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src/Advanced/Relativity/backups/dilation.tex Normal file
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\section{Length contraction and time dilation}
Up to now, we've been considering everything as points. We haven't considered how tall or wide you are when drawing these diagrams. Now we are going to.
\problem{}
Draw a stick of length $1$ at rest in a spacetime diagram.
\begin{solution}\begin{center}\begin{tikzpicture}[scale=2.0]
\message{Worldlines^^J}
\def\ymin{0.2}
\def\xmin{1.6}
\def\xmax{2}
\def\Nlines{4} % number of world lines (at constant x/t)
\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
\coordinate (O) at (0,0);
\coordinate (T) at (0,\xmax+0.2);
% WORLD LINES GRID
\message{ Making world lines...^^J}
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line] ( \x,-\ymin) -- ( \x,\xmax);
\draw[world line t] (-\xmin, \x) -- (\xmax, \x);
}
\draw[world line] (-\d,-\ymin) -- (-\d,\xmax);
\draw[world line] (-2*\d,-\ymin) -- (-2*\d,\xmax);
\draw[world line] (-3*\d,-\ymin) -- (-3*\d,\xmax);
% AXES
\draw[->,thick] (0,-\ymin) -- (T) node[left=-1] {$ct$};
\draw[->,thick] (-\xmin,0) -- (\xmax+0.2,0) node[below=0] {$x$};
% VECTORS
\draw[vector,myred] (O) -- (0,4*\d)
node[mydarkred,below left=0] {\contour{white}{stick: $x(t)=0$}};
\draw[vector,myred] (\d,0) -- (\d,4*\d);
%\node[right=8,above,mydarkpurple] at (T) {$x(t)=0$};
\fill[myred,opacity=1]
(0,0) -- (0,4*\d) -- (\d,4*\d) -- (\d,0) -- cycle;
\end{tikzpicture}\end{center}\end{solution}
\problem{}
Draw a stick of length $1$ sitting on a train moving at speed $c/2$.
{\em Hint: Begin by drawing the train's reference frame.}
\begin{solution}\begin{center}\begin{tikzpicture}[scale=2.0]
\message{Worldlines^^J}
\pgfmathsetmacro\ang{atan(1/2)} % angle between x and x' axes
\def\ymin{0.2}
\def\xmin{1.6}
\def\xmax{2}
\def\xmaxp{2.2} % maximum of rotated axis
\def\Nlines{4} % number of world lines (at constant x/t)
\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
\coordinate (O) at (0,0);
\coordinate (T) at (0,\xmax+0.2);
\def\xmax{2}
\def\xmaxp{2.2} % maximum of rotated axis
\def\Nlines{5} % number of world lines (at constant x/t)
\pgfmathsetmacro\ang{atan(1/2)} % angle between x and x' axes
\pgfmathsetmacro\d{0.9*\xmax/\Nlines}refer % grid size
\pgfmathsetmacro\D{\d/cos(\ang)/sqrt(1-tan(\ang)^2)} % grid size, boosted
\coordinate (O) at (0,0);
\coordinate (X) at (\xmax+0.2,0);
\coordinate (T) at (0,\xmax+0.2);
\coordinate (X') at (\ang:\xmaxp+0.2);
\coordinate (T') at (90-\ang:\xmaxp+0.2);
% WORLD LINES GRID
\message{ Making world lines...^^J}
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line] ( \x,-\ymin) -- ( \x,\xmax);
\draw[world line t] (-\xmin, \x) -- (\xmax, \x);
}
\draw[world line] (-\d,-\ymin) -- (-\d,\xmax);
\draw[world line] (-2*\d,-\ymin) -- (-2*\d,\xmax);
\draw[world line] (-3*\d,-\ymin) -- (-3*\d,\xmax);
\draw[vector,mydarkred] (O) -- (\ang:\xmaxp)
node[mydarkred,right=0] {\contour{white}{$x'$}};
\draw[vector,mydarkred] (O) -- (90-\ang:\xmaxp)
node[mydarkred,left=0] {\contour{white}{$ct'$}};
% AXES
\draw[->,thick] (0,-\ymin) -- (T) node[left=-1] {$ct$};
\draw[->,thick] (-\xmin,0) -- (\xmax+0.2,0) node[below=0] {$x$};
% BOOSTED WORLD LINE GRID
\message{ Making world lines for boosted frame...^^J}
\fill[mydarkred,opacity=0.05]
(O) --++ (\ang:\xmaxp) --++ (90-\ang:\xmaxp) --++ (\ang:-\xmaxp) -- cycle;
% \fill[mydarkred,opacity=0.05]
% (O) --++ (\ang:-\xmaxp) --++ (90-\ang:-\xmaxp) --++ (\ang:\xmaxp) -- cycle;
\foreach \i [evaluate={\x=\i*\D;}] in {1,...,4}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
% \draw[world line'] (\ang:-\x) --++ (90-\ang:-\xmaxp);
% \draw[world line'] (90-\ang:-\x) --++ (\ang:-\xmaxp);
\draw[world line'] (\ang:\x) --++ (90-\ang:\xmaxp);
\draw[world line'] (90-\ang:\x) --++ (\ang:\xmaxp);
}
\fill[myred,opacity=1]
(O) --++ (\ang:\D) --++ (90-\ang:\xmaxp) --++ (\ang:-\D) -- cycle;
\end{tikzpicture}\end{center}\end{solution}
\problem{}
In the rest frame, what is the length of the stick? This should require only some simple trigonometry.
{\em Hint: pick a particular time in the rest frame and look at where the stick is at that particular time.}
Generalize this to a general reference frame moving at speed $ck$.
Suppose that the stick has a length $\ell'$ in the reference frame of the train moving at speed $ck$. What is the length, $\ell$, of the stick in the rest frame?
Physically, what does this mean? What happens to the stick as it moves faster?
\begin{solution}
$\ell$ is the length of the stick on the ground, $\ell'$ is the length of the stick on the train. For a speed of $ck$, the $x'$ axis has a slope of $k$. We then calculate
\begin{align*}
\ell^2 + k^2 \ell^2 & = (\ell')^2\\
\ell & = \frac{\ell'}{\sqrt{1 + k^2}}\\
\end{align*}
\end{solution}
\problem{}
Now suppose that you measure out $t'$ seconds of time on the train which is moving at a speed of $kc$. When you start your clock, you yell "START" and when you stop your clock, you yell "STOP". How long between the two yells would someone at rest measure? What does this imply physically?
\begin{solution}
Let $t'$ be the seconds on the train and $t$ be the seconds in the rest frame. The same calculation as the length contraction implies that
\begin{align*}
(ct)^2 + (ckt)^2 & = (ct')^2 \\
t & = \frac{t'}{\sqrt{1 + k^2}} \\
\end{align*}
\end{solution}

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% use [nosolutions] flag to hide solutions.
% use [solutions] flag to show solutions.
\documentclass[
solutions,
singlenumbering,
shortwarning
]{../../../lib/tex/ormc_handout}
\def\emptydiagrams{0} % set this to 1 to generate empty diagrams, 0 removes all empty diagrams
\include{matt}
\usepackage{../../../lib/tex/macros}
\usepackage{tikz}
\usetikzlibrary{patterns}
\usepackage{hyperref}
\usepackage{graphicx}
\usepackage{caption}
\usepackage{etoolbox} % ifthen
\usepackage[outline]{contour} % glow around text
\usetikzlibrary{calc} % for adding up coordinates
\usetikzlibrary{decorations.markings,decorations.pathmorphing}
\usetikzlibrary{angles,quotes} % for pic (angle labels)
\usetikzlibrary{arrows.meta} % for arrow size
\usepackage{xfp} % higher precision (16 digits?)
\contourlength{1.1pt}
\tikzset{>=latex} % for LaTeX arrow head
\colorlet{myred}{red!85!black}
\colorlet{mydarkred}{red!55!black}
\colorlet{mylightred}{red!85!black!12}
\colorlet{myfieldred}{mydarkred!5} % for S' background
\colorlet{myredhighlight}{myred!20} % highlights simultaneity in ladder paradox
\colorlet{myblue}{blue!80!black}
\colorlet{mydarkblue}{blue!50!black}
\colorlet{mylightblue}{blue!50!black!30}
\colorlet{mylightblue2}{myblue!10}
\colorlet{mygreen}{green!80!black}
\colorlet{mypurple}{blue!40!red!80!black}
\colorlet{mydarkgreen}{green!50!black}
\colorlet{mydarkpurple}{blue!40!red!50!black}
\colorlet{myorange}{orange!40!yellow!95!black}
\colorlet{mydarkorange}{orange!40!yellow!85!black}
\colorlet{mybrown}{brown!20!orange!90!black}
\colorlet{mydarkbrown}{brown!20!orange!55!black}
\colorlet{mypurplehighlight}{mydarkpurple!20} % highlights simultaneity in ladder paradox
\tikzstyle{world line}=[myblue!40,line width=0.3]
\tikzstyle{world line t}=[mypurple!50!myblue!40,line width=0.3]
\tikzstyle{world line'}=[mydarkred!40,line width=0.3]
\tikzstyle{mysmallarr}=[-{Latex[length=3,width=2]},thin]
\tikzstyle{mydashed}=[dash pattern=on 3 off 3]
\tikzstyle{rod}=[mydarkbrown,draw=mydarkbrown,double=mybrown,double distance=2pt,
line width=0.2,line cap=round,shorten >=1pt,shorten <=1pt]
%\tikzstyle{rod'}=[rod,draw=mydarkbrown!80!red!85,double=mybrown!80!red!85]
\tikzstyle{vector}=[->,line width=1,line cap=round]
\tikzstyle{vector'}=[vector,shorten >=1.2]
\tikzstyle{particle}=[mygreen,line width=0.9]
\tikzstyle{photon}=[-{Latex[length=5,width=4]},myorange,line width=0.8,decorate,
decoration={snake,amplitude=1.0,segment length=5,post length=5}]
\def\tick#1#2{\draw[thick] (#1) ++ (#2:0.06) --++ (#2-180:0.12)}
\def\tickp#1#2{\draw[thick,mydarkred] (#1) ++ (#2:0.06) --++ (#2-180:0.12)}
\def\Nsamples{100} % number samples in plot
% COMMON AXES
\pgfdeclarelayer{back} % to draw on background
\pgfsetlayers{back,main} % set order
\def\xmin{0.23}
\def\xmax{2}
\def\Nlines{6} % number of world lines (at constant x/t)
\def\DNxp{0} % difference in number of world lines of x' axis
\def\DNyp{0} % difference in number of world lines of ct' axis
\def\DNy{0} % difference in number of world lines of ct axis
\def\ang{20} % angle between x and x' axes
\def\xplabelang{180} % anchor angle of x' axis label
%\pgfmathsetmacro\ang{atan(0.44)} % angle between x and x' axes
\def\axes{
\pgfmathsetmacro\d{\xmax/(\Nlines+0.4)} % grid size
\pgfmathsetmacro\D{\d/cos(\ang)/sqrt(1-tan(\ang)^2)} % grid size, boosted
\pgfmathsetmacro\ymax{\xmax+\DNy*\d} % maximum of y = ct axis
\pgfmathsetmacro\xmaxp{(\xmax/\d+\DNxp)*\D} % maximum of x' axis
\pgfmathsetmacro\ymaxp{(\xmax/\d+\DNyp)*\D} % maximum of y' = ct' axis
\pgfmathsetmacro\Nylines{\Nlines+\DNy} % number of world lines at constant ct'
\pgfmathsetmacro\Nxplines{\Nlines+\DNxp} % number of world lines at constant x'
\pgfmathsetmacro\Nyplines{\Nlines+\DNyp} % number of world lines at constant ct'
\coordinate (O) at (0,0);
\coordinate (X) at (\xmax+0.15,0);
\coordinate (T) at (0,\ymax+0.15);
\coordinate (X') at (\ang:\xmaxp+0.2);
\coordinate (T') at (90-\ang:\ymaxp+0.2);
% FILL
\begin{pgfonlayer}{back} % draw on back
\fill[myfieldred]
(\ang:-\xmin) -- (\ang:\xmaxp) --++ (90-\ang:\ymaxp) --++ (\ang:-\xmaxp)
-- (90-\ang:-\xmin) -- cycle;
\end{pgfonlayer}
% WORLD LINE GRID
\message{ Making world lines...^^J}
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
%\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line] (\x,0) -- (\x,\ymax);
}
\foreach \i [evaluate={\t=\i*\d;}] in {1,...,\Nylines}{
%\message{ Running i/N=\i/\Nlines, t=\t...^^J}
\draw[world line t] (0,\t) -- (\xmax,\t);
}
% BOOSTED WORLD LINE GRID
\message{ Making world lines for boosted frame...^^J}
\foreach \i [evaluate={\x=\i*\D;}] in {1,...,\Nxplines}{
