如何将方程式编号放在底部

如何将方程式编号放在底部

我希望我的等式编号 11.72 出现在最后一行。我该怎么做?

After a wick rotation we obtain


\begin{equation}
\begin{split}
\int \frac{d^4 \mathnormal{k}}{(2\pi)^4}\text{log}(-k^2+m^2)=& i\int 
  \frac{d^4 \mathnormal{k_E}}{(2\pi)^4}\text{log}(k_E^2+m^2)\\
  =&-i \frac{\partial}{\partial \alpha}\int \frac{d^4 
    \mathnormal{k_E}}{(2\pi)^4}\frac{1} 
    {(k_E^2+m^2)^{\alpha}}\Big\vert_{\alpha=0}\\
  =&-i \frac{\partial}{\partial \alpha} \Big(\frac{1} 
     {(4\pi)^{d/2}}\frac{\Gamma(\alpha-\frac{d}{2})} 
   {\Gamma(\alpha)}\frac{1}{(m^2)^{\alpha-\frac{d}{2}}} \Big) \Big 
    \vert_{\alpha=0}\\
       =& -i\frac{\Gamma(\frac{-d}{2})}{(4\pi)^{d/2}}\frac{1}{(m^2)^{- 
        d/2}}
       \end{split}
      \tag{11.72}
      \end{equation}

这是我获得的 pdf。我怎样才能将等式数字放在底部? enter image description here

答案1

像这样。

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{align}
\int \frac{d^4 \mathnormal{k}}{(2\pi)^4}\text{log}(-k^2+m^2)=& i\int 
  \frac{d^4 \mathnormal{k_E}}{(2\pi)^4}\text{log}(k_E^2+m^2)\notag\\
  =&-i \frac{\partial}{\partial \alpha}\int \frac{d^4 
    \mathnormal{k_E}}{(2\pi)^4}\frac{1} 
    {(k_E^2+m^2)^{\alpha}}\Big\vert_{\alpha=0}\notag\\
  =&-i \frac{\partial}{\partial \alpha} \Big(\frac{1} 
     {(4\pi)^{d/2}}\frac{\Gamma(\alpha-\frac{d}{2})} 
   {\Gamma(\alpha)}\frac{1}{(m^2)^{\alpha-\frac{d}{2}}} \Big) \Big 
    \vert_{\alpha=0}\notag\\
       =& -i\frac{\Gamma(\frac{-d}{2})}{(4\pi)^{d/2}}\frac{1}{(m^2)^{- 
        d/2}}
      \tag{11.72}
      \end{align}
\end{document}

enter image description here

但是,我至少会使用\log而不是\text{log},并使微分 d 与维度正则化中的维度可区分d,虚数 也是如此i。我也不会手动设置方程编号。

\documentclass{article}
\usepackage{amsmath}
\usepackage{cleveref}
\newcommand{\dd}[1]{\mathrm{d}#1}
\numberwithin{equation}{section}
\begin{document}
\section{Dimensional regularization}

\dots

After a Wick rotation we get
\begin{align}
\int\! \frac{\dd^4 \mathnormal{k}}{(2\pi)^4}\log(-k^2+m^2)&= \mathrm{i}\,\int\! 
  \frac{\dd^4 \mathnormal{k_E}}{(2\pi)^4}\log(k_E^2+m^2)\notag\\
  &=-\mathrm{i} \frac{\partial}{\partial \alpha}\int\! \frac{\dd^4 
    \mathnormal{k_E}}{(2\pi)^4}\frac{1} 
    {(k_E^2+m^2)^{\alpha}}\Big\vert_{\alpha=0}\notag\\
  &=-\mathrm{i}\, \frac{\partial}{\partial \alpha} \Big(\frac{1} 
     {(4\pi)^{d/2}}\frac{\Gamma(\alpha-\frac{d}{2})} 
   {\Gamma(\alpha)}\frac{1}{(m^2)^{\alpha-\frac{d}{2}}} \Big) \Big 
    \vert_{\alpha=0}\notag\\
       &= -\mathrm{i}\,\frac{\Gamma(\frac{-d}{2})}{(4\pi)^{d/2}}
       \frac{1}{(m^2)^{- 
        d/2}}
 \label{eq:DimRegInt1}
\end{align}

