### 物理代写|统计力学代写Statistical mechanics代考|PHYS 3006

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• 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|Convexity

The convexity of the internal energy function is an important property directly related to the stability of thermodynamic equilibrium. In this chapter we derive some properties of convex functions which will enable us to develop the theory of thermodynamics further. We state the theorems in general but give proofs for the one-dimensional case only. Proofs in the general $k$-dimensional case can be found in appendix C. In any case, the proofs are not essential for understanding the subsequent theory though the definitions, concepts, and theorems are.
If $g: \mathbb{R}^{k} \rightarrow \mathbb{R} \cup{+\infty}$ is a function which can take the value $+\infty$ then one defines its essential domain by
$$D(g)=\left{\vec{x} \in \mathbb{R}^{k} \mid g(\vec{x})<+\infty\right} .$$
Theorem 7.1 A convex function $g: \mathbb{R}^{k} \rightarrow \mathbb{R} \cup{+\infty}$ is continuous at every interior point of its essential domain.

Proof. It is clear that the essential domain of $g$ is convex. We give the proof of the theorem in one dimension, so that $D(g)=I$ is an interval. Without loss of generality we may assume that $I$ is an open interval. We shall prove that for all $x_{0} \in I, g\left(x_{0}\right) \leqslant \liminf {x \rightarrow x{0}} g(x)$ and $g\left(x_{0}\right) \geqslant \limsup {x \rightarrow x{0}} g(x)$.
To prove that $g\left(x_{0}\right) \leqslant \liminf {x \rightarrow x{0}} g(x)$, suppose to the contrary that $g\left(x_{0}\right)>\liminf {x \rightarrow x{0}} g(x)$. If there were to exist $x_{1}<x_{0}<x_{2}$ such that $g\left(x_{1}\right)<g\left(x_{0}\right)$ and $g\left(x_{2}\right)<g\left(x_{0}\right)$ then
$$g\left(x_{0}\right) \leqslant \frac{x_{2}-x_{0}}{x_{2}-x_{1}} g\left(x_{1}\right)+\frac{x_{0}-x_{1}}{x_{2}-x_{1}} g\left(x_{2}\right)<g\left(x_{0}\right)$$
which is a contradiction. We conclude that $g(x) \geqslant g\left(x_{0}\right)$, either for all $x \geqslant x_{0}$ or for all $x \leqslant x_{0}$. Both cases are similar. We consider only the first. We may assumẻ that therrê êxists a sęuuencê $\left{\bar{x}{n}\right}$ such that $x{n} \sim_{n} x_{0}$ and $g\left(x_{n}\right) \rightarrow$

$\liminf {x \rightarrow x{0}}[g(x)]x_{0}$ we have (see figure $7.1)$
$$g\left(x_{0}\right) \leqslant \frac{\tilde{x}-x_{0}}{\tilde{x} \quad x_{n}} g\left(x_{n}\right)+\frac{x_{0}-x_{n}}{\tilde{x} \quad x_{n}} g(\tilde{x}) .$$
As $x_{n} \rightarrow x_{0}$ the right hand side tends to $\liminf {x \rightarrow x{0}} g(x)$ which contradicts the assumption.

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

After the preliminaries about convex functions in chapter 7 , we can now define the free energy density $f(v, T)$ as minus the Legendre transform of $u(s, v)$ with respect to the variable $s$ :
$$f(v, T)=-\sup {s}{T s-u(s, v)} .$$ It is easy to see that the function $u(s, v)$ is convex so that we can use theorem $7.2$ to invert the Legendre transform and write $$u(s, v)=\sup {T}{T s+f(v, T)} .$$
Note that we can also write
$$f(v, T)=\inf {u}{u-T s(u, v)}$$ that is, $-(1 / T) f(v, T)$ is the Legendre transform of $-s(u, v)$. Inverting this relation we obtain $$s(u, v)=\inf {T>0} \frac{1}{T}{u-f(v, T)} .$$
The free energy density is an important quantity in thermodynamics because in many experimental situations it is easier to control the temperature than the internal energy. Note that the supremum in (8.1) is attained at the value of $s$ satisfying (6.8) so that it is legitimate to denote the variable $T$ in the free energy density as the temperature. We can therefore write
$$f(v, T)=u(s(v, T), v)-T s(v, T),$$
where $s(v, T)$ is defined as the solution of (6.8). Alternatively,
$$f(v, T)=u(v, T)-T s(u(v, T), v)$$

