这个暑假一直在重温热力学（因为一直都没怎么学明白过），突然想到我们一直在说平衡态时能量最低，但是为什么？而且这个能量是指什么能？根据我的知识结构，能量最低的原则是作为分析力学的最小作用量原理的推论给出的。不知道在大家的知识结构里能量最低原则是怎么确立的。但是单论我的这一角度，仔细想想存在很大问题。凭什么说‘’热力学里大家通过其他路径所定义出的内能等能量就能等同于分析力学里的拉格朗日量的路径积分‘’（1）？平衡态的熵最大是熵增加原理的推论，虽然后者对于前者来说是一个过强的条件。那能量最低，从热力学的角度，是如何阐述的呢？在读Luca Peliti, Simone Pigolotti的’‘Stochastic thermodynamics: an introduction’‘时注意到其中第14页（我弱爆了，本来想暑假仔细读这本书的，但却把主要时间放在了读另一本热力学的书上）有写到：’' W≥ΔE−TΔS Therefor, the equilibrium state can be identified as the one in which the expression E-TS attains its minimum.'' 其中W是指对系统所做的功，而E则是指内能，就是我们常用的U. 那么按照这个定义，E-TS最小时，任何d(E-TS)都大于等于零，因此一个等温系统，必然有W大于等于0，也就是说系统不可能再对外做功了。也就是说，一个等温系统的平衡态时，满足自由能最低，而这个平衡态的特征是不可能再对外做功。那么‘’这个定义和熵增加原理下的平衡态是否consistent‘’（2）呢？以及，我们使用Gibbs的系综语言来描述的平衡态，根据Ingo Muller的’‘A history of thermodynamics’'的第120页，是指玻尔兹曼熵最大时的态。由于玻尔兹曼熵和克劳修斯熵的等价性，那么只要证明了表述（2）的正确，就可以确认系综语言体系里可以安全使用等温系统（此时则是正则系综）平衡态时自由能最低的原则。那么表述（2）如何证明呢？