\dots

As shown in \cref{eq:DimRegInt1}, \dots

\end{document}

enter image description here

答案2

您可以简单地替换splitaligned使用[b]可选参数。

我擅自将大分隔符的大小更改为\bigg,在我看来,这样在这种情况下看起来更好。

\documentclass{article}
\usepackage{amsmath} 

\begin{document} 

\begin{equation}
    \begin{aligned}[b]
    \int \frac{d^4 \mathnormal{k}}{(2\pi)^4}\text{log}(-k^2+m^2)&= i\int
    \frac{d^4 \mathnormal{k_E}}{(2\pi)^4}\text{log}(k_E^2+m^2)\\
    &=-i \frac{\partial}{\partial \alpha}\int \frac{d^4
    \mathnormal{k_E}}{(2\pi)^4}\frac{1}
    {(k_E^2+m^2)^{\alpha}}\bigg\vert_{\alpha=0}\\
    &=-i \frac{\partial}{\partial \alpha} \biggl(\frac{1}
     {(4\pi)^{d/2}}\frac{\Gamma(\alpha-\frac{d}{2})}
   {\Gamma(\alpha)}\frac{1}{(m^2)^{\alpha-\frac{d}{2}}} \biggr) \bigg
    \vert_{\alpha=0}\\
      & = -i\frac{\Gamma(\frac{-d}{2})}{(4\pi)^{d/2}}\frac{1}{(m^2)^{-
        d/2}}
    \end{aligned}
  \tag{11.72}
\end{equation}. 

\end{document} 

enter image description here

答案3

tbtags该软件包中有一个选项amsmath可以为您完成此操作:

Sample output

\documentclass{article}

\usepackage[tbtags]{amsmath}

\begin{document}

\begin{equation}
  \begin{split}
    \int \frac{d^4 k}{(2\pi)^4}\log(-k^2+m^2)
    &= i\int \frac{d^4 k_E}{(2\pi)^4}\log(k_E^2+m^2)\\
    &=-i \frac{\partial}{\partial \alpha}\int \frac{d^4
       k_E}{(2\pi)^4}\frac{1}
       {(k_E^2+m^2)^{\alpha}}\Bigr\vert_{\alpha=0}\\
    &=-i \frac{\partial}{\partial \alpha} \Bigl(\frac{1}
       {(4\pi)^{d/2}}\frac{\Gamma(\alpha-\frac{d}{2})}
       {\Gamma(\alpha)}\frac{1}{(m^2)^{\alpha-\frac{d}{2}}} \Bigr) \Bigr
       \vert_{\alpha=0}\\
    &= -i\frac{\Gamma(\frac{-d}{2})}{(4\pi)^{d/2}}\frac{1}{(m^2)^{-
       d/2}}
  \end{split}
  \tag{11.72}
\end{equation}

\end{document}

其他变化

  • =&&= (如果您确实想对齐,=则需要写入={}&正确的间距)
  • \Big达到\Bigl\Bigr达到正确的间距
  • \text{log}\log
  • \mathnormal被移除,没有被持续使用,并且没有效果

此选项的反义词tbtags是默认选项centertags。其名称tbtags源自“顶部或底部标签”;如果方程式编号在左侧,则将标签放在顶部;如果方程式编号在右侧,则将标签放在底部。

答案4

下面的操作可以稍微清理一下,包括使用\nonumber(或\notag)的组合在必要时删除方程编号,并使用 指定硬编码方程编号\tag

enter image description here

\documentclass{article}

\usepackage{amsmath,mleftright}

\begin{document}

\begin{align}
  \int \frac{\mathrm{d}^4 k}{(2 \pi)^4} \log \bigl(-k^2 + m^2 \bigr)
    &=  i \int \frac{\mathrm{d}^4 k_E}{(2 \pi)^4} \log \bigl(k_E^2 + m^2 \bigr) \nonumber \\
    &= -i \mleft. \frac{\partial}{\partial \alpha} \int \frac{\mathrm{d}^4 k_E}{(2 \pi)^4}
      \frac{1}{\bigl(k_E^2 + m^2 \bigr)^\alpha} \mright\rvert_{\alpha = 0} \nonumber      \\
    &= -i \mleft. \frac{\partial}{\partial \alpha} \biggl(
      \frac{1}{(4 \pi)^{d / 2}} \frac{\Gamma(\alpha - d/2)}{\Gamma(\alpha)} 
      \frac{1}{(m^2)^{\alpha - d/2}} \biggr) \mright\rvert_{\alpha = 0} \nonumber         \\
    &= -i \frac{\Gamma(-d / 2)}{(4 \pi)^{d / 2}} \frac{1}{(m^2)^{-d / 2}} \tag{11.72}
\end{align}

\end{document}

一些指示:

  • \frac如果只是单个标记,则似乎没有必要在指数/上标(或下标)中使用。也就是说,a/b似乎比更干净\frac{a}{b}

  • 使用 (运算符)\log代替\text{log}

  • 使用...\rvert组合来拉伸可扩展分隔符(如上所示) (感谢\mleft\mrightmleftright,或者使用\left... \right,但间距可能会略有不同/更宽)。

  • align对齐=应该使用&=,而不是=&。您可以使用后者,但您需要={} &

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