OUT AND BACK AGAIN

The scenario explored here is extremely simple: an object goes out and comes back again. The return trip is important to ensure a “closed-loop”. Therefore, the scenario described can be thought of as an extension of the famous twin paradox to superluminal speeds. For the sake of clarity, to promote interest, and to place distances on scales where temporal effects correspond with common human time scales, the initial launching location will be called “Earth”, which can also be thought of as representing the twin that stays at home. The ballistic projectile will be referred to as a “spaceship”, which can also be thought of as representing the twin the travels away and then comes back. The turnaround location will be referred to as a “planet”. Furthermore, an example where the distance scale is on the order of light-years will be described concurrently.
The following conventions are observed. In general, unless stated otherwise, all times and distances will be given in the inertial frame of the Earth and from the location of Earth. All relative motion for the spaceship will take place in the line connecting the Earth to the planet, here defined as the \(x\) axis. All velocities are assumed constant. The planet is assumed moving away from the Earth at the subluminal speed \(u\) as measured in the Earth’s inertial frame. Times are given by the variable \(t\), and the standard time when the spaceship is scripted to leave Earth is set to \(t=0\). At this time the spaceship leaves Earth from a location called the Launch Pad, and aims to return to an Earth location called the Landing Pad. The distance on Earth between the Launch Pad and the Landing Pad is considered negligible. Velocities away from the Earth are considered positively valued, and velocities toward the Earth are considered negatively valued. The outbound velocity of the ship relative to the Earth is given by \(v\), the return velocity of the ship relative to the planet is given by \(-v\), and the return velocity of the ship relative to the Earth is designated \(w\). Keeping both spaceship speeds at magnitude \(v\) is a useful didactic simplification that demonstrates the logic of a much larger set of event sequences when the outgoing and incoming spaceship speeds are decoupled.
At time \(t=0\), the planet is designated to be a distance \(x_{po}\) and moving at positive velocity \(u\) away from the Earth, with respect to the Earth. Therefore, at time \(t\) the distance between Earth and the planet is simply
\begin{equation} x_{p}(t)=x_{po}+ut.\\ \end{equation}
Similarly, at time \(t=0\), the spacecraft leaves Earth from the Launch Pad. After launch and before reaching the planet, the distance between Earth and the spaceship is
\begin{equation} \label{xout}x_{out}(t)=vt.\\ \end{equation}
The x-coordinate will usually describe the spaceship and so, when it does, no subscript will be appended.
The spacecraft reaches the planet at the time when \(x(t)=x_{p}(t)\). Combining the above two equations shows that the amount of time it takes for the spacecraft to reach the planet is
\begin{equation} \label{tout}t_{out}=\Delta t_{out}={x_{po}\over v-u}.\\ \end{equation}
The distance to both the spacecraft and the planet at this time is
\begin{equation} \label{xturn}x_{turn}={vx_{po}\over v-u}.\\ \end{equation}
The spaceship turns around at the planet. For this calculation, the turnaround is considered instantaneous but the important point is that the turnaround duration is small compared to other time scales involved. After turnaround, the velocity of the spaceship relative to the Earth is
\begin{equation} \label{weq}w={u+v\over 1+uv/c^{2}},\\ \end{equation}
where \(u\) is measured relative to the Earth but \(v\) is measured relative to the planet. Only in Eq. (\ref{weq}) is \(v\) negative as it describes the spaceship returning back to Earth – in all other equations \(v\) is to be considered positive as it refers to the speed of the ship leaving Earth. Note that Eq. (\ref{weq}) is the standard equation of velocity addition in special relativity and one that has been consistently invoked even when superliminal speeds are assumed. \citep{Mermin2005}, \citep{Hill2012} Because of the centrality of this equation to physics, it is retained here in its classic form.
After the spaceship leaves the planet, its distance from the Earth is given by
\begin{equation} \label{xback}x_{back}(t)=x_{turn}+w(t-t_{out})={vx_{po}\over v-u}+{(u-v)(t-x_{po}/(v-u))\over 1-uv/c^{2}}.\\ \end{equation}
The scenario is defined so that once the spaceship returns to Earth, it lands on the Landing Pad and stays there. The spacecraft can only move between the Earth and the planet – it does not go past the Earth to negative \(x\) values.
The time it takes for the spaceship to return back to the Earth from the planet is
\begin{equation} \label{tback0}\Delta t_{back}=-x_{turn}/w.\\ \end{equation}
The negative sign leading this equation is necessary to make the amount of return time positive when \(w\) is negative. Writing \(\Delta t_{back}\) in terms of (positively valued) \(u\), \(v\), and \(x_{po}\) yields
\begin{equation} \label{tback}\Delta t_{back}={vx_{po}(1-uv/c^{2})\over(u-v)^{2}}.\\ \end{equation}
The total time that the ship takes for this trip is
\begin{equation} \label{ttot}\Delta t_{tot}=t_{tot}=t_{out}+\Delta t_{back}={x_{po}(2v-u-uv^{2}/c^{2})\over(v-u)^{2}},\\ \end{equation}
where, again, all interior speeds are defined as being positively valued and \(v>u\). Were the planet stationary with respect to Earth, parameterized by \(u=0\) then \(\Delta t_{out}=\Delta t_{back}=\Delta t_{tot}/2=x_{po}/v\) which agrees with the non-relativistic classical limit, no matter the (positive) value of \(v\).
It is also of interest to track how long it takes light signals to go from the spaceship to Earth, as measured on Earth. The time after launch that an Earth observer sees the spaceship at position \(x(t)\) will be labeled \(\tau(t)\). Since light moves at \(c\) in any frame, then Earth observers will see the outbound spaceship at position \(x(t)\) at time
\begin{equation} \label{iout}\tau_{out}(t)=x_{out}(t)/v+x_{out}(t)/c,\\ \end{equation}
where the first term is the time it takes for the spacecraft to reach the given position, and the second term is the time it takes for light to go from this position back to Earth. The time that Earth observers will see the spacecraft reach the planet is
\begin{equation} \label{iturn}\tau_{turn}=x_{turn}/v+x_{turn}/c.\\ \end{equation}
During the spaceship’s return back to Earth, Earth observers will see the spaceship at \(x_{back}(t)\) such that
\begin{equation} \label{iback}\tau_{back}(t)=x_{turn}/v+(x_{back}(t)-x_{turn})/w+x_{back}(t)/c.\\ \end{equation}
The first term is the time it takes for the spacecraft to reach the planet, the second term is the time it takes for the spacecraft to go from the planet to intermediate position \(x_{back}(t)\), and the third term is the time it takes for light to reach Earth from position \(x_{back}(t)\).
A series of threshold \(v\) values occur, which will be reviewed here in terms of increasing magnitude.

