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}
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}
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}
答案2
您可以简单地替换split
并aligned
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}
答案3
tbtags
该软件包中有一个选项amsmath
可以为您完成此操作:
\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
=&
至&=
(如果您确实想对齐,=
则需要写入={}&
正确的间距)\Big
达到\Bigl
或\Bigr
达到正确的间距\text{log}
到\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
:
\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
\mright
mleftright
,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={} &
。