Threshold Speed: \(v=c^{2}/u+c\sqrt{c^{2}/u^{2}-1}\)

The spacecraft speed \(v=c^{2}/u+c\sqrt{c^{2}/u^{2}-1}\) is a threshold value because here \(\Delta t_{back}=-\Delta t_{out}\), meaning that \(\Delta t_{tot}=0\) and that the total time it takes the spacecraft to go out to the planet and return back to Earth is zero. This speed is one of two formal solutions to setting \(\Delta t_{tot}\), as given in Eq. (\ref{ttot}), equal to zero. The other formal solution, \(v=c^{2}/u-c\sqrt{c^{2}/u^{2}-1}\) always results in a \(v<u\) and so is discarded because it describes a scenario where the spaceship never reaches the planet.
The general scenario has a spacecraft traveling in this speed range leave from the Launch Pad toward the planet. At the same time, a pair of spacecraft together materialize on the Landing Pad, one of which also immediately leaves for the planet, while the other spacecraft remains in place. The two simultaneously launched spacecraft both approach the planet, arrive simultaneously, and then de-materialize together at the planet.
In a specific example, the spacecraft is assigned the speed \(v=19.95c\). The return speed is computed from Eq. (\ref{weq}) to be, also, \(w=19.95c\). Therefore, at \(t=0\), one spacecraft leaves the Launch Pad, while another spacecraft leaves the Landing Pad. A third spacecraft remains in place on the Landing Pad. Both outgoing spacecraft reach the planet at \(t=0.5038\) years when the planet is \(x_{turn}=10.05\) light years from Earth. At this time, both spacecraft merge and dematerialize.
What is perceived as happening for spacecraft in this general speed range is qualitatively the same as what happens. Earth observers see a spacecraft image launch for the planet from the Launch Pad and, simultaneously, a second spacecraft image leave for the planet from the Landing Pad. The image of the other member of the spacecraft pair that appears on the Landing Pad stays put. Images of both outgoing spacecraft are seen to approach the planet, arrive at the planet at the same time, and disappear together when they reach the planet.
In the specific example of \(v=19.95c\), Earth observers see an image of the spacecraft leave from the Launch Pad – and another image leave from the Landing Pad – at \(t=0\). Images of both spaceships are seen to arrive at the planet at \(t=10.55\) years, whereafter the images merge and disappear.

Spaceship speeds \(v>c^{2}/u+c\sqrt{c^{2}/u^{2}-1}\)

