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# 物理代写|量子力学代写Quantum mechanics代考|PHYS4141 Symmetry in classical physics

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## 物理代写|量子力学代写Quantum mechanics代考|Symmetry in classical physics

Observers A and B use their own coordinate systems to keep track of the particles. For the particle labelled by the index $i$ let us define A’s coordinates by $\mathbf{r}i$ and B’s coordinates by $\mathbf{r}_i^{\prime}$. These are related to each other by coordinate transformations that involve several parameters. For example in the case of translations $\mathbf{r}_i^{\prime}=\mathbf{r}_i+\mathbf{a}_i$, where $\mathbf{a}_i$ are the parameters. It is useful to consider nearby observers which are related to each other by infinitesimal coordinate transformations. If the infinitesimal parameters for N symmetries are $\epsilon_a, a=1,2, \cdots, N$, we may expand the relation between $\mathbf{r}_i$ and $\mathbf{r}_i^{\prime}$ to first order in the $\epsilon_a$ ‘s, and write $$r_i^{\prime I}=r_i^I+\delta\epsilon r_i^I, \quad \delta_\epsilon r_i^I=\sum_a \epsilon_a f_i^{I a}(\mathbf{r}, \dot{\mathbf{r}})$$
where $I=1,2,3$ denotes the vector index.

If the two observers $\mathrm{A}$ and $\mathrm{B}$ see identical physical phenomena and measure the same results, it must be that the equations that they use in terms of $\mathbf{r}_i$ and $\mathbf{r}_i^{\prime}$ respectively have the same form. If one takes A’s equations and substitutes $\mathbf{r}_i^{\prime}$ instead of $\mathbf{r}_i$ the resulting equations are B’s equations. Only if there is a symmetry, B’s equations, rewritten in terms of $\mathbf{r}_i$, will yield A’s equations in identical form, not otherwise.

Instead of discussing the symmetries of the equations of motion, it is more efficient to consider the action from which they are derived by a variational principle. The action $S$ is constructed from a Lagrangian in the form $S\left(\mathbf{r}_i\right)=$ $\int_1^2 d t L\left(\mathbf{r}_i(t), \dot{\mathbf{r}}_i(t)\right)$. The Euler equations are then
$$\frac{\partial}{\partial t} \frac{\partial L}{\partial \dot{\mathbf{r}}_i(t)}-\frac{\partial L}{\partial \mathbf{r}_i(t)}=0$$
There will be a symmetry provided, under the substitution $\mathbf{r}_i \rightarrow \mathbf{r}_i^{\prime}$, the form of the action remains invariant up to a “constant”
$$S(\mathbf{r}) \rightarrow S\left(\mathbf{r}^{\prime}\right)=S(\mathbf{r})+\operatorname{constant}(1,2)$$

