2\delta_{ij}-x_ix_j) dV$$

then $$(\mathcal{I_{ij}})_{\tilde{e}}=\iiint_\tilde{V} \rho(r^2\delta_{ij}-\tilde{x}_i\tilde{x}_j) d\tilde{V}.$$

Note that $\mathbf{\tilde{x}}=\mathbf{Rx}.$


1 Answer 1


No, it won't be the same.

$$ (I_{\tilde e})_{ij} = \int_V \rho (\tilde r^2 \delta_{ij}-\tilde x_i \tilde x _j)$$ of course $r^2 = \tilde{\mathbf{r}} \cdot \tilde{\mathbf{r} } = (\mathbf{R}\mathbf{r})^{\mathrm{tr}}\cdot \mathbf{R} \mathbf{r} = \mathbf{r}\mathbf{R}^{-1}\mathbf{R}\mathbf{r}=\mathbf{r}\cdot\mathbf{r}$. This is true for any scalar, also the square of the angular momentum. Let us go back to the formula $$ (I_{\tilde e})_{ij} = \int_V \rho (\tilde r^2 \delta_{ij}-R_{ia}x_a\,R_{jb} x_b)$$ which differs from $ (I_{e})_{ij}$.


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