最大熵与平衡态最大熵导出热力学平衡态是在总能量守恒， \sum_i p_i E_i = const<script defer crossorigin="anonymous" integrity="sha384-X/XCfMm41VSsqRNQgDerQczD69XqmjOOOwYQvr/uuC+j4OPoNhVgjdGFwhvN02Ja" onload="katex.render(this.parentNode.innerText, this.parentNode, {"fleqn":false,"leqno":true,"output":"htmlAndMathml","throwOnError":false,"errorColor":"#cc0000","minRuleThickness":0.05,"maxSize":10,"maxExpand":1000,"macros":{},"colorIsTextColor":false,"displayMode":false})" src="https://cdn.jsdelivr.net/npm/katex@0.16.0/dist/katex.min.js"> ，的情况下最大化熵 H=-\sum_i p_i\ln p_i<script defer crossorigin="anonymous" integrity="sha384-X/XCfMm41VSsqRNQgDerQczD69XqmjOOOwYQvr/uuC+j4OPoNhVgjdGFwhvN02Ja" onload="katex.render(this.parentNode.innerText, this.parentNode, {"fleqn":false,"leqno":true,"output":"htmlAndMathml","throwOnError":false,"errorColor":"#cc0000","minRuleThickness":0.05,"maxSize":10,"maxExpand":1000,"macros":{},"colorIsTextColor":false,"displayMode":false})" src="https://cdn.jsdelivr.net/npm/katex@0.16.0/dist/katex.min.js"> 。这个用一个拉格朗日乘子的方法很容易证明最后得到的是玻尔兹曼分布，而再归一化一下就回到了最小的自由能。 \max (-\sum_ip_i\ln p_i) <script defer crossorigin="anonymous" integrity="sha384-X/XCfMm41VSsqRNQgDerQczD69XqmjOOOwYQvr/uuC+j4OPoNhVgjdGFwhvN02Ja" onload="katex.render(this.parentNode.innerText, this.parentNode, {"fleqn":false,"leqno":true,"output":"htmlAndMathml","throwOnError":false,"errorColor":"#cc0000","minRuleThickness":0.05,"maxSize":10,"maxExpand":1000,"macros":{},"colorIsTextColor":false,"displayMode":true})" src="https://cdn.jsdelivr.net/npm/katex@0.16.0/dist/katex.min.js"> \Rightarrow\partial_{p_i} \left(-\sum_i p_i \ln p _i +\lambda_1(1-\sum_i p_i)+\lambda_2(\overline{E}-\sum_i p_i E_i)\right)=0 <script defer crossorigin="anonymous" integrity="sha384-X/XCfMm41VSsqRNQgDerQczD69XqmjOOOwYQvr/uuC+j4OPoNhVgjdGFwhvN02Ja" onload="katex.render(this.parentNode.innerText, this.parentNode, {"fleqn":false,"leqno":true,"output":"htmlAndMathml","throwOnError":false,"errorColor":"#cc0000","minRuleThickness":0.05,"maxSize":10,"maxExpand":1000,"macros":{},"colorIsTextColor":false,"displayMode":true})" src="https://cdn.jsdelivr.net/npm/katex@0.16.0/dist/katex.min.js"> \Rightarrow-\ln p_i-(\underbrace{1+\lambda_1}_{\lambda_1'}+\lambda_2E_i)=0 <script defer crossorigin="anonymous" integrity="sha384-X/XCfMm41VSsqRNQgDerQczD69XqmjOOOwYQvr/uuC+j4OPoNhVgjdGFwhvN02Ja" onload="katex.render(this.parentNode.innerText, this.parentNode, {"fleqn":false,"leqno":true,"output":"htmlAndMathml","throwOnError":false,"errorColor":"#cc0000","minRuleThickness":0.05,"maxSize":10,"maxExpand":1000,"macros":{},"colorIsTextColor":false,"displayMode":true})" src="https://cdn.jsdelivr.net/npm/katex@0.16.0/dist/katex.min.js"> \Rightarrow p_i=\exp(\lambda_2 E_i+\lambda_1') <script defer crossorigin="anonymous" integrity="sha384-X/XCfMm41VSsqRNQgDerQczD69XqmjOOOwYQvr/uuC+j4OPoNhVgjdGFwhvN02Ja" onload="katex.render(this.parentNode.innerText, this.parentNode, {"fleqn":false,"leqno":true,"output":"htmlAndMathml","throwOnError":false,"errorColor":"#cc0000","minRuleThickness":0.05,"maxSize":10,"maxExpand":1000,"macros":{},"colorIsTextColor":false,"displayMode":true})" src="https://cdn.jsdelivr.net/npm/katex@0.16.0/dist/katex.min.js"> 利用约束条件 1-\sum_i p_i=0<script defer crossorigin="anonymous" integrity="sha384-X/XCfMm41VSsqRNQgDerQczD69XqmjOOOwYQvr/uuC+j4OPoNhVgjdGFwhvN02Ja" onload="katex.render(this.parentNode.innerText, this.parentNode, {"fleqn":false,"leqno":true,"output":"htmlAndMathml","throwOnError":false,"errorColor":"#cc0000","minRuleThickness":0.05,"maxSize":10,"maxExpand":1000,"macros":{},"colorIsTextColor":false,"displayMode":false})" src="https://cdn.jsdelivr.net/npm/katex@0.16.0/dist/katex.min.js"> 得到 e^{\lambda_1'} = \frac{1}{\sum_{i}e^{\lambda_2E_i}}\equiv\frac{1}{Z} <script defer crossorigin="anonymous" integrity="sha384-X/XCfMm41VSsqRNQgDerQczD69XqmjOOOwYQvr/uuC+j4OPoNhVgjdGFwhvN02Ja" onload="katex.render(this.parentNode.innerText, this.parentNode, {"fleqn":false,"leqno":true,"output":"htmlAndMathml","throwOnError":false,"errorColor":"#cc0000","minRuleThickness":0.05,"maxSize":10,"maxExpand":1000,"macros":{},"colorIsTextColor":false,"displayMode":true})" src="https://cdn.jsdelivr.net/npm/katex@0.16.0/dist/katex.min.js"> 利用另一个约束条件 \overline{E}-\sum_i p_i E_i<script defer crossorigin="anonymous" integrity="sha384-X/XCfMm41VSsqRNQgDerQczD69XqmjOOOwYQvr/uuC+j4OPoNhVgjdGFwhvN02Ja" onload="katex.render(this.parentNode.innerText, this.parentNode, {"fleqn":false,"leqno":true,"output":"htmlAndMathml","throwOnError":false,"errorColor":"#cc0000","minRuleThickness":0.05,"maxSize":10,"maxExpand":1000,"macros":{},"colorIsTextColor":false,"displayMode":false})" src="https://cdn.jsdelivr.net/npm/katex@0.16.0/dist/katex.min.js"> 得到 \overline{E} = \frac{\sum_iE_ie^{\lambda_2E_i}}{\sum_i e^{\lambda_2 E_i}} <script defer crossorigin="anonymous" integrity="sha384-X/XCfMm41VSsqRNQgDerQczD69XqmjOOOwYQvr/uuC+j4OPoNhVgjdGFwhvN02Ja" onload="katex.render(this.parentNode.innerText, this.parentNode, {"fleqn":false,"leqno":true,"output":"htmlAndMathml","throwOnError":false,"errorColor":"#cc0000","minRuleThickness":0.05,"maxSize":10,"maxExpand":1000,"macros":{},"colorIsTextColor":false,"displayMode":true})" src="https://cdn.jsdelivr.net/npm/katex@0.16.0/dist/katex.min.js"> 那么如何决定 \lambda_2<script defer crossorigin="anonymous" integrity="sha384-X/XCfMm41VSsqRNQgDerQczD69XqmjOOOwYQvr/uuC+j4OPoNhVgjdGFwhvN02Ja" onload="katex.render(this.parentNode.innerText, this.parentNode, {"fleqn":false,"leqno":true,"output":"htmlAndMathml","throwOnError":false,"errorColor":"#cc0000","minRuleThickness":0.05,"maxSize":10,"maxExpand":1000,"macros":{},"colorIsTextColor":false,"displayMode":false})" src="https://cdn.jsdelivr.net/npm/katex@0.16.0/dist/katex.min.js"> ? 这个由温度的定义得到。注意到系统的熵写作 H(p) = -\sum_ip_i\ln p_i = \ln Z + \lambda_2 \overline{E} <script defer crossorigin="anonymous" integrity="sha384-X/XCfMm41VSsqRNQgDerQczD69XqmjOOOwYQvr/uuC+j4OPoNhVgjdGFwhvN02Ja" onload="katex.render(this.parentNode.innerText, this.parentNode, {"fleqn":false,"leqno":true,"output":"htmlAndMathml","throwOnError":false,"errorColor":"#cc0000","minRuleThickness":0.05,"maxSize":10,"maxExpand":1000,"macros":{},"colorIsTextColor":false,"displayMode":true})" src="https://cdn.jsdelivr.net/npm/katex@0.16.0/dist/katex.min.js"> 使用温度的定义 \frac{1}{T}\equiv \frac{\partial H}{\partial \overline E} = \lambda_2 <script defer crossorigin="anonymous" integrity="sha384-X/XCfMm41VSsqRNQgDerQczD69XqmjOOOwYQvr/uuC+j4OPoNhVgjdGFwhvN02Ja" onload="katex.render(this.parentNode.innerText, this.parentNode, {"fleqn":false,"leqno":true,"output":"htmlAndMathml","throwOnError":false,"errorColor":"#cc0000","minRuleThickness":0.05,"maxSize":10,"maxExpand":1000,"macros":{},"colorIsTextColor":false,"displayMode":true})" src="https://cdn.jsdelivr.net/npm/katex@0.16.0/dist/katex.min.js"> 这整个过程和“热力学里大家通过其他路径所定义出的内能等能量就能等同于分析力学里的拉格朗日量的路径积分”没有关系。个人很认同这个说法：笼统来看作用量原理就是纯纯的一句废话。自由能与可用功如果从对外做功的角度思考，我们先写出广义自由能 \mathcal{F}(p) = \sum_i p_i E_i -TH(p)<script defer crossorigin="anonymous" integrity="sha384-X/XCfMm41VSsqRNQgDerQczD69XqmjOOOwYQvr/uuC+j4OPoNhVgjdGFwhvN02Ja" onload="katex.render(this.parentNode.innerText, this.parentNode, {"fleqn":false,"leqno":true,"output":"htmlAndMathml","throwOnError":false,"errorColor":"#cc0000","minRuleThickness":0.05,"maxSize":10,"maxExpand":1000,"macros":{},"colorIsTextColor":false,"displayMode":false})" src="https://cdn.jsdelivr.net/npm/katex@0.16.0/dist/katex.min.js"> , 热力学平衡态对应到广义自由能的最小值，或者说是 -\mathcal{F}<script defer crossorigin="anonymous" integrity="sha384-X/XCfMm41VSsqRNQgDerQczD69XqmjOOOwYQvr/uuC+j4OPoNhVgjdGFwhvN02Ja" onload="katex.render(this.parentNode.innerText, this.parentNode, {"fleqn":false,"leqno":true,"output":"htmlAndMathml","throwOnError":false,"errorColor":"#cc0000","minRuleThickness":0.05,"maxSize":10,"maxExpand":1000,"macros":{},"colorIsTextColor":false,"displayMode":false})" src="https://cdn.jsdelivr.net/npm/katex@0.16.0/dist/katex.min.js"> 的最大值，这里最大化其实可以回去对应到在能量守恒约束下最大化entropy。而不难证明，系统对外做功不能超过 W \leq \Delta \mathcal{F}=\mathcal{F}(p)-\mathcal{F}(p^{eq})<script defer crossorigin="anonymous" integrity="sha384-X/XCfMm41VSsqRNQgDerQczD69XqmjOOOwYQvr/uuC+j4OPoNhVgjdGFwhvN02Ja" onload="katex.render(this.parentNode.innerText, this.parentNode, {"fleqn":false,"leqno":true,"output":"htmlAndMathml","throwOnError":false,"errorColor":"#cc0000","minRuleThickness":0.05,"maxSize":10,"maxExpand":1000,"macros":{},"colorIsTextColor":false,"displayMode":false})" src="https://cdn.jsdelivr.net/npm/katex@0.16.0/dist/katex.min.js"> , 这个最直观的可以参考 Parrondo et al. Thermodyanmics of information, Nature physics, 2015. 在其中你会发现这个自由能差值正是非平初识态所蕴含的信息（KL distance）。这里或许可以参考上一篇文章里引用的Esposito在EPL上的工作。 \begin{split} \Delta \mathcal{F} &= \sum_i( p_i E_i + Tp_i\ln p_i - (p_i^{eq}E_i+Tp_i\ln p_i^{eq}))\\\ &= \sum_i\left(Tp_i\ln p _i - p_iT\ln e^{-\beta E_i} +T\ln Z\right)\\\ & = T\sum_i p_i\ln p_i-p_i\ln\frac{e^{-\beta E_i}}{Z}\\\ &=T\underbrace{\sum_ip_i\ln\frac{p_i}{p_i^{eq}}}\_\{\geq 0\} \end{split} <script defer crossorigin="anonymous" integrity="sha384-X/XCfMm41VSsqRNQgDerQczD69XqmjOOOwYQvr/uuC+j4OPoNhVgjdGFwhvN02Ja" onload="katex.render(this.parentNode.innerText, this.parentNode, {"fleqn":false,"leqno":true,"output":"htmlAndMathml","throwOnError":false,"errorColor":"#cc0000","minRuleThickness":0.05,"maxSize":10,"maxExpand":1000,"macros":{},"colorIsTextColor":false,"displayMode":true})" src="https://cdn.jsdelivr.net/npm/katex@0.16.0/dist/katex.min.js"> 大家还会发现，这个 \Delta \mathcal{F}<script defer crossorigin="anonymous" integrity="sha384-X/XCfMm41VSsqRNQgDerQczD69XqmjOOOwYQvr/uuC+j4OPoNhVgjdGFwhvN02Ja" onload="katex.render(this.parentNode.innerText, this.parentNode, {"fleqn":false,"leqno":true,"output":"htmlAndMathml","throwOnError":false,"errorColor":"#cc0000","minRuleThickness":0.05,"maxSize":10,"maxExpand":1000,"macros":{},"colorIsTextColor":false,"displayMode":false})" src="https://cdn.jsdelivr.net/npm/katex@0.16.0/dist/katex.min.js"> 正可以作为一个热力学consistent系统的Lyapunov函数。

