如果你也在 怎样代写理论力学Theoretical Mechanics这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。理论力学Theoretical Mechanics是一组密切相关的经典力学的替代公式。它是由许多科学家和数学家在18世纪及以后，在牛顿力学之后发展起来的。由于牛顿力学考虑的是运动的矢量，特别是系统中各组成部分的加速度、动量、力，因此由牛顿定律和欧拉定律所支配的力学的另一个名称是矢量力学。
理论力学Theoretical Mechanics使用代表系统整体的运动标量属性–通常是其总动能和势能–而不是牛顿对单个粒子的矢量力。运动方程是由标量通过一些关于标量变化的基本原理推导出来的。分析力学使用代表系统整体的运动标量属性–通常是其总动能和势能–而不是牛顿对单个粒子的矢量力。运动方程是由标量通过一些关于标量变化的基本原理推导出来的。
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物理代写|理论力学作业代写Theoretical Mechanics代考|Motivation
Classical Mechanics appears in four equivalent formulations:
- Newton (Vol. 1),
- Lagrange (Sect. 1),
- Hamilton (Sect. 2),
- Hamilton-Jacobi (Sect. 3).
The transition from the Lagrange to the Hamilton formalism was mathematically carried out by use of a Legendre transformation. In the next section the HamiltonJacobi theory will be based on the Hamilton mechanics, discussed in this section with the aid of a so-called ‘canonical transformation’. For this purpose, some preliminary considerations are certainly advisable.
We have shown previously that in the Lagrange formalism the choice of the generalized coordinates $q_{1}, \ldots, q_{S}$ is in principle arbitrary, only the total number $S$ is fixed. That is because the Lagrange equations
$$
\frac{d}{d t} \frac{\partial L}{\partial \dot{q}{j}}-\frac{\partial L}{\partial q{j}}=0, \quad j=1,2, \ldots, S
$$
are in the configuration space forminvariant with respect to point transformations. That we proved in Sect. 1.2.1. For the transformation
$$
\left(q_{1}, \ldots, q_{S}\right) \Longleftrightarrow\left(\bar{q}{1}, \ldots, \bar{q}{S}\right)
$$
with
$$
\bar{q}{j}=\bar{q}{j}(\mathbf{q}, t) ; \quad q_{j}=q_{j}(\overline{\mathbf{q}}, t), \quad j=1,2, \ldots, S
$$
formally unchanged Lagrange equations arise,
$$
\frac{d}{d t} \frac{\partial \bar{L}}{\partial \overline{\bar{q}}{j}}-\frac{\partial \bar{L}}{\partial \bar{q}{j}}=0, \quad j=1,2, \ldots, S
$$
where the new Lagrangian $\bar{L}$ emerges from the old one simply by insertion of the transformation formulas:
$$
\bar{L}=L(\mathbf{q}(\overline{\mathbf{q}}, t), \dot{\mathbf{q}}(\overline{\mathbf{q}}, \dot{\overline{\mathbf{q}}}, t) t)=\bar{L}(\overline{\mathbf{q}}, \dot{\overline{\mathbf{q}}}, t)
$$
In addition, the Lagrange equations are also invariant with respect to so-called mechanical gauge transformations (1.84):
$$
L \Rightarrow L+L_{0} ; \quad L_{0}=\frac{d}{d t} f(\mathbf{q}, t)
$$
Thereby $f$ is allowed to be an almost arbitrary function of $\mathbf{q}$ and $t$. The actual reason for these invariances stems from the action functional $S{\mathbf{q}(t)}(1.120)$, which becomes always extremal for the same path in $M$ (1.118) independently of the special choice of coordinates. On the other hand, the Lagrange equations of motion result from the requirement $\delta S=0$.
物理代写|理论力学作业代写Theoretical Mechanics代考|The Generating Function
Starting point for the following considerations is the modified Hamilton’s principle (2.55). This states that the motion of the system takes place in such a way that the action functional
$$
S{\mathbf{q}(t), \mathbf{p}(t)}=\int_{t_{1}}^{t_{2}} d t\left(\sum_{j=1}^{S} p_{j} \dot{q}{j}-H(\mathbf{p}, \mathbf{q}, t)\right) $$ becomes extremal for the actual path on the competing set $M$ of admitted phase paths: $$ M=\left{(\mathbf{q}(t), \mathbf{p}(t)): \mathbf{q}\left(t{1}\right)=\mathbf{q}{\mathrm{a}}, \mathbf{q}\left(t{2}\right)=\mathbf{q}{\mathrm{e}} ; \mathbf{p}\left(t{1}\right), \mathbf{p}\left(t_{2}\right) \text { arbitrary }\right}
$$
物理代写|理论力学作业代写THEORETICAL MECHANICS代考|Equivalent Forms of the Generating Function
The $(\mathbf{q}, \overline{\mathbf{q}})$-dependence of the generating function $F_{1}$ is in principle by no means selfevident. Applying Legendre transformations one can find three further types of generating functions:
$$
\begin{aligned}
&F_{2}=F_{2}(\mathbf{q}, \overline{\mathbf{p}}, t) \
&F_{3}=F_{3}(\mathbf{p}, \overline{\mathbf{q}}, t) \
&F_{4}=F_{4}(\mathbf{p}, \overline{\mathbf{p}}, t)
\end{aligned}
$$
The generating functions always combine a new and an old coordinate. The ongoing statement of the problem decides which form is most convenient. For all the three functions there exists a theorem as that for $F_{1}$ in (2.149) and (2.150) which we have proved in the last section. Let us inspect this point in the following in a little bit more detail.