%\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line'] (\ang:\x) --++ (90-\ang:\ymaxp);
}
\foreach \i [evaluate={\t=\i*\D;}] in {1,...,\Nyplines}{
%\message{ Running i/N=\i/\Nlines, t=\t...^^J}
\draw[world line'] (90-\ang:\t) --++ (\ang:\xmaxp);
}
% AXES
\draw[->,thick] (0,-\xmin) -- (T) node[left=-1] {$ct$};
\draw[->,thick] (-\xmin,0) -- (X) node[below=0] {$x$};
\draw[->,thick,mydarkred] (90-\ang:-\xmin) -- (T')
node[right=5,above=-1] {$ct'$};
\draw[->,thick,mydarkred] (\ang:-\xmin) -- (X')
node[anchor=\xplabelang,inner sep=2] {$x'$};
}
\uptitlel{Advanced 2}
\uptitler{\smallurl{}}
\title{Special Relativity}
\subtitle{
Prepared by Matthew Kowalski on \today{}\\
All diagrams are adapted from Izaak Neutelings, see \url{https://tikz.net/relativity_minkowski_diagram/}
}
\begin{document}
\maketitle
%\input{parts/relative velocity}
\input{parts/spacetime diagrams}
\input{parts/galilean}
\input{parts/special}
%\input{parts/relative velocity}
\input{parts/simultaneity}
\input{parts/proper}
\input{parts/contraction}
\end{document}

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\newcommand{\emptydiagram}[1]{
\if\emptydiagrams1
\begin{center}\begin{tikzpicture}[scale=2]
\message{Lorentz boost^^J}
\def\xmax{2}
\def\xmaxp{2.2} % maximum of rotated axis
\def\Nlines{6} % number of world lines (at constant x/t)
\pgfmathsetmacro\ang{atan(1/2)} % angle between x and x' axes
\pgfmathsetmacro\d{0.9*\xmax/\Nlines}refer % grid size
\pgfmathsetmacro\D{\d/cos(\ang)/sqrt(1-tan(\ang)^2)} % grid size, boosted
\coordinate (O) at (0,0);
\coordinate (X) at (\xmax+0.2,0);
\coordinate (T) at (0,\xmax+0.2);
\coordinate (X') at (\ang:\xmaxp+0.2);
\coordinate (T') at (90-\ang:\xmaxp+0.2);
% WORLD LINE GRID
\message{ Making world lines...^^J}
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line] (-\x,-\xmax) -- (-\x,\xmax);
\draw[world line] ( \x,-\xmax) -- ( \x,\xmax);
\draw[world line t] (-\xmax,-\x) -- (\xmax,-\x);
\draw[world line t] (-\xmax, \x) -- (\xmax, \x);
}
% AXES
\draw[->,thick] (0,-\xmax) -- (T) node[left=-1] {#1: $t$};
\draw[->,thick] (-\xmax,0) -- (X) node[below=0] {$x$};
\end{tikzpicture}\end{center}
\fi
}
\newcommand{\emptydiagramc}[1]{
\if\emptydiagrams1
\begin{center}\begin{tikzpicture}[scale=2]
\message{Lorentz boost^^J}
\def\xmax{2}
\def\xmaxp{2.2} % maximum of rotated axis
\def\Nlines{6} % number of world lines (at constant x/t)
\pgfmathsetmacro\ang{atan(1/2)} % angle between x and x' axes
\pgfmathsetmacro\d{0.9*\xmax/\Nlines}refer % grid size
\pgfmathsetmacro\D{\d/cos(\ang)/sqrt(1-tan(\ang)^2)} % grid size, boosted
\coordinate (O) at (0,0);
\coordinate (X) at (\xmax+0.2,0);
\coordinate (T) at (0,\xmax+0.2);
\coordinate (X') at (\ang:\xmaxp+0.2);
\coordinate (T') at (90-\ang:\xmaxp+0.2);
% WORLD LINE GRID
\message{ Making world lines...^^J}
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line] (-\x,-\xmax) -- (-\x,\xmax);
\draw[world line] ( \x,-\xmax) -- ( \x,\xmax);
\draw[world line t] (-\xmax,-\x) -- (\xmax,-\x);
\draw[world line t] (-\xmax, \x) -- (\xmax, \x);
}
% AXES
\draw[->,thick] (0,-\xmax) -- (T) node[left=-1] {#1: $ct$};
\draw[->,thick] (-\xmax,0) -- (X) node[below=0] {$x$};
\end{tikzpicture}\end{center}
\fi
}
\newcommand{\halfdiagramc}[1]{
\if\emptydiagrams1
\begin{center}\begin{tikzpicture}[scale=2]
\message{Lorentz boost^^J}
\def\xmax{2}
\def\xmaxp{2.2} % maximum of rotated axis
\def\Nlines{6} % number of world lines (at constant x/t)
\pgfmathsetmacro\ang{atan(1/2)} % angle between x and x' axes
\pgfmathsetmacro\d{0.9*\xmax/\Nlines}refer % grid size
\pgfmathsetmacro\D{\d/cos(\ang)/sqrt(1-tan(\ang)^2)} % grid size, boosted
\coordinate (O) at (0,0);
\coordinate (X) at (\xmax+0.2,0);
\coordinate (T) at (0,\xmax+0.2);
\coordinate (X') at (\ang:\xmaxp+0.2);
\coordinate (T') at (90-\ang:\xmaxp+0.2);
% WORLD LINE GRID
\message{ Making world lines...^^J}
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line] (-\x,-0.2) -- (-\x,\xmax);
\draw[world line] ( \x,-0.2) -- ( \x,\xmax);
%\draw[world line t] (-\xmax,-\x) -- (\xmax,-\x);
\draw[world line t] (-\xmax, \x) -- (\xmax, \x);
}
% AXES
\draw[->,thick] (0,-0.2) -- (T) node[left=-1] {#1: $ct$};
\draw[->,thick] (-\xmax,0) -- (X) node[below=0] {$x$};
\end{tikzpicture}\end{center}
\fi
}
\newcommand{\halfdiagramcwide}[1]{
\if\emptydiagrams1
\begin{center}\begin{tikzpicture}[scale=2]
\message{Lorentz boost^^J}
\def\xmax{3.5}
\def\xmaxp{2.2} % maximum of rotated axis
\def\Nlines{10} % number of world lines (at constant x/t)
\pgfmathsetmacro\ang{atan(1/2)} % angle between x and x' axes
\pgfmathsetmacro\d{0.9*\xmax/\Nlines}refer % grid size
\pgfmathsetmacro\D{\d/cos(\ang)/sqrt(1-tan(\ang)^2)} % grid size, boosted
\coordinate (O) at (0,0);
\coordinate (X) at (\xmax+0.2,0);
\coordinate (T) at (0,1.95);
\coordinate (X') at (\ang:\xmaxp+0.2);
\coordinate (T') at (90-\ang:\xmaxp+0.2);
% WORLD LINE GRID
\message{ Making world lines...^^J}
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line] (-\x,-0.2) -- (-\x,1.95);
\draw[world line] ( \x,-0.2) -- ( \x,1.95);
%\draw[world line t] (-\xmax,-\x) -- (\xmax,-\x);
%\draw[world line t] (-\xmax, \x) -- (\xmax, \x);
}
\message{ Making world lines...^^J}
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,5}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
% \draw[world line] (-\x,-0.2) -- (-\x,\xmax);
% \draw[world line] ( \x,-0.2) -- ( \x,\xmax);
%\draw[world line t] (-\xmax,-\x) -- (\xmax,-\x);
\draw[world line t] (-\xmax, \x) -- (\xmax, \x);
}
% AXES
\draw[->,thick] (0,-0.2) -- (T) node[left=-1] {#1: $ct$};
\draw[->,thick] (-\xmax,0) -- (X) node[below=0] {$x$};
\end{tikzpicture}\end{center}
\fi
}

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\section{Length Contraction}
With proper time and distance done, we can now tackle length contraction easily! Length contraction is weird because different parts of our object will now be experiencing different times.
\problem{}
Suppose that you (at rest) see a rod of length $L$ moving at speed $v$.
\halfdiagramc{you}
\begin{enumerate}
\item Using the provided grid, draw a spacetime diagram where the left side of the rod is at $x = 0$ at $t = 0$.
\item When we switch to the rod's reference frame, space gets rotated. Draw a line in the rod's reference frame which represents the rod at time $t' = 0$, when the left side is at $x' = 0$. Call the right side of the rod at this time $P$.
\item Switching back to your reference frame, what are the spacetime coordinates $(ct,x)$ of $P$?
\vfill
\item Use the formula for proper distance to compute the length $L'$ of the rod in its own reference frame.
\vfill
\item Which is larger, $L'$ or $L$? Do moving objects shrink?
\vspace{30pt}
\end{enumerate}
\begin{solution}
\begin{center}\begin{tikzpicture}[scale=1]
\message{Ladder paradox^^J}
%\def\R{2*\xmax} % radius of clip
%\clip (-\xmin,\R) |- (\R,-\xmin) arc(0:90:\xmin+\R);
% AXES
\def\xmin{0.2}
\def\xmax{2.6}
\def\ang{33.5} % angle between x and x' axes
\def\Nlines{12} % number of world lines (at constant x/t)
\def\DNy{1} % difference in number of world lines of y axis (lengthen)
\def\DNxp{-6} % difference in number of world lines of x' axis (shorten)
\def\DNyp{-2} % difference in number of world lines of y' axis (shorten)
\def\xplabelang{170} % anchor angle of x' axis label
\axes
% SETTINGS
\pgfmathsetmacro\Lz{4*\D} % proper/rest length L0 of ladder in S'
\pgfmathsetmacro\L{cos(2*\ang)/cos(\ang)*\Lz} % contracted length L in S
\pgfmathsetmacro\yminb{-0.7*\xmin} % ymin of barn in S
\pgfmathsetmacro\xb{4.96*\d} % x coordinate of barn in S
\pgfmathsetmacro\wb{3.08*\d} % width of barn in S
\pgfmathsetmacro\yA{(\xb+0.04*\d)/tan(\ang)} % y = ct coordinate when ladder is fully in barn in S
\coordinate (L) at (\L,0); % ladder end in S
\coordinate (L') at (\ang:\Lz); % ladder end in S'
\coordinate (A) at (90-\ang:{(\xb+0.04*\d)/sin(\ang)}); % left end of ladder when fully in barn
\coordinate (B) at ($(A)+(\L,0)$); % right end of ladder when fully in barn
\coordinate (C) at (90-\ang:{(\xb+\wb+0.08*\d)/sin(\ang)}); % left end of ladder when fully passed through barn
% FILL
\begin{pgfonlayer}{back} % draw on back (behind axes)
% \fill[mydarkblue!22] % barn frame
% (\xb,\yminb) rectangle (\xb+\wb,\ymax);
\fill[mylightred] % ladder frame
(90-\ang:-\xmin) -- (90-\ang:\ymaxp) --++ (\ang:\Lz) -- (L) --++ (90-\ang:-\xmin) -- cycle;
% \begin{scope}
% \clip (0,0) rectangle(1.3*\xmax,\ymax+0.2);
% \draw[myredhighlight,line width=3.1] % highlight simultaneity in S'
% (A)++(\ang:{-\xb/cos(\ang)-0.05}) --++ (\ang:\xmax+3.6*\D)
% (B)++(\ang:{-\xb/cos(\ang)-0.05-\Lz}) --++ (\ang:\xmax+3.8*\D);
% \draw[mypurplehighlight,line width=3.1] % highlight simultaneity in S
% (0,\yA) --++ (\xmax+0.8*\d,0);
% \end{scope}
\end{pgfonlayer}
% \draw[->,thick,mydarkblue] % barn left door
% (\xb,\yminb) -- (\xb,\ymax+0.15);
% \draw[->,thick,mydarkblue] % barn right door
% (\xb+\wb,\yminb) -- (\xb+\wb,\ymax+0.15);
\draw[->,thick,mydarkbrown] % rod left end
(L)++(90-\ang:-\xmin) -- (L) -- (L') --++ (90-\ang:\ymaxp+0.2);
% LADDER
\draw[rod] (O) -- (L')
node[pos=0.55,above=2,scale=0.8] {\contour{mylightred}{$L'$}};
\draw[rod] (O) -- (L)
node[pos=0.46,below=1,scale=0.8] {$L$};
% % LADDER IN BARN
% \draw[rod] (A) --++ (L');
% \draw[rod] (B) --++ (\ang:-\Lz);
% \draw[rod] (A) --++ (L);
% % LADDER RIGHT OF BARN
% \draw[rod] (C) --++ (L');
% \draw[rod] (C) --++ (L);
% LABELS
% \node[mydarkblue,below=0,align=center,scale=0.8,yshift=1] at (\xb+\wb/2,0)
% {barn\\$L$};
% \node[mydarkpurple,right,align=left,scale=0.65,yshift=1.2] at (\xb+3.6*\d,\yA)
% {both doors closed in S};
% %{both doors\\[-3]close in S};
% \node[mydarkred,right,scale=0.65,yshift=1.8,rotate=\ang] at ($(A)+(\ang:\Lz+0.8*\D)$)
% {left door closed in S$'$};
% \node[mydarkred,right,scale=0.65,yshift=0.7,rotate=\ang] at ($(B)+(\ang:0.8*\D)$)
% {right door closed in S$'$};
\end{tikzpicture}\end{center}
We want to calculate the spacetime position of the right side side of the rod when the left side is at the origin.
