### 物理代写|统计力学代写Statistical mechanics代考|PHYC30017

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• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 物理代写|统计力学代写Statistical mechanics代考|Hamilton’s Equations

It is often convenient to rewrite Newton’s equations (3.2.6) in Lagrangian form or in Hamiltonian form. We will only use the latter one. 4 To do so, we will introduce the phase space $\mathbb{R}^{6 N}$, and write a vector $\mathbf{x} \in \mathbb{R}^{6 \mathbf{N}}$ as a pair $\mathbf{x}=(\mathbf{q}, \mathbf{p})$, with $\mathbf{q}=$ $\left(\vec{q}{1}, \vec{q}{2}, \ldots, \vec{q}{N}\right) \in \mathbb{R}^{3 N}, \mathbf{p}=\left(\vec{p}{1}, \vec{p}{2}, \ldots, \vec{p}{N}\right) \in \mathbb{R}^{3 N}$
The Hamiltonian is a function $H: \mathbb{R}^{6 N} \rightarrow \mathbb{R}:$
$$H(\mathbf{q}, \mathbf{p})=K(\mathbf{p})+V(\mathbf{q})$$
with a kinetic energy
$$K(\mathbf{p})=\sum_{i=1}^{N} \frac{\left|\vec{p}{i}\right|^{2}}{2 m{i}}$$
and a potential energy $V(\mathbf{q})$ given by (3.2.5).
Then Hamilton’s equations are given by the following pair:
$$\frac{d \vec{q}{i}(t)}{d t}=\nabla{\bar{p}{i}} H(\mathbf{q}(t), \mathbf{p}(t))$$ and $$\frac{d \vec{p}{i}(t)}{d t}=-\nabla_{\bar{q}{i}} H(\mathbf{q}(t), \mathbf{p}(t)),$$ for $i=1, \ldots, N$ With $H$ defined by (3.3.1), (3.3.2), these equations are: $$\frac{d \vec{q}{i}(t)}{d t}=\frac{\vec{p}{i}(t)}{m{i}}$$
and
\begin{aligned} &\frac{d \vec{p}{i}(t)}{d t}=-\nabla{\overline{q_{i}}} V(\mathbf{q}(t)) . \ &x(t)=\frac{x(0)}{1-t x(0)}, \end{aligned}

## 物理代写|统计力学代写Statistical mechanics代考|The Hamiltonian Flow

Since (3.2.6) and (3.3.5), (3.3.6) are equivalent, if we assume that the potentials are such that a unique solution $\mathbf{q}(t)$ of (3.2.6) exist for all times, we also have, for the pair of (3.3.5), (3.3.6), for any $t_{0} \in \mathbb{R}$ and any initial conditions ( $\mathbf{q}{0}, \mathbf{p}{0}$ ), a unique solution $(\mathbf{q}(t), \mathbf{p}(t))$ satisfying (3.3.5), (3.3.6) for all times and such that $\left(\mathbf{q}\left(t_{0}\right), \mathbf{p}\left(t_{0}\right)\right)=\left(\mathbf{q}{0}, \mathbf{p}{0}\right)$.

It will be convenient to associate to such solutions a family of maps $T^{t}: \mathbb{R}^{6 N} \rightarrow$ $\mathbb{R}^{6 N}$, for $t \in \mathbb{R}$, defined by:
$$T^{t}(\mathbf{q}, \mathbf{p})=(\mathbf{q}(t), \mathbf{p}(t))$$

where $(\mathbf{q}(t), \mathbf{p}(t))$ is the unique solution of $(3.3 .5),(3.3 .6)$ satisfying $(\mathbf{q}(0), \mathbf{p}(0))=$ $(\mathbf{q}, \mathbf{p})$

So, the map $T^{t}$ associates to every pair $(\mathbf{q}, \mathbf{p}) \in \mathbb{R}^{6 N}$ the value at time $t$ of the unique solution $(\mathbf{q}(t), \mathbf{p}(t))$ of $(3.3 .5),(3.3 .6)$ that “passes” through the point (q, p) at time 0 . Since the solutions are assumed to exist for all $t \in \mathbb{R}, T^{t}$ is invertible: $T^{t} T^{-t}=$ Id, where Id is the identity operator.
The family of maps $\left(T^{t}\right)_{t \in \mathbb{R}}$ is called the Hamiltonian flow.

## 物理代写|统计力学代写Statistical mechanics代考|Conservation of Energy

The energy of a mechanical system of the type considered here is a function $E$ : $\mathbb{R}^{6 N} \rightarrow \mathbb{R}$ defined by
$$E(\mathbf{q}, \mathbf{p})=K(\mathbf{p})+V(\mathbf{q})$$
with the kinetic energy $K$ (p) defined in (3.3.2) and the potential energy $\mathrm{V}(\mathbf{q})$ defined in (3.2.5). This is of course identical to the Hamiltonian function and the energy, like the potential, is defined up to an additive constant. We have:

Theorem $3.1$ (Conservation of energy) Let $(\mathbf{q}(t), \mathbf{p}(t)$ ) be a solution of (3.3.5), (3.3.6). Then, $\forall t \in \mathbb{R}$ :
$$\frac{d E(t)}{d t}=0$$
where $E(t)=E(\mathbf{q}(t), \mathbf{p}(t))$
Proof It is enough to compute the time derivative of $E(t)$, using (3.3.7), (3.3.1), $(3.3 .2)$ :
$$\frac{d E(\mathbf{q}(t), \mathbf{p}(t))}{d t}=\sum_{i=1}^{N} \frac{\vec{p}{i} \cdot \frac{d \vec{p}{i}(t)}{d t}}{m_{i}}+\sum_{i=1}^{N} \nabla_{\bar{q}{i}} V(\mathbf{q}(t)) \cdot \frac{d \vec{q}{i}(t)}{d t}$$
By (3.3.5), (3.3.6), this equals 0 .
Remark 3.2 One may rewrite the energy in a more familiar form, using $\vec{p}{i}(t)=$ $\frac{1}{m{i}} \frac{d \vec{a}{u}(t)}{d t}:$ $$E(t)=\sum{i=1}^{N} \frac{m_{i}\left|\vec{v}{i}(t)\right|^{2}}{2}+V(\mathbf{q}(t))$$ with $\vec{v}{i}(t)=\frac{d \vec{q}_{.}(t)}{d t}$.

## 物理代写|统计力学代写Statistical mechanics代考|Hamilton’s Equations

H(q,p)=ķ(p)+在(q)

ķ(p)=∑一世=1ñ|p→一世|22米一世

dq→一世(吨)d吨=∇p¯一世H(q(吨),p(吨))和

dp→一世(吨)d吨=−∇q¯一世H(q(吨),p(吨)),为了一世=1,…,ñ和H由 (3.3.1), (3.3.2) 定义，这些方程是：

dq→一世(吨)d吨=p→一世(吨)米一世

dp→一世(吨)d吨=−∇q一世¯在(q(吨)). X(吨)=X(0)1−吨X(0),

## 物理代写|统计力学代写Statistical mechanics代考|Conservation of Energy

d和(吨)d吨=0

d和(q(吨),p(吨))d吨=∑一世=1ñp→一世⋅dp→一世(吨)d吨米一世+∑一世=1ñ∇q¯一世在(q(吨))⋅dq→一世(吨)d吨

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