Threshold Speed: \(v=c\)
The next threshold value for the spaceships outbound speed explored is \(v=c\). Generally when \(v=c\), however, then not only is the spaceships speed \(c\) relative to Earth on the way out, but it is \(c\) relative to the planet on the way out, it is \(-c\) relative to Earth on the way back, and it is \(-c\) relative to the planet on the way back. Here \(\Delta t_{out}=\Delta t_{back}\): the spaceship takes the same amount of time to reach the planet as it does to return.
In the example, the only quantity changed is the speed \(v\) of the spaceship. At \(v=c\), the spaceship catches up to the planet at \(x_{turn}=11.11\) light years at time \(t_{out}=11.11\) years after launch. The speed of return is \(w=-c\). The time it takes for the ship to return is \(\Delta t_{back}=11.11\) years, meaning the total time for the trip, as measured on Earth, is \(\Delta t_{tot}=22.22\) years.
What Earth observers see, in general, is quite different from the classical non-relativistic cases. Neglecting redshifting effects, the spaceship would appear to travel out to the planet normally, but would arrive back on Earth at the same time that the spacecraft appears to arrive at the planet. In fact, light from the entire journey back to Earth would arrive at the same time the spacecraft itself arrived back on Earth. This is because as perceived from Earth, the spacecraft, returning at speed \(c\), rides alongside all of the photons it releases toward Earth on the way back.
In the specific example, the \(v=c\) spaceship is seen to arrive back on Earth when \(x_{back}(t_{tot})=0\), which Eq. (\ref{iback}) gives as \(\tau_{back}=\tau_{turn}=\) 22.22 years. The solid lines in Figures \ref{superslow} and \ref{superfast} depict the world lines of this ship.
Spaceship speeds \(c<v<c^{2}/u\)
To explore the question as to how faster-than-light motion can lead to backward time travel, it is now supposed that superluminal speeds are possible for material spaceships. In general, as greater spacecraft speeds \(v\) are considered in the range \(c<v<c^{2}/u\), the magnitude of the speed of the spacecraft’s return \(w\) increases without bound. Here, in general, \(\Delta t_{back}<\Delta t_{out}\).
In the specific example, the spaceship speed is now taken to be \(v=5c\). The spaceship catches up with the planet at \(x_{turn}=10.20\) light years at time \(t_{out}=2.041\) years after launch. The return speed is \(w=-9.800c\) so that it takes the spaceship \(\Delta t_{back}=1.041\) years to get back to Earth. Earth receives the spaceship back after 3.082 years away.
What Earth observers see, in general, is perhaps surprising, as tracked by Eqs. (\ref{iout}) and (\ref{iback}). First the spacecraft appears to leave for the planet as normal. Next, however, two additional images of the spacecraft appear on Earth on the Landing Pad, one of which stays on the Landing Pad, while the other image immediately appears to leave for the planet. The underlying reason for these strange apparitions is that spacecraft itself returns to Earth before two images of the spacecraft return to Earth. Therefore, after the spacecraft returns, the Earth observer sees not only the returned spacecraft, but an image of the spacecraft on the way out, and an image of the spacecraft on the way back, all simultaneously.
In the specific example, an image of the \(v=5c\) spacecraft is seen moving toward the planet and arriving, as determined by Eq. (\ref{iturn}) at \(\tau_{turn}=12.24\) years, even though the spacecraft itself arrived back on Earth earlier – after only 3.082 years. The light dotted line in Figure \ref{superfast} describes the world line for this trip’s journey.
It is particularly illuminating to consider what is visible from Earth five years after the spacecraft left the Launch Pad, after the actual spacecraft has arrived back on Earth but before the spacecraft appears to have arrived at the planet. First, assuming the returned spacecraft has remained on the Landing Pad, there is the image of the returned spacecraft remaining on the Landing Pad. Next, an image of the spacecraft going out to the planet is visible back on Earth. Focusing on this image of the outbound craft, one can solve Eq.(\ref{iout}) for \(x_{out}\) to find that \(x_{out}=4.167\) light years from Earth when this image of the outbound ship arrives back on Earth. Last, a third image – an image of the spacecraft on its return back from the planet – is simultaneously visible back on Earth. At the five-year mark, one can solve Eq. (\ref{iback}) to find that \(x_{back}(5\ {\rm years})=2.135\) light years. Therefore, Earth sees this third image as the spaceship is returning to Earth, but still 2.135 light years distant.
