此代码分为两部分。首先是数字和随后的计算。(由两个注释表示)。但是当我运行它们时,它们以相反的顺序出现。
\subsection*{Figures}
\begin{center}
%should appear first
\begin{figure}[ht]
\begin{subfigure}[b]{0.5\linewidth}
\centering
\includegraphics[width=0.9\linewidth]{../figures/Ptime.jpg}
\caption{Pressure vs Step}
\vspace{4ex}
\end{subfigure}
\begin{subfigure}[b]{0.5\linewidth}
\centering
\includegraphics[width=0.9\linewidth]{../figures/temptime.jpg}
\caption{Temperature vs Step}
\vspace{4ex}
\end{subfigure}
\begin{subfigure}[b]{0.5\linewidth}
\centering
\includegraphics[width=0.9\linewidth]{../figures/KEtime.jpg}
\caption{Kinetic Energy vs Step}
\end{subfigure}
\begin{subfigure}[b]{0.5\linewidth}
\centering
\includegraphics[width=0.9\linewidth]{../figures/PEtime.jpg}
\caption{Potential Energy vs Step}
\end{subfigure}
\caption{}
\end{figure}
\end{center}
%should appear next
\textit{Calculation of Self Diffusion Coefficient}
\begin{equation*}
\begin{split}
\text{slope} & = 0.0021 \\
\text{Self Diffusion Coeff} & =\frac{1}{6}\times \text{(Slope of MSD at long time)}\\
& = \frac{1}{6} \times \text{0.0021}\\
& = \SI{3.5e-4} {\text{LJ units}}
\end{split}
\end{equation*}
\text{Considering the LJ liquid to be liquid Ar, we have $\sigma = 0.34 \textit{ nm}$, $\frac{\epsilon}{k_B}=120 \textit{ K}$. }\\
\text{$\therefore \frac{{(L^*)}^2}{t^*}=\frac{{L^2}/{\sigma^2}}{t\sqrt{\frac{\epsilon}{m\sigma^2}}}\implies \frac{L^2}{t}= (\sigma^3\sqrt{m/\epsilon}) \frac{(L^*)^2}{t^*} \implies \frac{L^2}{\epsilon}=\num{2.49e-31}\frac{(L^*)^2}{t^*}$}
\linebreak
\begin{equation*}
\begin{split}
\text{Diffusion Coefficient in SI Units} &= \num{3.5e-4}\times \SI{2.489e-31}{\meter^2 /\second}\\
&=\SI{8.7e-34}{\meter^2 /\second}
\end{split}
\end{equation*}
答案1
一旦将第二个方程上方的文本缩短为一个文本单词considering
,就可以正常工作 - 方程中可能存在一些错误 - 除了正确排列之外
\documentclass[]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{mathtools}
\usepackage{amssymb,booktabs}
\usepackage{caption,subcaption}
\begin{document}
\subsection*{Figures}
\begin{center}
%should appear first
\begin{figure}[ht]
\begin{subfigure}[b]{0.5\linewidth}
\centering
\includegraphics[width=0.9\linewidth]{example-image-a}
\caption{Pressure vs Step}
\vspace{4ex}
\end{subfigure}
\begin{subfigure}[b]{0.5\linewidth}
\centering
\includegraphics[width=0.9\linewidth]{example-image-a}
\caption{Temperature vs Step}
\vspace{4ex}
\end{subfigure}
\begin{subfigure}[b]{0.5\linewidth}
\centering
\includegraphics[width=0.9\linewidth]{example-image-a}
\caption{Kinetic Energy vs Step}
\end{subfigure}
\begin{subfigure}[b]{0.5\linewidth}
\centering
\includegraphics[width=0.9\linewidth]{example-image-a}
\caption{Potential Energy vs Step}
\end{subfigure}
\caption{}
\end{figure}
\end{center}
%should appear next
\textit{Calculation of Self Diffusion Coefficient}
\begin{equation}
\begin{split}
TCU & = \frac{\text{purchasing cost} + \text{setup cost} + \text{holding cost}}{time} \\
& = \frac{cy+K+h\frac{y}{2}t_0}{t_0}
\end{split}
\end{equation}
\text{Considering the LJ liquid to be liquid Ar, we have $\sigma = 0.34 \textit{ nm}$, $\frac{\epsilon}{k_B}=120 \textit{ K}$. }\\
\text{$\therefore \frac{{(L^*)}^2}{t^*}=\frac{{L^2}/{\sigma^2}}{t\sqrt{\frac{\epsilon}{m\sigma^2}}}\implies \frac{L^2}{t}= (\sigma^3\sqrt{m/\epsilon}) \frac{(L^*)^2}{t^*} \implies \frac{L^2}{\epsilon}=\num{2.49e-31}\frac{(L^*)^2}{t^*}$}
\linebreak
\begin{equation}
\begin{split}
TCU & = \frac{\text{purchasing cost} + \text{setup cost} + \text{holding cost}}{time} \\
& = \frac{cy+K+h\frac{y}{2}t_0}{t_0}
\end{split}
\end{equation}
\end{document}