# Physics Quiz: Accelerated Twins

Just about everybody has heard of the Twin Paradox in relativity: one twin becomes as astronaut and sets off for Alpha Centauri, the other remains on Earth at mission control. Thanks to time dilation, the two age at different rates, and the one who made the trip out and back ends up younger than the one who stayed behind.

Of course, the paradox is not that the two twins have different ages– rather, it’s that from a simple approach to special relativity, you would think that each twin should see the other’s clock running slow, since it seems like getting into a rocket and flying off into space should be equivalent to sitting still in the rocket, and having the entire Earth go zipping off in the opposite direction. This is resolved by noting that the twin in the rocket experiences significant acceleration during the trip, while the other twin does not, and so the two frames of reference are not equivalent.

So, with that in mind, here’s a more subtle question:

Two twins, named Alice and Bob in keeping with convention, get into identical rocket ships separated by a distance L, with Alice in front and Bob behind her. At a pre-arranged time, they each start their rocket, and accelerate for a pre-determined time. At the end of the acceleration, they are each moving at a relativistic speed– 4/5ths the speed of light, say. Which of these twins is older at the end of the acceleration?

You can find the answer using Google, but that would be cheating. We’ll do this as a poll first, and I’ll give the answer probably tomorrow:

A second question, which may or may not help you think about the answer: At the end of their acceleration, what is the spacing between their ships as measured by Alice and Bob?

As weird as relativity is, it does not allow you to choose more than one answer to these questions. Think carefully, choose wisely, and come back tomorrow for the answer.

1. #1 Eric Lund
February 2, 2010

The defect in the second question is that you haven’t specified which frame will be used for the measurement. An observer in an inertial frame will measure the same distance as the initial distance, since she will see both Alice and Bob accelerate in the same way and therefore keep their relative displacement constant.

The first one is easy. Apart from a sign reversal in the displacement vector, the trajectory of Bob in Alice’s reference frame must be identical to the trajectory of Alice in Bob’s reference frame. So by symmetry they must experience the same elapsed time.

2. #2 John Novak
February 2, 2010

What does this mean, “At a pre-arranged time?”

February 2, 2010

What does this mean, “At a pre-arranged time?”

Alice and Bob synchronize their clocks in their rest frame, and each agree that at exactly noon on their shared birthday, they will start the engines of their rockets.

February 2, 2010

The defect in the second question is that you haven’t specified which frame will be used for the measurement. An observer in an inertial frame will measure the same distance as the initial distance, since she will see both Alice and Bob accelerate in the same way and therefore keep their relative displacement constant.

Sorry. The question is asking for the distance according to Alice and Bob. You are correct that there is some observer in some frame who sees the distance being the same. The question is, is that observer in the same final frame as Alice and Bob?

5. #5 Uncle Al
February 2, 2010

The four-vector is conserved. The longest accumulated path will have the shortest accumulated interval. Running clocks can only be compared when local (touching). Pick your POV.

6. #6 Sili
February 2, 2010

Does it make sense to compare times at different points in space? If not, we need to know how A and B are brought together to answer the question.

(That is to say, I think I answered wrong …)

7. #7 Alan E.
February 2, 2010

The second question seems as if you would need to specify where the observer is standing to be able to take a measurement. Outside of the system, the distance L would appear to be the same. But within the system, wouldn’t the distance between the two actually seem longer? The exact location of the measurement would be different depending on which ship you are measuring from, but the distance would be the same, but longer than outside the system.

Here is a crude drawing of what I think I am getting at (ships are A and B, and the dashes are the measurement):

Outside the system (20 dashes between):
A——————–B

Measure from A’s POV (21 dashes between):
A———————B–

Measure from B’s POV (21 dashes between):
–A———————B

8. #8 Alex
February 2, 2010

As far as I can tell, there’s nothing different between Alice and Bob*. You say that they are accelerating from the perspective of an observer in the frame where they were initially at rest, but if they both accelerate at the same rate with respect to that observer then as far as they are concerned they haven’t moved at all with respect to each other. The observer from their original frame is the one accelerating and encountering whatever weirdness happens when you accelerate to large speeds, but they are standing still with respect to each other the whole time. They might have some weird observations concerning the guy who’s accelerating with respect to them, but they themselves see nothing weird.