In the rest frame, the equation for the right edge of the rod is $x = vt + L$.
The equation for the spatial axis of the rod's rest frame is $c^2t = vx$.
This implies that they intersect at $ct = \frac{cvL}{c^2 - v^2}$ and $x = \frac{L}{1-v^2/c^2}$.
Calculating the proper length between the left and right side of the rod, we find that
\begin{align*}
L' = \chi = \sqrt{\frac{c^4L^2}{(c^2 - v^2)^2} - \frac{c^2 v^2 L^2}{(c^2 - v^2)}} = L\sqrt{\frac{c^2(c^2 - v^2)}{(c^2 - v^2)^2}} = \frac{L}{\sqrt{1 - v^2/c^2}}
\end{align*}
We note that this implies $L' > L$. i.e., moving objects get shorter
\end{solution}
\pagebreak
\problem{Ladder paradox}
Aiden and Matt have a ladder of length $2L$ that they are trying to squeeze into a barn of length $L$. Suppose that the barn has a front door and back door which can be open/closed simultaneously.
Now, Aiden is particular smart, so he gives Matt the ladder and has Matt run at the barn at speed $v = \sqrt{3}c/2$. Does the ladder fit in the barn? We'll analyze this in the new few questions.
\begin{enumerate}
\item
What is the length of the ladder from Aiden's perspective when Matt is running? Does the ladder fit in the barn?
\begin{solution}
$L$
\end{solution}
\vfill
\item As soon as Matt and the ladder are inside the barn, Aiden quickly closes and opens the doors of the barn. Success!
However, consider this from Matt's perspective. From Matt's perspective, he's holding a ladder of length $2L$ and a barn is flying at him at speed $v = \sqrt{3}c/2$. By length contraction, what is the length of the barn?
\begin{solution}
$L/2$
\end{solution}
\vfill\pagebreak
\item Despite the barn being too short, we know that the ladder has to fit! Using the provided grid, draw a spacetime diagram of the situation. Include Matt's reference frame on your diagram.
\emptydiagramc{Aiden}
\vfill
\item From Matt's perspective, why don't the doors of the barn crush the ladder?
\end{enumerate}
\begin{solution}
In short, from Matt's perspective, the barn doors do not close at the same time. The back door closes right when the front of the ladder reaches it, then opens again. Later, the front door closes right when the back of the ladder passes it, then opens again.
% SPACETIME DIAGRAM - LADDER PARADOX
\begin{tikzpicture}[scale=2.5]
\message{Ladder paradox^^J}
%\def\R{2*\xmax} % radius of clip
%\clip (-\xmin,\R) |- (\R,-\xmin) arc(0:90:\xmin+\R);
% AXES
\def\xmin{0.2}
\def\xmax{2.6}
\def\ang{33.5} % angle between x and x' axes
\def\Nlines{12} % number of world lines (at constant x/t)
\def\DNy{1} % difference in number of world lines of y axis (lengthen)
\def\DNxp{-6} % difference in number of world lines of x' axis (shorten)
\def\DNyp{-2} % difference in number of world lines of y' axis (shorten)
\def\xplabelang{170} % anchor angle of x' axis label
\axes
% SETTINGS
\pgfmathsetmacro\Lz{4*\D} % proper/rest length L0 of ladder in S'
\pgfmathsetmacro\L{cos(2*\ang)/cos(\ang)*\Lz} % contracted length L in S
\pgfmathsetmacro\yminb{-0.7*\xmin} % ymin of barn in S
\pgfmathsetmacro\xb{4.96*\d} % x coordinate of barn in S
\pgfmathsetmacro\wb{3.08*\d} % width of barn in S
\pgfmathsetmacro\yA{(\xb+0.04*\d)/tan(\ang)} % y = ct coordinate when ladder is fully in barn in S
\coordinate (L) at (\L,0); % ladder end in S
\coordinate (L') at (\ang:\Lz); % ladder end in S'
\coordinate (A) at (90-\ang:{(\xb+0.04*\d)/sin(\ang)}); % left end of ladder when fully in barn
\coordinate (B) at ($(A)+(\L,0)$); % right end of ladder when fully in barn
\coordinate (C) at (90-\ang:{(\xb+\wb+0.08*\d)/sin(\ang)}); % left end of ladder when fully passed through barn
% FILL
\begin{pgfonlayer}{back} % draw on back (behind axes)
\fill[mydarkblue!22] % barn frame
(\xb,\yminb) rectangle (\xb+\wb,\ymax);
\fill[mylightred] % ladder frame
(90-\ang:-\xmin) -- (90-\ang:\ymaxp) --++ (\ang:\Lz) -- (L) --++ (90-\ang:-\xmin) -- cycle;
\begin{scope}
\clip (0,0) rectangle(1.3*\xmax,\ymax+0.2);
\draw[myredhighlight,line width=3.1] % highlight simultaneity in S'
(A)++(\ang:{-\xb/cos(\ang)-0.05}) --++ (\ang:\xmax+3.6*\D)
(B)++(\ang:{-\xb/cos(\ang)-0.05-\Lz}) --++ (\ang:\xmax+3.8*\D);
\draw[mypurplehighlight,line width=3.1] % highlight simultaneity in S
(0,\yA) --++ (\xmax+0.8*\d,0);
\end{scope}
\end{pgfonlayer}
\draw[->,thick,mydarkblue] % barn left door
(\xb,\yminb) -- (\xb,\ymax+0.15);
\draw[->,thick,mydarkblue] % barn right door
(\xb+\wb,\yminb) -- (\xb+\wb,\ymax+0.15);
\draw[->,thick,mydarkbrown] % rod left end
(L)++(90-\ang:-\xmin) -- (L) -- (L') --++ (90-\ang:\ymaxp+0.2);
% LADDER
\draw[rod] (O) -- (L')
node[pos=0.55,above=2,scale=0.8] {\contour{mylightred}{$2L$}};
\draw[rod] (O) -- (L)
node[pos=0.46,below=1,scale=0.8] {$L$};
% LADDER IN BARN
\draw[rod] (A) --++ (L');
\draw[rod] (B) --++ (\ang:-\Lz);
\draw[rod] (A) --++ (L);
% LADDER RIGHT OF BARN
\draw[rod] (C) --++ (L');
\draw[rod] (C) --++ (L);
% LABELS
\node[mydarkblue,below=0,align=center,scale=0.8,yshift=1] at (\xb+\wb/2,0)
{barn\\$L$};
\node[mydarkpurple,right,align=left,scale=0.65,yshift=1.2] at (\xb+3.6*\d,\yA)
{both doors closed in S};
%{both doors\\[-3]close in S};
\node[mydarkred,right,scale=0.65,yshift=1.8,rotate=\ang] at ($(A)+(\ang:\Lz+0.8*\D)$)
{left door closed in S$'$};
\node[mydarkred,right,scale=0.65,yshift=0.7,rotate=\ang] at ($(B)+(\ang:0.8*\D)$)
{right door closed in S$'$};
\end{tikzpicture}
% SPACETIME DIAGRAM - LADDER PARADOX from perspective of S' (i.e. in the S' frame)
% \begin{tikzpicture}[scale=2.5]
% \message{Ladder paradox from the perspective of S'^^J}
% % SETTINGS
% \def\ang{-33.5} % angle between x and x' axes
% \def\Nxlines{9} % number of world lines (at constant x)
% \def\Nylines{13} % number of world lines (at constant t)
% \def\Nxplines{6} % number of world lines (at constant x')
% \def\Nyplines{10} % number of world lines (at constant t')
% \def\xmin{0.2}
% \pgfmathsetmacro\D{2.6/13} % grid size
% \pgfmathsetmacro\d{\D/cos(\ang)/sqrt(1-tan(\ang)^2)} % grid size, boosted
% \pgfmathsetmacro\xmax{(\Nxlines+0.4)*\d} % maximum of x axis in S
% \pgfmathsetmacro\ymax{(\Nylines+0.4)*\d} % maximum of y = ct axis in S
% \pgfmathsetmacro\xmaxp{(\Nxplines+0.4)*\D} % maximum of x' axis in S'
% \pgfmathsetmacro\ymaxp{(\Nyplines+0.4)*\D} % maximum of y' = ct' axis in S'
% \pgfmathsetmacro\Lz{4*\D} % proper/rest length L0 of ladder in S'
% \pgfmathsetmacro\L{\Lz/cos(\ang)} % contracted length L in S
% \pgfmathsetmacro\xb{4.96*\d} % x coordinate of barn in S
% \pgfmathsetmacro\wb{3.08*\d} % width of barn in S
% \pgfmathsetmacro\yAp{(\xb+0.04*\d)*sin(\ang)*(1-cot(\ang)^2)} % y' = ct' coordinate when ladder is fully in barn in S
% \pgfmathsetmacro\yBp{\yAp+\L*sin(\ang)} % y' = ct' coordinate when ladder is fully in barn in S
% \coordinate (O) at (0,0);
% \coordinate (X) at (\ang:\xmax+0.05);
% \coordinate (T) at (90-\ang:\ymax+0.05);
% \coordinate (X') at (\xmaxp+0.15,0);
% \coordinate (T') at (0,\ymaxp+0.15);
% \coordinate (L) at (\ang:\L); % ladder end in S
% \coordinate (L') at (\Lz,0); % ladder end in S'
% \coordinate (A) at (0,\yAp); % left end of ladder when fully in barn in S
% \coordinate (B) at ($(A)+(L)$); % right end of ladder when fully in barn in S
% \coordinate (C) at (0,{(\xb+\wb+0.08*\d)*sin(\ang)*(1-cot(\ang)^2)}); % left end of ladder when fully passed through barn
% % FILL
% \fill[myfieldred]
% (-\xmin,0) -| (\xmaxp,\ymaxp) -| (0,-\xmin) -| cycle;
% \fill[mylightred] % ladder frame
% (0,-\xmin) |- (\Lz,\ymaxp) -- ($(L)+(0,-\xmin)$) -- cycle;
% \fill[mydarkblue!22] % barn frame
% (\ang:\xb)++(90-\ang:-\xmin) --++ (90-\ang:\xmin+\ymax)
% --++ (\ang:\wb) --++ (90-\ang:-\xmin-\ymax) -- cycle;
% % HIGHLIGHT DOORS OPEN/CLOSED
% \begin{scope}
% \clip (0,0) --++ (90-\ang:\ymax) -- (\xmaxp+1.8*\d,\ymaxp) --++ (0,-1.1*\ymaxp) -- cycle;
% \draw[myredhighlight,line width=3.1] % highlight simultaneity in S'
% ({\yAp*tan(\ang)-0.1},\yAp) -- (\xmaxp+1.6*\d,\yAp)
% ({\yBp*tan(\ang)-0.1},\yBp) -- (\xmaxp+1.8*\d,\yBp);
% \draw[mypurplehighlight,line width=3.1] % highlight simultaneity in S
% (A)++(\ang:-\xb-0.1) --++ (\ang:{\xb+\L+3.75*\d});
% \end{scope}
% % BOOSTED WORLD LINE GRID
% \message{ Making world lines for boosted frame...^^J}
% \foreach \i [evaluate={\x=\i*\D;}] in {1,...,\Nxplines}{
% \draw[world line] (\x,0) -- (\x,\ymaxp);
% }
% \foreach \i [evaluate={\t=\i*\D;}] in {1,...,\Nyplines}{
% \draw[world line t] (0,\t) -- (\xmaxp,\t);
% }
% % WORLD LINE GRID
% \message{ Making world lines...^^J}
% \foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nxlines}{
% \draw[world line'] (\ang:\x) --++ (90-\ang:\ymax);
% }
% \foreach \i [evaluate={\t=\i*\d;}] in {1,...,\Nylines}{
% \draw[world line'] (90-\ang:\t) --++ (\ang:\xmax);
% }
% % WORLD LINES BARN & ROD
% \draw[->,thick,mydarkblue] % barn left door
% (\ang:\xb+\wb)++(90-\ang:-\xmin) --++ (90-\ang:\xmin+\ymax+0.2);
% \draw[->,thick,mydarkblue] % barn right door
% (\ang:\xb)++(90-\ang:-\xmin) --++ (90-\ang:\xmin+\ymax+0.2);
% \draw[->,thick,mydarkbrown] % rod right end
% (L)++(0,-\xmin) -- (\Lz,\ymaxp+0.15);
% % AXES
% \draw[->,thick] (90-\ang:-\xmin) -- (T) node[below left=-1] {$ct$};
% \draw[->,thick] (\ang:-\xmin) -- (X) node[below left=0] {$x$};
% \draw[->,thick,mydarkred] (0,-\xmin) -- (T')
% node[right=3,above=-1] {$ct'$};
% \draw[->,thick,mydarkred] (-\xmin,0) -- (X')
% node[anchor=140,inner sep=0.5] {$x'$};
% % LADDER
% \draw[rod] (O) -- (L)
% node[pos=0.45,below=2,scale=0.8] {$L$};
% \draw[rod] (O) -- (L')
% node[pos=0.45,above=0.6,scale=0.8] {\contour{mylightred}{$2L$}};
% \draw[rod] (O) -- (L');
% % LADDER IN BARN
% \draw[rod] (A) --++ (L);
% \draw[rod] (A) --++ (L');
% \draw[rod] (B) --++ (-\Lz,0);
% % LADDER RIGHT OF BARN
% \draw[rod] (C) --++ (L');
% \draw[rod] (C) --++ (L);r
% % LABELS
% \node[mydarkbrown,above=1,scale=0.8] at (\Lz/2,\ymaxp)
% {rod};
% \node[mydarkblue,below=0,scale=0.8]
% at ({(\xb+\wb/2)*cos(\ang)*(1-tan(\ang)^2)+0.07},0)
% {\contour{mydarkblue!22}{barn}};
% %\node[mydarkblue,anchor=90-\ang,inner sep=2,scale=0.8,rotate=\ang] at (\ang:\xb+\L/2)
% % {barn}; %{barn\\$L$};
% \node[mydarkpurple,right,align=left,scale=0.65,yshift=1.6,rotate=\ang]
% at ($(B)+(\ang:0.4*\d)$) {both doors closed in S};
% \node[mydarkred,right,scale=0.65,yshift=1.8] at ($(A)+(\Lz+0.3*\d,0)$)
% {left door closed in S$'$};
% \node[mydarkred,right,scale=0.65,yshift=0.7] at ($(B)+(0.3*\d,0)$)
% {right door closed in S$'$};
% \end{tikzpicture}
\end{solution}
\vfill\pagebreak

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\section{Galilean Relativity}
Much like you can watch your pets scatter from the perspective of a train, we can watch the world from anyone's perspective. When we shift perspective like this, just using our normal intuition, we call this {\em Galilean relativity}.
