PX285 - E4 - rotation of space

consider the isotropy of space (rotational symmetry)
same dynamics expected, regardless of rotation, so there should be no 'special direction'
consider a position vector, $\vec{r}$ , rotated by a small angle, $\vec{δ ϕ} :$ $\vec{r} \to \vec{r} + \vec{δ r}$ , where, $\vec{δ r} = \vec{δ ϕ} \times \vec{r}$
the rotation also acts on velocities: $\vec{δ v} = \vec{δ ϕ} \times \vec{v}$
the lagrangian:

L (\vec{r}, \vec{v}) \to L (\vec{r} + \vec{δ r}, \vec{v} + \vec{δ v}) = L (\vec{r}, \vec{v}) + δ L

if space is isotropic, $δ L = 0 :$

L (\vec{r}, \vec{v}) = L (\vec{r} + \vec{δ r})

δ L = \vec{δ r} \cdot \vec{\nabla} L + \vec{δ v} \cdot {\vec{\nabla}}_{v} L = (\vec{δ ϕ} \times \vec{r}) \cdot \vec{\nabla} L + (\vec{δ ϕ \times \vec{v}}) \cdot {\vec{\nabla}}_{v} L

{\vec{\nabla}}_{v} L = (\begin{matrix} \frac{\partial L}{\partial v_{x}} \\ \frac{\partial L}{\partial v_{y}} \\ \frac{\partial L}{\partial v_{z}} \end{matrix}) = \vec{p}

\vec{\nabla} L = (\begin{matrix} \frac{\partial L}{\partial x} \\ \frac{\partial L}{\partial y} \\ \frac{\partial L}{\partial z} \end{matrix}) = \dot{\vec{p}}

\begin{aligned} δ L & = (\vec{δ ϕ} \times \vec{r}) \cdot \dot{\vec{p}} + (\vec{δ ϕ} \times \vec{v}) \cdot \vec{p} \\ = \vec{δ ϕ} \cdot (\vec{r} \times \dot{\vec{p}}) + \vec{δ ϕ} \cdot (\vec{v} \times \vec{p}) \\ = \vec{δ ϕ} \cdot [\vec{r} \times \dot{\vec{p}} + \vec{v} \times \vec{p}] \\ = \vec{δ ϕ} \cdot [\vec{r} \times \dot{\vec{p}} + \dot{\vec{r}} \times \vec{p}] \\ = \vec{δ ϕ} \cdot \frac{d}{d t} (\vec{r} \times \vec{p}) \end{aligned}

if space is isotropic, $δ L = 0 :$

\frac{d}{d t} \vec{M} = \frac{d}{d t} (\vec{r} \times \vec{p}) = 0

the angular momentum, $\vec{M}$ , is an invariant, ie: a constant of the motion
$\vec{M}$ contains $d$ invariant components, each arising from the rotational symmetries in $d$ -dimensional space

general case

more generally, for $N$ interacting particles, eg: bound in a solid body, there will be $3 N$ coordinates and $3 N$ velocities, giving the lagrangian:

L (x_{1}, \dots z_{n}, {\dot{x}}_{1}, \dots {\dot{z}}_{n})

for a rotation, $L \to L + δ L :$

δ L = \sum_{i = 1}^{N} [(\begin{matrix} \frac{\partial L}{\partial x_{i}} \\ \frac{\partial L}{\partial y_{i}} \\ \frac{\partial L}{\partial z_{i}} \end{matrix}) \cdot (\begin{matrix} δ x_{1} \\ δ y_{1} \\ δ z_{1} \end{matrix}) + (\begin{matrix} \frac{\partial L}{\partial {\dot{x}}_{i}} \\ \frac{\partial L}{\partial {\dot{y}}_{i}} \\ \frac{\partial L}{\partial {\dot{z}}_{i}} \end{matrix}) \cdot (\begin{matrix} δ {\dot{x}}_{1} \\ δ {\dot{y}}_{1} \\ δ {\dot{z}}_{1} \end{matrix})]

this is the first order taylor expansion in $6 N$ dependent variables

\begin{aligned} δ L & = \sum_{i} (\dot{{\vec{p}}_{i}} \cdot \vec{δ r_{i}} + {\vec{p}}_{i} \cdot \vec{δ v_{i}}) \\ = \sum_{i} \vec{δ ϕ} \cdot \frac{d}{d t} ({\vec{r}}_{i} \times {\vec{p}}_{i}) \\ = \vec{δ ϕ} \cdot \frac{d}{d t} \sum_{i} ({\vec{r}}_{i} \times {\vec{p}}_{i}) \end{aligned}

for an arbitrary rotation, $δ L = 0$ only if $\frac{d}{d t} \sum_{i} ({\vec{r}}_{i} \times {\vec{p}}_{i}) = 0$
define the angular momentum, $\vec{M} = \sum_{i} ({\vec{r}}_{i} \times {\vec{p}}_{i})$
$\frac{d}{d t} \vec{M} = 0$ , so it is a conserved quantity
$\vec{M}$ has $d = 3$ components, one for each rotational symmetry