# 物理代写|分析力学代写Analytical Mechanics代考|PHYS3351

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

Hamilton’s principle guarantees only that the action $S=\int_{t_1}^{t_2} L d t$ is stationary for the actual path described by the system. In general, one can only ensure that the value of the action for the true motion is a local minimum – for comparison paths sufficiently close to the physical path – and as long as the time interval $\left[t_1, t_2\right]$ is sufficiently small. In particular cases its is possible to obtain stronger results by elementary arguments. For a particle in a uniform gravitational field, the action is an absolute minimum without restriction on the time interval (Problem 2.2). In the case of a harmonic oscillator, the action is an absolute minimum provided the time interval is shorter than half the period of the oscillator (Problem 2.3).

In the calculus of variations (Gelfand \& Fomin, 1963; Elsgoltz, 1969; Fox, 1987; Kot, 2014) sufficient conditions for a minimum are formulated which can be applied to variational problems such as Hamilton’s principle. Here, however, we settle upon a direct approach which, for the sake of simplicity, will be restricted to the case of a particle in one-dimensional motion under a force with potential energy $V$. The admission of a general potential energy comes with a price: in order to prove a weaker result – compared with the results valid for the projectile and the harmonic oscillator – one has to resort to rather sophisticated mathematical reasonings. Therefore, let the reader be duly warned that the understanding of the proof of the theorem that follows requires a degree of mathematical maturity somewhat above the one presupposed in the rest of this book.

## 物理代写|分析力学代写Analytical Mechanics代考|Hamilton’s Principle in the Non-Holonomic Case

In the previous section we assumed that the mechanical system was described by mutually independent coordinates, an independence which was crucial for the derivation of (2.48) from (2.47). Such independent generalised coordinates always exist if all of the constraints imposed on the system are holonomic. When non-holonomic constraints are present it is impossible to introduce generalised coordinates such that the constraint conditions are identically satisfied. It is possible, nevertheless, to deduce the equations of motion from Hamilton’s principle in the special case in which the non-holonomic constraints are differential equations of the form
$$\sum_{k=1}^n a_{l k} d q_k+a_{l t} d t=0, \quad l=1, \ldots, p$$
whose coefficients $a_{l k}$ and $a_{l t}$ are functions solely of $q_1, \ldots, q_n$ and $t$. Let us suppose, therefore, that the system is described by $n$ coordinates $q_1, \ldots, q_n$ and is subject to the $p$ independent differential constraints $(2.66)$, where the index $l$ distinguishes the constraint equations from one another. In spite of their seemingly restricted form, constraints of the type (2.66) encompass virtually all non-holonomic constraints of physical interest.

Let $L=T-U$ be the Lagrangian for the system written as if there were no constraints. The generalised potential in $L$ refers to the applied forces alone, so that the constraint forces responsible for the enforcement of Eqs. (2.66) are not taken into account by the Lagrangian. Hamilton’s principle $\delta S=0$ implies
$$\int_{t_1}^{t_2} d t \sum_k\left[\frac{\partial L}{\partial q_k}-\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}_k}\right)\right] \delta q_k=0$$
but now we cannot infer that the coefficient of each $\delta q_k$ is zero because the $\delta q$ s are not mutually independent. Indeed, Eqs. (2.43) show that each $\delta q_k$ is a virtual displacement since time remains fixed as one performs the variation that leads from $q_k(t)$ to $\bar{q}_k(t)=$ $q_k(t)+\delta q_k(t)$. But, according to Eqs. (2.66), for there to be compatibility with the constraints the virtual displacements must obey $$\sum_{k=1}^n a_{l k} \delta q_k=0, \quad l=1, \ldots, p,$$
where we have used $d t=0$ for virtual displacements. Since the $n$ variations $\delta q_1, \ldots, \delta q_n$ have to satisfy the $p$ equations (2.68), only $n-p$ variations of the $q$ s are independent of each other. ${ }^{10}$ Thus, we are faced with the problem of determining a conditional extremum for the functional $S=\int_{t_1}^{t_2} L d t$, and the treatment imitates the method of Lagrange multipliers of differential calculus.

## 物理代写|分析力学代写分析力学代考|非完整情况下的汉密尔顿原理

$$\sum_{k=1}^n a_{l k} d q_k+a_{l t} d t=0, \quad l=1, \ldots, p$$
，其系数$a_{l k}$和$a_{l t}$仅是$q_1, \ldots, q_n$和$t$的函数，则有可能从汉密尔顿原理推导出运动方程。因此，让我们假设系统由$n$坐标$q_1, \ldots, q_n$描述，并服从$p$独立微分约束$(2.66)$，其中索引$l$将约束方程彼此区分开来。尽管(2.66)型约束的形式看似有限，但它实际上包含了物理兴趣的所有非完整约束

$$\int_{t_1}^{t_2} d t \sum_k\left[\frac{\partial L}{\partial q_k}-\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}k}\right)\right] \delta q_k=0$$
，但现在我们不能推断每个$\delta q_k$的系数为零，因为$\delta q$ s不是相互独立的。的确，eq。(2.43)表明每个$\delta q_k$都是一个虚位移，因为当人们执行从$q_k(t)$到$\bar{q}_k(t)=$$q_k(t)+\delta q_k(t)的变化时，时间是固定的。但是，根据equs。(2.66)，为了与约束兼容，虚拟位移必须服从$$ \sum{k=1}^n a_{l k} \delta q_k=0, \quad l=1, \ldots, p,$$，其中我们使用$d t=0$作为虚拟位移。由于$n$的变化$\delta q_1, \ldots, \delta q_n$必须满足$p$方程(2.68)，因此$q$的变化只有$n-p$是相互独立的。${ }^{10}$因此，我们面临着确定函数$S=\int_{t_1}^{t_2} L d t\$条件极值的问题，处理方法模仿微分学的拉格朗日乘子方法

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