*I vaguely recall some of these examples having Al and Bert, once upon a time, but that would be less inclusive.

February 2, 2010

I don’t see anything in the initial question that states Alice and Bob are accelerating in the same direction. Same time same amount of time, and same final speed, yes, but I notice it does NOT say “same velocity.” Which is why I went with “not enough information to answer” on question 2.

10. #10 Robert
February 2, 2010

Assuming Bob is not L closer to the bottom of a gravity well, or some other complication…

Put a big paper bag around both ships (or a bit of pink ribbon to tie them together if you prefer), it accelerates at the same rate etc.. IE: accelerate the bag, with two ships, Bod & Alice in it; rather than thinking of two separate ships.

End result is the bag/ribbon, Alice, and Bod have all aged by the same amount; and the bag size hasn’t changed by the end, all measured from (a frame) within the moving bag. No changes relative to each other.

11. #11 Alex
February 2, 2010

Electric Landlady makes a good point, but if that’s the case then this is more a test of reading comprehension than on relativity. The part where Chad specifies that one person is in front and the other is behind sort of implies something about direction, so I think that the assumption most of us are making is a valid one. Still, it would be nice if this were phrased a bit more carefully.

February 2, 2010

They are headed in the same direction, with Alice starting out a distance L in front of Bob.

13. #13 Alan Coles
February 2, 2010

Here’s an interesting variation on the twin question.

Suppose you first slowly separate the two rockets with symmetrical and opposite thrusts. No time difference caused. Then accelerate only Alice’s ship to 4/5 c by using the gravitational field of a large moving mass close to her ship (not using the rocket engine). Don’t apply any force to Bob’s ship. Bob is far enough from Alice that the mass near her ship gives him negligible acceleration; that’s the purpose of separating them as the first step. Rocket engines on the huge mass keep it slightly in front of her rocket as she speeds up, so it continually accelerates her gravitationally, but there is no non-gravitational effect of the mass on her ship.

Artfully contrive to reverse Alice’s ship with another gravitational acceleration to -4/5 c velocity. Bring her back to zero speed at her starting point with a third gravitational acceleration. Then use their rockets to slowly reunite the 2 ships.

How will their clocks compare?

Apart from the initial and final use of rocket thrust to separate and reunite the ships, neither ship has experienced any detectable acceleration, since both are in free fall the entire time. External observers can see Alice accelerate several times, but no instruments in her ship can detect it. She remains in an inertial frame the whole trip. (Inertial and gravitational mass are the same; Dicke-Eotvos experiment. Their world-lines are also geodesics of the metric tensor – straight lines in local Lorentz frames; this is the equivalence principle.)

Both ships have identical human experiences and instrument records. Neither think they have shifted reference frames (as is caused by experiencing acceleration). But outside observers plainly see the ref frame changes in Alice, so they know who speeded up and slowed down.

So, is the time-lapse difference a consequence of their different velocity histories, an external relative property (since only accelerations can be observed locally, but not velocities)? Or is it a consequence of change in inertial frames (an internal property)? With gravitational accelerations there is no locally observable consequence of the acceleration; that is the essence of free fall (in the freely falling Lorentz frame the gradient of the metric tensor is zero, and this is always possible). And if tidal effects from the big mass are a worry, make it a large slab and position her ship infinitesimally close to it, so no geodesic deviation is detectable

Or do Alice and Bob find identical time-lapses when they hold hands upon their reunion?

14. #14 Rhett
February 2, 2010

I like the third question: “Which twin was happier?”

15. #15 Buddha Buck
February 2, 2010

I’ll add a third twin/triplet, Chuck, who helps synchronize everything. I’m assuming that initially Alice, Bob, and Chuck are in the same reference frame and situated such that Alice and Bob are equidistant from Chuck. All three share a sense of “simultaneous”. At T=-L/2c, Chuck fires a starter signal such that at T=0 (in the shared starting frame) Alice and Bob receive it and start their engines.

At a later T, both Alice and Bob turn off their engines. In Chuck’s frame, this turn-off is simultaneous for both Alice and Bob. Both Alice and Bob report the same time on their clocks, albeit a different time than what is reported on Chuck’s clock.