\example{}
Consider the situation of Example $1$ again, but now from the perspective of your cat.
From your cat's perspective, she's the one staying still and you're the one walking away, only now you're walking away at speed $1$ to the left. We'll denote our new spatial variable with $x'$.
\begin{center}
% SPACETIME DIAGRAM with WORLD LINES
\begin{tikzpicture}[scale=2.0]
\message{Worldlines^^J}
\def\ymin{0.2}
\def\xmin{2}
\def\xmax{2}
\def\Nlines{4} % number of world lines (at constant x/t)
\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
\coordinate (O) at (0,0);
\coordinate (T) at (0,\xmax+0.2);
% WORLD LINES GRID
\message{ Making world lines...^^J}
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line] ( \x,-\ymin) -- ( \x,\xmax);
\draw[world line t] (-\xmin, \x) -- (\xmax, \x);
}
\draw[world line] (-\d,-\ymin) -- (-\d,\xmax);
\draw[world line] (-2*\d,-\ymin) -- (-2*\d,\xmax);
\draw[world line] (-3*\d,-\ymin) -- (-3*\d,\xmax);
\draw[world line] (-4*\d,-\ymin) -- (-4*\d,\xmax);
% AXES
\draw[->,thick] (0,-\ymin) -- (T) node[left=-1] {$t$};
\draw[->,thick] (-\xmin,0) -- (\xmax+0.2,0) node[below=0] {$x'$};
% VECTORS
\draw[vector,myred, very thick] (O) -- (0,4*\d)
node[mydarkred,above right] {\contour{white}{cat: $x'(t)=0$}};
\draw[vector,myblue, very thick] (O) -- (-4*\d,4*\d)
node[mydarkblue,below left=0] {\contour{white}{you: $x'(t)=-t$}};
%\node[right=8,above,mydarkpurple] at (T) {$x(t)=0$};
\end{tikzpicture}
\end{center}
\problem{}
Draw the situation from \ref{pets scatter} in your cat's perspective.\\
What if we drew the situation from \ref{pets scatter train} in your cat's perspective?\\
Would there be any change when the cat is on the train? Why or why not?
\emptydiagram{Cat}
\begin{solution}
\begin{center}
\begin{tikzpicture}[scale=2.0]
\message{Worldlines^^J}
\def\ymin{0.2}
\def\xmin{4}
\def\xmax{2}
\def\Nlines{4} % number of world lines (at constant x/t)
\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
\coordinate (O) at (0,0);
\coordinate (T) at (0,\xmax+0.2);
% WORLD LINES GRID
\message{ Making world lines...^^J}
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line] ( \x,-\ymin) -- ( \x,\xmax);
\draw[world line t] (-\xmin, \x) -- (\xmax, \x);
}
\draw[world line] (-\d,-\ymin) -- (-\d,\xmax);
\draw[world line] (-2*\d,-\ymin) -- (-2*\d,\xmax);
\draw[world line] (-3*\d,-\ymin) -- (-3*\d,\xmax);
\draw[world line] (-4*\d,-\ymin) -- (-4*\d,\xmax);
\draw[world line] (-5*\d,-\ymin) -- (-5*\d,\xmax);
\draw[world line] (-6*\d,-\ymin) -- (-6*\d,\xmax);
\draw[world line] (-7*\d,-\ymin) -- (-7*\d,\xmax);
\draw[world line] (-8*\d,-\ymin) -- (-8*\d,\xmax);
% AXES
\draw[->,thick] (0,-\ymin) -- (T) node[left=-1] {$t$};
\draw[->,thick] (-\xmin,0) -- (\xmax+0.2,0) node[below=0] {$x$};
% VECTORS
\draw[vector,myred] (O) -- (-8*\d,4*\d)
node[mydarkred,below left=0] {\contour{white}{you: $x(t)=-2t$}};
\draw[vector,myblue] (O) -- (0,4*\d)
node[mydarkblue,above left=0] {\contour{white}{cat: $x(t)=0$}};
\draw[vector,mygreen] (O) -- (-9*\d,3*\d)
node[mydarkgreen,below left=0] {\contour{white}{dog: $x(t)=-3t$}};
\draw[vector,black] (O) -- (-2*\d,1*\d) -- (-6*\d, 2*\d) -- (-4*\d, 4*\d)
node[black,below right=0] {\contour{white}{hamster}};
% \draw[vector,myblue]
% (O) to[out=35,in=-100] (O)
% to[out=80,in=-80,looseness=1.5] (0.3*\xmax,4*\d)
% node[mydarkblue,above=-3] {\contour{white}{cat: $x(t)$}};
%\node[right=8,above,mydarkpurple] at (T) {$x(t)=0$};
\end{tikzpicture}
\end{center}
\end{solution}
\newpage
\definition{Reference Frame}
When we view the world from the perspective of different objects, we say that we are working in different {\em reference frames}. The original Example 1, where you are stationary, is your reference frame. The new plot in Example 5, where your cat is stationary, is your cat's reference frame.
If we want to compare what is happening in multiple reference frames at once, we can graph multiple spacetime grids on one plot. If we overlay the cat's reference frame onto your reference frame, we can visualize everything in Example 1 as:
% SPACETIME DIAGRAM - GALILEAN TRANSFORMATION
\begin{center}\begin{tikzpicture}[scale=1.8]
\message{Galilean transformation^^J}
\def\xmax{2}
\def\xmaxp{2.1} % maximum of rotated axis
\def\Nlines{4} % number of world lines (at constant x/t)
\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
\pgfmathsetmacro\ang{atan(1)} % angle
\coordinate (O) at (0,0);
\coordinate (X) at (\xmax+0.2,0);
\coordinate (T) at (0,\xmax+0.2);
\coordinate (X') at (\ang:\xmaxp+0.2);
\coordinate (T') at (90-\ang:\xmaxp+0.2);
% WORLD LINES GRID
\message{ Making world lines...^^J}
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line] (-\x,-\xmax) -- (-\x,\xmax);
\draw[world line] ( \x,-\xmax) -- ( \x,\xmax);
\draw[world line t] ({-\xmax-tan(\ang)*\x},-\x) -- (\xmax,-\x);
\draw[world line t] (-\xmax,\x) -- ({\xmax+tan(\ang)*\x},\x);
}
% AXES
\draw[->,thick] (0,-\xmax) -- (T) node[left=0] {$t$};
\draw[->,thick] (-\xmax,0) -- (X) node[right=6,below=-1] {$x={\color{mydarkred}x'}$};
\draw[->,thick,mydarkred, very thick] (90-\ang:-\xmaxp) -- (T')
node[left=-1] {$t'$}
node[right=2,below right=-2] {cat: $x(t) = t$};
% VECTORS
\draw[vector,myblue, very thick] (O) -- (0,4*\d)
node[mydarkblue,below left=0] {\contour{white}{you: $x(t)=0$}};
% WORLD LINES GRID - BOOSTED
\message{ Making world lines, boosted...^^J}
\fill[mydarkred,opacity=0.05]
(O) --++ (90-\ang:\xmax) --++ (\xmax,0) --++ (90-\ang:-\xmax) -- cycle;
\fill[mydarkred,opacity=0.05]
(O) --++ (90-\ang:-\xmax) --++ (-\xmax,0) --++ (90-\ang:\xmax) -- cycle;
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line'] (\x,0) --++ (90-\ang:\xmax);
\draw[world line'] (-\x,0) --++ (90-\ang:-\xmax);
}
\node[right] at (5.6*\d, \d) {$x=\color{mydarkred}x' + 1$};
%\draw pic[<-,"$\theta$",draw=black,angle radius=34,angle eccentricity=1.2] {angle = T'--O--T};
\end{tikzpicture}\end{center}
Here $x,t$ are the spacetimes coordinates in your perspective and $x',t'$ are the spacetime coordinates in your cat's perspective. Note that $t = t'$ for any point while $x = x' + t$.
\problem{}
What does it mean for two events to lie on the same vertical (blue) line from your perspective?
What does it mean for two events to lie on the same slanted (red) line from your cat's perspective?
\vfill
\begin{solution}
Two events lie on the same vertical line if they occur at the same location in your reference frame.
Two events lie on the same slanted line if they occur at the same location in your cat's reference frame.
\end{solution}
% \remark{}
% Here we are forcing time in the cat's reference frame to behave the same as in our reference frame. As in, one second for the cat is one second for us and vice versa. However, looking at the plot and measuring distances, it almost looks like one second for us is longer than one second for the cat... Suspicious...
\problem{}
In the situation from Problem 2, when will your hamster catch up to your cat? \\
Choose the most convenient reference frame to work in, you shouldn't have to do much math.
\begin{solution}
Using the cat's reference frame drawn in Problem 6, the hamster will catch up to the cat at $t = 8$.
\end{solution}
\vfill
\pagebreak

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\section{Proper Time and Proper Length}
We mentioned that the space and time axes also get rescaled during a Lorentz boost. Let's figure out by how much! To do this, we're going to use the only consistent tool at our disposal, the speed of light.
\problem{Time dilation}
Suppose that Alice remains stationary. In her reference frame, Bob moves to the right at speed $v$. Every quantity in Alice's reference frame will be denoted normally, i.e.~$x,t$. Every quantity in Bob's reference frame will be given a prime, i.e.~$x',t'$.
Bob holds onto a very special clock. The clock consists of a light bulb, a mirror distance $L$ away from the light (perpendicular to the direction Bob is moving) and a photosensor.
\begin{enumerate}
\item
Draw the experimental setup. Note again that the light bulb and mirror are separated in the $y$ direction, not the $x$.
\vfill
\item In Bob's reference frame (where the clock isn't moving), how long does it take the light to be emitted, travel to the mirror, bounce back, and be reabsorbed? Write this as $t'$.
\begin{solution}
$ct' = 2L \implies t' = 2L/c$
\end{solution}
\vfill\pagebreak
\item In Alice's reference frame, she still sees the light move, reflect off the mirror, then come back. How long does this take?
{\em Note: The mirror and photosensor are moving at speed $v$ during this process. Also, the light is now moving at an angle, but it still moves at $c$.}
\begin{solution}
Pythagorean theorem:
$$(ct)^2 = (vt)^2 + 4L^2 \implies t^2 = \frac{4L^2}{c^2 - v^2} \implies t = \frac{2L}{\sqrt{c^2 - v^2}}$$
\end{solution}
\vfill
\item Combine these two formulas to find a relationship between $t$ and $t'$.
If one unit of time elapses in Bob's reference frame, does more or less time elapse in Alice's reference frame? Who ages faster?
\begin{solution}
$$ct' = \sqrt{c^2 - v^2}t \implies t' = \sqrt{1 - v^2/c^2} t$$
Since $v < c$, $0 \leq v^2/c^2 <1$. Then $t' < t$.
Alice ages faster. The time that Bob experiences is less than Alice.
\end{solution}
\end{enumerate}
\vfill \pagebreak
\problem{Space dilation}
Since the speed of light is always constant, we can use time dilation to see what happens to our notion of distance. Suppose that a photon is emitted from one location and absorbed in another.
\begin{enumerate}
\item Suppose that in Alice's reference frame, the photon takes time $t$ to be emitted and absorbed. In this time, what distance $x$ does the photon travel?
\begin{solution}
$x = ct$
\end{solution}
\vfill
\item Suppose that Bob is traveling at speed $v$ relative to Alice. Using time dilation, how long does the photon take to be emitted and absorbed in Bob's reference frame? What distance $x'$ does the photon travel in that time?