S-L 很有道理，作用量原理确实…emmm 但自由能一块儿有点没明白，计算 \Delta F<script defer crossorigin="anonymous" integrity="sha384-X/XCfMm41VSsqRNQgDerQczD69XqmjOOOwYQvr/uuC+j4OPoNhVgjdGFwhvN02Ja" onload="katex.render(this.parentNode.innerText, this.parentNode, {"fleqn":false,"leqno":true,"output":"htmlAndMathml","throwOnError":false,"errorColor":"#cc0000","minRuleThickness":0.05,"maxSize":10,"maxExpand":1000,"macros":{},"colorIsTextColor":false,"displayMode":false})" src="https://cdn.jsdelivr.net/npm/katex@0.16.0/dist/katex.min.js"> 时第一行第四项为什么不是 pi^{eq}<script defer crossorigin="anonymous" integrity="sha384-X/XCfMm41VSsqRNQgDerQczD69XqmjOOOwYQvr/uuC+j4OPoNhVgjdGFwhvN02Ja" onload="katex.render(this.parentNode.innerText, this.parentNode, {"fleqn":false,"leqno":true,"output":"htmlAndMathml","throwOnError":false,"errorColor":"#cc0000","minRuleThickness":0.05,"maxSize":10,"maxExpand":1000,"macros":{},"colorIsTextColor":false,"displayMode":false})" src="https://cdn.jsdelivr.net/npm/katex@0.16.0/dist/katex.min.js"> 而是 p_i<script defer crossorigin="anonymous" integrity="sha384-X/XCfMm41VSsqRNQgDerQczD69XqmjOOOwYQvr/uuC+j4OPoNhVgjdGFwhvN02Ja" onload="katex.render(this.parentNode.innerText, this.parentNode, {"fleqn":false,"leqno":true,"output":"htmlAndMathml","throwOnError":false,"errorColor":"#cc0000","minRuleThickness":0.05,"maxSize":10,"maxExpand":1000,"macros":{},"colorIsTextColor":false,"displayMode":false})" src="https://cdn.jsdelivr.net/npm/katex@0.16.0/dist/katex.min.js"> 呢？（不知道为啥我输入公式总是渲染不明白会崩掉…）

