# 1.3: A Trip to Alpha Centauri

- Page ID
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Let us take a trip to Alpha Centauri and leave our twin behind. This is a G-type star that is *4.37 light-years* from Earth’s Sun,* so if we travel at \(v\) = 0.6 \(c\), then the one-way Earth coordinate time will be \(t=\frac{d}{v}=\frac{4.37 c \ yr}{.6 c}=7.283 \ yr\), or a round-trip Earth coordinate time of *14.56 years*. (Note that we have written the distance unit *light-year* as \(c\)* year*, since this allows us to manipulate the units correctly; note also that \(c / c\) = *1* just as *2/2* = *1* or *pig/pig* = *1*, and multiplying an expression by one gives us that expression back.)

Now you and your twin agree to communicate using light pulses the whole time. Your twin will send 10 pulses of light over the *14.56 years*, or one pulse every *1.456 years*. Since you are trying to outrun the pulses (though failing) on the outbound trip, you will see fewer pulses, at longer intervals, than your twin sends. On the return trip, the pulses you see will pile up faster than they are sent.

We wish to actually see what happens on this trip, so I will show screenshots from a 1993 Macintosh OS 9 application called RelLab, which allows us to program in the motion of our flying saucer between these stars as well as the propagation of the light pulses we use to signal with the twin left on Earth. The initial setup is shown in Figure \(\PageIndex{1}\). In the upper left-hand corner, one sees “Frame: Earth,” which indicates that this is the frame of reference for our twin left on Earth. That means that what we are seeing is what is transpiring in the Earth coordinate system, courtesy of an omnipotent observer (defined as a collection of observers all at rest with respect to the Earth and with watches synchronized). No individual observer at rest with respect to the Earth, such as our twin, “sees” this omnipotent view, but she can construct it from all such observers if each of them sent her messages detailing the arrival times of the pulses of light at their locality.

Note

*Due to the extreme brightness of Alpha Centauri, the new Gaia space telescope has, as of 2019, not given a parallax measurement of its distance. So we must rely on other instruments. P. Kervella, F. Mignard, A. Mérand, and F. Thévenin, *A&A* **594**, A107 (2016), used the Very Large Telescope (VLT) and the New Technology Telescope (NTT) to find the parallax. Their result was 747.17 ± 0.61 milli-arc-seconds (mas), giving a distance in parsecs (from the phrase “parallax arcseconds,” the means by which distances are found) and light-years of

*d* = *1/(0.74717 ± 0.00061)"* = *(1.3384 ± 0.0011 pc)* × *3.2616 \(*c\)* yr/pc* = *4.3653* ± *0.0036 *\(c\)* yr*.

We will round this up to *4.37* and hope that their error bars will survive the test of time. Actually, an error of even a few percentage points would not alter this story in any significant way.

I have added RelLab to an application called WPMacApp, which may be downloaded from the same page where you downloaded this book, https:// doi .org/ 10 .15760/ pdxopen -29. Open the file “Alpha Centauri Trip 3c 10Eflash” and set “Frame: Earth” in the upper left-hand corner if you wish to step through it as you read, though hopefully the screenshots below will give you a sufficient experience without needing RelLab.

You agreed to also send out pulses at the same frequency of *1.456 years*, but did you? Your twin indeed sent out a pulse at *1.456 years* that can be seen (from our omniscient view) well spread out in space in Figure \(\PageIndex{2}\), *1.90 years* after you left, but yours is a tiny ring (just visible in Figure \(\PageIndex{2}\)) to the saucer’s left, about a month after emission at *1.82 years*. Why did you delay it after promising to be faithful?

Oh, that’s right, this is simply the time-dilation factor working, with our \(\gamma\) = *1.25* for \(v\) = *0.6* \(c\): 1.25 × *1.456 yr* = *1.82 yr*. This light pulse reaches Earth at *2.912 years*, shown two months afterward in Figure \(\PageIndex{3}\). It will be easiest to do the time comparisons if we simply count such pulses returning to Earth. Let us label this **red** one “count 1” of the outbound trip. That the **red** ring of light from the saucer and the small **orange** one just emitted by the Earth continue to share a wave-front means that the light emitted by the moving saucer and from the still Earth indeed move at the same speed, c.