{\em This will give you a formula relating distance $x'$ in Bob's reference frame to distance $x$ in Alice's reference frame.}
\begin{solution}
$x' = ct' = ct \sqrt{1 - v^2/c^2} = x \sqrt{1 - v^2/c^2}$
\end{solution}
\vfill
\end{enumerate}
\vfill \pagebreak
\remark{} Switching back and forth between reference frames gets confusing quickly, especially when you are consideration time and space dilation. To avoid this, we will define a new notion of time and space that doesn't care about which reference frame we're in. Since this will be universal, we'll call these {\em proper time}, denoted by $\tau$, and {\em proper distance}, denoted by $\chi$.
Recall our notion of past and future from earlier:
% SPACETIME DIAGRAM - LIGHT CONE
\begin{center}\begin{tikzpicture}[scale=1.8]
\message{Light cone^^J}
\def\xmax{2}
\def\xmaxp{2.2} % maximum of rotated axis
\def\Nlines{5} % number of world lines (at constant x/t)
\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
\pgfmathsetmacro\ang{atan(1/3)} % angle between x and x' axes
\coordinate (O) at (0,0);
\coordinate (X) at (\xmax+0.2,0);
\coordinate (T) at (0,\xmax+0.2);
% WORLD LINE GRID
\message{ Making world lines...^^J}
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line] (-\x,-\xmax) -- (-\x,\xmax);
\draw[world line] ( \x,-\xmax) -- ( \x,\xmax);
\draw[world line t] (-\xmax,-\x) -- (\xmax,-\x);
\draw[world line t] (-\xmax, \x) -- (\xmax, \x);
}
% AXES
\draw[->,thick] (0,-\xmax) -- (T) node[left=-1] {$ct$};
\draw[->,thick] (-\xmax,0) -- (X) node[below=0] {$x$};
% LABELS
%\draw pic[->,"$45^\circ$",draw=black,angle radius=23,angle eccentricity=1.38] {angle = X--O--C};
\node[mydarkorange,above right] at (0.1*\xmax,\xmax) {future light cone};
\node[mydarkorange,below] at (0,-\xmax) {past light cone};
% FILLS
\fill[myblue,opacity=0.05] % SPACELIKE
(\xmax,\xmax) -- (-\xmax,-\xmax) -- (-\xmax,\xmax) -- (\xmax,-\xmax) -- cycle;
\fill[myorange,opacity=0.05] % TIMELIKE
(\xmax,\xmax) -- (-\xmax,\xmax) -- (\xmax,-\xmax) -- (-\xmax,-\xmax) -- cycle;
\node[mydarkblue,right,align=center] at (-\xmax,0.18*\xmax)
{\contour{myblue!5}{present}\\[-2]\contour{myblue!5}{}};
\node[mydarkblue,left,align=center] at (\xmax,0.18*\xmax)
{\contour{myblue!5}{present}\\[-2]\contour{myblue!5}{}};
\node[mydarkorange,align=center] at (-0.22*\xmax,0.67*\xmax)
{\contour{myorange!5}{future}\\[-2]\contour{myorange!5}{}};
\node[mydarkorange,align=center] at (0.22*\xmax,-0.67*\xmax)
{\contour{myorange!5}{past}\\[-2]\contour{myorange!5}{}};
\node at (2*\d, 4*\d){\textbullet};
\node[black, above] at (2*\d, 4*\d){future event};
\node at (4*\d, 2*\d){\textbullet};
\node[black, above right] at (4*\d, 2*\d){present event};
% PHOTON
\draw[photon] ( \xmax,-\xmax) -- ( 0.02*\xmax,-0.02*\xmax);
\draw[photon] (-\xmax,-\xmax) -- (-0.02*\xmax,-0.02*\xmax);
\draw[photon] ( 0.02*\xmax,0.02*\xmax) -- ( \xmax,\xmax)
node[mydarkorange,above right] {$x=ct$};
\draw[photon] (-0.02*\xmax,0.02*\xmax) -- (-\xmax,\xmax);
% % PARTICLE WORLDLINE
% \draw[particle,decoration={markings,mark=at position 0.27 with {\arrow{latex}},
% mark=at position 0.76 with {\arrow{latex}}},postaction={decorate}]
% (-0.5*\xmax,-\xmax) to[out=80,in=-110] (O) to[out=70,in=-100] (0.45*\xmax,\xmax);
% \fill[mydarkgreen] (O) circle(0.04); % event
\end{tikzpicture}\end{center}
\problem{Proper Time}
For any event $(x,ct)$ in the past or future, we can choose a reference frame to make it occur at $x' = 0$.
\begin{enumerate}
\item What velocity do we need to boost to so that the event $(x,ct)$ now occurs at $x' = 0$?
\begin{solution}
We need to boost to $v = x/t$.
\end{solution}
\vfill
\item After boosting to this reference frame, what time $ct'$ does the event occur at? \\
This will be our proper time $c\tau$.
\begin{solution}
Using time dilation, the event now occurs at $$c\tau = ct' = ct\sqrt{1 - x^2/c^2 t^2} = \sqrt{c^2 t^2 - x^2}$$
\end{solution}
\vfill
\end{enumerate}
\pagebreak
\problem{Proper Distance}
Similarly, for any event $(x,ct)$ in the present, we can choose a reference frame to make it occur at $t' = 0$.
\begin{enumerate}
\item What velocity do we need to boost to so that the event $(x,ct)$ now occurs at $t' = 0$?
\begin{solution}
We need to boost to $v = c^2t/x$.
\end{solution}
\vfill
\item After boosting to that velocity, what spatial position $x'$ does the event occur at? This will be our proper distance $\chi$.
\begin{solution}
Using space dilation, the event now occurs at $$\chi = x' = x\sqrt{1 - c^2t^2/x^2} = \sqrt{x^2 - c^2t^2}$$.
\end{solution}
\end{enumerate}
\vfill
Defined in this way, proper time and proper distance give a quick formula to figure out how old an object is, or how far an object traveled, in its own reference frame!
We'll now use this proper time and proper space to quickly solve one of the most famous paradoxes in special relativity.
\pagebreak
% \remark{}
% To visualize the consequences of this change, let's see what lines of constant proper time and constant proper space look like.
% Lines in green are constant proper time and lines in blue are constant proper space.
% % SPACETIME DIAGRAM - MULTIPLE INVARIANT HYPERBOLOIDS
% % Inspiration: https://commons.wikimedia.org/wiki/File:Spacelike_and_Timelike_Invariant_Hyperbolas.png
% \begin{center}\begin{tikzpicture}[scale=1.8]
% \message{Multiple invariant hyperboloids^^J}
% \def\xmax{2}
% \def\Nlines{4} % number of world lines (at constant x/t)
% \pgfmathsetmacro\w{\xmax/(\Nlines+1)}
% % AXES
% \draw[->,thick] (0,-\xmax) -- (0,\xmax+0.2) node[left=-1] {$ct$};
% \draw[->,thick] (-\xmax,0) -- (\xmax+0.2,0) node[right=-1] {$x$};
% % LIGHTCONE
% \draw[myorange,thick] (-\xmax,-\xmax) -- (\xmax, \xmax);
% \draw[myorange,thick] (-\xmax, \xmax) -- (\xmax,-\xmax);
% \foreach \i [evaluate={\s=\xmax*\i/(\Nlines+1); \xm=sqrt(\xmax^2-\s^2);}] in {1,...,\Nlines}{
% % SPACELIKE HYPERBOLOIDS
% \draw[mygreen,thick,samples=\Nsamples,smooth,variable=\x,domain=-\xm:\xm]
% plot(\x,-{sqrt(\s^2+(\x)^2)})
% plot(\x,{sqrt(\s^2+(\x)^2)});
% % TIMELIKE HYPERBOLOIDS
% \draw[myblue,thick,samples=\Nsamples,smooth,variable=\y,domain=-\xm:\xm]
% plot(-{sqrt(\s^2+(\y)^2)},\y)
% plot({sqrt(\s^2+(\y)^2)},\y);
% }
% % LABELS
% \node[mydarkgreen,above left=2,align=center] at (-0.2*\xmax,\xmax)
% {timelike separation\\[-1]$s^2 = c^2t^2 - x^2 > 0$};
% \node[mydarkorange,left=2,above right=-2,align=center] at (\xmax,\xmax)
% {lightlike separation\\[-1]$s^2 = c^2t^2 - x^2 = 0$};
% \node[mydarkblue,right=0,align=center] at (0.88*\xmax,-0.24*\xmax)
% {spacelike separation\\[-1]$s^2 = c^2t^2 - x^2 < 0$};
% % % VECTORS
% % \def\xa{0.5}
% % \def\xb{2.7}
% % \def\ta{-0.7}
% % \def\tb{1.7}
% % \draw[mydarkgreen,very thick,decoration={markings,mark=at position 0.55 with {\arrow{latex}}},
% % postaction={decorate},samples=20,variable=\x,domain=\xa:\xb]
% % plot({\w*\x},{\w*sqrt((\x)^2+3^2)});
% % \draw[mydarkblue,very thick,decoration={markings,mark=at position 0.6 with {\arrow{latex}}},
% % postaction={decorate},samples=20,variable=\x,domain=\ta:\tb]
% % plot({\w*sqrt((\x)^2+3^2)},{\w*\x});
% % \fill[mydarkgreen] ({\w*\xa},{\w*sqrt(\xa^2+3^2)}) coordinate (A) circle(0.03);
% % \fill[mydarkgreen] ({\w*\xb},{\w*sqrt(\xb^2+3^2)}) coordinate (A') circle(0.03);
% % \fill[mydarkblue] ({(\w*sqrt((\ta)^2+3^2)},{\w*\ta}) coordinate (B) circle(0.03);
% % \fill[mydarkblue] ({(\w*sqrt((\tb)^2+3^2)},{\w*\tb}) coordinate (B') circle(0.03);
% % \draw[vector',mydarkgreen] (0,0) -- (A)
% % node[pos=0.53,right=-2] {$s$};
% % \draw[vector',mydarkgreen] (0,0) -- (A')
% % node[pos=0.57,right=-2] {$s$};
% % \draw[vector',mydarkblue] (0,0) -- (B)
% % node[pos=0.53,below=-1] {$s$};
% % \draw[vector',mydarkblue] (0,0) -- (B')
% % node[pos=0.53,above=-1] {$s$};
% \end{tikzpicture}\end{center}
\problem{Twin paradox}
Suppose that you have two twins, Mark and Lucas, born at the exact same instant and location. \\
One day, Mark sends Lucas off in a spaceship traveling at $c/2$ to a planet that is one lightyear away ($c*(\text{1 year})$). Upon reaching the planet, Lucas realizes that he's all alone! Saddened, he turns around immediately, traveling back at speed $c/2$.
The question is: Which twin is older? We'll break this down step by step.
\begin{enumerate}
\item From Mark's perspective, what time does Lucas reach the planet? What time does he return?
\begin{solution}
Lucas reaches the planet at $t = 2$ years.
Lucas reaches home at $t = 4$ years.
\end{solution}
\vfill
\item From Lucas' perspective, how long does it take him to reach the planet? Hint: use proper time.
\begin{solution}
The proper time of Lucas arriving at the planet is $$c\tau = \sqrt{4 - 1}\text{ lightyears} \implies \tau = \sqrt{3}\text{ years}.$$
\end{solution}
\vfill
\item How long does it take him to return?
\begin{solution}
$\tau = \sqrt{3}\text{ years}.$
\end{solution}
\vfill
\item Which twin is older?
\begin{solution}
Mark
\end{solution}
\end{enumerate}
\vfill\pagebreak
\problem{Twin paradox (continued)}
The weird part is, from Lucas' perspective, it was Mark who flew off and then came back! So why is Mark not younger than Lucas?