$$
F_{2}=F_{2}(\mathbf{q}, \overline{\mathbf{p}}, t)
$$
$F_{2}$ one obtains from $F_{1}$ by a Legendre transformation with respect to $\overline{\mathbf{q}}$ :
$$
F_{2}(\mathbf{q}, \overline{\mathbf{p}}, t)=F_{1}(\mathbf{q}, \overline{\mathbf{q}}, t)-\sum_{j=1}^{S} \frac{\partial F_{1}}{\partial \bar{q}{j}} \bar{q}{j}=F_{1}(\mathbf{q}, \overline{\mathbf{q}}, t)+\sum_{j=1}^{S} \bar{p}{j} \bar{q}{j}
$$
From the relation, already used to find (2.151),
$$
d F_{1}=\sum_{j=1}^{S}\left(p_{j} d q_{j}-\bar{p}{j} d \bar{q}{j}\right)+(\bar{H}-H) d t
$$
it follows for $F_{2}$ :
$$
d F_{2}=d F_{1}+\sum_{j=1}^{S}\left(\bar{p}{j} d \bar{q}{j}+\bar{q}{j} d \bar{p}{j}\right)=\sum_{j=1}^{S}\left(p_{j} d q_{j}+\bar{q}{j} d \bar{p}{j}\right)+(\bar{H}-H) d t
$$
That means:
$$
p_{j}=\frac{\partial F_{2}}{\partial q_{j}} ; \quad \bar{q}{j}=\frac{\partial F{2}}{\partial \bar{p}{j}} ; \quad \bar{H}=H+\frac{\partial F{2}}{\partial t}
$$
By inverting and resolving one shows, as explicitly demonstrated in the last section for $F_{1}$, that from (2.161) the transformation formulas $(\mathbf{q}, \mathbf{p}) \longrightarrow(\overline{\mathbf{q}}, \overline{\mathbf{p}})$ are uniquely derivable.
理论力学代写
物理代写|理论力学作业代写THEORETICAL MECHANICS代考|MOTIVATION
经典力学以四种等效的形式出现：
- 牛顿五这一世.1,
- 拉格朗日小号和C吨.1,
- 汉密尔顿小号和C吨.2,
- 汉密尔顿-雅可比小号和C吨.3.
从拉格朗日到汉密尔顿形式的转换是通过使用勒让德变换在数学上进行的。在下一节中，HamiltonJacobi 理论将基于 Hamilton 力学，在本节中借助所谓的“规范变换”进行讨论。为此目的，一些初步的考虑当然是可取的。
我们之前已经表明，在拉格朗日形式主义中，广义坐标的选择q1,…,q小号原则上是任意的，只有总数小号是固定的。这是因为拉格朗日方程
$$
\frac{d}{dt} \frac{\partial L}{\partial \dot{q} {j}}-\frac{\partial L}{\partial q {j} }=0, \quad j=1,2, \ldots, S
一种r和一世n吨H和C这nF一世G你r一种吨一世这nsp一种C和F这r米一世nv一种r一世一种n吨在一世吨Hr和sp和C吨吨这p这一世n吨吨r一种nsF这r米一种吨一世这ns.吨H一种吨在和pr这v和d一世n小号和C吨.1.2.1.F这r吨H和吨r一种nsF这r米一种吨一世这n
\剩下q_{1}, \ldots, q_{S}\rightq_{1}, \ldots, q_{S}\right\Longleftrightarrow\left(\bar{q} {1}, \ldots, \bar{q} {S}\right)
在一世吨H
\bar{q} {j}=\bar{q} {j}q,吨; \quad q_{j}=q_{j}q¯,吨, \quad j=1,2, \ldots, S
$$
形式上不变的拉格朗日方程出现，
$$
\frac{d}{d t} \frac{\partial \bar{L}}{\partial \overline{\bar{q}}{j}}-\frac{\partial \bar{L}}{\partial \bar{q}{j}}=0, \quad j=1,2, \ldots, S
$$
where the new Lagrangian $\bar{L}$ emerges from the old one simply by insertion of the transformation formulas:
$$
\bar{L}=L(\mathbf{q}(\overline{\mathbf{q}}, t), \dot{\mathbf{q}}(\overline{\mathbf{q}}, \dot{\overline{\mathbf{q}}}, t) t)=\bar{L}(\overline{\mathbf{q}}, \dot{\overline{\mathbf{q}}}, t)
$$
In addition, the Lagrange equations are also invariant with respect to so-called mechanical gauge transformations (1.84):
$$
L \Rightarrow L+L_{0} ; \quad L_{0}=\frac{d}{d t} f(\mathbf{q}, t)
$$
从而F被允许是一个几乎任意的函数q和吨. 这些不变性的实际原因源于作用泛函小号q(吨)(1.120)，对于相同的路径，它总是极值的米 1.118独立于坐标的特殊选择。另一方面，拉格朗日运动方程由要求产生d小号=0.