In Figure \(\PageIndex{3}\) and following, I have overlain colored halos around the black rings of light shown in the RelLab screenshots. Their order in time is indicated by their place in the spectrum from **red** to **lavender** to help track the rings in subsequent figures.

Figure \(\PageIndex{4}\) shows *5.096 years* into the trip, when the first (**red**) pulse sent from the moving saucer reaches Alpha Centauri. How long does it take for the second one to arrive? Just 3/4 of a year later, at *5.826 years*, the **orange** pulse seen in Figure \(\PageIndex{5}\), which is coincident with the Earth’s first (**red**) pulse’s arrival there.

Note that this is one pulse per *5.824 yr* − *5.096 yr* = *0.728 yr*, so the saucer pulses arrive at Alpha Centauri at twice the frequency they were supposed to be emitted, one per *1.456 years*. * This is the relativistic Doppler shift. The relativistic Doppler shift can also be seen by measuring the distance between the wave front of this pulse and the one just emitted by the saucer. It is about 3/4 of the one-light-year grid. (We can also double-check that the travel time of Earth’s first [**red**] pulse was \(t\) = \(d / v\) = *4.37* \(c\)* yr/c* = *4.37* yr, which, since it was emitted at *1.456 years*, gives us an arrival time of *5.826 years*.)

This increase in the frequency of the light waves is called a *blue shift* because visual colors shifted to higher frequencies move toward the blue end of the visual spectrum from their emission frequencies. For instance, red light might be blue-shifted to orange, orange light to yellow, yellow to green, green to blue, and blue to violet. One would also say that infrared light that is seen as ultraviolet light is extremely blue-shifted. (Please note that the colors in the figures are *labels* for the sequence of pulses and not the actual colors [shifted or not] of the pulses of light.)

The left edge of the third (**yellow**) pulse of light just emitted by the saucer in Figure \(\PageIndex{5}\) can be measured to be about *2.9 *\(c\) *years* away from the left edge of the second (**orange**) return pulse emitted by the saucer, which has incidentally also just reached Earth (count 2). Thus, the Earth sees the light from the saucer at half the frequency the pulses were supposed to be emitted— one per *2.912 years* instead of one per *1.456 years*, which is the relativistic Doppler red shift.

At *7.28 years*, you reach Alpha Centauri and immediately turn around (Figure \(\PageIndex{6}\)). At *8.736 years*, the third (**yellow**) saucer pulse reaches Earth (count 3), as seen in Figure \(\PageIndex{7}\).

Note

* Formally, the relativistic Doppler shift is given by

\[f=f_{0} \sqrt{\frac{1+v / c}{1-v / c}}=f_{0} \sqrt{\frac{5 / 5+3 / 5}{5 / 5+3 / 5}}=f_{0} \sqrt{\frac{8 / 5}{2 / 5}}=2 f_{0} \nonumber\]

where \(f_{0} \mathrm{i}\) is the emission frequency of light from the spaceship or star. We get the reciprocal one-half \(f_{0} \mathrm{i}\) if we change the signs of the velocities.

At *11.648 years* (Figure \(\PageIndex{8}\)), the fourth (**yellow-green**) return pulse from the outbound trip reaches Earth (count 4). You can see that the next (**blue-green**) one comes close on its heels, following in *0.728 years*. We see the frequency blue-shifted to twice the emitting frequency as the saucer now approaches the Earth. The first three pulses from the inbound trip arrive at *12.376 years* (**blue-green**) in Figure \(\PageIndex{9}\) (count 1); *13.104 years* (**cyan**), next in line in Figure \(\PageIndex{9}\) (count 2); and *13.832 years* (**blue**), seen having passed the Earth in Figure \(\PageIndex{10}\) (count 3). At *14.56 years*, the saucer reaches Earth and would have emitted its fourth pulse had it been needed at this point (count 4).

