物理代写|统计力学代写Statistical mechanics代考|PHYS3034

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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|The Zeroth Law

The zeroth law states:
There exists a function of state $T$, the temperature, such that two systems in thermal contact are in thermal equilibrium if and only if their temperatures are equal.
Note that this law does not fix the temperature scale! On the Celsius scale, one defines $0^{\circ} \mathrm{C}$ as the melting point of ice and $100^{\circ} \mathrm{C}$ as the boiling point of water, both under a pressure of $1 \mathrm{~atm}$. The temperature of a given system can then be determined by bringing it into thermal contact with a thermometer and measuring some property of the thermometer system using linear interpolation between $0^{\circ} \mathrm{C}$ and $100^{\circ} \mathrm{C}$ (and extrapolation beyond this range).
EXAMPLE 1.1: Mercury vs. Resistance Thermometer.
In a mercury thermometer for example, the temperature is determined by the length of a column of mercury as it expands due to heating. Thus:
$$\theta_{m}=\frac{\ell_{\theta}-\ell_{0}}{\ell_{100}-\ell_{0}} \times 100^{\circ} \mathrm{C}$$
is the empirical temperature as it is measured by a mercury thermometer. In a resistance thermometer, the temperature is determined by the electrical resistance of a metal wire:
$$\theta_{R}=\frac{R_{\theta}-R_{0}}{R_{100}-R_{0}} \times 100^{\circ} \mathrm{C} .$$
We now encounter the following problem: Suppose that the electrical resistance $R$ varies nonlinearly with temperature as measured on the mercury scảlé:
$$R_{\theta}=R_{0}\left(1+b \theta_{m}+c \theta_{m}^{2}\right)$$
Then
$$\theta_{R}=\frac{b \theta_{m}+c \theta_{m}^{2}}{b+100 c} \neq \theta_{m}$$

Fortunately, it turns out that the nonlinearity in equation (1.3) is rather small over a not too large interval of temperatures. (i.e. $c<<1$.) Nevertheless, for an accurate definition of the temperature scale one needs a standard thermometer with which all other thermometers can then be gauged. For this we can use an inert gas. Indeed, using one of the above types of thermometer, it was discovered that most rare gases satisfy the ideal gas law (1) mentioned in the introduction:
$$p V=n R_{0} T,$$
where the constant $R_{0}$ is independent of the gas and where $T=\theta+273$ if $\theta$ is measured in degrees Celsius.

物理代写|统计力学代写Statistical mechanics代考|The First Law

The first law of thermodynamics is an extension of the law of conservation of energy to include thermal processes. It can be formulated as follows:
There exists a function of state $U$, the internal energy, such that if an amount of energy $E$ is supplied to an otherwise isolated system, bringing it from an equilibrium state $\alpha$ to an equilibrium state $\beta$, then
$$E=U(\beta)-U(\alpha)$$
irrespective of the way in which this energy was supplied.

EXAMPLE 2.1: The sledge.
Consider a sledge of mass $M$ sliding down a hill of height $h$ (figure 2.1). Assume that the speed of the sledge at the top of the hill is negligible. By conservation of energy, the potential energy $E_{p o t}=M g h$ must have been transformed into kinetic energy: $E_{k i n}=\frac{1}{2} M v^{2}$ at the bottom of the hill. The speed at the bottom of the hill must be $v=\sqrt{2 g h}$. In fact the speed will be smaller due to friction. Where has the energy gone? The answer is, of course, that it was transformed into heat.

The first precise experiments about heat flow were done by James Joule in the $1840 \mathrm{~s}$. He did measurements on a thermally insulated container of gas to which he supplied energy in different ways: mechanical stirring, electrical heating, and compression. Figure $2.2$ shows the three processes schematically. Joule found that the final temperature of the gas depends only on the amount of energy supplied; not on the way it was supplied, whether it was heat or work. Note that it was important that the container was well-insulated so that no heat could escape. In that case all the energy supplied must have been absorbed by the gas. In each of the different ways of heating the gas, it is possible to determine the amount of energy supplied, and it turns out that this determines the final state of the gas completely.

物理代写|统计力学代写Statistical mechanics代考|Differentials

In this chapter we derive some c

onsequences of the first law. The first law is used most often in differential form. This is particularly useful for quasi-static processes, which can be subdivided into small stages during which the change in the various state functions is only small (infinitesimal). In infinitesimal form, the equation (2.10) reads:
$$\delta Q+\delta W=\mathrm{d} U$$
One can give a precise meaning to the differential quantities appearing in this equation by defining the concept of a differential form. We shall discuss this at the end of this chapter, but first let us do some formal calculations with the differential forms appearing in (3.1).

In example $2.2$ we already mentioned that the work differential $\delta W$ cannot be written as a small change in a function of state $\mathcal{W}$. We shall use the notation $\delta$ in such a case. On the other hand, the differential $\mathrm{d} U$ does mean: ‘a small change in $U^{\prime}$, and we write $\mathrm{d}$ instead of $\delta$. One says that the differential $\mathrm{d} U$ is exact. If we assume as basic coordinates of the state space $V$ and $T$ we can write $\mathrm{d} U$ in terms of the coordinate differentials $\mathrm{d} V$ and $\mathrm{d} T$ :
$$\mathrm{d} U=\left(\frac{\partial U}{\partial T}\right){V} \mathrm{~d} T+\left(\frac{\partial U}{\partial V}\right){T} \mathrm{~d} V=C_{V} \mathrm{~d} T+\left(\frac{\partial U}{\partial V}\right){T} \mathrm{~d} V$$ (The subscript on partial derivatives indicates the variable to be kept fixed.) We have already seen that the work differential can also be written in terms of $\mathrm{d} T$ and $\mathrm{d} V$ (in fact only the latter appears): $$\delta W=-p \mathrm{~d} V .$$ In general, one can write any arbitrary differential in terms of the coordinate differentials. Denoting an arbitrary differential by $\omega$, one has $$\omega=f{1}(T, V) \mathrm{d} T+f_{2}(T, V) \mathrm{d} V$$

Using equation (3.1) we have for the heat differential
$$\delta Q=C_{V} \mathrm{~d} T+\left(p+\left(\frac{\partial U}{\partial V}\right){T}\right) \mathrm{d} V$$ Alternatively, one can take $T$ and $p$ as independent variables and write $$\mathrm{d} U=\left(\frac{\partial U}{\partial T}\right){p} \mathrm{~d} T+\left(\frac{\partial U}{\partial p}\right)_{T} \mathrm{~d} p$$

物理代写|统计力学代写Statistical mechanics代考|The Zeroth Law

θ米=ℓθ−ℓ0ℓ100−ℓ0×100∘C

θR=Rθ−R0R100−R0×100∘C.

Rθ=R0(1+bθ米+Cθ米2)

θR=bθ米+Cθ米2b+100C≠θ米

p在=nR0吨,

物理代写|统计力学代写Statistical mechanics代考|Differentials

d问+d在=d在

d在=(∂在∂吨)在 d吨+(∂在∂在)吨 d在=C在 d吨+(∂在∂在)吨 d在（偏导数上的下标表示要保持固定的变量。）我们已经看到，功微分也可以写成d吨和d在（实际上只有后者出现）：

d在=−p d在.一般来说，可以根据坐标微分写出任意微分。表示任意微分ω, 一个有

ω=F1(吨,在)d吨+F2(吨,在)d在

d问=C在 d吨+(p+(∂在∂在)吨)d在或者，可以采取吨和p作为自变量并写

d在=(∂在∂吨)p d吨+(∂在∂p)吨 dp

有限元方法代写

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MATLAB代写

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