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.
\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.
% 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}{
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.
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?
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.