However, because they are travelling relative to Chuck, their sense of “simultaneous” are canted relative to Chuck (albeit with the same angle with each other) and there is an offset in space of their sense of “simultaneous”. When Alice turns off her engine, Bob will have yet to do so. When Bob turns off his engine, Alice will have already done so.

So from Chuck’s view point, they are the same age. From Alice or Bob’s view point, Alice got to the shut-off point first, so she’s older, from both their points of view.

As far as the distance… For chuck, the interval between the two shutdown events is -L^2, since the time difference is 0 and the spacial separation is L. For Alice and Bob, the interval between the two shutdown events is t’^2-L’^2 = -l^2, so L’^2 = L^2 + t’^2. Since t’ is non-zero, L’^2 > L^2, so L’ > L. The distance between the two ships, from Alice or Bob’s view point, is greater than when they started.

16. #16 Anton
February 2, 2010

From an inertial reference frame, one trajectory will simply be a space-translated version of the other trajectory. Thus, the proper time for each trajectory as calculated in this inertial frame will be identical. Since the proper time is the same as the time measure by an observer along the trajectory, both twins should be the same age at the end. Thus, for question 1, I vote “They are exactly the same age.”

Also, relative to this inertial frame, the distance between the twins will remain constant. However, this means that in the reference frame of the twins, the distance between them will now be greater than L. This is because the greater-than-L distance will be Lorentz-contracted to L in the inertial frame. Thus, for question 2, I vote “Greater than the initial separation L.”

17. #17 bcooper
February 2, 2010

I agree with Anton’s post #16. Pretty much the same reasoning I was using.

All you need is to tie a string between the rockets and you have John Bell’s question from one of his articles in Speakable and Unspeakable in Quantum Mechanics. He was apparently able to generate a fair bit of controversy with that, so I’ll be curious to see how this shakes out.

18. #18 Anton
February 2, 2010

I like Buddha Buck’s answer better than mine. Relative to an inertial frame in which Alice and Bob start out stationary, they are both the same age and their separation is still equal to L. Relative to an inertial frame in which Alice and Bob end up stationary, Alice is older than Bob and their separation is greater than L.

19. #19 Neil Ostrove
February 2, 2010

Given (12) the principle of equivalence should come into play. It’s as if they were in a constant gravitational field with Alice higher than Bob. IIRC (which I may not) gravitational red shift would make Alice age more slowly and be younger.

20. #20 bcooper
February 2, 2010

I like Buddha Buck’s answer better than mine. Relative to an inertial frame in which Alice and Bob start out stationary, they are both the same age and their separation is still equal to L. Relative to an inertial frame in which Alice and Bob end up stationary, Alice is older than Bob and their separation is greater than L.

I suppose it depends on whether their flying for an agreed upon amount of time before shutting off the engines involves their proper time or Chuck’s time. If it is the former, then the only difference between the two trajectories is that one is translated in space, as you pointed out, and then I can’t possibly see how they are aged any different. If this is what they do, then to Chuck it will look like they flew for different amounts of time, I guess. To me this is the natural interpretation. If they are the ones deciding whether to turn the engines on and off, it seems natural to specify the amount of time they should fly in terms of their rest frame.

21. #21 Monimonika
February 2, 2010

I went with not enough info on the first choice because I do not know if Alice and Bob started off being the exact same age (wouldn’t birth times be different?). Heck, I’m not sure if they’re even twins to each other.

Went with “same distance” on the second question.

22. #22 Anton
February 2, 2010

They stop accelerating at the same time relative to Chuck but not relative to themselves. Let’s say Alice is 1 year older when she stops accelerating. Then Bob will also be one year older when he stops accelerating. However, relative to Alice and Bob, Alice stops earlier. Thus, when Bob finally stops, Alice will be older in their reference frame.

Chuck would think that they are crazy because to him, it would look like they are comparing Alice’s age at one point in time to Bob’s age at an earlier point in time.

23. #23 miller
February 2, 2010

Okay, I can think of four ways to explain the answer. I could go into math, but that limits my audience to the people who already get it. I could make some pictures, but that takes too much work for a comment. I could reason intuitively, but that kind of reasoning is unreliable. So I will go for the fourth option, which is to simply assert without proof. :)

If we stick to Alice and Bob’s initial reference frame, they start the same age, and start accelerating simultaneously. They remain the same age and maintain a constant distance L throughout.

