Brief Review of Undergraduate Classical Mechanics (3) - Noether's Theorem & Generators of symmetry

이 글은 2020.10.14에 작성됨.

This post was written on Oct 14, 2020.

 

Sometimes it is the case that $\delta S$ vanishes for certain limited variations of the path without imposing any condition at all. When this happens, we say that $S$ has a symmetry. A symmetry of an action functional $S[x]$ is an infinitesimal transformation of the path, $x^i(t) \rightarrow x^i+\epsilon^i(x^j)$ that leaves the action invariant :

$$S[x^i+\epsilon^i(x^j(t),t)]=S[x^i(t)]$$

or, up to constant :

$$S[x^i+\epsilon^i(x^j(t),t)]=S[x^i(t)]+C$$

which means action is transformed up to total derivative of some function. $$L(x^j)=L(x^j+\epsilon^j)+\frac{d}{dt} G(x^j,t) $$

in any case, physics remains unchanged.

 

(1) For some symmetry transformation, regardless of the choice of path, suppose 

$$\delta L \equiv \frac{dG(x^j,t)}{dt}$$

 

(2) Also suppose path is transformed like below :

$$x^i \rightarrow x^i+\epsilon^i(x^j(t))$$

 

(3) Then Noether current (more natural concept; charge) is conserved.

$$\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{x}^i} \epsilon^i(x^j) - G(x^j,t) \right)_{Classical\:path} = 0$$

 

[PROOF] The action has a symmetry. so identically $\delta S \equiv G \mid ^{t_2}_{t_1}$. No equation of motion has been used. Since path transformed $x^i \rightarrow x^i+\epsilon^i(x^j(t))$, $\delta S$ can be represented also like

$$\begin{align} \delta S & \equiv \int_{t_1}^{t_2} dt \left( \frac{\partial L}{\partial x^i} \epsilon^i + \frac{\partial L}{\partial \dot{x}^i} \dot{\epsilon}^i \right) \\ & = \left. \frac{\partial L}{\partial \dot{x}^i} \epsilon^i \right|^{t_2}_{t_1} + \int_{t_1}^{t_2} dt \left( \frac{\partial L}{\partial x^i} - \frac{d}{dt} \frac{\partial L}{\partial \dot{x}^i} \right) \end{align}$$

This equation satisfies for all paths. Regardless of path selected, we here only imposed symmetry condition. Let's now select our path clasically. For the classical path, Lagrange equation holds. Thus 2nd term vanishes. Then we get

$$ \left. \left( \frac{\partial L}{\partial \dot{x}^i} \epsilon^i - G \right) \right|_{t_1}^{t_2} = 0$$

for any two times $t_1, t_2$.

 

Now let's discuss about the generators. In Hamiltonian mechanics, Poisson bracket is defined as 
$$\{ A,B \}=\frac{\partial A}{\partial q^i} \frac{\partial B}{\partial p_i} - \frac{\partial A}{\partial p_i} \frac{\partial B}{\partial q^i}$$ For some symmetry transformation $q^i \to q^i+\delta q^i, \;\; p_i \to p_i +\delta p_i$, $\delta A$ is 
$$\delta A = \delta q^i \frac{\partial A}{\partial q^i} + \delta p_i \frac{\partial A}{\partial p_i}$$ If there exists some generator $G$ such that $$\delta A = \delta a ^{ijk\cdots} \{A,G_{ijk\cdots}\}$$ then $G$ is called a 'generaor' of this symmetry. Here's some examples.

 

(1) Generator of time translation : for the symmetry of $t\to t+\delta t$, $$\delta A = \delta t \dot{q}^i \frac{\partial A}{\partial q^i} + \delta t \dot{p}_i \frac{\partial A}{\partial p_i}$$ In this symmetry, the corresponding generator is Hamiltonian since $$\begin{align} \delta t \{A, H\} & = \delta t \left( \frac{\partial A}{\partial q^i} \frac{\partial H}{\partial p_i} - \frac{\partial A}{\partial p_i} \frac{\partial H}{\partial q^i} \right) \\ &= \delta t \left(\dot{q}^i \frac{\partial A}{\partial q^i} +  \dot{p}_i \frac{\partial A}{\partial p_i}\right)\end{align}$$
(2) Generator of spatial translation : for the symmetry of $q^i \to q^i + \delta q^i$, $$\delta A = \delta q^i \frac{\partial A}{\partial q^i}$$ In this symmetry, the corresponding generator is canonical momentum since $$\begin{align} \delta q^i \{A, p_i\}& = \delta q^i \left( \frac{\partial A}{\partial q^j} \frac{\partial p_i}{\partial p_j} - \frac{\partial A}{\partial p_j} \frac{\partial p_i}{\partial q^j} \right) \\ &= \delta q^i \frac{\partial A}{\partial q^i} \end{align}$$ 
(3) Generator of a rotation : for the symmetry of $x^i \to x^i + \delta \theta_k \epsilon^{kli} x_l$, $p_i \to p_i + \delta \theta_k \epsilon^{kli} p_l$, $$\delta A = \delta \theta^k \epsilon_{kli} \left(  x^l \frac{\partial}{\partial x_i} + p^l \frac{\partial}{\partial p_i} \right) A$$  In this symmetry, the corresponding generator is angular momentum since  $$\begin{align} \delta A &= \delta \theta^k \{ A, L_k\} = \delta \theta^k \epsilon_{klm} \{ A, x^l p^m\} \\ &= \delta \theta^k \epsilon_{klm} \left( \frac{\partial A}{\partial x^i}\right. \underbrace{\frac{\partial (x^l p^m)}{\partial p_i}}_{x^l g^{mi}} - \underbrace{\frac{\partial (x^l p^m)}{\partial x^i}}_{\delta^l_i p^m} \left. \frac{\partial A}{\partial p_i}  \right) \\ &= \delta \theta^k \left( \epsilon_{kl}^{\;\;\;i} x^l \frac{\partial }{\partial x^i} - \epsilon_{kim} p^m \frac{\partial}{\partial p_i} \right) A \\ &= A(x^i + \delta \theta_k \epsilon^{kli} x_l, p_i + \delta \theta_k \epsilon^{kli} p_l) - A(x^i, p_i) \end{align}$$