## 物理代写|量子力学代写Quantum mechanics代考|Symmetry and classical conservation laws

The above examples illustrate some simple physical systems with symmetries. Now consider a general Lagrangian describing an arbitrary system of particles located at $\mathbf{r}_i(t)$ at time $t$. Suppose the Lagrangian has a symmetry under the infinitesimal transformation of (8.1) with some specific functions $f_i^{I a}\left(\mathbf{r}_j, \dot{\mathbf{r}}_j\right)$. According to Noether’s theorem, that we will prove below, corresponding to every symmetry parameter $\epsilon_a$ there exists a conserved quantity $Q_a\left(\mathbf{r}_i, \dot{\mathbf{r}}_i\right)$ that is time independent. That is, even though the location and velocities of the particles may be changing with time, the conserved quantities $Q_a$, which are constructed from them, remain unchanged, i.e. $d Q_a / d t=0$. The conservation of energy, momentum and angular momentum are some examples of consequences of symmetry. There are many more interesting cases in specific physical systems.
To construct the explicit form of $Q_a\left(\mathbf{r}_i, \dot{\mathbf{r}}_i\right)$ and prove Noether’s theorem, first note that the symmetry of the action (8.3) is satisfied most generally provided the Lagrangian behaves as follows
$$L\left(\mathbf{r}_i^{\prime}(t), \dot{\mathbf{r}}_i^{\prime}(t)\right)=\frac{\partial t^{\prime}}{\partial t} L\left(\mathbf{r}_i\left(t^{\prime}\right), \dot{\mathbf{r}}_i\left(t^{\prime}\right)\right)+\frac{\partial}{\partial t} \alpha\left(\mathbf{r}_i(t), \dot{\mathbf{r}}_i(t)\right) .$$
Here $t^{\prime}(t, \epsilon)$ is a change of variables that generally may depend on the parameters of the symmetry transformation, and $\frac{\partial t^{\prime}}{\partial t}$ is the Jacobian for the change of variables. $\alpha$ is some function of the dynamical variables and the parameters, which vanishes as $\epsilon_a \rightarrow 0$. The function $\alpha$ is zero in most cases, but not for every case, as will be seen in examples below. Also, in most cases $t^{\prime}(t, \epsilon)=t$, otherwise the infinitesimal expansion gives $t^{\prime}=t+\sum_a \epsilon_a \gamma^a\left(\mathbf{r}_i(t), \dot{\mathbf{r}}_i(t)\right)$ with some functions $\gamma^a$. When equation (8.10) is integrated, the left side yields $\int_1^2 L\left(\mathbf{r}_i^{\prime}(t), \dot{\mathbf{r}}_i^{\prime}(t)\right)=S\left(\mathbf{r}_i^{\prime}\right)$, and the right side gives
\begin{aligned} & \int_1^2 d t \frac{\partial t^{\prime}}{\partial t} L\left(\mathbf{r}_i\left(t^{\prime}\right), \dot{\mathbf{r}}_i\left(t^{\prime}\right)\right)+\int_1^2 d t \frac{\partial}{\partial t} \Lambda\left(\mathbf{r}_i(t), \dot{\mathbf{r}}_i(t)\right) \ & =\int d t^{\prime} L\left(\mathbf{r}_i\left(t^{\prime}\right), \dot{\mathbf{r}}_i\left(t^{\prime}\right)\right)+\Lambda(1)-\Lambda(2) \ & =S\left(\mathbf{r}_i\right)+\text { constant }(1,2) \end{aligned}
Thus, the condition of symmetry (8.3) is equivalent to (8.10).

## 物理代写|量子力学代写Quantum mechanics代考|Symmetry in classical physics

$$\frac{\partial}{\partial t} \frac{\partial L}{\partial \dot{\mathbf{r}}_i(t)}-\frac{\partial L}{\partial \mathbf{r}_i(t)}=0$$

$$S(\mathbf{r}) \rightarrow S\left(\mathbf{r}^{\prime}\right)=S(\mathbf{r})+\text { constant }(1,2)$$

## 物理代写|量子力学代写Quantum mechanics代考|Symmetry and classical conservation laws

$$L\left(\mathbf{r}_i^{\prime}(t), \dot{\mathbf{r}}_i^{\prime}(t)\right)=\frac{\partial t^{\prime}}{\partial t} L\left(\mathbf{r}_i\left(t^{\prime}\right), \dot{\mathbf{r}}_i\left(t^{\prime}\right)\right)+\frac{\partial}{\partial t} \alpha\left(\mathbf{r}_i(t), \dot{\mathbf{r}}_i(t)\right) .$$

$$\int_1^2 d t \frac{\partial t^{\prime}}{\partial t} L\left(\mathbf{r}_i\left(t^{\prime}\right), \dot{\mathbf{r}}_i\left(t^{\prime}\right)\right)+\int_1^2 d t \frac{\partial}{\partial t} \Lambda\left(\mathbf{r}_i(t), \dot{\mathbf{r}}_i(t)\right) \quad=\int d t^{\prime} L\left(\mathbf{r}_i\left(t^{\prime}\right), \dot{\mathbf{r}}_i\left(t^{\prime}\right)\right)+\Lambda(1)-\Lambda(2)=S\left(\mathbf{r}_i\right)+\operatorname{constant}(1,2)$$

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