\begin{frame}{Calcolo classico}
\vspace{-1cm}
\begin{equation*}
mv = \dfrac{E_{\gamma}}{c} + m(v-v_{R}); \qquad
E_{e} + \dfrac{1}{2}mv^{2} = E_{f} + E_{\gamma} + \dfrac{1}{2}m(v-v_{R})^{2}\\
\vspace{0.1cm}
\textcolor{blue}{Ponendo:}\\
E_{R} = \dfrac{1}{2}mv_{R}^{2} = \dfrac{E_{\gamma}^{2}}{2mc^{2}} \approx \dfrac{E_{0}^{2}}{2mc^{2}}; \qquad
\vspace{0.1cm}
E_{D} = mvv_{R} = \dfrac{E_{\gamma}v}{c} \approx \dfrac{E_{0}v}{c}\\
\vspace{0.1cm}
\textcolor{blue}{ Otteniamo:}\\
E_{\gamma}= E_{0} -E_{R} +E_{D}\\
\vspace{0.3cm}
\textcolor{blue}{ Per v=0, per avere fluorescenza di risonanza si deve avere:}\\
2E_{R} \lesssim \Gamma \\ \textcolor{black}{(condizione non verificata per nuclei con E_{0}= 100KeV; A=100)}\\
\end{equation*}
\vspace{-1.5cm}
\begin{equation*}
\textcolor{blue}{Agitazione termica} \Rightarrow \delta v= \sqrt{\frac{k_{B}T}{m}} \Rightarrow \Gamma_{D} = \dfrac{E_{0}\delta v}{c}= \dfrac{E_{0}}{c}\sqrt{\frac{k_{B}T}{m}}\\ \textcolor{black}{(caso tipico nucleare \Gamma_{D}=10^{-3} eV)} \\
\end{equation*}
\end{frame}
答案1
这是多个问题的组合:
不要在方程式中留下空行
将每个方程式放入其自己的
equation*
环境中不要滥用
\\
换行符,如果你处于文本模式,请保留一个空行
\documentclass{beamer}
\begin{document}
\begin{frame}{Calcolo classico}
\begin{equation*}
mv = \dfrac{E_{\gamma}}{c} + m(v-v_{R}); \qquad
E_{e} + \dfrac{1}{2}mv^{2} = E_{f} + E_{\gamma} + \dfrac{1}{2}m(v-v_{R})^{2}
\end{equation*}
\textcolor{blue}{Ponendo:}
\begin{equation*}
E_{R} = \dfrac{1}{2}mv_{R}^{2} = \dfrac{E_{\gamma}^{2}}{2mc^{2}} \approx \dfrac{E_{0}^{2}}{2mc^{2}};
\end{equation*}
% just the same for the remaining text
\end{frame}
\end{document}
答案2
对于需要多行数学运算的情况,例如,不能使用\\
in , from 。这里有一个带有和from 的版本(包 loads 和 extends ),以及带有单位的数字的from 。equation*
gather*
amsmath
gather*
\shortintertext
mathtools
mathtools
amsmath
\SI{<number>}{<unit>}
siunitx
\documentclass{beamer}
\usepackage{mathtools}
\usepackage{siunitx}
\begin{document}
\begin{frame}
\frametitle{Calcolo classico}
\begin{gather*}
mv = \frac{E_{\gamma}}{c} + m(v-v_{R}); \qquad
E_{e} + \frac{1}{2}mv^{2} = E_{f} + E_{\gamma} + \frac{1}{2}m(v-v_{R})^{2}\\
\shortintertext{\textcolor{blue}{Ponendo:}}
E_{R} = \frac{1}{2}mv_{R}^{2} = \frac{E_{\gamma}^{2}}{2mc^{2}} \approx \frac{E_{0}^{2}}{2mc^{2}}; \qquad
E_{D} = mvv_{R} = \frac{E_{\gamma}v}{c} \approx \frac{E_{0}v}{c}\\
\shortintertext{\textcolor{blue}{ Otteniamo:}}
E_{\gamma}= E_{0} -E_{R} +E_{D}\\
\shortintertext{\textcolor{blue}{Per $v=0$, per avere fluorescenza di risonanza si deve avere:}}
2E_{R} \lesssim \Gamma \\
\shortintertext{\textcolor{black}{(condizione non verificata per nuclei con $E_{0}= \SI{100}{\kilo\eV}$; $A=100$)}}
\textcolor{blue}{\text{Agitazione termica}} \Rightarrow \delta v= \sqrt{\frac{k_{B}T}{m}} \Rightarrow \Gamma_{D} = \frac{E_{0}\delta v}{c}= \frac{E_{0}}{c}\sqrt{\frac{k_{B}T}{m}}
\end{gather*}
(caso tipico nucleare $\Gamma_{D}=\SI{e-3}{\eV}$)
\end{frame}
\end{document}