\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.

\makeatletter
\if@solutions\else
	\emptydiagramc{Alice}
\fi
\makeatother

\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?

	\makeatletter
	\if@solutions\else
		\halfdiagramcwide{Ground}
	\fi
	\makeatother

	\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}


\makeatletter
\if@solutions\else
	\halfdiagramcwide{Ground}
\fi
\makeatother


\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]
	\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
	\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
		\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}


\vfill
\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?

\makeatletter
\if@solutions\else
	\emptydiagramc{Rest}
\fi
\makeatother
\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. \par
	In the boosted frame, the spaceship on the right accelerates first,
	stretching the floss, causing it to break. \par
	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