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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|统计物理代写Statistical Physics of Matter代考|Canonical Ensemble and the Boltzmann Distribution

What is the probability that the system in the canonical ensemble will be at a certain microstate $\mathcal{M}$ ? To find this probability we suppose that the composite of the system and the heat bath or reservoir $(B)$ surrounding it is an isolated system with total energy $E_{T}$, as depicted in Figs. $2.4$ and 3.6. Then the number of all the accessible microstates in the total system is
$$W_{T}\left(E_{T}\right)=\sum_{\mathcal{M}} W\left(\mathcal{E}{\mathcal{M}}\right) W{B}\left(E_{T}-\mathcal{E}{\mathcal{M}}\right)$$ where $\sum{\mathcal{M}}$ signifies the summation over all accessible microstates of the system, each having the energy $\mathcal{E}{\mathcal{M}}, W\left(\mathcal{E}{\mathcal{M}}\right)$ is its number of microstates. $W_{B}\left(E_{T}-\mathcal{E}{\mathcal{M}}\right)$ is the number of the microstates of the heat bath, given that the system has the energy $\mathcal{E}{M}$. In the each one of the microstates counted in (3.26) is equally probable a priori by the postulate (3.1), so that the probability that the system will be in a specific state $\mathcal{M}\left(W\left(\mathcal{E}{\mathcal{M}}\right)=1\right)$ is $$P{\mathcal{M}}=\frac{W{B}\left(E_{T}-\mathcal{E}{\mathcal{M}}\right)}{\sum{\mathcal{M}} W_{B}\left(E_{T}-\mathcal{E}{\mathcal{M}}\right)}$$ To go further, we note that $$W{B}\left(E_{T}-\mathcal{E}{\mathcal{M}}\right)=\exp \left[\frac{1}{k{B}} S_{B}\left(E_{T}-\mathcal{E}{\mathcal{M}}\right)\right]$$ and the system’s energy $\mathcal{E}{\mathcal{M}}$ is much smaller than the total energy $E_{T}$ or the reservoir energy $E_{T}-\mathcal{E}{\mathcal{M}}$. Consequently the exponent above is expanded as \begin{aligned} \exp \left[\frac{1}{k{B}} S_{B}\left(E_{T}-\mathcal{E}{\mathcal{M}}\right)\right] & \cong \exp \left[\frac{1}{k{B}}\left(S_{B}\left(E_{T}\right)-\mathcal{E}{\mathcal{M}} \frac{\partial}{\partial E{T}} S_{B}\left(E_{T}\right)\right)\right] \ &=\exp \left[\frac{1}{k_{B}}\left(S_{B}\left(E_{T}\right)-\frac{\mathcal{E}_{\mathcal{M}}}{T}\right)\right] \end{aligned}

## 物理代写|统计物理代写Statistical Physics of Matter代考|The Energy Fluctuations

The energy distribution of macroscopic systems in canonical ensemble is a sharp Gaussian around the average energy. To show this, consider that values of the microstate energy $\mathcal{E}$ are continuously distributed with density of states $w(\mathcal{E})$ over a range $d \mathcal{E}$, so that the partition function (3.33) can be written as
$$Z=\int d \mathcal{E} w(\mathcal{E}) e^{-\beta \mathcal{E}}$$
which implies that probability distribution of the energy within the range $d \mathcal{E}$ is

