\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.}

	\makeatletter
	\if@solutions\else
		\halfdiagramc{Mark}
	\fi
	\makeatother


	\begin{solution}
		% SPACETIME DIAGRAM of TWIN PARADOX
		\begin{tikzpicture}[scale=2.0]
			\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
			\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