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TL;DR: OP's reference to Noether's theorem is exactly right. Angular momentum $x^{[i}p^{j]}$ is the Noether charge for rotational symmetry.

  1. The Noether charge in point mechanics reads $$ Q~=~ p_i X^i - h T, \tag{1}$$ where $X^i$ and $T$ are spatial and temporal generators, and where $p_i=\frac{\partial L}{\partial v^i}$ is the momentum and $h=p_iv^i-L$ is the energy.

  2. The $n$-dimensional rotation group $SO(n)$ has a Lie algebra $$so(n)~=~\{\omega\in {\rm Mat}_{n\times n}(\mathbb{R}) \mid \omega^T=-\omega\} \tag{2}$$ consisting of antisymmetric matrices. It generates infinitesimal rotations.

  3. An infinitesimal spatial rotation $$ \delta x^i ~=~\epsilon X^i, \qquad \delta t~=~\epsilon T, \tag{3}$$ (where $\epsilon$ is an infinitesimal parameter), has generatorgenerators$^1$ $$ X^i = \omega^i{}_jx^j, \qquad T~=~0,\tag{4}$$ so$$ X^i = \omega^i{}_jx^j, \qquad T~=~0.\tag{4}$$

  4. So the corresponding Noether charge is $$ Q~\stackrel{(1)+(4)}{=}~p_i \omega^i{}_jx^j~=~\underbrace{x^{[j}p^{i]}}_{\text{ang. mom.}}\omega_{ij}.\tag{5} $$

See also e.g. this related Phys.SE post.

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$^1$ We raise and lower spatial indices with the spatial metric $\delta_{ij}$.

TL;DR: OP's reference to Noether's theorem is exactly right. Angular momentum $x^{[i}p^{j]}$ is the Noether charge for rotational symmetry.

  1. The Noether charge in point mechanics reads $$ Q~=~ p_i X^i - h T, \tag{1}$$ where $X^i$ and $T$ are spatial and temporal generators, and where $p_i=\frac{\partial L}{\partial v^i}$ is the momentum and $h=p_iv^i-L$ is the energy.

  2. The $n$-dimensional rotation group $SO(n)$ has a Lie algebra $$so(n)~=~\{\omega\in {\rm Mat}_{n\times n}(\mathbb{R}) \mid \omega^T=-\omega\} \tag{2}$$ consisting of antisymmetric matrices. It generates infinitesimal rotations.

  3. An infinitesimal spatial rotation $$ \delta x^i ~=~\epsilon X^i, \qquad \delta t~=~\epsilon T, \tag{3}$$ has generator$^1$ $$ X^i = \omega^i{}_jx^j, \qquad T~=~0,\tag{4}$$ so the corresponding Noether charge is $$ Q~\stackrel{(1)+(4)}{=}~p_i \omega^i{}_jx^j~=~\underbrace{x^{[j}p^{i]}}_{\text{ang. mom.}}\omega_{ij}.\tag{5} $$

See also e.g. this related Phys.SE post.

--

$^1$ We raise and lower spatial indices with the spatial metric $\delta_{ij}$.

TL;DR: OP's reference to Noether's theorem is exactly right. Angular momentum $x^{[i}p^{j]}$ is the Noether charge for rotational symmetry.

  1. The Noether charge in point mechanics reads $$ Q~=~ p_i X^i - h T, \tag{1}$$ where $X^i$ and $T$ are spatial and temporal generators, and where $p_i=\frac{\partial L}{\partial v^i}$ is the momentum and $h=p_iv^i-L$ is the energy.

  2. The $n$-dimensional rotation group $SO(n)$ has a Lie algebra $$so(n)~=~\{\omega\in {\rm Mat}_{n\times n}(\mathbb{R}) \mid \omega^T=-\omega\} \tag{2}$$ consisting of antisymmetric matrices. It generates infinitesimal rotations.

  3. An infinitesimal spatial rotation $$ \delta x^i ~=~\epsilon X^i, \qquad \delta t~=~\epsilon T, \tag{3}$$ (where $\epsilon$ is an infinitesimal parameter), has generators$^1$ $$ X^i = \omega^i{}_jx^j, \qquad T~=~0.\tag{4}$$

  4. So the corresponding Noether charge is $$ Q~\stackrel{(1)+(4)}{=}~p_i \omega^i{}_jx^j~=~\underbrace{x^{[j}p^{i]}}_{\text{ang. mom.}}\omega_{ij}.\tag{5} $$

See also e.g. this related Phys.SE post.

--

$^1$ We raise and lower spatial indices with the spatial metric $\delta_{ij}$.

Added explanation
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Qmechanic
  • 206.6k
  • 48
  • 566
  • 2.3k

TL;DR: OP's reference to Noether's theorem is exactly right. Angular momentum $x^{[i}p^{j]}$ is the Noether charge for rotational symmetry.

  1. Noether's theorem is exactly right. The Noether charge in point mechanics reads $$ Q~=~ p_i X^i - h T, \tag{1}$$ where $X^i$ and $T$ are spatial and temporal generators, and where where $p_i=\frac{\partial L}{\partial v^i}$ is the momentum and $h=p_iv^i-L$ is the energy.