Voyager 这个可能是markdown和mathjax的冲突，目前的暂时解决方法是关掉富文本模式图中那个钢笔小标签（确实有点麻烦看看后续怎么优化一下😅 e.g. \Delta F<script defer crossorigin="anonymous" integrity="sha384-X/XCfMm41VSsqRNQgDerQczD69XqmjOOOwYQvr/uuC+j4OPoNhVgjdGFwhvN02Ja" onload="katex.render(this.parentNode.innerText, this.parentNode, {"fleqn":false,"leqno":true,"output":"htmlAndMathml","throwOnError":false,"errorColor":"#cc0000","minRuleThickness":0.05,"maxSize":10,"maxExpand":1000,"macros":{},"colorIsTextColor":false,"displayMode":false})" src="https://cdn.jsdelivr.net/npm/katex@0.16.0/dist/katex.min.js"> and p_i^{eq}<script defer crossorigin="anonymous" integrity="sha384-X/XCfMm41VSsqRNQgDerQczD69XqmjOOOwYQvr/uuC+j4OPoNhVgjdGFwhvN02Ja" onload="katex.render(this.parentNode.innerText, this.parentNode, {"fleqn":false,"leqno":true,"output":"htmlAndMathml","throwOnError":false,"errorColor":"#cc0000","minRuleThickness":0.05,"maxSize":10,"maxExpand":1000,"macros":{},"colorIsTextColor":false,"displayMode":false})" src="https://cdn.jsdelivr.net/npm/katex@0.16.0/dist/katex.min.js">