Let us sum up what we know. Your twin on Earth sees four pulses from the outbound trip at *2.912 years* between pulses for a subtotal of *11.648 years*. She also sees four pulses from the inbound trip at *0.728 years* between pulses for a subtotal of *2.912 years*. Thus, the total trip took *11.648* + *2.912* = *14.56 years*, as we expected.

What about you? You sent eight pulses at *1.456-year* intervals for a total of *11.648 years*. Note that if we multiply your proper-time interval by \(\gamma\) = *1.25*, we get the *14.56 years* that your twin experienced! All this seems to work!

One may account for the time in a third way. The (**blue**) saucer pulse having just passed the Earth in Figure \(\PageIndex{10}\) is the seventh and last (the eighth “pulse” not being emitted because the saucer has returned). The corresponding seventh (**blue**) Earth-pulse is also shown in Figure \(\PageIndex{10}\) just arriving at Alpha Centauri. Concentric and inward from that one are two more pulses—colored **purple** and **lavender**, respectively—together accounting for an additional *2* × *1.456* = *2.912 years* to add to the saucer’s *11.648 years* to give *14.56 years*. That is, the Earth has sent 10 pulses, one at the end of every *1.456 years*, for a total of *14.56 years*, with the final one not shown, since the saucer has come to rest at the time it would normally have been emitted. All three methods are consistent.

## SAUCER FRAME

We turn now to the saucer frame of reference in which the saucer is at rest and the Sun and Alpha Centauri move off to the left for almost six years and then move back to the right to return the Earth to the saucer’s position. Note that in Figure \(\PageIndex{11}\), the distance between these two stars is now about 3.5 of the *1* \(c\) *year* grid spacings. Indeed, if we apply length contraction to the moving pair of stars, the distance from the Sun to Alpha Centauri should contract to

\[\ell=\frac{L}{\gamma}=\frac{4.37 c \ y r}{1.25}=3.49 c \ y r.\]

So we would expect a travel time of

\[\tau=\frac{L}{v}=\frac{3.496 c \ yr}{.6 c}=5.827 yr\]

each way, or *11.648 years* total.

If you are running RelLab, set the clock to zero in the file “Alpha Centauri Trip 3c 10Eflash” and set “Frame: Saucer 1” in the upper left-hand corner.

Figure \(\PageIndex{12}\) shows the two stars and their planets having moved to the left for a span of *1.6 years* and the emission of the first pulse expanding almost two months after the agreedupon time of *1.456 years*. You were not lying after all about sending the pulses on time! But what about your twin left on Earth? Why did she not send off a pulse?

Figure \(\PageIndex{13}\) shows the configuration *1.92 years* into the trip. We see a small expanding ring of (**red**) light that was emitted by the Earth a little over a month prior at *1.82 years*. From an omnipotent observer at rest relative to the saucer, the Earth appears to have sent its first pulse late. What could cause this? Time dilation would seem to be the cause, since *1.25* × *1.456 yr* = *1.82 yr*. But that would mean that the saucer at rest sees the moving Earth clocks slowed and the Earth at rest sees the moving saucer clocks slowed by the same time-dilation factor. This is why Einstein called this a theory of “relativity.”

Despite the fact that one’s view is relative in this case, somehow the twin on the saucer experiences less time for the overall round trip than does the twin on the Earth. This is called the Twin Paradox. Read on for its resolution.

Figure \(\PageIndex{14}\) shows the configuration at *2.912 years*. The receding Earth has its (**red**) pulse received by the saucer at a half the emission frequency, as one would expect with a relativistic red shift: count 1 of Earth-pulses received in the outward interval.