\begin{aligned} P(\mathcal{E}) &=\frac{w(\mathcal{E}) e^{-\beta \mathcal{E}}}{Z} \ &=e^{-\beta{\mathcal{E}-T \mathcal{S}(\mathcal{E})}} / Z=e^{-\beta \mathcal{F}(\mathcal{E})} / Z \end{aligned}
where $\mathcal{F}(\mathcal{E})=\mathcal{E}-T \mathcal{S}(\mathcal{E})=\mathcal{E}-k_{B} T \ln w(\mathcal{E})$ is the free energy given as a function of an energy $\mathcal{E}$. Because $e^{-\beta T \mathcal{S}(\mathcal{E})}$ increases and $e^{-\beta \mathcal{E}}$ decreases with $\mathcal{E}$, we expect that $P(\mathcal{E})$ is peaked at $\mathcal{E}^{}$, where $\mathcal{F}(\mathcal{E})$ is minimum. Around the minimum, $\mathcal{F}(\mathcal{E})$ can be expanded: \begin{aligned} \mathcal{F}(\mathcal{E}) & \cong \mathcal{E}^{}-T \mathcal{S}\left(\mathcal{E}^{}\right)+\frac{1}{2} T\left(\frac{\partial^{2} \mathcal{S}\left(\mathcal{E}^{}\right)}{\partial \mathcal{E}^{* 2}}\right)\left(\mathcal{E}-\mathcal{E}^{}\right)^{2} \ &=\mathcal{F}\left(\mathcal{E}^{}\right)-\frac{1}{2 T C_{V}}\left(\mathcal{E}-\mathcal{E}^{}\right)^{2} \end{aligned} In the above, we used $\partial^{2} \mathcal{S}\left(\mathcal{E}^{}\right) / \partial \mathcal{E}^{+2}=\partial / \partial \mathcal{E}^{}(1 / T)=-1 /\left(T^{2} C_{V}\right)$, along with $\partial \mathcal{S}\left(\mathcal{E}^{}\right) / \partial \mathcal{E}^{}=1 / T$ and $\partial T /\left(\partial \mathcal{E}^{}\right)=1 /\left(\partial \mathcal{E}^{} / \partial T\right)=1 / C_{V}$ (2. 38). Finally, we obtain $$P(\mathcal{E}) \propto e^{-\beta \mathcal{F}(\mathcal{E})} \cong \exp \left[-\frac{1}{2 k_{B} T^{7} C_{V}}\left(\mathcal{E}-\mathcal{E}^{}\right)^{2}\right]$$

# 统计物理代考

## 物理代写|统计物理代写Statistical Physics of Matter代考|Canonical Ensemble and the Boltzmann Distribution

$$W_{T}\left(E_{T}\right)=\sum_{\mathcal{M}} W(\mathcal{E M}) W B\left(E_{T}-\mathcal{E M}\right)$$

$$P \mathcal{M}=\frac{W B\left(E_{T}-\mathcal{E M}\right)}{\sum \mathcal{M} W_{B}\left(E_{T}-\mathcal{E M}\right)}$$

$$W B\left(E_{T}-\mathcal{E M}\right)=\exp \left[\frac{1}{k B} S_{B}\left(E_{T}-\mathcal{E M}\right)\right]$$

$$\exp \left[\frac{1}{k B} S_{B}\left(E_{T}-\mathcal{E M}\right)\right] \cong \exp \left[\frac{1}{k B}\left(S_{B}\left(E_{T}\right)-\mathcal{E M} \frac{\partial}{\partial E T} S_{B}\left(E_{T}\right)\right)\right] \quad=\exp \left[\frac{1}{k_{B}}\left(S_{B}\left(E_{T}\right)-\frac{\mathcal{E}_{\mathcal{M}}}{T}\right)\right]$$

## 物理代写|统计物理代写Statistical Physics of Matter代考|The Energy Fluctuations

$$Z=\int d \mathcal{E} w(\mathcal{E}) e^{-\beta \mathcal{E}}$$

$$P(\mathcal{E})=\frac{w(\mathcal{E}) e^{-\beta \mathcal{E}}}{Z} \quad=e^{-\beta \mathcal{E}-T \mathcal{S}(\mathcal{E})} / Z=e^{-\beta \mathcal{F}(\mathcal{E})} / Z$$ 值。在最小值附近， $\mathcal{F}(\mathcal{E})$ 可以扩展:
$$\mathcal{F}(\mathcal{E}) \cong \mathcal{E}-T \mathcal{S}(\mathcal{E})+\frac{1}{2} T\left(\frac{\partial^{2} \mathcal{S}(\mathcal{E})}{\partial \mathcal{E}^{* 2}}\right)(\mathcal{E}-\mathcal{E})^{2} \quad=\mathcal{F}(\mathcal{E})-\frac{1}{2 T C_{V}}(\mathcal{E}-\mathcal{E})^{2}$$

$$P(\mathcal{E}) \propto e^{-\beta \mathcal{F}(\mathcal{E})} \cong \exp \left[-\frac{1}{2 k_{B} T^{7} C_{V}}(\mathcal{E}-\mathcal{E})^{2}\right]$$

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