To recap, five years after leaving the Launch Pad, three images of the spacecraft are visible on Earth. One image is emitted by the spacecraft as it remains sitting on the Landing Pad after its return, another image is emitted by the spacecraft on its way to the planet, and a third image is emitted by the spacecraft on its way back from the planet.
It is further illuminating to check on the spacecraft images still visible on Earth eight years after it left the Launch Pad, three years after the previous check. At eight years, assuming the spaceship remains on the Landing Pad, an image of the returned spaceship remains visible on the Landing Pad. Solving Eq. (\ref{iout}) for \(x_{out}\) now shows the “outbound image” of the ship at location 6.667 light years from Earth, further out than it was before. The image of the returning spaceship shows from Eq. (\ref{iback}) the ship at a distance 5.477 light years from Earth. This might appear odd as the return image arriving at Earth eight years out shows the ship as further away from Earth – not closer – than the image of the returning craft that arrived after five years. It therefore appears that this return spaceship is moving backward in time, as seen from Earth.
For clarity, to recap again, eight years after leaving the Launch Pad, three images of the spacecraft remain visible from Earth: one image on the Landing Pad, one image on the way out, and one image on the way back. Both the outbound image and the return image show the ship appearing further away after year eight than at year five.
Right at the time the spaceship returns to Earth, the number of spaceship images visible on Earth jumps from one to three. Before this, Eq. (\ref{iback}) shows that both the image of the spaceship on the Landing Pad and an image of the spaceship returning to the Landing Pad have yet to reach Earth. Therefore for \(v\) in this interval, the spaceship reaching the Landing Pad marks an image pair creation event.
Similarly, when the spacecraft reaches the planet, both the outbound and the return images of this event arrive back at Earth simultaneously, as can be seen from Eqs. (\ref{iout}) and (\ref{iback}). Because no further images of the spacecraft going out or returning exist, these images then both disappear, leaving only the spacecraft image on the Landing Pad. This disappearance is an image pair annihilation event. These image pair events are conceptually similar to spot pair events seen for non-material illumination fronts moving superluminally. \citep{Nemiroff2015} Image events are entirely perceptual – the actual location of the spaceship is given at any time by Eqs. (\ref{xout}) and (\ref{xback}). Observers at other vantage points – or in other inertial frames – may see things differently, including possible different relative timings of image pair creation and annihilation events. Also, in this velocity range, since the location of the spacecraft \(x(t)\) in the Earth frame is unique, it is clear that only one spaceship ever exists at any given time.
Threshold Speed: \(v=c^{2}/u\)
At the threshold speed \(v=c^{2}/u\), the speed of return \(w\) of the spacecraft to Earth diverges as the denominator of Eq. (\ref{weq}) goes to zero. Although slight variations of \(v\) below and above this threshold speed will yield \(w\) divergences to positive or negative infinity, the negative infinity realization will be discussed. At formally infinite return speed \(w\), the time it takes for the spacecraft to return to Earth is \(\Delta t_{back}=0\). Therefore, as soon as the spaceship reaches the planet it arrives back on Earth. To be clear, what returns to Earth immediately is not only an image of the spaceship as perceived on Earth, but the actual physical spaceship itself.
In a specific example, the spaceship speed is now taken to be \(v=c^{2}/u=10c\). This spaceship catches up with the planet at about \(x_{turn}=10.10\) light years at time \(t_{out}=1.010\) years after launch. With zero return time, Earth receives the spaceship back after \(\Delta t_{tot}=1.010\) years away. The dashed line in Figure \ref{superfast} describes the world line for this trip’s journey.
In general, the Earth-bound observer sees the same series of events as perceived when \(c<v<c^{2}/u\), they just happen a bit more compact in time. First, the spaceship is seen leaving. Next, a pair of spaceships appears on Earth on the Landing Pad. One ship from this image pair immediately leaves for the planet, while the other spaceship – and its image – remain on Earth. Both outbound spaceship images appear to reach the planet at the same time, and both then disappear from view.
In the specific example when \(v=10c\), the Earth observer first sees the spaceship leave the Launch Pad at \(t=0\). At \(t=1.010\) years, the Earth observer suddenly sees two images appear on the Landing Pad, one of which immediately takes off – time reversed – toward the planet, while the other image stays put. At \(t=11.11\) years, the Earth observer sees both the outbound and return spacecraft images reach the planet, and both disappear.
