整体思路
$$\begin{align} \frac{\Delta F}{NE_n} = \left[ \tilde{\tilde{\Omega}}_{\mathrm{int}} - \tilde{\Omega}_{0}^{\mathrm{M}} + \tilde{\mu} \right]\left( \frac{n_{\varepsilon}}{n} \right)^{2/3} \end{align}$$ $$\begin{align} \frac{n}{n_{\varepsilon}} =& - \frac{\partial}{\partial\tilde{\mu}}\left[ \tilde{\tilde{\Omega}}_{\mathrm{int}} + \tilde{\Omega}_{0}^{\mathrm{B}} \right] \end{align}$$ $$\begin{align} \tilde{\tilde{\Omega}}_{\mathrm{int}} = \int \mathrm{d}\tilde{q}\cdot \tilde{q}^2 \int_{-\infty}^{+\infty}\mathrm{d}\tilde{\omega} \cdot \frac{3}{\pi}\cdot \frac{1}{e^{\tilde{\beta}\tilde{\omega}}-1} \delta^p(\vec{q},z) \end{align}$$ $$\begin{align} \tilde{\Omega}_0^{\mathrm{B}} = \frac{3}{\tilde{\beta}} \int_0^{\infty} \mathrm{d} \tilde{k} \cdot \tilde{k}^2 \ln \left[1 - e^{- \tilde{\beta} \tilde{\xi}_k}\right] \end{align}$$ $$\begin{align} \delta^p(\vec{q}, z) =& \mathrm{Arg}\left[ \frac{1}{4\pi}\cdot \frac{2R}{k_{\varepsilon}^2 v} + \tilde{z}\cdot\frac{1}{4\pi} + \frac{2 R}{M k_{\varepsilon}^2}\Pi_r(\vec{q},z + \mathrm{i}0^+) \right] \end{align}$$ $$\begin{align} \frac{2 R}{M k_{\varepsilon}^2}\Pi_r(\vec{q},z) =& \frac{2}{\pi^2} \cdot k_{\varepsilon}R\cdot \int \mathrm{d}\tilde{k} \left\{ \int_{-1}^1 \mathrm{d}x \cdot x^2\left[ 1+n(\xi_{\vec{k}+\vec{q}/2}) + n(\xi_{-\vec{k}+\vec{q}/2}) \right]\frac{3}{2}\frac{\tilde{k}^4} {2\tilde{k}^2 - \tilde{z}} -\frac{1}{2}\tilde{k}^2 - \frac{1}{4}\tilde{z} \right\} \end{align}$$带 tilde 的都是以 $\varepsilon$ 或 $k_{\varepsilon}$ 为单位的无量纲量.
最终画的是 $(1)$ 式. $(1)$ 式左边以 $E_n$ 为单位, 右边乘上 $\left( \frac{n_{\varepsilon}}{n} \right)^{2/3}$ 进行单位转换.
将 $(2)~(5)$ 式依次代入可得结果
还有一些关系: $\tilde{\xi} = \tilde{k}^2 - \tilde{\mu} ,\quad n(\xi) = \frac{1}{e^{\tilde{\beta} \tilde{\xi}}-1}, \quad \tilde{z} = \tilde{\omega} - \frac{\tilde{q}^2}{2} + 2 \tilde{\mu}$
$\delta^p(\vec{q}, z)$ 的近似的解析表达式
$\delta^p(\vec{q}, z)$ 的近似的解析表达式为
$$\begin{align} -\pi \theta(\omega - a) \end{align}$$其中
$$\begin{align} a = \frac{\tilde{q}^2}{2} - 2 \tilde{\mu} - \frac{2R}{k_{\varepsilon}^2 v} \end{align}$$以 $\tilde{q}$ 和 $\tilde{\omega}$ 为变量, 数值上画出的 $\delta^p(\vec{q}, z)$ 的二维图为
file:./2019-09-08-physics-NSRcalv2/2Dfig.png
是一个阶跃函数. 其边界和近似的解析表达式对比
file:./2019-09-08-physics-NSRcalv2/stepLine.png
计算 density I
$$\begin{align} \frac{n}{n_{\varepsilon}} =& - \frac{\partial}{\partial\tilde{\mu}}\left[ \tilde{\tilde{\Omega}}_{\mathrm{int}} + \tilde{\Omega}_{0}^{\mathrm{B}} \right] \end{align}$$density $\frac{n}{n_{\varepsilon}}$ 与温度 $\tilde{\beta}$ , $k_{\varepsilon}R$ , $\frac{2R}{k_{\varepsilon}^2 v}$ , 化学势 $\tilde{\mu}$ 有关.
将温度 $\tilde{\beta}1$ 和 $k_{\varepsilon}R1/30$ 固定, 以
$\frac{2R}{k_{\varepsilon}^2 v}$ 为横坐标, 画出不同化学势
$\tilde{\mu}$ 时积分部分的多条曲线.
取 $\frac{2R}{k_{\varepsilon}^2 v}\in [-2, 2]$ , $\mu \in [-2, -.0.1]$
计算 density II
用 $\delta^p(\vec{q}, z)$ 的近似的解析表达式求 density 的解析表达式
$$\begin{align} \frac{n}{n_{\varepsilon}} =& - \frac{\partial}{\partial\tilde{\mu}}\left[ \tilde{\tilde{\Omega}}_{\mathrm{int}} + \tilde{\Omega}_{0}^{\mathrm{B}} \right] \\ =& \int_0^{\infty}\mathrm{d}\tilde{q}\cdot 3 \tilde{q}^2\left[ \frac{2}{e^{\tilde{\beta}a}-1} + \frac{1}{e^{\tilde{\beta}(\tilde{q}^2 - \tilde{\mu})}-1} \right] \end{align}$$最终结果
将结果与原文中的结果对比
file:./2019-09-08-physics-NSRcalv2/co-result.png
- 蓝线: 原文结果
- num 数值结果.
- ana1 只把 $\delta^p$ 的边界的边界用数值解代替(analytically calculate $\delta^p$ ).
- ana2 直到积分的结果都是解析的, 微分是数值微分(analytically calculate the integral F).
- ana3 对 $\mu$ 的微分也用解析结果(analytically calculate $\mu$ ).
计算 $T_C$
weak coupling
算出 $\mu = 0$ 时的 density .
对应的纵坐标为
$$\begin{align} \frac{k_BT_C}{E_n} = \frac{1}{\beta E_n} = \frac{1}{\varepsilon \beta} \frac{\varepsilon}{E_n} = \frac{1}{\tilde{\beta}} \cdot \left(\frac{n_{\varepsilon}}{n}\right)^{2/3} \end{align}$$横坐标为
$$\begin{align} \frac{2 R_p}{k_n^2v_p} = \frac{2 \tilde{R_p}}{k_{\varepsilon}^2 \tilde{v_p}}\cdot \left( \frac{n_{\varepsilon}}{n}\right)^{2/3} \end{align}$$file:./2019-09-08-physics-NSRcalv2/Tc.png
file:./2019-09-08-physics-NSRcalv2/TcNum.png