# 物理代写|分析力学代写Analytical Mechanics代考|PHYS330

## 物理代写|分析力学代写Analytical Mechanics代考|Hamilton’s Principle and Lagrange’s Equations

The mere change of notation
$$x \rightarrow t, \quad y \rightarrow q, \quad y^{\prime} \equiv \frac{d y}{d x} \rightarrow \dot{q} \equiv \frac{d q}{d t}, \quad f \rightarrow L, \quad J \rightarrow S$$
shows that Euler’s equation takes the form of Lagrange’s equation
$$\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}}\right)-\frac{\partial L}{\partial q}=0,$$
which, therefore, follows from the variational principle
$$\delta S \equiv \delta \int_{t_1}^{t_2} L(q, \dot{q}, t) d t=0$$
with $\delta q\left(t_1\right)=\delta q\left(t_2\right)=0$. The generalisation for holonomic systems with any number of degrees of freedom is straightforward. Let $S$ be the action defined by
$$S=\int_{t_1}^{t_2} L\left(q_1, \ldots, q_n, \dot{q}_1, \ldots, \dot{q}_n, t\right) d t$$
and consider
$$\begin{gathered} \bar{q}_1(t)=q_1(t)+\delta q_1(t), \ \vdots \ \bar{q}_n(t)=q_n(t)+\delta q_n(t), \end{gathered}$$
with the variations $\delta q_1, \ldots, \delta q_n$ mutually independent and arbitrary except for the endpoint conditions $\delta q_k\left(t_1\right)=\delta q_k\left(t_2\right)=0, k=1, \ldots, n$. It is worth stressing that it is the mutual independence of the generalised coordinates that ensures that each can be varied independently of the others.

## 物理代写|分析力学代写Analytical Mechanics代考|Equivalent Lagrangians

Hamilton’s principle, also known as the principle of least action, ${ }^7$ entails that the same equations of motion are generated by two Lagrangians that differ by a total time derivative of an arbitrary function of the generalised coordinates and time.

Definition 2.3.1 Two Lagrangians $\bar{L}(q, \dot{q}, t)$ and $L(q, \dot{q}, t)$ are said to be equivalent if they differ by the total time derivative of an arbitrary function $f(q, t)$ of the generalised coordinates and time:
$$\bar{L}(q, \dot{q}, t)=L(q, \dot{q}, t)+\frac{d}{d t} f(q, t) .$$
Theorem 2.3.1 Equivalent Lagrangians give rise to the same equations of motion. Proof The action $\bar{S}$ associated with $\bar{L}$ is
$$\bar{S}=\int_{t_1}^{t_2} \bar{L}(q, \dot{q}, t) d t=\int_{t_1}^{t_2} L(q, \dot{q}, t) d t+\int_{t_1}^{t_2} \frac{d f}{d t} d t=S+f\left(q\left(t_2\right), t_2\right)-f\left(q\left(t_1\right), t_1\right) .$$
Since the variation of the action leaves the endpoints $q\left(t_1\right)$ and $q\left(t_2\right)$ fixed, $\delta \bar{S}=\delta S$. Therefore, the conditions $\delta \bar{S}=0$ and $\delta S=0$ are identical, showing that $\bar{L}$ and $L$ engender exactly the same equations of motion.

Exercise 2.3.1 Prove, by direct substitution into Lagrange’s equations, that equivalent Lagrangians give rise to the same equations of motion.

## 物理代写|分析力学代写分析力学代考|汉密尔顿原理和拉格朗日方程

$$x \rightarrow t, \quad y \rightarrow q, \quad y^{\prime} \equiv \frac{d y}{d x} \rightarrow \dot{q} \equiv \frac{d q}{d t}, \quad f \rightarrow L, \quad J \rightarrow S$$

$$\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}}\right)-\frac{\partial L}{\partial q}=0,$$
，因此，遵循变分原理
$$\delta S \equiv \delta \int_{t_1}^{t_2} L(q, \dot{q}, t) d t=0$$
with $\delta q\left(t_1\right)=\delta q\left(t_2\right)=0$。对于具有任意数量自由度的完整系统的推广是直接的。假设$S$是
$$S=\int_{t_1}^{t_2} L\left(q_1, \ldots, q_n, \dot{q}_1, \ldots, \dot{q}_n, t\right) d t$$

$$\begin{gathered} \bar{q}_1(t)=q_1(t)+\delta q_1(t), \ \vdots \ \bar{q}_n(t)=q_n(t)+\delta q_n(t), \end{gathered}$$
，其变化$\delta q_1, \ldots, \delta q_n$是相互独立和任意的，端点条件$\delta q_k\left(t_1\right)=\delta q_k\left(t_2\right)=0, k=1, \ldots, n$除外。值得强调的是，正是广义坐标的相互独立性保证了每个广义坐标可以独立于其他广义坐标变化

## 物理代写|分析力学代写分析力学代考|等效拉格朗日量

$$\bar{L}(q, \dot{q}, t)=L(q, \dot{q}, t)+\frac{d}{d t} f(q, t) .$$

$$\bar{S}=\int_{t_1}^{t_2} \bar{L}(q, \dot{q}, t) d t=\int_{t_1}^{t_2} L(q, \dot{q}, t) d t+\int_{t_1}^{t_2} \frac{d f}{d t} d t=S+f\left(q\left(t_2\right), t_2\right)-f\left(q\left(t_1\right), t_1\right) .$$

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