Figure \(\PageIndex{15}\) shows the configuration at *5.824 years* when Alpha Centauri just reaches the saucer: count 2 on the outbound trip. This matches the time, \(t=d / v\), that the saucer thinks is required for Alpha Centauri to move a distance of *3.496 light-years* (to reach the saucer) at speed \(v\) = *0.6* \(c\). Note that the second (**orange**) pulse emitted by the moving Earth reaches the saucer simultaneously with Alpha Centauri, a simultaneity also seen in the other reference frame in Figure \(\PageIndex{6}\).

As we move forward by just 3 seconds (10^{−7} years), notice the huge shift in perspective of Figure \(\PageIndex{16}\). From the point of view of the saucer at rest, the Sun and Alpha Centauri have instantaneously screeched to a halt from a velocity leftward (which we write as \(v\) = *−0.6* \(c\)), reversed course, and instantaneously ramped their velocity up to \(v\) = *+0.6* \(c\) (rightward at the same speed).

But—and this is the crucial clarification—however much you on the saucer may assert the privilege of claiming to be at rest while the universe moves back and forth around you, no one on the Earth feels any acceleration as they stop moving leftward and start moving rightward. On the other hand, your body *does* feel acceleration.

What we see in our omniscient perspective in Figure \(\PageIndex{15}\) is that in the saucer’s outgoing rest frame, there is a (**red**) Earth-pulse having been received by the saucer and having moved on to the right, an (**orange**) Earth-pulse currently being received, and a (**yellow**) Earthpulse in flight yet to be received, having barely left the Earth. In the saucer’s incoming rest frame 3 seconds later, our omniscient perspective in Figure \(\PageIndex{16}\) shows the same (**red**) Earthpulse having been received, and having moved on to the right, and the same (**orange**) Earth-pulse currently being received just as in Figure \(\PageIndex{15}\). But it also shows four Earth-pulses in flight (**yellow**, **green**, **cyan**, and **blue**), yet to be received, and a fifth is about to be emitted. Those extra three (**green**, **cyan**, and **blue**), almost four, in-flight pulses *each* mark the passage of about *0.728 years* of Earth time in 3 seconds of saucer time. We should thus expect the Earth to have experienced almost *2.9 year*s more than the saucer during those 3 seconds. The saucer must be shifting from one reference frame to another reference frame for this to be possible. Note that exactly two pulses (**red** and then **orange**) have already been received in both the outgoing and incoming saucer frames: the saucer cannot go back and change its own reality by such frame shifting under acceleration.

The symmetry of the trip allows us to infer that the initial instantaneous jump from stillness to motion at the start of the saucer’s outbound trip added to the instantaneous jump from motion to stillness at the end of the saucer’s return trip would speed up the Earth’s clocks by another *2.9 years* as seen by the saucer, for a total acceleration-induced extension of Earth’s clocks of *5.8 years*. But time dilation in the unaccelerated portion of the trip slows the Earth’s clocks down, as perceived by the saucer, precisely negating this extra *2.9 years*. We should thus have Earth clocks reading *2.9* + *2.9* − *2.9* = *2.9* more years than the saucer, and we have seen that they do.

At *6.552 years*, the first (count 1) of Earth-pulses is received by the saucer on the return trip. The others come at *7.28*, *8.008*, *8.736*, *9.464*, *10.192*, and *10.92* (counting 2 , 3, . . . and 7), and at *11.64 years*, the planets come to rest and need not send the eighth pulse. The twin on the traveling Earth is back. We will not display the corresponding figures, as not that much changes between them, but you should play the full movie of the trip, available at https:// doi .org/10.15760/ pdxopen-29.

The saucer sees two pulses from Earth on the outbound trip, at *2.912 years* between pulses, for a subtotal of *5.824 years*. The saucer sees eight pulses on the inbound trip, at *0.7284 years* between pulses, for a subtotal of *5.824 years*. Thus, the total trip took *2* × *5.824 year* = *11.648 years*, the duration we expected from time dilation. The difference between the Earth and saucer clocks is *14.56* − *11.648* = *2.912 years*, just about the factor we predicted from the new Earth pulses that appeared as the saucer shifted through a series of accelerated frames of reference as it changed directions at the Alpha Centauri end of the trip (since the Earth-end acceleration and deceleration time effect and the time-dilation effect while coasting essentially nullify each other).

