# What does the accelerated twin see in the twin paradox?

I am referring to the usual twin "paradox" where one twin remains on Earth while the other one takes a round trip journey on a rocket. I understand how the experience of the twins is not symmetrical, because one is accelerating, and that the solution involves changing reference frames. What I haven't seen in common answers, and what I am curious about, is what would one twin (on the rocket) see if she could constantly look at the other twin's clock. Imagine that the travelling twin had a super telescope and watched the Earth-bound twin's clock constantly during her trip. This scenario includes the following stages:

1. Positive acceleration, +a, away from Earth
2. Floating freely for a while at velocity +v
3. Negative acceleration, -a, to slow to a stop relative to Earth
4. Briefly motionless relative to Earth (v=0)
5. Accelerating towards Earth at -a until reaching velocity -v
6. Floating freely towards earth at velocity -v
7. Final positive acceleration, +a, to land on Earth (v=0).

What would the twin on the rocket see on Earth, looking from afar:

1. While accelerating away?
2. While floating freely before starting the deceleration?
3. While decelerating but still moving away?
4. During transition (turn around)?
5. While decelerating but now moving back towards Earth?
6. While floating freely back towards Earth?
7. While accelerating to a stop on Earth?

Imagine the traveling twin constantly looking through the telescope and comparing the Earth clock to her rocket clock. I'm looking for an answer like "...now the Earth-clock appears to be ticking slower than her own clock on the rocket". And "...at this point the clocks are momentarily showing the same time." You can keep the math to the bare minimum, in fact no math is best! Thank you in advance.

• As usual the best possible answer is "draw the space-time diagram, already", but in this instance that isn't just because it covers the hyperbolic geometry, but also because it builds the retardation of signals into the diagram. – dmckee Feb 25 '18 at 0:36
• I've never drawn a spacetime diagram, but I'll give it a try. – Tom B. Feb 25 '18 at 1:23
• My answer here contains the full answer to your question: physics.stackexchange.com/questions/307573/… – WillO Feb 25 '18 at 3:05
• Words like "see" and "looking" need to be examined carefully. People learning relativity have to rid themselves of some basic Newtonian misconceptions, one of which is that we can "see" what the universe is doing, like an instantaneous snapshot at a moment in time. When you see, you are receiving optical signals, which take time to propagate to you at $c$. The propagation time is different for sources at different distances from you. The Lorentz transformation doesn't describe what you see optically. There is also no guarantee that it makes sense to use successive Lorentz frames to [...] – Ben Crowell Feb 25 '18 at 18:56
• [...] describe some single frame of reference of an accelerated observer. For example, two different events on an observer's world-line can both be simultaneous (in the sense of instantaneous Lorentz frames) with the same distant event. I have some detailed discussion of this in section 3.9.3 of my SR book, lightandmatter.com/sr . – Ben Crowell Feb 25 '18 at 18:58

I made some videos about this a few years back, on YouTube. It shows what the traveler sees on a variety of clocks in the scene, for the entire duration of the journey.

I would suggest "Earth gravity front view" to start with ("Earth gravity" represents the value of acceleration in this context!)

The terse explanatory text is reproduced here:

The twin "paradox" (in quotes because it is NOT a real paradox!) is a basic teaching scenario for Special Relativity: http://math.ucr.edu/home/baez/physics...

This is a series of visualizations of the journey from the point of view of the traveling twin, who flies 20 light years away from his home station then returns. The journey consists of four parts, joined together. The first quarter is an acceleration away from the station. During the second and third quarters the ship accelerates towards the station (so that at the half way point the ship is stationary 20 light years away). In the fourth quarter the ship accelerates away from the station in order to come to rest there.

The total coordinate travel time (shown as a red dot in the top left HUD clock) is 43.711/58.918 years for acceleration at earth/moon gravity levels, whilst proper time (green dot) is 12.101/38.694 years. The yellow dot represents the time the traveler would see on the station clock face through a very powerful telescope! The octahedral stations (spaced one light year apart and one light year to the left of the flight path) are all synchronized to coordinate time and rotate once over the course of the whole journey. There is a 2x2 light year wall one light year beyond the far end of the journey, and large rectangular frames every 5 light years. Where a floor is shown it is 1 light year per stripe, and there are small 1 ly milestone spheres along the way, with a larger one every five light years.

