整体思路

$$\begin{align} \frac{\Delta F}{NE_n} = \left[ \tilde{\tilde{\Omega}}_{\mathrm{int}} - \tilde{\Omega}_{0}^{\mathrm{M}} + \tilde{\mu} \right]\left( \frac{n_{\varepsilon}}{n} \right)^{2/3} \end{align}$$ $$\begin{align} \frac{n}{n_{\varepsilon}} =& - \frac{\partial}{\partial\tilde{\mu}}\left[ \tilde{\tilde{\Omega}}_{\mathrm{int}} + \tilde{\Omega}_{0}^{\mathrm{B}} \right] \end{align}$$ $$\begin{align} \tilde{\tilde{\Omega}}_{\mathrm{int}} = \int \mathrm{d}\tilde{q}\cdot \tilde{q}^2 \int_0^{\pi}\sin\theta_q\mathrm{d}\theta_q \int_{-\infty}^{+\infty}\mathrm{d}\tilde{\omega} \cdot \frac{3}{2\pi}\cdot \frac{1}{e^{\tilde{\beta}\tilde{\omega}}-1} \delta^p(\vec{q},z) \end{align}$$ $$\begin{align} \tilde{\Omega}_0^{\mathrm{B}} = \frac{3}{\tilde{\beta}} \int_0^{\infty} \mathrm{d} \tilde{k} \cdot \tilde{k}^2 \ln \left[1 - e^{- \tilde{\beta} \tilde{\xi}_k}\right] \end{align}$$ $$\begin{align} \delta^p(\vec{q}, z) =& \mathrm{Arg}\left[ \frac{1}{4\pi}\cdot \frac{2R}{k_{\varepsilon}^2 v} + \tilde{z}\cdot\frac{1}{4\pi} + \frac{2 R}{M k_{\varepsilon}^2}\Pi_r(\vec{q},z + \mathrm{i}0^+) \right] \end{align}$$ $$\begin{align} \frac{2 R}{M k_{\varepsilon}^2}\Pi_r(\vec{q},z) =& \frac{2}{\pi^2} \cdot k_{\varepsilon}R\cdot \int \mathrm{d}\tilde{k}\\ & \left\{ \int_0^{2\pi}\mathrm{d}\phi_k \int_{0}^{\pi} \sin\theta \mathrm{d}\theta_k \cdot \cos^2 \theta_k \left[ 1+n(\xi_{\vec{k}+\vec{q}/2}) + n(\xi_{-\vec{k}+\vec{q}/2}) \right]\frac{3}{4\pi}\frac{\tilde{k}^4} {2\tilde{k}^2 - \tilde{z}} -\frac{1}{2}\tilde{k}^2 - \frac{1}{4}\tilde{z} \right\} \end{align}$$

而

$$\begin{align} n(\xi_{\vec{k}+\vec{q}/2}) = \frac{1}{e^{\beta(\tilde{k}^2 + \tilde{q}^2/4 + \tilde{k} \tilde{q} x-2\mu)} -1} \end{align}$$

其中

$$\begin{align} x = \cos\theta_{kq} =& \sin\theta_k \cos \phi_k \sin \theta_q \cos \phi_q + \sin \theta_k \sin\phi_k \sin \theta_q \sin \phi_q + \cos \theta_k \cos \theta_q \\ =& \sin\theta_k \sin \theta_q \cos(\phi_k-\phi_q) + \cos \theta_k \cos \theta_q \end{align}$$

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: $\tilde{\xi} = \tilde{k}^2 - \tilde{\mu} ,\quad n(\xi) = \frac{1}{e^{\tilde{\beta} \tilde{\xi}}-1}, \quad \tilde{z} = \tilde{\omega} - \frac{\tilde{q}^2}{2} + 2 \tilde{\mu}$

Fig

file:./2019-10-09-physics-NSRcalv3/numGiven.png

file:./2019-10-09-physics-NSRcalv3/TcAll-ana.png

file:./2019-10-09-physics-NSRcalv3/fig3-p-wave.png

file:./2019-10-09-physics-NSRcalv3/fig4-pwave.png

file:./2019-10-09-physics-NSRcalv3/fig5-a.png

file:./2019-10-09-physics-NSRcalv3/fig5-b.png

fig3-p-nc file:./2019-10-09-physics-NSRcalv3/fig3-p-nc.png

file:./2019-10-09-physics-NSRcalv3/Tc-nc.png