Voyager 有一个typo，应该是下面这样 \begin{split} \Delta \mathcal{F} &= \sum_i( p_i E_i + Tp_i\ln p_i - (p_i^{eq}E_i+Tp_i^{\color{red}eq}\ln p_i^{eq}))\\\ &= \sum_i\left(Tp_i\ln p _i - p_iT\ln e^{-\beta E_i} +T\ln Z\right)\\\ & = T\sum_i p_i\ln p_i-p_i\ln\frac{e^{-\beta E_i}}{Z}\\\ &=T\underbrace{\sum_ip_i\ln\frac{p_i}{p_i^{eq}}}_{\geq 0} \end{split} <script defer crossorigin="anonymous" integrity="sha384-X/XCfMm41VSsqRNQgDerQczD69XqmjOOOwYQvr/uuC+j4OPoNhVgjdGFwhvN02Ja" onload="katex.render(this.parentNode.innerText, this.parentNode, {"fleqn":false,"leqno":true,"output":"htmlAndMathml","throwOnError":false,"errorColor":"#cc0000","minRuleThickness":0.05,"maxSize":10,"maxExpand":1000,"macros":{},"colorIsTextColor":false,"displayMode":true})" src="https://cdn.jsdelivr.net/npm/katex@0.16.0/dist/katex.min.js">