The flights are rendered without relativistic effects and then for two values of acceleration (currently earth gravity and moon gravity). The distortion artifacts are due to aberration of light, the Doppler effect and the headlight effect. The earth gravity videos exhibit some nice penrose-terrel "rotation" effects. Magenta markers in a circle show where you really are in the scene (in the sense that things outside the circle are behind you, whilst those inside the circle are in front of you. Doppler shift = gamma for this circle), and grey markers show the circle where the doppler shift is 1 (gamma is in theory directly observable on this circle).

Let's have Bob travel out for 30 seconds and back for 30 seconds at 4/5 the speed of light while Alice stays at home. They each have clocks that (according to them) tick once per second.

I. When Bob leaves, his clock and Alice's both show time $0$.

II. As Bob travels and looks back through his telescope, he sees Alice's clock tick once every 3 seconds. But he figures that this is partly because the distance between him and Alice is increasing, so the light from each tick takes longer to reach him than the light from the previous tick. After correcting for this, he concludes that Alice's clock is ticking once every 1.67 seconds. That is, her clock is running slow, but not by as much as you'd think if you just measured the interval between arriving ticks and failed to correct for travel times.

III. When Bob's clock reads 30 seconds, he receives the light from Alice's tenth tick (that is, the image in his telescope shows Alice's clock reading 10), and calculates that Alice's clock now reads "18 seconds" (though the light from the most recent 8 ticks hasn't reached him yet).

IV. Now Bob turns around and decides that he and Alice are (and always have been) moving closer together, not farther apart. His telescope still shows Alice's clock reading "10", but now, unlike before, Bob believes that each tick has traveled less distance than the one before it. This causes him to recalculate that Alice's clock currently reads "82 seconds".

V. Bob now travels home, during which time his clock ticks 30 times, and he sees Alice's clock tick 90 times (advancing from 10 to 100). But (according to Bob) 72 of these ticks had already occurred before he turned around, though he had not yet received the light from those ticks. Therefore, according to Bob, the remaining 18 of those 90 ticks take place during his 30-second return journey. So he sees 3 ticks a second, but calculates that Alice's clock is ticking once every 1.67 seconds.

VI. Bob arrives home. His clock reads 60 seconds. Alice's clock reads 100 seconds.

Where did I get all this? I drew the spacetime diagram. So should you.

• It appears I've thrown a wrench in it by specifying a particular position, in one reference frame, from which to observe distant events in another reference frame. I see I've used the word "observe" differently than is standard. By doing so, I've added Doppler and delay. I assume that ordinarily, I should imagine an observer stationed at all points on a given reference frame, ready to make local observations. (And yes, I should draw the space-time diagram.) – Tom B. Feb 25 '18 at 6:59
• @TomB. : There is, I think, no well-established standard use for the word "observe" in this context. If Bob's telescope shows Alice's clock reading $10$ and he calculates that it currently reads $18$, some writers will say he "observes" a reading of 10 and others that he "observes" a reading of 18. (I think the latter is somewhat more common, but not universal.) So one always needs to be careful about clarifying what one means. – WillO Feb 25 '18 at 7:02

The short version of the answer, which is really helped by the, ahem, spacetime diagram, is as follows: as the twin rockets off into space, she sees her brother's clock ticking slowly (they are fraternal). This is because her definition of simultaneous looks 'back' w.r.t to his definition of simultaneous. He, of course, see the same thing and thinks her clock is slow. (again, spacetime diagrams really help visualize this symmetry).

Now when she turns the ships around she has a whole new reference frame, and her hyper-slice of simultaneity looks 'forward'--it is at this point the brother has gotten much older. On the trip home, each sees the other aging more slowly.

So, during constant motion, each thinks the other is aging at a slower rate--it's when the astro-twin turns around that the paradox happens: bro on Earth doesn't notice anything different, while sis says "yikes! you aged real fast."

I left out the acceleration phases because-they don't make anything clearer--when you understand that the astronaut's definition of "when is now back on Earth" leaps forward when she flips the U-turn, the paradox is resolved.