*Note that we are required to have the saucer see the Earth’s proper time dilated as Earth moves relative to the saucer at rest (Figure \(\PageIndex{13}\))*.* Without the Earth’s clocks s lowed d own when seen from the saucer, the forward skipping of Earth time as the saucer shifts from one accelerated reference frame to another accelerated reference frame would have produced too much Earth time having passed. *Mutual time dilation is not a paradox of relativity but a requirement*.

Note

* Had the saucer not seen the Earth’s proper time dilated, the distance the Earth would move in the saucer frame during the \(\tau\) = *1.456 yr* until the first pulse would be sent would be *0.6 *\(c\) × *1.456 yr* = *0.874 *\(c\)* yr*. The return trip for the light would be *0.874 years*. Add \(\tau\) = *1.456 yr* for the Earth’s outbound proper time and you get a return time for the pulse of *2.33 yr* after the Earth left. This is the wrong frequency. It is *4/5 (1/1.25)* of the correct result of *2.912 years *!

## A SMOOTHLY ACCELERATED SAUCER FRAME

Given the huge reality shift when the saucer instantaneously shifted direction, we would like to redo this simulation with moderate acceleration away from Earth at the beginning, deceleration at Alpha Centauri, reacceleration back to Earth, and deceleration at Earth. In between the periods of acceleration, we will travel at an unaccelerated glide of \(v\) = *0.6* \(c\).

We want to keep the rocket’s acceleration at a reasonable level. As the Earth tries to pull us down through the floor (gravitationally accelerating us at “one Earth gravity,” or 1 \(g\)), the floor resists that intrusion, of the electrons in our shoes into the electrons of the floor, with an equal force upward. If we use thrusters to accelerate the saucer at 1 \(g\), the floor of the saucer will push up on your feet with exactly the force you are used to, and your mass will try to resist the change in velocity and push against the floor. (You have experienced this effect if you have turned a corner too sharply in a car and have felt the door pushing against you as your mass tries to maintain its forward momentum.) You will have the illusion that you are walking on Earth. This is the *principle of equivalence* that we will study in the next chapter. If the saucer were to accelerate at 2 \(g\), you would feel twice as heavy—those whose bathroom scale on Earth reads 150 pounds would read 300 pounds on a scale in your saucer-board cabin. This would put a strain on your heart so we will stick with a 1 \(g\) acceleration.

As might be expected, RelLab does not allow for accelerated motion, but we can nevertheless model a smooth 1 \(g\) acceleration by a series of steps of increasing velocity, as seen in Figure \(\PageIndex{17}\). One sees that it takes about *3/4 of a year* to boost up to \(v\) = *0.6* \(c\).

The saucer would accelerate at 1 \(g\) until reaching the velocity of *0.6* \(c\), then coast for the bulk of the trip at this constant velocity before flipping over and decelerating at 1 \(g\) for the remainder of the outbound trip. It would immediately start the boost up to speed back toward the Earth, coast for the bulk of the return trip, and then flip over and decelerate as it approaches Earth.

At this acceleration, the saucer will reach \(\gamma\) = 1.25 at a distance of \(r=\frac{c^{2}}{A}(\gamma-1) \).* If you are running RelLab, open the file “Alpha Centauri Trip g acc 7c10” and set “Frame: Saucer 1” in the upper left-hand corner. You can also play the linked movie “Alpha Centauri Trip g acc 7c10.mp4.” In either case, you will see that I have imposed a grid on it whose small squares are *1/4 of a light-year* apart. The effects of acceleration will come into play only while the saucer is within that distance of the Earth or Alpha Centauri.