Cases where faster-than-light motion leads to closed-loop backward time travel can finally be explained as a logical extension of previously discussed results. In cases of increasing \(v\) where \(v>c^{2}/u+c\sqrt{c^{2}/u^{2}-1}\), in general, the “return back” speed \(w\) is not only positive – and so indicating motion away from the Earth – but decreasing – and so indicating slower motion toward the planet. The result is that the spacecraft “returns back” to the Landing Pad before it leaves from the Launch Pad. Surprisingly, this scenario does not give the straightforward out-and-back sequence of events commonly assumed for superluminal time-travel. On the contrary, this scenario relies on pair-creation and pair-annihilation events.
The first scenario event for spacecraft speeds in this general speed range is that two spacecraft appear on the Landing Pad, one of which immediately sets off for the planet. The next event, as described in the Earth frame, is that the initial spacecraft takes off from the Launch Pad and heads out toward the planet. Both the spacecraft that left from the Landing Pad and the spaceship that left from the Launch Pad reach the planet at the same time and de-materialize.
In the specific example of \(v=30c\), Eq. (\ref{weq}) now yields a “return back” speed of \(w=14.95c\), a positive value that describes movement away from the Earth. Note also that \(w<v\), so that \(w\) describes less rapid outward motion. Further, Eq. (\ref{tout}) shows \(\Delta t_{out}=0.3344\) years, while Eq. (\ref{tback}) shows \(\Delta t_{back}=-0.6711\) years, so that their sum has \(\Delta t_{tot}=-0.3367\) years, meaning that the spaceship “returns” 0.3367 years before it left. The dot-dashed line in Figure \ref{superfast} describes the world line for this trip’s journey.
Eq. (\ref{xout}) describes the spaceship that left from the Launch Pad at \(t=0\). Eq. (\ref{xturn}) shows that this \(v=30c\) spacecraft catches up with the planet at \(x_{turn}=10.03\) light years from Earth. The time this spaceship catches up to the planet is at \(t_{out}=0.3344\) years.
Eq. (\ref{xback}) describes the spaceship that left from the Landing Pad at \(t=-0.3367\) years. This spacecraft also catches up to the planet at \(t=0.3344\) years. Times greater than \(t=0.3344\) years yield distance values in Eq. (\ref{xback}) larger than that of the planet, but these are considered unphysical because the scenario gives the boundary condition that the spacecraft turns around at the planet. Similarly, times less than \(t=-0.3367\) years yield negative distance values in Eq. (\ref{xback}), but these are also considered unphysical because the scenario states a boundary condition that the spacecraft travels only between Earth and the planet. One might argue that all times before \(t=0\) are similarly unphysical because the scenario dictated that the spaceship launched at \(t=0\), but no such temporal boundary condition was placed on the time of return.
It is educational to query the locations of the \(v=30c\) outbound and “return back” spaceships during their journeys to see how they progress. At time \(t=-0.3367\) years, Eq. (\ref{xback}) indicates that the return spaceship has a location of \(x_{back}(-0.3367)=0\) light years, meaning the “return back” spaceship is still on Earth. At time \(t=-0.1\) years, this equation indicates that the return spaceship has a location \(x_{back}=3.538\) light years from Earth in the direction of the planet. At time \(t=0\) years, Eq. (\ref{xout}) indicates that \(x_{out}(0\ \rm{years})=0\) light years, meaning that the “outbound” spaceship is still on Earth, while Eq. (\ref{xback}) indicates that \(x_{back}=5.033\) light years from Earth. At time \(t=0.1\) years the equations hold that \(x_{out}(0.1\ \rm{years})=3.000\) light years, while \(x_{back}(0.1\ \rm{years})=6.528\) light years. Next, at \(t=0.2\) years, the equations hold that \(x_{out}(0.2\ \rm{years})=6.000\) light years, while \(x_{back}(0.2\ \rm{years})=8.023\) light years. Finally, at \(t=0.3344\) years, both \(x_{out}(0.3344\ \rm{years}\)) and \(x_{back}\ \rm{years}\) yield 10.03 light years.
What is seen on Earth in general for speeds in this range is again qualitatively similar to what physically happens. First, two spacecraft images are seen to appear on the Landing Pad, one of which is seen to launch immediately for the planet, while the other appears to stay put. Next, an image of the spacecraft on the Launch Pad launches for the planet. Both the image of the spacecraft that left from the Landing Pad and the image of the spacecraft that left from the Launch Pad appear to reach the planet at the same time. These two images merge and disappear.
In the specific example where \(v=30c\), Earth sees the spaceships that materialized on the Landing Pad at \(t=-0.3367\) years with no time delay because this Landing Pad is on Earth. The spaceship that stays on the Landing Pad is always seen on Earth to be on the Landing Pad with no time delay, from \(t=-0.3367\) years onward. The spaceship that left the Landing Pad for the planet at \(t=-0.3367\) is seen with increasing time delay due to the travel time of the light between the spaceship and Earth. At \(t=-0.1\) years, the “return back” spaceship is at \(x_{back}=3.538\) light years from Earth but because of light travel time, is seen when it was only at \(x_{back}=0.2218\) light years away. At \(t=0\) years, a spaceship is seen to take off from the Launch Pad, while the spaceship that left the Landing Pad while actually at \(x_{back}(0)=5.033\) light years distant, appears as it did when at \(x_{back}=0.3156\) light years distant. At \(t=0.2\) years \(x_{out}(0.2\ \rm{years})\) = 6.000 light years, but due to light travel time, this “outbound” spaceship appears as it did when at \(x_{out}=0.1935\) light years. Similarly, at \(t=0.2\) years, the “return back” spaceship is at \(x_{back}(0.2\ \rm{years})=8.023\) light years distant, but due to light travel time appears as it did when at \(x_{back}=0.5030\) light years. Images of both spaceships arriving at the planet are received back on Earth at \(t=10.367\) years. At this time, both images merge and disappear.
Even before the outbound spaceship leaves from the Launch Pad, astronauts on the spaceship that appeared and remained on the Landing Pad may come out, recount their journey, and even watch the subsequent launch of their spacecraft on the nearby Launch Pad. A physical conundrum occurs, for example, if these astronauts go over to the Launch Pad and successfully interfere with the initial spacecraft launch. This would create a causal paradox that may reveal any time travel to the past to be unphysical. \citep{Hossenfelder2012} Alternatively, however, the presented scenario may proceed but the disruption event may be disallowed by the Novikov Chronology Protection Conjecture \citep{Novikov1992} – or a similarly acting physical principle. Then, try as they might, the astronauts could find that they just cannot disrupt the launch. \citep{Lloyd2011} In a different alternative, such disruption actions may be allowed were the universe to break into a sufficiently defined multiverse, with the disruption just occurring in a different branch of the multiverse \citep{Everett1957} than the one where the spacecraft initially launched. Then, life for the disrupting astronauts would continue on normally even after they disrupted the launch, even though they could remember this launch. It is not the purpose of this work, however, to review or debate causal paradoxes created by backward time travel, but rather to show how some superluminal speeds do not lead to closed-loop backward time-travel, while other speeds do, but by incorporating non-intuitive pair creation and annihilation events.

When \(v\) Diverges

Perhaps counter-intuitively, an infinite amount of backward time travel does not result when \(v\) diverges. In the general case, as \(v\) approaches infinity, the time it takes for the spacecraft to reach the planet, \(\Delta t_{out}\), approaches zero. However, Eq. (\ref{tback}) shows that the time it takes for the spacecraft to return to Earth, \(\Delta t_{back}\), does not approach negative infinity but rather \(\Delta t_{back}\sim-x_{po}u/c^{2}\). The reason for this is that, in Eq. (\ref{weq}), \(w\) only approaches \(c^{2}/u\) as \(v\) diverges, which is always superluminal and never zero. Therefore, the faster the planet is moving away from the Earth, and the further the planet is initially from the Earth, the further back in time the returning spaceship may appear on the Landing Pad.
In the specific example, diverging \(v\) leads to a maximum backward time travel according to Eq. (\ref{tback}) of \(\Delta t_{back}=1.000\) year. The return speed \(w\) approaches 10 c. Therefore, in this scenario, the earliest a spaceship pair could appear on the Landing Pad would be \(t=-1.000\) year, one year before the outbound spaceship leaves the Launch Pad. After this, events would unfold qualitatively as described in the last section. The triple dot-dashed line in Figure \ref{superfast} describes the world line for this trip’s journey.