• Right. Then there is the other thing that the diagram helps with. Though the sudden aging happens during the acceleration phase at the far end, the light from the that time arrives during an period of time part-way through the in-bound leg of the trip. – dmckee Feb 25 '18 at 3:11
• This account of relativity is unreal. Why would the space-sister's turnaround, have any effect on the earth-brother's age (or anything else in the universe, other than the space-sister's observation of it)? – Steve Feb 25 '18 at 3:24
• @Steve I failed the OP by only answering "how does the asto-twin see her bro in what she defines as simultaneous on Earth"--which differs from what she sees 'now on her ship', as there's retardation. During travel, she always sees the twin "as he was", it's that after the turn around she is looking much further into his past. – JEB Feb 25 '18 at 3:50
• @Steve This account of relativity is correct. You've entirely misunderstood it. The traveling twin's acceleration does not affect the stay at home in a physical sense, it affects her concept of simultaneity: after the acceleration her time slice is simultaneous with her brother at a much later Earth date. – dmckee Feb 25 '18 at 4:27
• @TomB. Well, to get what she 'sees', you first have to understand when is her "now"--which is what I described. From that, you add in propagation delay (because she always sees his past) and you are almost there. Throw in time-dilation and Doppler (which are 2 parts of the same effect), and you done. – JEB Feb 26 '18 at 2:10

There is something in relativity that people don't really mention: when someone says, for example, that the lengths of objects contract, that doesn't necessarily mean little Johnny looking through a telescope will see a contracted object. In fact, approaching objects will appear lengthened.

https://en.wikipedia.org/wiki/Terrell_rotation

I suspect there is a similar effect for time dilation. I will gladly accept the name "Goling Onner's effect" for this phenomenon.

When someone says that length contracts or time dilates, what they mean is that measured lengths will be shorter and measured times will be longer. A typical definition for measurement in relativity is that you measure the distance and time of an event by firing a photon at the event and waiting for it to bounce back to you. Using the roundtrip time and the formula $distance=c\cdot time$ you can determine when and how far the event was.

So the answer to your question is unilluminating: I suspect both twins will see the other's clock slow down on the outbound trip and then speed up on the inbound trip, if they look through a telescope. But special relativity is not about pointing a telescope around and looking at stuff, despite the way all the pop-sci books talks about.

What special relativity says is that lengths are contracted and times are dilated when you compute them using the above procedure. So the real question is

If the travelling twin uses this procedure to measure time and distance, what will they observe?

The brief answer is: as they decelerate at the midpoint of the trip, and then accelerate back towards earth, they will measure earth-twin's clock ticking faster than the spaceship clock. This is what distinguishes the two scenarios. Earth-twin will never measure the spaceship clock to be ticking faster than earth clocks; earth-twin will measure space-twin's clock slowing down as the ship accelerates away, speeding up to normal as the ship slows down at the far end of the trip, slowing down again as it accelerates back to earth, and finally speeding up to normal as the ship slows down at earth.

• "and then accelerate back towards earth, they will measure earth-twin's clock speeding up" You presumably have in mind that they will see the light of the stay-at-home- clock ticking coming in fast, which is true. But that doesn't mean that they see the clock peed up. After correcting for the propagation delay both twins see the clock of their opposite number running slow. That's the basic fact of time-dilation that twists everyone's brains into jelly. And the thing that you have to grapple with when the twin paradox comes up. – dmckee Feb 25 '18 at 5:07
• @dmckee, that statement cannot be consistently maintained across a full journey from start (when they sychronise their clocks) to finish (when they compare them). – Steve Feb 25 '18 at 5:15
• @Steve It can be maintained precisely because simultaneity is a relative concept. But then it can't make any sense on any part of the journey without the relativity of simultaneity, so making it work for this cases dones't require any further assumptions. – dmckee Feb 25 '18 at 5:19
• @dmckee, it cannot be maintained across the full span of the journey in the manner I described, because then you'd have two twins standing in front of each other, somehow arguing that only the other clock is slow (and heaven knows how a third observer could describe the two clocks put side-by-side). I suspect you are simply failing to fully correct for the delay despite claiming to do so (an error that naturally has it's largest impact at the greatest distances or largest discrepancies in speed, and dissolves to nothing as the twins come back together to finally compare their clocks). – Steve Feb 25 '18 at 5:28
• @dmckee no that is wrong. What you said is true for inertial observers. An accelerating spaceship is not inertial. "You presumably have in mind that they will see the light of the stay-at-home- clock ticking coming in fast, which is true.", no, that's why I mentioned that relativity is not about what you see (like through a telescope), but what you measure. Draw the spacetime diagram for an accelerating spaceship, use the operational definitions, and you will see that the ship measures earth's clock ticking faster. – goling onner Feb 25 '18 at 14:55