  2. Next theThe $n$-dimensional rotation group $SO(n)$ has a Lie algebra $$so(n)~=~\{\omega\in {\rm Mat}_{n\times n}(\mathbb{R}) \mid \omega^T=-\omega\} \tag{2}$$ consisting of antisymmetric matrices. It generates infinitesimal rotations.

  3. An infinitesimal spatial rotation $$ \delta x^i ~=~\epsilon X^i, \qquad \delta t~=~\epsilon T, \tag{3}$$ has generator$^1$ $$ X^i = \omega^i{}_jx^j, \qquad T~=~0,\tag{4}$$ so the corresponding Noether charge is $$ Q~\stackrel{(1)+(4)}{=}~p_i \omega^i{}_jx^j~=~\underbrace{x^{[j}p^{i]}}_{\text{ang. mom.}}\omega_{ij}.\tag{5} $$

See also e.g. this related Phys.SE post.

--

$^1$ We raise and lower spatial indices with the spatial metric $\delta_{ij}$.

  1. Noether's theorem is exactly right. The Noether charge in point mechanics reads $$ Q~=~ p_i X^i - h T, \tag{1}$$ where $X^i$ and $T$ are spatial and temporal generators, and where $p_i=\frac{\partial L}{\partial v^i}$ is the momentum and $h=p_iv^i-L$ is the energy.

  2. Next the $n$-dimensional rotation group $SO(n)$ has a Lie algebra $$so(n)~=~\{\omega\in {\rm Mat}_{n\times n}(\mathbb{R}) \mid \omega^T=-\omega\} \tag{2}$$ consisting of antisymmetric matrices. It generates infinitesimal rotations.

  3. An infinitesimal spatial rotation $$ \delta x^i ~=~\epsilon X^i, \qquad \delta t~=~\epsilon T, \tag{3}$$ has generator$^1$ $$ X^i = \omega^i{}_jx^j, \qquad T~=~0,\tag{4}$$ so the corresponding Noether charge is $$ Q~\stackrel{(1)+(4)}{=}~p_i \omega^i{}_jx^j~=~\underbrace{x^{[j}p^{i]}}_{\text{ang. mom.}}\omega_{ij}.\tag{5} $$

--

$^1$ We raise and lower spatial indices with the spatial metric $\delta_{ij}$.

TL;DR: OP's reference to Noether's theorem is exactly right. Angular momentum $x^{[i}p^{j]}$ is the Noether charge for rotational symmetry.

  1. The Noether charge in point mechanics reads $$ Q~=~ p_i X^i - h T, \tag{1}$$ where $X^i$ and $T$ are spatial and temporal generators, and where $p_i=\frac{\partial L}{\partial v^i}$ is the momentum and $h=p_iv^i-L$ is the energy.

  2. The $n$-dimensional rotation group $SO(n)$ has a Lie algebra $$so(n)~=~\{\omega\in {\rm Mat}_{n\times n}(\mathbb{R}) \mid \omega^T=-\omega\} \tag{2}$$ consisting of antisymmetric matrices. It generates infinitesimal rotations.

  3. An infinitesimal spatial rotation $$ \delta x^i ~=~\epsilon X^i, \qquad \delta t~=~\epsilon T, \tag{3}$$ has generator$^1$ $$ X^i = \omega^i{}_jx^j, \qquad T~=~0,\tag{4}$$ so the corresponding Noether charge is $$ Q~\stackrel{(1)+(4)}{=}~p_i \omega^i{}_jx^j~=~\underbrace{x^{[j}p^{i]}}_{\text{ang. mom.}}\omega_{ij}.\tag{5} $$

See also e.g. this related Phys.SE post.

--

$^1$ We raise and lower spatial indices with the spatial metric $\delta_{ij}$.

Source Link
Qmechanic
  • 206.6k
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  • 566
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  1. Noether's theorem is exactly right. The Noether charge in point mechanics reads $$ Q~=~ p_i X^i - h T, \tag{1}$$ where $X^i$ and $T$ are spatial and temporal generators, and where $p_i=\frac{\partial L}{\partial v^i}$ is the momentum and $h=p_iv^i-L$ is the energy.

  2. Next the $n$-dimensional rotation group $SO(n)$ has a Lie algebra $$so(n)~=~\{\omega\in {\rm Mat}_{n\times n}(\mathbb{R}) \mid \omega^T=-\omega\} \tag{2}$$ consisting of antisymmetric matrices. It generates infinitesimal rotations.

  3. An infinitesimal spatial rotation $$ \delta x^i ~=~\epsilon X^i, \qquad \delta t~=~\epsilon T, \tag{3}$$ has generator$^1$ $$ X^i = \omega^i{}_jx^j, \qquad T~=~0,\tag{4}$$ so the corresponding Noether charge is $$ Q~\stackrel{(1)+(4)}{=}~p_i \omega^i{}_jx^j~=~\underbrace{x^{[j}p^{i]}}_{\text{ang. mom.}}\omega_{ij}.\tag{5} $$

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$^1$ We raise and lower spatial indices with the spatial metric $\delta_{ij}$.