Let's break this down step by step.
\begin{enumerate}
\item[\bf E:]
To help answer this, draw a spacetime diagram of this situation. In Lucas' reference frames (one for going out to the planet and one for coming back), draw lines of constant time. What happens to Mark from Lucas' perspective?\\
{\em Hint: changing speed is weird. Focus on where Lucas changes speed.}
\halfdiagramc{Mark}
\begin{solution}
% SPACETIME DIAGRAM of TWIN PARADOX
\begin{tikzpicture}[scale=2.0]
\message{Twin paradox^^J}
\def\xmin{0.2}
\def\xmax{2}
\def\ymax{2.3}
\def\Nlines{5} % number of world lines (at constant x/t)
\def\ang{60} % angle between ct and ct' axes
\pgfmathsetmacro\d{0.94*\xmax/\Nlines} % grid size
\pgfmathsetmacro\dt{3*\d} % time of half trip
\pgfmathsetmacro\D{\dt/tan(\ang)} % distance between observers
\pgfmathsetmacro\h{\dt-\D/tan(\ang)} % half time gap of return
\coordinate (A) at (0,0); % observer A at t=0
\coordinate (B) at (\D,0); % observer B at t=0
\coordinate (C) at (\D,\dt); % point of return
\coordinate (T1) at (0,\dt); % time of return
\coordinate (T2) at (0,2*\dt); % time of arrival
% WORLD LINES GRID
\message{ Making world lines...^^J}
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line] ( \x,-\xmin) -- ( \x,\ymax);
\draw[world line t] (-\xmin, \x) -- (\xmax, \x);
}
\draw[world line t] (-\xmin,{(\Nlines+1)*\d}) -- (\xmax,{(\Nlines+1)*\d});
% AXES
\draw[->,thick] (0,-\xmin) -- (0,\ymax+0.2) node[above left=-2] {$ct$};
\draw[->,thick] (-\xmin,0) -- (\xmax+0.2,0) node[below=0] {$x$};
% VECTORS
\draw[vector,myred,shorten >=1] (A) -- (T2);
\draw[vector,mygreen,shorten >=2] (A) -- (C);
\draw[vector,mygreen,shorten >=1] (C) -- (T2);
% PLANES OF SIMULTANEITY
\fill[mydarkred,opacity=0.06]
(0,\h) -- (C) -- (0,2*\dt-\h) -- cycle;
\pgfmathsetmacro\ystep{\h/3}
\foreach \i [evaluate={\dy=(\i-1)*\ystep; \ya=\i*\ystep; \yb=2*\dt-\i*\ystep;}] in {1,...,3}{
\draw[mydarkred,dashed,line width=0.6]
(0,\ya)++(90-\ang:-0.8*\xmin) --++ (90-\ang:{1.2*\xmin+\D/sin(\ang)});
\draw[mydarkblue,dashed,line width=0.6]
(0,\yb)++(\ang-90:-0.8*\xmin) --++ (\ang-90:{1.2*\xmin+\D/sin(\ang)});
\fill[mydarkred] (0,\ya) circle(0.02);
\fill[mydarkblue] (0,\yb) circle(0.02);
%\fill[mydarkblue] ({\D-\dy*cot(\ang)},\dt+\dy) circle(0.02);
%\fill[mydarkred] ({\D-\dy*cot(\ang)},\dt-\dy) circle(0.02);
\fill[mydarkblue] (C)++(-\ang:{\dy*sin(\ang)/cos(2*\ang)}) circle(0.02);
\fill[mydarkred] (C)++( \ang:{\dy*sin(\ang)/cos(2*\ang)}) circle(0.02);
}
\fill[mydarkred] (A) circle(0.04) node[below left=-1] {Mark}; % observer A
\fill[mydarkgreen] (C) circle(0.04)
node[right=4] {\contour{white}{Lucas turns around}}; % observer B returns
\node[mydarkblue,above right=0,align=left] at (2*\d,1.15*\dt)
{\contour{white}{planes of}\\[-2]\contour{white}{simultaneity}};
\node[mydarkred,below right=0,align=left] at (2*\d,0.85*\dt)
{\contour{white}{planes of}\\[-2]\contour{white}{simultaneity}};
% TICKS
\node[fill=white,inner sep=1,above=1,left=3] at (T1) {$\dfrac{ct_2}{2}=ct_1$};
\node[fill=white,inner sep=1,above=2,left=3] at (T2) {$ct_2$};
\tick{T1}{0};
\tick{T2}{0};
\end{tikzpicture}
\end{solution}
\vfill
\item[\bf F:] We said that when Lucas landed on the planet, he immediately took off back towards Earth. From Mark's perspective, how long was Lucas on that planet?
\begin{solution}
$4 - 2 \sqrt{3}$ years.
\end{solution}
\vfill
\item[\bf G:] What is physically different between Lucas' perspective and Mark's perspective?
\begin{solution}
Lucas is the one who accelerates. His reference frame is non-inertial (not defined in this packet) and would require additional care.
\end{solution}
\vfill
\end{enumerate}
\pagebreak

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\section{Relative Velocity}
The first weirdness occurs if we consider multiple people running in multiple directions and speeds. Let's explore this.
You shouldn't have to do any math here besides thinking about angles.
\problem{}
Using the given diagram, verify that if, in Bob's reference frame, Alice is running away from Bob at speed $v$, then in Alice's reference frame, Bob is running away at speed $v$.
{\em Hint: slopes}
% SPACETIME DIAGRAM - LORENTZ BOOST
\begin{center}\begin{tikzpicture}[scale=1.8]
\message{Lorentz boost^^J}
\def\xmax{2}
\def\xmaxp{2.2} % maximum of rotated axis
\def\Nlines{5} % number of world lines (at constant x/t)
\pgfmathsetmacro\ang{atan(1/2)} % angle between x and x' axes
\pgfmathsetmacro\d{0.9*\xmax/\Nlines}refer % grid size
\pgfmathsetmacro\D{\d/cos(\ang)/sqrt(1-tan(\ang)^2)} % grid size, boosted
\coordinate (O) at (0,0);
\coordinate (X) at (\xmax+0.2,0);
\coordinate (T) at (0,\xmax+0.2);
\coordinate (X') at (\ang:\xmaxp+0.2);
\coordinate (T') at (90-\ang:\xmaxp+0.2);
% WORLD LINE GRID
\message{ Making world lines...^^J}
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line] (-\x,-\xmax) -- (-\x,\xmax);
\draw[world line] ( \x,-\xmax) -- ( \x,\xmax);
\draw[world line t] (-\xmax,-\x) -- (\xmax,-\x);
\draw[world line t] (-\xmax, \x) -- (\xmax, \x);
}
% BOOSTED WORLD LINE GRID
\message{ Making world lines for boosted frame...^^J}
\fill[mydarkred,opacity=0.05]
(O) --++ (\ang:\xmaxp) --++ (90-\ang:\xmaxp) --++ (\ang:-\xmaxp) -- cycle;
\fill[mydarkred,opacity=0.05]
(O) --++ (\ang:-\xmaxp) --++ (90-\ang:-\xmaxp) --++ (\ang:\xmaxp) -- cycle;
\foreach \i [evaluate={\x=\i*\D;}] in {1,...,4}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line'] (\ang:-\x) --++ (90-\ang:-\xmaxp);
\draw[world line'] (90-\ang:-\x) --++ (\ang:-\xmaxp);
\draw[world line'] (\ang:\x) --++ (90-\ang:\xmaxp);
\draw[world line'] (90-\ang:\x) --++ (\ang:\xmaxp);
}
% AXES
\draw[->,thick] (0,-\xmax) -- (T) node[left=-1] {Alice: $ct$};
\draw[->,thick] (-\xmax,0) -- (X) node[below=0] {$x$};
\draw[->,thick,mydarkred] (90-\ang:-\xmaxp) -- (T')
node[right=5,above=-1] {Bob: $ct'$};
\draw[->,thick,mydarkred] (\ang:-\xmaxp) -- (X') node[right=-1] {$x'$};
% ANGLES
\draw pic[->,"$\theta$",draw=black,angle radius=34,angle eccentricity=1.2] {angle = X--O--X'};
\draw pic[<-,"$\theta$",draw=black,angle radius=34,angle eccentricity=1.2] {angle = T'--O--T};
% % PHOTON
% \draw[photon] (O) --++ (4*\d, 4*\d);
\end{tikzpicture}\end{center}
\begin{solution}
Extend the graph to the left and look at the slope of Alice's world line in Bob's reference frame. You'll find that Alice's slope in Bob's reference frame is the opposite of Bob's slope in Alice's reference frame.
\end{solution}
\problem{}
What if Alice runs to the left at speed $v$, Bob stays still, and Charlie runs to the right at speed $v$?
From Charlie's reference frame, is Alice running away at speed $2v$? Why or why not?
{\em Think about the consequences.}
\begin{solution}
If $v > c/2$, then this poses an immediate problem. Alice has to move at a speed below $2v$. So velocities do not just add.
\end{solution}
\vfill
\pagebreak

216
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\section{Simultaneity}
Let's look now at the consequences of a rotated spatial coordinate. To help ourselves switch between different perspectives, we'll bring in some professionals: Alice and Bob.
\ref{simultaneity setup} to \ref{Bob overreacted} can all be completed on one spacetime diagram. Feel free to use the one provided below, or draw your own if it gets too crowded.
\emptydiagramc{Alice}
\problem{}<simultaneity setup>
Draw a spacetime diagram from Alice's reference frame or use the one provided.
What do the horizontal gridlines represent?
If two events (remember, points in spacetime) lie on the same horizontal line, what does that imply about the events?
\begin{solution}
Horizontal lines have constant $t$ value. Therefore, if two events lie on the same horizontal line then they occur at the same time.
\end{solution}
\vfill
\pagebreak
\problem{}
Suppose that Bob is walking to the right at speed $c/3$, relative to Alice. Add Bob's worldline to your spacetime diagram. Superimpose Bob's reference frame onto the diagram
What do the "horizontal" lines in Bob's reference frame represent to Bob?
\begin{solution}
Same solution, horizontal lines in Bob's reference frame represent events at a particular time.
\end{solution}
\vfill
\problem{}
Suppose that Alice is passing time, snapping to some music with her arms out.\\
Both of her arms are length $1$ and she snaps both hands every unit time. So there is a snap at location $x = \pm 1$ at $ct = 1$, $ct = 2$, and so on.
Draw this situation on your spacetime diagram.
Consider this from Bob's perspective. What does Bob hear?
(We'll assume that sound propagates instantly, so if a sound occurs at $ct' = 3.5$ then Bob hears it instantly at $ct' = 3.5$.)
\begin{solution}
The snaps will be out of time in Bob's reference frame. He will hear Alice's right hand first and then her left hand after.
\end{solution}
\vfill
\problem{} <Bob overreacted>
Bob is deeply annoyed at Alice because she {keeps}.~{\em snapping}.~{\em out}.~{\bf\em of}.~{\bf\em TIME}.
In a fit of frustration he starts running faster and faster to the right. Eventually, he notices that the snaps are changing. What happens to the timing of the snaps? Do the snaps ever come back into time? At what speed is Bob running if they do?
\begin{solution}
We can line up new sets of snaps with lines of slope $1/2$. So if Bob is running at speed $c/2$ then he will hear the snaps in time again.
\end{solution}
\vfill
\pagebreak
\problem{}
Done with music, Alice and Bob decide to have a race and they agree to race from $x = 0$ to $x = 2$.
Suppose that Alice can run at an impressive speed of $c/2$ while Bob can only run at a measly speed of $c/4$.
\begin{enumerate}
\item Who wins the race?
\begin{solution}
Alice
\end{solution}
\vspace{20pt}
\item Is there any reference frame in which Bob wins?
\halfdiagramcwide{Ground}
\begin{solution}
No
\end{solution}
\vfill
\item Suppose instead that Alice starts at $x = 0$ and is racing to $x = 2$, while Bob starts at $x = 8$ and is racing to $x = 10$. Is there now a reference frame where Bob wins? Why?\\
{\em Hint: Plot these events on a spacetime diagram.}
\begin{solution}
Bob finishes at $(2, 8)$ and Alice finishes at $(10,4)$.
If you consider a reference frame moving to the left at speed $c/2$, then Alice and Bob will tie. Anything faster, Bob will win. Anything slower, Alice will win.
\end{solution}
\end{enumerate}
\halfdiagramcwide{Ground}
\vfill
\pagebreak
% \problem{}
% We've seen that by changing your reference frame, events that were simultaneous may now occur at different times.
% Which of the following pairs of events (points in spacetime $(x,ct)$) are simultaneous in {\em some} inertial reference frame?
% % Remember, we can only consider reference frames which are moving less than the speed of light.
% \begin{enumerate}
% \item $(1,1)$ and $(0,1)$
% \item $(2,1)$ and $(0,0)$
% \item $(2,3)$ and $(0,1)$
% \item $(1,2)$ and $(0,0)$
% \item $(1,1)$ and $(1,3)$
% \end{enumerate}
% \begin{solution}
% \begin{enumerate}
% \item $(1,1)$ and $(0,1)$ - yes
% \item $(2,1)$ and $(0,0)$ - yes
% \item $(2,3)$ and $(0,1)$ - no, would require a reference frame at the speed of light
% \item $(1,2)$ and $(0,0)$ - no, would require a reference frame moving faster than the speed of light
% \item $(1,1)$ and $(1,3)$ - definitely not
% \end{enumerate}
% \end{solution}
\problem{}
This weirdness with simultaneity might make us question what the past, present, and future are.
To make sense of this, suppose that we want to define the {\em present} to be every event which is simultaneous to you, right here, right now --- a.k.a.~the spacetime point $(0,0)$ --- in at least one reference frame.
On the given spacetime diagram, draw the region which represents the present.
Draw a region which represents the future: events that are later in time in every reference frame.
