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# 物理代写|热力学代写Thermodynamics代考|MECH337 Discontinuous Systems

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## 物理代写|热力学代写Thermodynamics代考|What is the Linear Non-equilibrium Thermodynamics

Here we focus our attention on discontinuous systems. ${ }^1$ Let $S=S\left(x_1, \ldots x_N\right)$ and $x_{i-e q}=0$ with no loss of generality. ${ }^2$ We have shown that $\frac{d S}{d t}=X_i J_i, J_i=-\frac{d x_i}{d t}$, $X_i=-\frac{\partial S}{\partial x_i}=\beta_{i k} x_k, S-S_{e q}=-\frac{1}{2} \beta_{i k} x_i x_k$ and $\beta_{i k}=\beta_{k i}$. We introduce the probability $w\left(x_1, \ldots x_N\right) d x_1 \ldots d x_N$ that the $1^{s t}$ quantity takes a value between $x_1$ and $x_1+d x_1$ and that the $2^{\text {nd }}$ quantity takes a value between $x_2$ and $x_2+d x_2 \ldots$ and that the $N^{t h}$ quantity takes a value between $x_N$ and $x_N+d x_N \cdot{ }^3$ With the help of statistical mechanics, it is possible to show that ( $k_B$ Boltzmann’s constant):

$$w \propto \exp \left[\frac{\left(S-S_{e q}\right)}{k_B}\right]$$
This relationship is referred to as ‘Einstein’s formula’ in the literature. We define also the average $\equiv \int a w d x_1, \ldots d x_N$ of the generic quantity $a$ and the correlation function $$of the generic quantities a and b. We recollect some relevant results below. Firstly, the symmetry X_i \leftrightarrow x_i (Sect. 2.5) ensures that:$$
=
$$Secondly, with the help of statistical mechanics it is possible to show that:$$
=k_B \delta_{i k}
$$where \delta_{i k}=1 if i=k and \delta_{i k}=0 if i \neq k (Kronecker’ delta). Thirdly, time invariance of the laws of physics implies that:$$
=
$$for arbitrary times t and \tau. Physically, this means that physics does not depend on the choice of the origin of time. ## 物理代写|热力学代写Thermodynamics代考|Onsager’s Symmetry The following, fundamental result of LNET is due to Onsager. We multiply both sides of =\tau \cdot \frac{d x_i}{d t} by x_l(t), take the average, invoke the results of Sect.4.1.1 and obtain:$$
\begin{gathered}
>== \
=-\cdot \tau \cdot J_i>=-\tau \cdot L_{i k}= \
=-\tau \cdot k_B \cdot L_{i k} \cdot \delta_{l k}=-\tau \cdot k_B \cdot L_{i l}
\end{gathered}
$$Now, there is nothing special about the indices i and l. We may swap them and obtain:$$
>=-\tau \cdot k_B \cdot L_{l i}
$$Term-by-term subtraction leads to:$$
L_{l i}-L_{i l}=\left(\tau k_B\right)^{-1}\left[>->\right]
$$Time invariance and microscopic reversibility imply that the R.H.S. vanishes, { }^8 hence:$$
L_{l i}=L_{i l} \quad ; \quad \text { Onsager’s symmetry }
$$If =-x_i(t) x_k(t+\tau) then term-by term sum of the relationships >=-\tau \cdot k_B \cdot L_{i l} \quad and >=-\tau \cdot k_B \cdot L_{l i} leads to L_{i l}=-L_{l i}. However, no ambiguity ever arises: since entropy is even in \mathbf{v}, if no magnetic field and no vorticity occur then S-S_{e q} is the sum of two quadratic forms, the former and the latter containing only quantities even and odd in \mathbf{v} respectively, and L_{i l}=L_{l i}. Moreover, if L_{i l} depends on magnetic field (vorticity) then nothing changes provided that the sign of the magnetic field (vorticity) is changed as we swap i and l. Finally, we stress the point that Onsager’s symmetry is not equivalent to linearity of relaxation-ruling laws and to lack of self-organisation; see, e.g. the counterexamples in [4,5] respectively. ## 热力学代写 ## 物理代写|热力学代写Thermodynamics代考|What is the Linear Non-equilibrium Thermodynamics 在这里，我们将注意力集中在不连续系统上。 { }^1 让 S=S\left(x_1, \ldots x_N\right) 和 x_{i-e q}=0 不失一般性。 { }^2 我们已经证明 \frac{d S}{d t}=X_i J_i, J_i=-\frac{d x_i}{d t}, X_i=-\frac{\partial S}{\partial x_i}=\beta_{i k} x_k, S-S_{e q}=-\frac{1}{2} \beta_{i k} x_i x_k 和 \beta_{i k}=\beta_{k i}. 我们引入概率 w\left(x_1, \ldots x_N\right) d x_1 \ldots d x_N 那个 1^{s t \text { th }} 数量取值之间 x_1 和 x_1+d x_1 并且 2^{\text {nd }} 数量取值之间 x_2 和 x_2+d x_2 \ldots 并且 N^{t h} 数量取值之间 x_N 和 x_N+d x_N \cdot{ }^3 在统计力学的帮助下，可以证明 ( k_B 玻尔兹曼常数) :$$
w \propto \exp \left[\frac{\left(S-S_{e q}\right)}{k_B}\right]
$$—个andthecorrelation functionofthegenericquantities—个and \mathrm{b} .Werecollectsomerelevantresultsbelow. Firstly, thesymmetry \mathrm{x} _ \mathrm{i} \backslash leftrightarrow \mathrm{x}{-} \mathrm{i} (Sect. 2.5)ensuresthat := Secondly, withthehelpofstatisticalmechanicsitispossibletoshowthat :=k_B \delta{i k} where \backslash delta_{ik }=1 i f 我 = kand \backslash delta_{ik }=0 i f 我 \backslash neq \mathrm{k} (Kronecker’delta). Thirdly, timeinvarianceofthelawsofphysicsimpliesthat :=forarbitrarytimes吨and |taus。在物理上，这意味着物理学不依赖于时间起源的选择。 ## 物理代写|热力学代写Thermodynamics代考|Onsager’s Symmetry LNET 的以下其本结果归功于 Onsager。我们将两边相乘 =\tau \cdot \frac{d x_i}{d t} 经过 x_l(t) ，取平均值，调用 Sect.4.1.1 的结果，得到:$$

==-\cdot \tau \cdot J_i>=-\tau \cdot L_{i k}=-\tau \cdot k_B \cdot L_{i k} \cdot \delta_{l k}=-\tau \cdot k_B \cdot L_{i l}
$$现在，䒺引没有什么特别之处 i 和 l. 我们可以交换它们并获得:$$
=-\tau \cdot k_B \cdot L_{l i}
$$逐项咸法导致:$$
L_{l i}-L_{i l}=\left(\tau k_B\right)^{-1}[>->]
$$时间不变性和微观可逆性意味着 RHS 消失， { }^8 因此:$$
L_{l i}=L_{i l} \quad ; \quad \text { Onsager’s symmetry }


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