Note

* A derivation for the formulas for the time dilation and such under acceleration is given in the appendix of this book. It is helpful to have simplified the quantity \(c\)*/A = 0.968715 yr*, which we then multiply by \(c\) and by \(\gamma\)−*1* = 0.25 to get *0.242179* \(c\).

Since we are taking a while to get up to speed, the round trip will take a bit longer than before, \(t\) = *15.8426 Earth yr*, as will be the time between pulses so that *10* are emitted by the Earth. The ratio of the round trip Earth time to saucer time is *1.21539* rather than *1.25*, but all else is pretty much the same.

We will start by having 2 pulses of light be emitted by Earth before the saucer leaves so that we can see the acceleration (rapid frame shift) more easily on the first part of the trip. But it is the period when the saucer decelerates into Alpha Centauri and immediately starts the boost back up to speed toward the Earth that the saucer’s perspective shift from frame to frame becomes clear. Please note that the language of this paragraph is chosen to match the reality: it is the saucer that accelerates and decelerates. However, in subsequent paragraphs and in the figures, we will talk about and show the saucer frame, in which the saucer is at rest and it is Earth and Alpha Centauri that move.

Stop the movie at 56 seconds, when the clock shows *5.1 years*, just before deceleration. This is also shown in Figure \(\PageIndex{18}\). There are 12.2 grid lines between the Earth pulse that is just touching the right-hand frame and the one catching up to the saucer. Watch what happens to the distance between pulses seen from the saucer frame when we start to slow down. Clearly, we are rapidly changing frames of reference.

Pause the movie again at 1:11, when the clock shows *6.5 years*. Now pulses are 6.5 grid lines apart, and there are two additional (**orange** and **yellow**) Earth-pulse emissions, showing that an additional *2.9 Earth years* have elapsed during the *3/4-year* saucer deceleration.

The saucer reverses direction, and we should pause the movie again at 1:19, when the clock shows *7.2 years* and we have returned to coasting, shown in Figure \(\PageIndex{20}\). The pulses have continued to bunch up with the relativistic Doppler shift and are now 3 grid lines apart, 1/4 of the 12.2 spacing on the outbound coast, just as expected.

Back the movie up to *5.8 years*, and note that two pulses of light are in flight from the Earth since the saucer left (the **orange** one has yet to reach the saucer). Run the movie up to *7.2 years* again, and note that six pulses of light are now in flight from the Earth since the saucer left, a difference of four (**yellow**, **green**, **cyan**, and **blue**). The Earth was due to send pulses at *1.58-year* intervals, so about *4* × *1.58 yr* = *6.3 yr* have passed on Earth during the deceleration, and reacceleration phase, which took *1.4 saucer years*.

Over the *5 years* that the saucer saw the Earth moving away at a constant \(v\) = *0.6* \(c\), after the initial acceleration, it would see Earth clocks slow down, showing that only *4 Earth years* would have passed. But during the deceleration and reacceleration phase near Alpha Centauri, *6.3 years* passed on Earth while the saucer experienced only *1.4 years*. Combining these two eras shows that *10.3 Earth years* pass for every *7.7 saucer years*, for a ratio of 1.3. This is quite close to the 1.22 overall time-dilation factor for the trip given by our crude counting of rings on a grid. (As with the earlier trip using instantaneous jumps from stillness to motion, and the reverse, the initial acceleration-induced time effects at the beginning of the trip and the final deceleration-induced time effects at the end of the trip will essentially cancel the time-dilation effects of the unaccelerated portion of the trip.)

This sequence of events shows two things. The first is that the seeming “paradox” of each of us seeing our moving counterpart’s clocks running slowly is not only real but required to compensate for the much larger reverse effects of acceleration on the clocks of the person who actually feels the effects of that acceleration. Without the former, the latter would be too large. The second takeaway is that it is the person experiencing the physical effects of acceleration whose clocks run slower over a round trip and who is therefore the younger twin.