Draw a region which represents the past: events that are earlier in time in every reference frame.
\begin{center}\begin{tikzpicture}[scale=2]
\message{Light cone^^J}
\def\xmax{2}
\def\xmaxp{2.2} % maximum of rotated axis
\def\Nlines{5} % number of world lines (at constant x/t)
\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
\pgfmathsetmacro\ang{atan(1/3)} % angle between x and x' axes
\coordinate (O) at (0,0);
\coordinate (X) at (\xmax+0.2,0);
\coordinate (T) at (0,\xmax+0.2);
% WORLD LINE GRID
\message{ Making world lines...^^J}
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line] (-\x,-\xmax) -- (-\x,\xmax);
\draw[world line] ( \x,-\xmax) -- ( \x,\xmax);
\draw[world line t] (-\xmax,-\x) -- (\xmax,-\x);
\draw[world line t] (-\xmax, \x) -- (\xmax, \x);
}
% AXES
\draw[->,thick] (0,-\xmax) -- (T) node[left=-1] {$ct$};
\draw[->,thick] (-\xmax,0) -- (X) node[below=0] {$x$};
% LABELS
%\draw pic[->,"$45^\circ$",draw=black,angle radius=23,angle eccentricity=1.38] {angle = X--O--C};
% \node[mydarkorange,above right] at (0.1*\xmax,\xmax) {future light cone};
% \node[mydarkorange,below] at (0,-\xmax) {past light cone};
% FILLS
% \fill[myblue,opacity=0.05] % SPACELIKE
% (\xmax,\xmax) -- (-\xmax,-\xmax) -- (-\xmax,\xmax) -- (\xmax,-\xmax) -- cycle;
% \fill[myorange,opacity=0.05] % TIMELIKE
% (\xmax,\xmax) -- (-\xmax,\xmax) -- (\xmax,-\xmax) -- (-\xmax,-\xmax) -- cycle;
% \node[mydarkblue,right,align=center] at (-\xmax,0.18*\xmax)
% {\contour{myblue!5}{present}\\[-2]\contour{myblue!5}{}};
% \node[mydarkblue,left,align=center] at (\xmax,0.18*\xmax)
% {\contour{myblue!5}{present}\\[-2]\contour{myblue!5}{}};
% \node[mydarkorange,align=center] at (-0.22*\xmax,0.67*\xmax)
% {\contour{myorange!5}{future}\\[-2]\contour{myorange!5}{}};
% \node[mydarkorange,align=center] at (0.22*\xmax,-0.67*\xmax)
% {\contour{myorange!5}{past}\\[-2]\contour{myorange!5}{}};
% PHOTON
\draw[photon] ( \xmax,-\xmax) -- ( 0.02*\xmax,-0.02*\xmax);
\draw[photon] (-\xmax,-\xmax) -- (-0.02*\xmax,-0.02*\xmax);
\draw[photon] ( 0.02*\xmax,0.02*\xmax) -- ( \xmax,\xmax)
node[mydarkorange,above right] {$x=ct$};
\draw[photon] (-0.02*\xmax,0.02*\xmax) -- (-\xmax,\xmax);
% % PARTICLE WORLDLINE
% \draw[particle,decoration={markings,mark=at position 0.27 with {\arrow{latex}},
% mark=at position 0.76 with {\arrow{latex}}},postaction={decorate}]
% (-0.5*\xmax,-\xmax) to[out=80,in=-110] (O) to[out=70,in=-100] (0.45*\xmax,\xmax);
% \fill[mydarkgreen] (O) circle(0.04); % event
\end{tikzpicture}\end{center}
Another way to think of this is thinking of causality. In a reference frame, the {\em future} is every event that can be affected by an event at $(0,0)$. The {\em past} is every event that could have affected an event at $(0,0)$. The {\em present} is every event that is causally independent of $(0,0)$.
\begin{solution}
% SPACETIME DIAGRAM - LIGHT CONE
The photons split the diagram into four regions.
The top region is the future, the bottom region is the past, and the left/right regions are the present.
\end{solution}
\pagebreak
\problem{Bell's Spaceship Paradox}
Suppose that we have two spaceships, distance $L$ apart, tied together with floss of length $L$. The floss is so weak that any stretching at all will cause it to disintegrate.
The spaceships are at rest and then simultaneously accelerate to speed $c/2$.
Draw the spacetime diagram for this situation. Include the reference frame of the spaceships {\em after} they start moving (i.e. the reference frame moving at speed $c/2$).\\
What happens to the floss in the boosted frame? Does it break? Why?
\emptydiagramc{Rest}
\vfill
\problem{Bell's Spaceship Paradox (continued)}
The same outcome has to occur in all reference frames, so we know that the floss breaks in the stationary reference frame. However, in the stationary reference frame, the two spaceships accelerate at the same time, so our explanation no longer seems accurate.
Can you come up with a hypothesis for why the floss breaks in the stationary reference frame?
{\em Hint: The floss breaks if the ships are farther apart than the length of the floss.}
\begin{solution}
The floss will break as soon as the spaceships accelerate.
In the boosted frame, the spaceship on the right accelerates first, stretching the floss, causing it to break.
In the rest frame, this paradox motivates length contraction. Either the ships get farther apart (doesn't happen) or the floss gets shorter. This implies that the length of moving objects must get smaller.
\end{solution}
\vfill
\pagebreak

175
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\section{Spacetime Diagrams}
We are going to derive the principles and consequences of special relativity using basic geometry.
\\ To help with our visualization, we will be using spacetime diagrams (called {\em Minkowski diagrams}).
To make our models simpler, we will only be considering {\em one spatial dimension}.
We plot space, which we denote by $x$, as the horizontal axis, and time, which we denote by $t$ as the vertical axis. For a given object, we can then plot its position at any given time. \\
This will give a (potentially curvy) line that we call the object's {\em world line}.
\example{}
Suppose that at time $t = 0$, you are standing still with your cat at your feet. Your cat walks away from you at speed $1$. We can represent this with a spacetime diagram:
\begin{center}
% SPACETIME DIAGRAM with WORLD LINES
\begin{tikzpicture}[scale=2.0]
%\message{Worldlines^^J}
\def\ymin{0.2}
\def\xmin{1.6}
\def\xmax{2}
\def\Nlines{4} % number of world lines (at constant x/t)
\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
\coordinate (O) at (0,0);
\coordinate (T) at (0,\xmax+0.2);
% WORLD LINES GRID
%\message{ Making world lines...^^J}
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line] ( \x,-\ymin) -- ( \x,\xmax);
\draw[world line t] (-\xmin, \x) -- (\xmax, \x);
}
\draw[world line] (-\d,-\ymin) -- (-\d,\xmax);
\draw[world line] (-2*\d,-\ymin) -- (-2*\d,\xmax);
\draw[world line] (-3*\d,-\ymin) -- (-3*\d,\xmax);
% AXES
\draw[->,thick] (0,-\ymin) -- (T) node[left=-1] {$t$};
\draw[->,thick] (-\xmin,0) -- (\xmax+0.2,0) node[below=0] {$x$};
% VECTORS
\draw[vector,myred, very thick] (O) -- (4*\d,4*\d)
node[mydarkred,right=10,above] {\contour{white}{cat: $x(t)=t$}};
\draw[vector,myblue, very thick] (O) -- (0,4*\d)
node[mydarkblue,below left=0] {\contour{white}{you: $x(t)=0$}};
%\node[right=8,above,mydarkpurple] at (T) {$x(t)=0$};
\end{tikzpicture}
\end{center}
{\em NOTE:} The horizontal axis is space and the vertical axis is time. We are only working with one spatial dimension.
\problem{}<pets scatter>
Suppose that you are standing still at time $t = 0$ and your many pets lie at your feet.
\begin{itemize}
\item Your cat, unhappy that she is not fed, begins walking away to your right at speed $2$.
\item Your dog, distracted by a squirrel, walks away to your left at speed $1$.
\item Your hamster, just wanting to feel included, waits a second and then follows the dog at speed $2$. After reaching your dog, your hamster turns around and sprints after the cat at speed $3$.
\end{itemize}
Draw this situation in the provided spacetime diagram.
\emptydiagram{Alice}
\begin{solution}
\begin{center}
\begin{tikzpicture}[scale=2.0]
\message{Worldlines^^J}
\def\ymin{0.2}
\def\xmin{2}
\def\xmax{2}
\def\Nlines{4} % number of world lines (at constant x/t)
\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
\coordinate (O) at (0,0);
\coordinate (T) at (0,\xmax+0.2);
% WORLD LINES GRID
\message{ Making world lines...^^J}
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line] ( \x,-\ymin) -- ( \x,\xmax);
\draw[world line t] (-\xmin, \x) -- (\xmax, \x);
}
\draw[world line] (-\d,-\ymin) -- (-\d,\xmax);
\draw[world line] (-2*\d,-\ymin) -- (-2*\d,\xmax);
\draw[world line] (-3*\d,-\ymin) -- (-3*\d,\xmax);
\draw[world line] (-4*\d,-\ymin) -- (-4*\d,\xmax);
% AXES
\draw[->,thick] (0,-\ymin) -- (T) node[left=-1] {$t$};
\draw[->,thick] (-\xmin,0) -- (\xmax+0.2,0) node[below=0] {$x$};
% VECTORS
\draw[vector,myred] (O) -- (0,4*\d)
node[mydarkred,below left=0] {\contour{white}{you: $x(t)=0$}};
\draw[vector,myblue] (O) -- (4*\d,2*\d)
node[mydarkblue,above left=0] {\contour{white}{cat: $x(t)=2t$}};
\draw[vector,mygreen] (O) -- (-4*\d,4*\d)
node[mydarkgreen,below left=0] {\contour{white}{dog: $x(t)=-t$}};
\draw[vector,black] (O) -- (0,\d) -- (-2*\d, 2*\d) -- (4*\d, 4*\d)
node[black,below right=0] {\contour{white}{hamster}};
% \draw[vector,myblue]
% (O) to[out=35,in=-100] (O)
% to[out=80,in=-80,looseness=1.5] (0.3*\xmax,4*\d)
% node[mydarkblue,above=-3] {\contour{white}{cat: $x(t)$}};
%\node[right=8,above,mydarkpurple] at (T) {$x(t)=0$};
\end{tikzpicture}
\end{center}
\end{solution}
\problem{Event}
Any single point $(x,t)$ on a spacetime diagram is considered an {\em event} because it describes a time and place. For instance, what is the event that corresponds to your hamster catching up to your dog?
\begin{solution}
$(-2,2)$
\end{solution}
\vfill
\begin{problem}{}<pets scatter train>
Suppose that the situation of \ref{pets scatter} occurred while you were riding on a train moving to the right at speed $1$.
Everything occurs relative to you in the same way.
Draw the same diagram in this new situation.
Are any of your pets staying still in this new situation?
\emptydiagram{Train}
\begin{solution}
\begin{center}
\begin{tikzpicture}[scale=2.0]
\message{Worldlines^^J}
\def\ymin{0.2}
\def\xmin{2}
\def\xmax{2}
\def\Nlines{4} % number of world lines (at constant x/t)
\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
\coordinate (O) at (0,0);
\coordinate (T) at (0,\xmax+0.2);
% WORLD LINES GRID
\message{ Making world lines...^^J}
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line] ( \x,-\ymin) -- ( \x,\xmax);
\draw[world line t] (-\xmin, \x) -- (\xmax, \x);
}
\draw[world line] (-\d,-\ymin) -- (-\d,\xmax);
\draw[world line] (-2*\d,-\ymin) -- (-2*\d,\xmax);
\draw[world line] (-3*\d,-\ymin) -- (-3*\d,\xmax);
\draw[world line] (-4*\d,-\ymin) -- (-4*\d,\xmax);
% AXES
\draw[->,thick] (0,-\ymin) -- (T) node[left=-1] {$t$};
\draw[->,thick] (-\xmin,0) -- (\xmax+0.2,0) node[below=0] {$x$};
% VECTORS
\draw[vector,myred] (O) -- (4*\d,4*\d)
node[mydarkred,below left=0] {\contour{white}{you: $x(t)=t$}};
\draw[vector,myblue] (O) -- (6*\d,2*\d)
node[mydarkblue,above left=0] {\contour{white}{cat: $x(t)=3t$}};
\draw[vector,mygreen] (O) -- (0,4*\d)
node[mydarkgreen,below left=0] {\contour{white}{dog: $x(t)=0$}};
\draw[vector,black] (O) -- (\d,\d) -- (0, 2*\d) -- (8*\d, 4*\d)
node[black,below right=0] {\contour{white}{hamster}};
% \draw[vector,myblue]
% (O) to[out=35,in=-100] (O)
% to[out=80,in=-80,looseness=1.5] (0.3*\xmax,4*\d)
% node[mydarkblue,above=-3] {\contour{white}{cat: $x(t)$}};
%\node[right=8,above,mydarkpurple] at (T) {$x(t)=0$};
\end{tikzpicture}
\end{center}
The dog remains stationary in this reference frame.
\end{solution}
\end{problem}
\vfill
\pagebreak

270
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\section{Special Relativity}
Galilean relativity is nice until we start going really, {\em really} fast. Since most of us are terribly slow, we can use it without any issues. However, in reality, things are much weirder. In particular, there is a maximum speed: the speed of light,
$$c = 299,792,458 \tfrac{m}{s}.$$
Nothing can move faster than the speed of light and {\bf in every reference frame, light will move at this speed}.
Let's see if this is consistent with Galilean relativity.
We are going to making things easier for ourselves now and change units. Instead of measuring time $t$, we are now going to measure $ct$.
\problem{}<photon diagram>
Suppose you are sitting still and you send one photon to your right. Draw this photon on a spacetime diagram, with horizontal axis $x$ and vertical axis $ct$.