物理代写|理论力学作业代写THEORETICAL MECHANICS代考|THE GENERATING FUNCTION
以下考虑的出发点是修改后的汉密尔顿原理2.55. 这表明系统的运动发生在动作泛函
$$
S{\mathbf{q}(t), \mathbf{p}(t)}=\int_{t_{1}}^{t_{2}} d t\left(\sum_{j=1}^{S} p_{j} \dot{q}{j}-H(\mathbf{p}, \mathbf{q}, t)\right) $$ becomes extremal for the actual path on the competing set $M$ of admitted phase paths: $$ M=\left{(\mathbf{q}(t), \mathbf{p}(t)): \mathbf{q}\left(t{1}\right)=\mathbf{q}{\mathrm{a}}, \mathbf{q}\left(t{2}\right)=\mathbf{q}{\mathrm{e}} ; \mathbf{p}\left(t{1}\right), \mathbf{p}\left(t_{2}\right) \text { arbitrary }\right}
$$
物理代写|理论力学作业代写THEORETICAL MECHANICS代考|EQUIVALENT FORMS OF THE GENERATING FUNCTION
这(q,q¯)- 生成函数的依赖性F1原则上绝不是不言而喻的。应用勒让德变换可以找到另外三种类型的生成函数：
F2=F2(q,p¯,吨) F3=F3(p,q¯,吨) F4=F4(p,p¯,吨)
生成函数总是结合新坐标和旧坐标。问题的持续陈述决定了哪种形式最方便。对于所有三个函数，都存在一个定理，如F1在2.149和2.150我们在上一节中已经证明了这一点。让我们在下面更详细地检查这一点。
F2=F2(q,p¯,吨)
F2一个从F1由勒让德变换相对于q¯:
$$
F_{2}=F_{2}(\mathbf{q}, \overline{\mathbf{p}}, t)
$$
$F_{2}$ one obtains from $F_{1}$ by a Legendre transformation with respect to $\overline{\mathbf{q}}$ :
$$
F_{2}(\mathbf{q}, \overline{\mathbf{p}}, t)=F_{1}(\mathbf{q}, \overline{\mathbf{q}}, t)-\sum_{j=1}^{S} \frac{\partial F_{1}}{\partial \bar{q}{j}} \bar{q}{j}=F_{1}(\mathbf{q}, \overline{\mathbf{q}}, t)+\sum_{j=1}^{S} \bar{p}{j} \bar{q}{j}
$$
From the relation, already used to find (2.151),
$$
d F_{1}=\sum_{j=1}^{S}\left(p_{j} d q_{j}-\bar{p}{j} d \bar{q}{j}\right)+(\bar{H}-H) d t
$$
it follows for $F_{2}$ :
$$
d F_{2}=d F_{1}+\sum_{j=1}^{S}\left(\bar{p}{j} d \bar{q}{j}+\bar{q}{j} d \bar{p}{j}\right)=\sum_{j=1}^{S}\left(p_{j} d q_{j}+\bar{q}{j} d \bar{p}{j}\right)+(\bar{H}-H) d t
$$
That means:
$$
p_{j}=\frac{\partial F_{2}}{\partial q_{j}} ; \quad \bar{q}{j}=\frac{\partial F{2}}{\partial \bar{p}{j}} ; \quad \bar{H}=H+\frac{\partial F{2}}{\partial t}
$$
通过反转和解析一个显示，如上一节中明确展示的F1，即从2.161转换公式(q,p)⟶(q¯,p¯)是唯一可导出的。
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