But I’m assuming you’re asking about the final reference frame of Alice and Bob. In this reference frame, Alice and Bob start out with a velocity of 4/5 times the speed of light, with Bob ahead of Alice. Their initial distance is less than L. Alice starts out older than Bob. And then they start to decelerate, with Alice decelerating first. Because she decelerates first, Bob experiences more time dilation than she, and the age difference becomes even greater. Furthermore, the distance between the two increases to something greater than L.

So my final answer: Alice ends up older than Bob, and their distance ends up greater than L. More precisely (if I did my math right), the distance is (5/3)*L and the age difference is (4/3)*(L/c).

24. #24 bcooper
February 2, 2010

However, relative to Alice and Bob, Alice stops earlier. Thus, when Bob finally stops, Alice will be older in their reference frame.

If Alice is stopping at an earlier point than Bob, then talking about “their reference frame” doesn’t make much sense because from their perspective they aren’t always in the same frame. At the end of the journey after all the engines are off everybody should be in the same frame again anyway.

I’m assuming the point of the separation question is to suggest the intuitive viewpoint that if Bob sees the separation between himself and Alice grow on the way out, then he needs to see it shrink back down to L before they arrive in the final rest frame, thus suggesting that she stops earlier. (Of course this same viewpoint also suggests that she started sooner in order to build up that distance; the two effects should be a wash. I don’t have a good image in my head for what this really looks like to Bob and would have to probably do some calculations to build intuition.)

It’s been years since my brain has done anything formal with relativity, so I haven’t worked out the details of what happens if we assume that Chuck back on Earth is sending them a message telling them when to stop. I can at least see how that would break the symmetry of the situation and consider that it might result in an age difference. But in the case where they agree to stop the flight after an hour passes on their onboard clocks, the argument that they are the exact same flight started at two different points in space is so powerful that I can’t really consider anything else.

25. #25 Anton
February 2, 2010

If Alice is stopping at an earlier point than Bob, then talking about “their reference frame” doesn’t make much sense because from their perspective they aren’t always in the same frame.
I mean the inertial reference frame in which Alice and Bob are at rest once they both stop accelerating. Relative to this frame, Alice will be older once both of them stop accelerating.

At the end of the journey after all the engines are off everybody should be in the same frame again anyway.
Not Chuck. Chuck will be in a different inertial reference frame and that is the key to the answer. Relative to Chuck’s frame, they are the same age when Alice and Bob stop.

26. #26 neil344
February 2, 2010

I think, alice will be older. When Bob starts his engine at the prearranged time, he will see himself as catching up to alice (since it takes L/c time for light to reach him from Alice), so by the Doppler effect he sees A’s light clock as ticking faster during the acceleration. When Alice starts her engine at the prearranged time, she looks in her rear view mirror and sees herself pulling away from Bob, so she see’s Bob’s light clock as ticking more slowly. Thus Alice ages faster than Bob.

27. #27 neil344
February 2, 2010

Hmmm. I should add that when they stop accelerating, it would seem that the reverse Doppler effect should happen (Bob will see himself as falling back and Alice will see Bob as catching up.) However, I don’t think their clocks stay synchronized during acceleration. Alice stops before Bob because she reaches the “agreed on time to stop” earlier than Bob does. So my answer stands.

28. #28 johan couder
February 2, 2010

I was happy to answer both questions, then I started reading the comments, and now my brain hurts! BTW, I’m surprised acceleration is mentioned to resolve the original twin paradox. Surely, in special theory of relativity acceleration is irrelevant?

29. #29 Matt Leifer
February 2, 2010

Imagining that the rockets are connected by a taut piece of string a la John Bell definitely helps with the second question. According to an observer at rest in the inertial frame that Alice and Bob start off in before accelerating, the distance between the rockets is constant but the string undergoes length contraction so it snaps. According to Alice and Bob, the string does not undergo length contraction because it is always instantaneously at rest in the same frame that they are. Therefore, because the string breaks they must be further apart than they were to start with.

I agree with niel344 on the age question. I am sure it can be derived from the same sort of stringy reasoning, but I can’t see how at the moment.

30. #30 dr. dave
February 2, 2010

I’m going with the equivalence principle on this one. I think if we don’t tell them they are in spaceships, then B is just at the bottom of a tall building and A is at the top. I think even in a uniform gravitational field, Bob’s clock runs slower.