学术吹水试水：关于平衡态的一点讨论

Voyager

这个暑假一直在重温热力学（因为一直都没怎么学明白过），突然想到我们一直在说平衡态时能量最低，但是为什么？而且这个能量是指什么能？

根据我的知识结构，能量最低的原则是作为分析力学的最小作用量原理的推论给出的。不知道在大家的知识结构里能量最低原则是怎么确立的。但是单论我的这一角度，仔细想想存在很大问题。凭什么说‘’热力学里大家通过其他路径所定义出的内能等能量就能等同于分析力学里的拉格朗日量的路径积分‘’（1）？

平衡态的熵最大是熵增加原理的推论，虽然后者对于前者来说是一个过强的条件。那能量最低，从热力学的角度，是如何阐述的呢？

在读Luca Peliti, Sim...

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S-L

最大熵与平衡态

最大熵导出热力学平衡态是在总能量守恒，\sum_i p_i E_i = const，的情况下最大化熵 H=-\sum_i p_i\ln p_i。这个用一个拉格朗日乘子的方法很容易证明最后得到的是玻尔兹曼分布，而再归一化一下就回到了最小的自由能。
\max (-\sum_ip_i\ln p_i) 时第一行第四项为什么不是pi^{eq}而是p_i呢？

（不知道为啥我输入公式总是渲染不明白会崩掉…）...

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Charlie

Voyager 这个可能是markdown和mathjax的冲突，目前的暂时解决方法是关掉富文本模式
图中那个钢笔小标签（确实有点麻烦看看后续怎么优化一下😅

e.g. \Delta F and p_i^{eq}

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S-L

Voyager

有一个typo，应该是下面这样

\begin{split} \Delta \mathcal{F} &= \sum_i( p_i E_i + Tp_i\ln p_i - (p_i^{eq}E_i+Tp_i^{\color{red}eq}\ln p_i^{eq}))\\\ &= \sum_i\left(Tp_i\ln p _i - p_iT\ln e^{-\beta E_i} +T\ln Z\right)\\\ & = T\sum_i p_i\ln p_i-p_i\ln\frac{e^{-\beta E_i}}{Z}\\\ &=T\under

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