Spaceship speeds \(c^{2}/u<v<c^{2}/u+c\sqrt{c^{2}/u^{2}-1}\)
For spaceship velocities \(c^{2}/u<v<c^{2}/u+c\sqrt{c^{2}/u^{2}-1}\), the return velocity \(w\), in general, becomes formally positive. Since the spacecraft never moves toward the Earth in this scenario, how can it return to Earth? Although such a conundrum may seem like an end to a physically reasonable scenario, a physically consistent sequence of events does exist that is compatible with the formalism. This sequence is as follows. A spaceship leaves the Launch Pad on Earth for the planet. At a later time, as measured on Earth, a physical pair of spaceships materializes on the Landing Pad on Earth. One of these spaceships immediately goes off to the planet, while the other spaceship remains in place on the Landing Pad. Eventually both the spaceship that initially left the Launch Pad and the spaceship that later left the Landing Pad arrive at the planet simultaneously and dematerialize into nothing.
There is a fundamental difference between this sequence of events and the events when \(c<v\leq c^{2}/u\). When the spacecraft has speeds in this range, a real pair creation event occurs on the Landing Pad, and a real pair annihilation event occurs at the planet. These are not images, but are consistent with the location(s) of the actual spacecraft(s) to any observer in the inertial frame of Earth, as computed by Eqs. (\ref{xout}) and (\ref{xback}). Interpreting these equations as describing physical spaceship pair events is a natural extension of the image pair events that occurs at lower speeds. Although observers in Earth’s inertial frame that are located off the Earth may perceive events and sequences of events differently, they all must use the actual locations of the spacecraft as computed in the inertial frame of the Earth as the basis for what they see.
In a specific example, the spaceship speed is now taken to be \(v=15c\). The spaceship takes off from the Launch Pad at \(t=0\). Eq. (\ref{tout}) gives the outbound time as \(\Delta t_{out}=0.6711\) years and Eq. (\ref{tback}) gives the return time as \(\Delta t_{back}=-0.3378\) years, so that the total time the spaceship is away from Earth is \(\Delta t_{tot}=0.3333\) years. Therefore the next thing that happens is that a pair of spaceships materialize on the Landing Pad at \(t=0.3333\) years. One spacecraft stays on the Landing Pad. The speed of the spaceship that leaves the Landing Pad is from Eq. (\ref{weq}) \(w=29.80c\). Therefore, even though this ship left later, it is just the right amount faster to arrive at the planet at the same time as the spacecraft that left the Launch Pad. Both spacecraft catch up to the planet when it is, from Eq. (\ref{xturn}), \(x_{turn}=10.07\) light years distant. This occurs at is \(t=0.6711\) years. At this time, both outbound spacecraft merge and dematerialize.
A potential point of confusion is that the equation for the location of the spacecraft that left the Landing Pad, Eq. (\ref{xback}), formally returns a negative valued location for the spaceship when \(t<0.3333\) years. It is claimed here that such locations are outside of the described scenario and so do not occur. The two spaceships that materialize at \(t=0.3333\) years on the Landing Pad do not have previous positions described in this scenario. The situation is similar to the equation for the spaceship that left the Launch Pad, Eq. (\ref{xout}). This equation also does not indicate that the outbound spaceship that leaves the Launch Pad occupied negative valued locations before \(t=0\), as such positions are outside the described scenario and do not occur.
Surprisingly, perhaps, what Earth observers see, in general, is not conceptually different from events perceived when the spacecraft has \(c<v\leq c^{2}/u\) as described in the previous two sections. Still the first event witnessed is the launching of the spacecraft from the Launch Pad. Next, Earth observers see a pair of spacecraft appear on the Landing Pad, one of which stays put and the other goes off to the planet. Last, the observers see both spacecraft arrive at the planet at the same time and disappear.
In the specific example, a spaceship is seen from Earth to leave the Launch Pad at \(t=0\). Suddenly, at \(t=0.3378\) years, a pair of spaceships appear on the Landing Pad, one of which stays there, and the other leaves for the planet. The spaceship that leaves the Landing Pad appears time reversed and faster than the spaceship that left Launch Pad. The images of the spacecraft arriving at the planet, described by Eq. (\ref{iout}) and Eq. (\ref{iback}), arrive back on Earth at \(t=10.74\) years, where the images appear to merge and disappear.