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I've got a frame that first has some itemize text and then want to split it in two columns to have some more itemized text on each column.

The problem is I can't align the top of the columns and the result doesn't look great as is shown in the attached image.

Also, I've read beamer: Top-aligning columns within a top-aligned frame but was wondering if I'm doing something wrong with the t, T before I start messing with things I don't quite understand (ie the \topsep, \partopsep

\documentclass{beamer}
\usepackage[utf8]{inputenc}
\usepackage[english]{babel}
\usepackage{graphicx}
\usepackage{subfig}
\usepackage{braket}
\usepackage{geometry}
\usepackage{empheq}
\usepackage{amsmath}
\usepackage{amssymb}

\begin{document}

\begin{frame}[T]{Angular Spectrum Method}{Free Space Propagation}
    A constant index medium such as free space is: 
    \begin{itemize}
        \item isotropic \pause $\Rightarrow$ diagonal permittivity, permeability tensors \pause
        \item spatially uniform \pause $\Rightarrow$ relative permittivity and permeability are constants equal to 1 \pause
        \item Source free \pause $\Rightarrow$ $\mathbf{J} = 0$, $\rho = 0$
    \end{itemize}
    \pause
    \begin{columns}[T]
    \begin{column}[T]{.5\textwidth}
        \begin{itemize}
            \item<7-> Polarisation Independent!
            \item<8-> Modes are plane waves
        \end{itemize}
    \end{column}
    \hfill
    \begin{column}[T]{.5\textwidth}
        \begin{itemize}
            \item<7-> $ \nabla^2 \mathbf{E} - \mu_0 \mu_r \epsilon_0 \epsilon_r \dfrac{ \partial^2 \mathbf{E} }{ \partial t } = 0 $
            \item<8-> $ E_i = \sum_\mathbf{k} A_\mathbf{k} e^{ \mathbf{k} \cdot \mathbf{r} } $
            \item<8-> $ \left( \dfrac{ 2 \pi n }{ \lambda } \right)^2 = k^2 = k_x^2 + k_y^2 + k_z^2 $
        \end{itemize}
    \end{column}
    \end{columns}
\end{frame}

\end{document}

enter image description here

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  • Your code has errors. Please fix them! Mar 4, 2015 at 16:46

1 Answer 1

2

Use some \vphantoms in the itemize to the left so items on both environments have the same height:

\documentclass{beamer}
\usepackage[utf8]{inputenc}
\usepackage[english]{babel}
\usepackage{graphicx}
\usepackage[caption=false]{subfig}
\usepackage{braket}
\usepackage{geometry}
\usepackage{empheq}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage[backend=biber,url=false,doi=false,isbn=false,firstinits=true]{biblatex}

\begin{document}

\begin{frame}{Angular Spectrum Method}{Free Space Propagation}
    A constant index medium such as free space is: 
    \begin{itemize}
        \item isotropic \pause $\Rightarrow$ diagonal permittivity, permeability tensors \pause
        \item spatially uniform \pause $\Rightarrow$ relative permittivity and permeability are constants equal to 1 \pause
        \item Source free \pause $\Rightarrow$ $\mathbf{J} = 0$, $\rho = 0$
    \end{itemize}
    \pause
    \begin{columns}[T]
    \begin{column}[T]{.5\textwidth}
        \vspace{0pt}
        \begin{itemize}
            \item<7-> Polarisation Independent!\vphantom{$ \dfrac{ \partial^2 \mathbf{E} }{ \partial t }$}
            \item<8-> Modes are plane waves\vphantom{$A_\mathbf{k} e^{ \mathbf{k} \cdot \mathbf{r} }$}
        \end{itemize}
    \end{column}
    \hfill
    \begin{column}[T]{.5\textwidth}
        \begin{itemize}
            \item<7-> $ \nabla^2 \mathbf{E} - \mu_0 \mu_r \epsilon_0 \epsilon_r \dfrac{ \partial^2 \mathbf{E} }{ \partial t } =  0 $
            \item<8-> $ E_i = \sum_\mathbf{k} A_\mathbf{k} e^{ \mathbf{k} \cdot \mathbf{r} } $
            \item<8-> $ \left( \dfrac{ 2 \pi n }{ \lambda } \right)^2 = k^2 = k_x^2 + k_y^2 + k_z^2 $
        \end{itemize}
    \end{column}
    \end{columns}
\end{frame}

\end{document}

enter image description here

Remarks

  • Your code had a spurious & character that I deleted and [T] was not a valid option for frame.

    • You need to load subfig with caption=false, as in my example code.
  • \bold is deprecated; you should use \mathbf instead.

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  • hm... the subfig doesn't create problems for me whether I have caption=false or not. (I suspect because I'm not using the latest version of beamer) Otherwise perfect thank you.
    – evan54
    Mar 4, 2015 at 17:05

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