(I think? Even uniform? Yes???)

31. #31 dr. dave
February 2, 2010

28) John – acceleration isn’t “IRRELAVENT” in SR, it’s not allowed! The paradox arises because it describes an impossible situation – that A & B stand next to each other and compare ages. They can not do so completely within the framework of SR, because in order to get in the same place again, the rocket-twin must at some point have slowed down, stopped, and accelerated back.

32. #32 neil344
February 2, 2010

The separation question (whether the string snaps) is less clear to me. The string snaps in the Bell paradox because a stationary observer sees a Lorentz contraction in the length of the spaceships so the space between them increases. But from Bob’s perspective, the spaceships are getting closer during acceleration, while from Alice’s perspective the spaceships are getting farther apart. The string can’t both snap for Alice and go slack for Bob, so I still can’t figure it out.

33. #33 miller
February 2, 2010

@Johan Couder #28

“BTW, I’m surprised acceleration is mentioned to resolve the original twin paradox. Surely, in special theory of relativity acceleration is irrelevant?”

The fact that one of the twins accelerates is important because an accelerating reference frame is not an inertial reference frame. The twin who stays home is in an inertial reference frame the entire time, but the traveling twin is not.

It’s sometimes said that Special Relativity can’t describe accelerating reference frames. This is not true. You simply treat it as a smoothly boosting frame. But this is very difficult, whether we’re doing it mathematically or intuitively, so I would take the easier way by assuming the acceleration is instantaneous. That is, we just start in one reference frame and switch instantaneously to another.

34. #34 Anonymous Coward
February 2, 2010

I’m still not sure I follow the setup of the problem. They use their watches (synchronized) to start at the same time as explained in comment #3. But the other measurements aren’t clear to me:

1) How do they coordinate the “accelerate for a pre-determined time”? By their own watches again?

2) And how do we compare their ages? By their individual reference frames?

If so, then they are both the same age at the end of their accelerations. I expect that – because that seems too mundane an answer – my assumptions to #1 and/or #2 are wrong.

35. #35 Danil
February 2, 2010

Are the acceleration programs of the two rocket ships supposed to be independent of each other?

Because if they are the same, and the string between them snaps, doesn’t it necessarily follow that the rocket ships themselves also rip in half (for the same reason).

The problem does state that the rockets are accelerating for the same amount of time (presumably each in their own frame), which I think means that if they have a constant acceleration, they must have the same constant acceleration (haven’t confirmed that with a calculation). If the ships can vary their accelerations differently, my intuition segfaults….

36. #36 Bee
February 3, 2010

Well, absent a universal time what sense does the question make when they’re not at the same place?

37. #37 johan couder
February 3, 2010

@dr dave & miller. Thanks for enlightening me. I find spacetime diagrams the easiest way to resolve the (original) twin paradox, but the role of acceleration wasn’t very clear to me.
(there are some very nice animated spacetime diagrams at http://www.phys.unsw.edu.au/einsteinlight/jw/module4_twin_paradox.htm, but that’s probably old news, and they didn’t particularly help me with this problem, although they probably should have).

38. #38 Pete J
February 3, 2010

I we ignore the Sagnac effect we get a diffrent answer, but we can’t as it’s real, and just confirmed in NASA’s Laser Lunar Ranging analysis (See arXiv; D Gezari); It also works longitudinally the same as rotationally; light travels at ‘c’ in the local frame it’s in. (See Wang et al). You just need to use the DFM (discrete field) correction.

From Alices pov Bob will be much younger as the light from him will take longer to reach her. From Bob’s point of view Alice will be a little older.
(The time it took the light to travel distance ‘l’ before they left will be reduced as it moves through the rest frame at ‘c’ and Bob is moving towards it).

That’s the facts, but for us? It depends entirely on our point of view!
Peter J

39. #39 Arrow
February 3, 2010

Alice is accelerating in front of Bob, according to equivalence principle it’s as if they were both accelerated by the massive body in front of them. In both cases Alice is closer to this equivalent massive body then Bob is and the closer one is the more slowly clocks run. This means clock will run slower for Alice and she will end up younger to some outside observer.

As for distance both Alice and Bob see Alice as accelerating faster due to being closer to the hypothetical massive body so they will see the distance between them increase.