Threshold Speed: \(v=u\)

The first threshold speed explored is \(v=u\). Below this speed, the spaceship is moving too slowly to reach the planet. When \(v=u\), the spaceship has the same outward speed as the planet and only reaches the planet after an infinite time has passed. This is shown by the denominators going to zero in Eqs. (\ref{tout}, \ref{tback}, and \ref{ttot}). Earth observers will see both the spaceship and the planet moving away in tandem forever.

Spaceship speeds \(u<v<c\)

When \(u<v<c\), and when both the spaceship and the planet are moving much less than \(c\), then Earth observers see the spaceship move out to the planet and return back to Earth in a normal fashion that is expected classically.
When \(u<v<c\) generally, then \(\Delta t_{back}>\Delta t_{out}\). This results from the magnitude of the spacecraft’s speed coming back to Earth, \(w\), being less than the magnitude of the spacecraft’s speed going out to the planet, \(v\), even though the distance traveled by the spacecraft is the same in both cases: \(x_{turn}\). In this speed range, the occurrence of events in the Earth’s inertial frame proceeds as expected in non-relativistic classical physics. As tracked from the Earth, the spaceship simply goes out to the planet, turns around, and returns.
For clarity, a series of specific numerical examples are given, with values echoed in Table \ref{table1}, and world lines depicted in the Minkowski spacetime diagrams of Figures \ref{superslow} and \ref{superfast}. In all of these examples, the planet starts at distance \(x_{po}=10\) light years from Earth when \(t=0\), and the planet’s speed away from the Earth is \(u=0.1c\). For the speed range being investigated in this sub-section, the spaceship has speed \(v=0.5c\). Then by Eq. (\ref{xturn}) the spaceship reaches the planet when both are \(x_{turn}=12.50\) light years from Earth. The duration of this outbound leg is given by Eq. (\ref{tout}) as \(\Delta t_{out}=25.00\) years. For clarity, the speed of the ship’s return is computed from Eq. (\ref{weq}) as \(w\sim-0.4211c\), to four significant digits. The duration of the ship’s return back to Earth is given by Eqs. (\ref{tback0}, \ref{tback}) as \(\Delta t_{back}\sim 29.69\) years. The total time that the spaceship is away is the addition of the “out” time and the “back” time, which from Eq. (\ref{ttot}) is \(\Delta t_{tot}=54.69\) years. The world line of this ship’s trip is given by the dashed line in Figure 1.
Due to the finite speed of light, Earth observers perceive the spacecraft as arriving at the planet only after the equations indicate that it has already started back toward Earth. The closer \(v\) is to \(c\), the closer the spacecraft is to the Earth, as defined by Eq. (\ref{iturn}), when Earth observers see the spaceship arrive at the planet.
In the concurrent example of \(v=0.5c\), the spaceship reaches the planet at time \(t=25.00\) years, but light from this event does not reach Earth until the time given by Eq. (\ref{iturn}) – \(\tau_{turn}=37.50\) years, well after the spaceship has actually left the planet. However, the ship is first seen on Earth to arrive back at Earth when it actually arrives back on Earth – 54.69 years after it left.