\halfdiagramc{You}
\begin{solution}\begin{center}
\begin{tikzpicture}[scale=2.0]
\message{Worldlines^^J}
\def\ymin{0.2}
\def\xmin{1.6}
\def\xmax{2}
\def\Nlines{4} % number of world lines (at constant x/t)
\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
\coordinate (O) at (0,0);
\coordinate (T) at (0,\xmax+0.2);
% WORLD LINES GRID
\message{ Making world lines...^^J}
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line] ( \x,-\ymin) -- ( \x,\xmax);
\draw[world line t] (-\xmin, \x) -- (\xmax, \x);
}
\draw[world line] (-\d,-\ymin) -- (-\d,\xmax);
\draw[world line] (-2*\d,-\ymin) -- (-2*\d,\xmax);
\draw[world line] (-3*\d,-\ymin) -- (-3*\d,\xmax);
% AXES
\draw[->,thick] (0,-\ymin) -- (T) node[left=-1] {$ct$};
\draw[->,thick] (-\xmin,0) -- (\xmax+0.2,0) node[below=0] {$x$};
% VECTORS
\draw[vector,myred] (O) -- (0,4*\d)
node[mydarkred,below left=0] {\contour{white}{you: $x(t)=0$}};
\draw[photon,shorten >=2] (O) -- (4*\d, 4*\d)
node[black, above right] {\contour{white}{photon}};
%\node[right=8,above,mydarkpurple] at (T) {$x(t)=0$};
\end{tikzpicture}
\end{center}
\end{solution}
\problem{}
Suppose you are now sitting on a train that is moving to the right at $c/2$ and again send one photon to your right. Draw this diagram in the reference frame of the ground.
Draw your (Galilean) reference frame on top of this diagram. What is the speed of the photon in your reference frame? Is that a problem?
\halfdiagramc{Train}
\begin{solution}\begin{center}\begin{tikzpicture}[scale=1.5]
\message{Worldlines^^J}
\pgfmathsetmacro\ang{atan(1/2)} % angle
\def\ymin{0.2}
\def\xmin{1.6}
\def\xmax{2}
\def\Nlines{4} % number of world lines (at constant x/t)
\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
\coordinate (O) at (0,0);
\coordinate (T) at (0,\xmax+0.2);
% WORLD LINES GRID
\message{ Making world lines...^^J}
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line] ( \x,-\ymin) -- ( \x,\xmax);
\draw[world line t] (-\xmin, \x) -- (\xmax, \x);
}
\draw[world line] (-\d,-\ymin) -- (-\d,\xmax);
\draw[world line] (-2*\d,-\ymin) -- (-2*\d,\xmax);
\draw[world line] (-3*\d,-\ymin) -- (-3*\d,\xmax);
% AXES
\draw[->,thick] (0,-\ymin) -- (T) node[left=-1] {$ct$};
\draw[->,thick] (-\xmin,0) -- (\xmax+0.2,0) node[below=0] {$x=\color{mydarkred}x'$};
% VECTORS
\draw[vector,myred] (O) -- (2*\d,4*\d)
node[mydarkred,below left=0] {\contour{white}{you: $x(t)=ct/2$}};
\draw[photon,shorten >=2] (O) -- (4*\d, 4*\d)
node[black, above right] {\contour{white}{photon}};
%\node[right=8,above,mydarkpurple] at (T) {$x(t)=0$};
% WORLD LINES GRID - BOOSTED
\message{ Making world lines, boosted...^^J}
\fill[mydarkred,opacity=0.05]
(O) --++ (90-\ang:\xmax) --++ (\xmax,0) --++ (90-\ang:-\xmax) -- cycle;
% \fill[mydarkred,opacity=0.05]
% (O) --++ (90-\ang:-\xmax) --++ (-\xmax,0) --++ (90-\ang:\xmax) -- cycle;
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line'] (\x,0) --++ (90-\ang:\xmax);
% \draw[world line'] (-\x,0) --++ (90-\ang:-\xmax);
}
\end{tikzpicture}\end{center}
In your reference frame, the light is only moving at speed $c/2$. Uh oh.
\end{solution}
\pagebreak
\problem{}
Clearly, Galilean relativity and the absolute speed of light do not mix well together. Having noticed this, you are now in the same boat as early 20th century physicists.
Can you brainstorm any ways to fix Galilean relativity to account for this absolute speed of light?
{\em Hint 1}: Try different methods of drawing the axes of your reference frame that would maintain the speed of light in both the rest frame and in your reference frame.
{\em Hint 2}: The photon worldline always bisects the angle between the space and time axes. \\ Is there a way that you can make that happen in your reference frame?
{\em \color{gray} Don't worry if you don't have any ideas! It took physicists a while to figure this out. Whenever you want to move on, we have the solution on the new page.}
\emptydiagramc{Train}
\pagebreak
\definition{Lorentz Boost}
Looking at our spacetime diagram from \ref{photon diagram}, we see that the photon worldline bisects the angle between the $ct$ axis and $x$ axis. So if we want to maintain this speed in all reference frames, we just need to make sure that photons bisect our new time axis $ct'$ and our new space axis $x'$.
In order to do this, we're going to to rotate our space axis $x'$ by the same angle that our $ct'$ axis is rotated. Rotating both axes like this is called a {\em Lorentz boost} and is best visualized in the following diagram:
% SPACETIME DIAGRAM - LORENTZ BOOST
\begin{center}\begin{tikzpicture}[scale=1.8]
\message{Lorentz boost^^J}
\def\xmax{2}
\def\xmaxp{2.2} % maximum of rotated axis
\def\Nlines{5} % number of world lines (at constant x/t)
\pgfmathsetmacro\ang{atan(1/2)} % angle between x and x' axes
\pgfmathsetmacro\d{0.9*\xmax/\Nlines}refer % grid size
\pgfmathsetmacro\D{\d/cos(\ang)/sqrt(1-tan(\ang)^2)} % grid size, boosted
\coordinate (O) at (0,0);
\coordinate (X) at (\xmax+0.2,0);
\coordinate (T) at (0,\xmax+0.2);
\coordinate (X') at (\ang:\xmaxp+0.2);
\coordinate (T') at (90-\ang:\xmaxp+0.2);
% WORLD LINE GRID
\message{ Making world lines...^^J}
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line] (-\x,-\xmax) -- (-\x,\xmax);
\draw[world line] ( \x,-\xmax) -- ( \x,\xmax);
\draw[world line t] (-\xmax,-\x) -- (\xmax,-\x);
\draw[world line t] (-\xmax, \x) -- (\xmax, \x);
}
% BOOSTED WORLD LINE GRID
\message{ Making world lines for boosted frame...^^J}
\fill[mydarkred,opacity=0.05]
(O) --++ (\ang:\xmaxp) --++ (90-\ang:\xmaxp) --++ (\ang:-\xmaxp) -- cycle;
\fill[mydarkred,opacity=0.05]
(O) --++ (\ang:-\xmaxp) --++ (90-\ang:-\xmaxp) --++ (\ang:\xmaxp) -- cycle;
\foreach \i [evaluate={\x=\i*\D;}] in {1,...,4}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line'] (\ang:-\x) --++ (90-\ang:-\xmaxp);
\draw[world line'] (90-\ang:-\x) --++ (\ang:-\xmaxp);
\draw[world line'] (\ang:\x) --++ (90-\ang:\xmaxp);
\draw[world line'] (90-\ang:\x) --++ (\ang:\xmaxp);
}
% AXES
\draw[->,thick] (0,-\xmax) -- (T) node[left=-1] {$ct$};
\draw[->,thick] (-\xmax,0) -- (X) node[below=0] {$x$};
\draw[->,thick,mydarkred] (90-\ang:-\xmaxp) -- (T')
node[right=5,above=-1] {you: $ct'$};
\draw[->,thick,mydarkred] (\ang:-\xmaxp) -- (X') node[right=-1] {$x'$};
% ANGLES
\draw pic[->,"$\theta$",draw=black,angle radius=34,angle eccentricity=1.2] {angle = X--O--X'};
\draw pic[<-,"$\theta$",draw=black,angle radius=34,angle eccentricity=1.2] {angle = T'--O--T};
% PHOTON
\draw[photon] (O) --++ (5*\d, 5*\d)
node[black, above right] {\contour{white}{photon}};
\end{tikzpicture}\end{center}
As before, the slanted (red) axes are your reference frame as you're moving. \\
Note that you (and anything moving the same speed as you) are stationary in this reference frame.
\remark{}
In this diagram, we've not only rotated your space axis ($x'$), we've also adjusted the scale of $ct'$ and $x'$ relative to the rest frame. This scaling comes from physical experiments which we will conduct later.
\problem{}
Please verify that in the diagram above, if you shoot a photon behind you, it still moves at speed $c$ in both your reference frame and the ground's reference frame. You can do this by drawing directly on the diagram.
\begin{solution}
This follows by just extending the boosted axes to the second quadrant and drawing the photon's worldline.
\end{solution}
\pagebreak
% \begin{solution}\begin{center}\begin{tikzpicture}[scale=1.8]
% \message{Lorentz boost^^J}
% \def\xmax{2}
% \def\xmaxp{2.2} % maximum of rotated axis
% \def\Nlines{5} % number of world lines (at constant x/t)
% \pgfmathsetmacro\ang{atan(1/2)} % angle between x and x' axes
% \pgfmathsetmacro\d{0.9*\xmax/\Nlines}refer % grid size
% \pgfmathsetmacro\D{\d/cos(\ang)/sqrt(1-tan(\ang)^2)} % grid size, boosted
% \coordinate (O) at (0,0);
% \coordinate (X) at (\xmax+0.2,0);
% \coordinate (T) at (0,\xmax+0.2);
% \coordinate (X') at (\ang:\xmaxp+0.2);
% \coordinate (T') at (90-\ang:\xmaxp+0.2);
% % WORLD LINE GRID
% \message{ Making world lines...^^J}
% \foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
% \message{ Running i/N=\i/\Nlines, x=\x...^^J}
% \draw[world line] (-\x,-\xmax) -- (-\x,\xmax);
% \draw[world line] ( \x,-\xmax) -- ( \x,\xmax);
% \draw[world line t] (-\xmax,-\x) -- (\xmax,-\x);
% \draw[world line t] (-\xmax, \x) -- (\xmax, \x);
% }
% % BOOSTED WORLD LINE GRID
% \message{ Making world lines for boosted frame...^^J}
% \fill[mydarkred,opacity=0.05]
% (O) --++ (\ang:\xmaxp) --++ (90-\ang:\xmaxp) --++ (\ang:-\xmaxp) -- cycle;
% \fill[mydarkred,opacity=0.05]
% (O) --++ (\ang:-\xmaxp) --++ (90-\ang:-\xmaxp) --++ (\ang:\xmaxp) -- cycle;
% \foreach \i [evaluate={\x=\i*\D;}] in {1,...,4}{
% \message{ Running i/N=\i/\Nlines, x=\x...^^J}
% \draw[world line'] (\ang:-\x) --++ (90-\ang:-\xmaxp);
% \draw[world line'] (90-\ang:-\x) --++ (\ang:-\xmaxp);
% \draw[world line'] (\ang:\x) --++ (90-\ang:\xmaxp);
% \draw[world line'] (90-\ang:\x) --++ (\ang:\xmaxp);
% }
% % AXES
% \draw[->,thick] (0,-\xmax) -- (T) node[left=-1] {$ct$};
% \draw[->,thick] (-\xmax,0) -- (X) node[below=0] {$x$};
% \draw[->,thick,mydarkred] (90-\ang:-\xmaxp) -- (T')
% node[right=5,above=-1] {you: $ct'$};
% \draw[->,thick,mydarkred] (\ang:-\xmaxp) -- (X') node[right=-1] {$x'$};
% % ANGLES
% \draw pic[->,"$\theta$",draw=black,angle radius=34,angle eccentricity=1.2] {angle = X--O--X'};
% \draw pic[<-,"$\theta$",draw=black,angle radius=34,angle eccentricity=1.2] {angle = T'--O--T};
% % PHOTON
% \draw[photon] (O) --++ (4*\d,4*\d);
% \end{tikzpicture}\end{center}\end{solution}
\problem{}
A caveat to Lorentz boosts is that we cannot boost to reference frames which are at the speed of light or faster. Based on the diagram given, why can't we do that? \\
{\em "If my calculations are correct, when this baby hits $c$, you're gonna see some serious stuff."}
\begin{solution}
If boosted to $c$, our axes would overlap, compressing time and space into one. If we boosted past $c$, our axes would flip, making the past the future and the future the past.
\end{solution}
\vfill
\problem{}
This diagram implies some strange things. We'll spend the next sections discussing some consequences of this, but take a minute to note anything weird that you notice.
In particular, look at a unit of time (or a unit of length) in your frame vs the rest frame.
Which is longer, one unit of time in your reference frame or in the rest frame?
Which is longer, one unit of distance in your reference frame or in the rest frame?
Consider the implications of a slanted space line.
What does it mean if two events both lie on this line?
\vfill
\newpage