Let's say that we have two twins, the EarthBound Twin ($$EBT$$), and the Traveling Twin ($$TT$$). Each twin has a clock that ticks once/second, and that the other can see (telescope technology having advanced along with transportation technology). $$TT$$ takes a trip, traveling away at $$3/5c$$ for a total of three light years, stopping, and then returning home at the same velocity of $$3/5c$$. So:

• $$TT$$'s velocity of $$3/5c$$ gives a time dilation factor $$\gamma$$ of $$4/5$$
• In $$EBT$$'s reference frame, the trip takes five years out and five years back, while $$EBT$$ ages ten years
• $$TT$$ experiences traveling only four years out and four years back, aging only eight years
• $$EBT$$ ends up two years older than $$TT$$.

While $$TT$$ is traveling outward, $$TT$$ will see $$EBT$$'s clock ticking at half speed (due to both time dilation and the steadily increasing distance). While $$TT$$ is traveling home, $$TT$$ will see $$EBT$$'s clock ticking at double speed. So, here are your stages of travel, and what $$TT$$ will see during each:

1. "Positive acceleration, +a, away from Earth": $$TT$$ will see $$EBT$$'s clock smoothly slowing down
2. "Floating freely for a while at velocity +v": $$TT$$ will see $$EBT$$'s clock running at half speed, ticking once every two seconds
3. "Negative acceleration, -a, to slow to a stop relative to Earth": $$TT$$ will see $$EBT$$'s clock smoothly speeding up
4. "Briefly motionless relative to Earth (v=0)": $$TT$$ will see $$EBT$$'s clock running at normal speed, ticking once a second
5. "Accelerating towards Earth at -a until reaching velocity -v": $$TT$$ will see $$EBT$$'s clock gradually speeding up
6. "Floating freely towards earth at velocity -v": $$TT$$ will see $$EBT$$'s clock running at double speed, ticking twice a second
7. "Final positive acceleration, +a, to land on Earth (v=0)": $$TT$$ will see $$EBT$$'s clock gradually slowing down until it reaches normal speed.

Here's the brain twister: $$EBT$$ will see exactly the same sequence of clock changes in $$TT$$'s clock. That is, during $$TT$$'s trip $$EBT$$ will see $$TT$$'s clock slow down, run at half speed, speed up during the turnaround, run at double speed, and then slow back to normal as $$TT$$ arrives home.

If they each see the other's clock go through the same sequence of changes, why do they agree that $$TT$$ end up younger than $$EBT$$? Here's the key: when $$TT$$ turns around halfway through the trip, it takes another three years for that information to reach $$EBT$$. Thus, although $$TT$$ sees $$EBT$$'s clock run slow for half the trip and then fast for half the trip; $$EBT$$ sees $$TT$$'s clock run slow for 80% of the trip and then fast for 20% of the trip.

$$TT$$ will see $$EBT$$'s clock move half-speed for four years, and double-speed for four years, so $$TT$$ will see $$EBT$$ age for $$(1/2) * 4 + 2 * 4 = 10$$ years. $$EBT$$ will see $$TT$$'s clock move at half-speed for eight years, and double-speed for two years, so $$EBT$$ will see $$TT$$ age for $$(1/2) * 8 + 2 * 2 = 8$$ years. In other words, they both agree that $$EBT$$ ends up two years older than $$TT$$.

Note that I'm basing these numbers on A. John Mallinckrodt's page on the twin "paradox". The page includes spacetime diagrams from the point of view of each twin, which may be helpful.

I believe I have the broad strokes of this answer correct, but as I am no expert I likely at least have some details wrong. In any case, corrections, small or large, are welcome.

