$$ \begin{aligned}\text{U}& =\mathrm{e}^{-\alpha}\sum_{l}\varepsilon_{l}\omega_{l}\mathrm{e}^{-\beta\varepsilon_{l}}=\mathrm{e}^{-\alpha}\bigg(-\frac{\partial}{\partial\beta}\bigg)\sum_{l}\omega_{l}\mathrm{e}^{-\beta\varepsilon_{l}}=\frac{N}{Z_{1}}\Big(-\frac{\partial}{\partial\beta}\Big)Z{1}=-N\frac{\partial}{\partial\beta}\mathrm{ln}Z{_1}\end{aligned} $$
广义力 :粒子能量一般式外参量的函数,例如三维粒子的能量
$\varepsilon_{l}=\frac{2\pi^{2}\hbar^{2}}{mL^{2}}\big(n_{x}^{2}+n_{y}^{2}+n_{z}^{2}\big)=\frac{2\pi^{2}\hbar^{2}}{mV^{2/3}}\big(n_{x}^{2}+n_{y}^{2}+n_{\bar{z}}^{2}\big)$
根据功能原理,外力对质点说做的功等于动能的增量$f=\frac{\partial\varepsilon}{\partial y}$
$\begin{aligned} \text{Y}& =\sum_{l}\frac{\partial\varepsilon_{l}}{\partial y}a_{l}=\sum_{l}\frac{\partial\varepsilon_{l}}{\partial y}\omega_{l}\mathrm{e}^{-\alpha-\beta\cdot\varepsilon_{l}} \\ &=\mathrm{e}^{-\alpha}\left(-\frac1\beta\frac\partial{\partial y}\right)\sum_l\omega_l\mathrm{e}^{-\beta-\varepsilon_l} \\ &=\frac N{Z_1}\Big(-\frac1\beta\frac\partial{\partial y}\Big)Z_1 \\ &=-\frac N\beta\frac\partial{\partial y}\mathrm{ln~}Z_1 \end{aligned}$广义力公式
功和热的统计意义
$\mathrm{d}U=\sum_la_l\mathrm{d}\varepsilon_l(Ydy)+\sum_l\varepsilon_l\mathrm{d}a_l(dQ)$
前者功使粒子能级变化而粒子分布不变,后者从外界吸收的热量式能级分布发生变化能级不变
熵的表达式
$\begin{aligned}\text{dQ}& =\operatorname{d}U-Y\operatorname{d}y =-N\mathrm{d}\left(\frac{\partial\ln\left.Z_1\right)}{\partial\beta}\right)+\frac{N}\beta\frac{\partial\ln\left.Z_1\right)}{\partial\gamma}\mathrm{d}y\end{aligned}$
$\beta\left(\text{d}U-Y\text{d}y\right)=-N\text{B}\text{d}\left(\frac{\partial \ln Z_1}{\partial\beta}\right)+N\frac{\partial\ln Z_1}{\partial y}\text{d}y$
$\beta\left(\text{d}U-Y\text{d}y\right)=N\text{d}\bigg(\ln Z_1-\beta\frac{\partial}{\partial\beta}\text{ln}Z_1\bigg)$
$\beta=\frac1{kT}$怎么假定的我也不知道(doge)$k=1.381\times10^{-23}J\cdot K^{-1}=R/N_A$
$\beta$系统的温度因子
$\mathrm{d}S=Nk\mathrm{d}\Big(\ln Z_{1}-\beta\frac{\partial}{\partial\beta}\mathrm{ln}Z{_1}\Big)$
玻尔兹曼关系
$\left.S=k\left(N{ln}\\N+\alpha N+\beta U\right.\right)$,$S=k \ln \Omega$
熵宏观与围观的桥梁,系统无序度的表现。给普朗克绝对熵以有力的支持
T→0时粒子往低能级排列。
若简并度不为1,则为有限数,k很小,熵有限
$\begin{aligned}S&=Nk\Bigg[\ln Z_1-\beta\frac{\partial}{\partial\beta}\ln Z_1\Bigg]-k\ln N!\end{aligned}$
上面的式子符合玻色费米系统,满足对广延量的需求
自由能
玻尔兹曼系统:$F=-NkT\\operatorname{ln}Z_{1}=-kT\\operatorname{ln}\\biggl(Z_{1}^{N}\\biggr)$
满足经典极限条件的玻色(费米)系统:$\\begin{aligned}F=-NkT\\ln Z_1+kT\\ln N!=-kT\\ln\\!\\left(Z_1^N/N!\\right)\\end{aligned}$
首先考虑单原子分子气体
能量:$\\varepsilon=\\frac{p^2}{2m}=\\frac1{2m}\\left(p_x^2+p_y^2+p_z^2\\right)$ 像空间体积元:$d\\omega=dxdydzdp_xdp_ydp_z$
可以得出配分函数 :$\\int_{-\\infty}^\\infty e^{-\\alpha x^2}dx=\\sqrt{\\pi/\\alpha}\\begin{array}{c}\\Longrightarrow Z_1=\\frac V{h^3}(\\frac{2\\pi m}\\beta)^{3/2}=\\frac V{h^3}(2\\pi mkT)^{3/2}\\end{array}$
$\\ln Z_1=\\frac32\\ln(\\frac{2\\pi m}{h^2})-\\frac32\\ln\\beta+\\ln V$
物态方程$P=\\frac N\\beta\\frac{\\partial\\ln Z_1}{\\partial V}=\\frac N\\beta\\frac{\\partial\\ln V}{\\partial V}=\\frac{NkT}V$
双原子分子或者多原子分子问题:有转动或者振动能量,但是不改变配分函数对体积V 的依赖 关系,因此理想气体的物态方程不变**。**