where $u(v, T)$ is the solution of
$$\frac{1}{T}=\left(\frac{\partial s}{\partial u}\right){v}$$ Differentiating (8.5) w.r.t. to $v$ we obtain, using equations (6.8) and $(6.9)$, $$\left(\frac{\partial f}{\partial v}\right){T}(v, T)=\left(\frac{\partial u}{\partial v}\right){s}(s(v, T), v)=-p(v, T)$$ Similarly we have $$\left(\frac{\partial f}{\partial T}\right){v}=-s(v, T)$$
so that we can write
$$\mathrm{d} f=-p \mathrm{~d} v-s \mathrm{~d} T$$
Note that this implies in particular that for an isothermal process from an equilibrium state $\alpha$ to an equilibrium state $\beta$, the work done per particle $w$ equals
$$w=f(\beta)-f(\alpha)$$
Thus, the free energy plays a role similar to the potential energy in mechanics: given the work done in an isothermal process, (8.11) determines the new equilibrium state.

In some cases it is convenient to work in terms of the variables $s$ and $p$. We then take a Legendre transform with respect to $v$ and define the enthalpy $h(s, p)$ by
$$h(s, p)=\inf _{v}{p v+u(s, v)}$$
It is easy to prove that
$$\mathrm{d} h=\delta q+v \mathrm{~d} p$$

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

Consider again the $p-V$ diagram for a general fluid given as figure 1 in the introduction. In the region under the broken curve, liquid and vapour (gas at temperatures below $T_{c}$ is usually called vapour) coexist. As the volume is increased in this region the pressure remains constant and liquid vaporizes (figure 9.1).

It follows that the pressure at which liquid and vapour coexist is a function of the temperature alone. If the temperature of the liquid is raised while keeping the pressure above the liquid constant, it will no longer be in equilibrium and will evaporate. If we want to keep it in equilibrium, we must increase the pressure. The equilibrium pressure is therefore an increasing function of the temperature. The graph of this function is the line of coexistence of liquid and vapour shown in figure $9.2$. Above this line, only liquid exists in equilibrium while below the curve only vapour exists in equilibrium. This argument also shows that by lowering the pressure above a liquid its boiling point decreases. This is an important method for obtaining low temperatures (refrigeration). For example, by pumping away the vapour above liquid helium-3 $\left({ }^{3} \mathrm{He}\right.$ ) (a rare isotope of helium), one can reduce the temperature by an order of magnitude, from approximately $3 \mathrm{~K}$ to $0.3 \mathrm{~K}$.

Another look at the $p-V$ diagram shows that the coexistence curve must end at a maximum temperature $T_{c}$ called the critical temperature. This point of the coexistence curve $p_{e q}(T)$ corresponds to one single point $\left(p_{c}, V_{c}\right)$ in the $p-V$ diagram, the critical point. At this point remarkable things happen: the so called critical phenomena. For example, the isothermal compressibility $\kappa_{T}$ (see equation (4.14)) diverges at the critical point. Also the specific heat $c_{V}$ diverges. This means that the free energy is not twice differentiable at the critical point. One therefore speaks of a second-order phase transition. The degree of divergence can be expressed in terms of critical exponents. It turns out that $\kappa_{T}$ and $c_{V}$ behave near the critical point asymptotically as
$$\kappa_{T} \sim \mathcal{K}\left|’ I^{\prime}-H_{c}^{\prime}\right|^{-\gamma}$$

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

D(g)=\left{\vec{x} \in \mathbb{R}^{k} \mid g(\vec{x})<+\infty\right} 。D(g)=\left{\vec{x} \in \mathbb{R}^{k} \mid g(\vec{x})<+\infty\right} 。

G(X0)⩽X2−X0X2−X1G(X1)+X0−X1X2−X1G(X2)<G(X0)

G(X0)⩽X~−X0X~XnG(Xn)+X0−XnX~XnG(X~).

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

F(在,吨)=−支持s吨s−在(s,在).很容易看出函数在(s,在)是凸的，所以我们可以使用定理7.2反转勒让德变换并写入

F(在,吨)=信息在在−吨s(在,在)那是，−(1/吨)F(在,吨)是勒让德变换−s(在,在). 反转这个关系，我们得到

s(在,在)=信息吨>01吨在−F(在,吨).

F(在,吨)=在(s(在,吨),在)−吨s(在,吨),

F(在,吨)=在(在,吨)−吨s(在(在,吨),在)

1吨=(∂s∂在)在将 (8.5) 微分到在我们得到，使用方程（6.8）和(6.9),

(∂F∂在)吨(在,吨)=(∂在∂在)s(s(在,吨),在)=−p(在,吨)同样我们有

(∂F∂吨)在=−s(在,吨)

dF=−p d在−s d吨

H(s,p)=信息在p在+在(s,在)

dH=dq+在 dp

ķ吨∼ķ|′我′−HC′|−C

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## MATLAB代写

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