@TomB., I doubt the spacetime diagram will give you much insight - I never found them much help.

There's basically three components to consider.

The first is that, by virtue of being at a distance from the clock you're observing, there is a distance-dependent delay in the signal reaching you.

Second, if you're moving, then there's a direction-dependent Doppler effect (which is basically the dynamics of the distance-dependent delay).

Thirdly, if you're moving (not necessarily accelerating), there is a speed-dependent dilation - which, if you're the moving twin, will (despite the name) mean the clock is going faster on Earth.

So putting it all together, when you accelerate from Earth, you see an increasing redshift in the image, and the Earth clock proceeds at an increasingly slow rate - but not quite as slowly as you would expect, because your clock rate (and in fact, your entire local environment) is also slowing down slightly owing to dilation.

When you start to cruise at a steady speed, the observed rate of the Earth clock settles into the groove that you'd established when you were accelerating - and it's moving noticeably slower than your clock.

When you decelerate at the far end, it's basically the effects of the acceleration in reverse. The redshift attenuates, the dilation attenuates, and the clock returns to appearing to proceed at a normal rate.

But of course, we still have two factors to account for.

First, you're now a substantial distance away from the Earth clock and it takes some time for the signal to reach you, so you're reading it substantially in arrears, compared to what the clock face is currently displaying on Earth. That's why you experienced all that redshift on the way out, because you were establishing more and more delay in the signal, based on moving further and further away.

Secondly, because you've been experiencing dilation during the trip, the Earth clock is not quite as much in arrears as you would expect, based simply on your distance from Earth. The Earth clock seems to be "ahead" of you (once you've offset the expected delay based on distance), even though it is now ticking at a rate that is consistent with your clock.

Now, accelerating toward Earth on the return trip, some effects are reversed. Instead of redshift, we have blueshift, so the Earth clock starts appearing to move forward like the clappers - which is to be expected, as you're purging the "delay" that you'd established on the outbound journey. But in fact, it's going even more like the clappers than you would expect, because you're suffering dilation again.

Once the acceleration is over and you've reached steady cruising velocity, the clock remains blueshifted, and it's gaining on you steadily.

Finally, we decelerate to land on Earth. At this point, the blueshift attenuates, the dilation attenuates, and finally at a point very close to Earth, the Earth clock finally overtakes yours.

So you're back on Earth, looking at a clock that has gained substantially on you, and a twin whose beard has grown noticeably more than yours.

• @TomB., the Earth clock overtakes yours immediately as soon as you set off, because that is when the dilation kicks in for you. That fact is only masked by the fact that, because you're moving away from the clock on the outbound journey, you're introducing a constantly increasingly delay in your "reception" of the clock reading - which more than offsets the gain. Obviously, I'm making a distinction between the "real" reading of the clock (i.e. the reading it would show if it could convey the information across space instantaneously), and the apparent reading. (1/2) – Steve Feb 25 '18 at 3:11
• The delays and the Doppler effects are merely apparent effects due to the limited speed of light and the dynamics of how the delay in reception changes according to whether you move away or toward the clock that you're observing. The dilation on the other hand is real - that's why your clock is in fact behind when you finally return to Earth. (2/2) – Steve Feb 25 '18 at 3:13
• @TomB., watch out though, because I will soon be pilloried with a series of unexplained downvotes (I've already received one)! – Steve Feb 25 '18 at 3:16
• "which, if you're the moving twin, will (despite the name) mean the clock is going faster on Earth. " Is simply wrong. Both twins observe the counterpart's clock as running slow on both legs. @TomB. For the "How can that work" see the space-time diagram provided by Takeuchi reproduce in physics.stackexchange.com/a/111089/520. – dmckee Feb 25 '18 at 4:40
• Steve, you can't workout the difference of the clocks without doing one of a few things: (a) drawing the space-time diagram (the easy thing, low math way), (b) computing intervals for the participants (the easy thing, mathier way), or (c) computing (to use Einsteins phrase) "the moment of acceleration" for each participant (a royal pain). But they all rely on the hyperbolic geometry embedded in the theory. And the last one tells you that just knowing the speeds ins't enough, you have to know where the world-lines bend, as well as how much. – dmckee Feb 25 '18 at 5:03

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