0
$\begingroup$

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

$$\boldsymbol{\omega}=\omega_1\mathbf{e}_1+\omega_2\mathbf{e}_2+\omega_3\mathbf{e}_3=\tilde{\omega}_1\mathbf{\tilde{e}}_1+\tilde{\omega}_2\mathbf{\tilde{e}}_2+\tilde{\omega}_3\mathbf{\tilde{e}}_3,$$

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

$\endgroup$

1 Answer 1

0
$\begingroup$

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}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.