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Twin Paradox Without Acceleration ?

  1. Dec 9, 2012 #1


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    What would be the example of the twin paradox where there is no acceleration at all ?
    Only different periods of constant velocity for both twins. Or of periods with different velocities for both twins.
  2. jcsd
  3. Dec 9, 2012 #2


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    The common easy example of "no acceleration" is to have the travelling twin turn around by doing a hyperbolic orbit around a massive gravitating body (zooms in at greater than escape velocity, does a hairpin turn around the object, heads back out with (almost) the same speed but in a different direction. The traveling twin can be in free fall for the entire turnaround.

    This version has the advantage of avoiding the red herring of acceleration completely; it's clear that the proper times on the two paths are different and that's where the "paradox" comes from. The disadvantage is that (as far as I know) there's no way of computing traveller's proper time without using general relativity or making some shaky assumptions... so it's hard to convincingly resolve the "paradox" without doing some serious math.

    Different speeds, and/or different times traveling at those speeds, is much easier. Just assume instantaneous changes in speed (unphysical, but simplifies the math no end and doesn't change the result) and calculate the flat Minkowski spacetime distance on each segment of both paths. It's generally easiest and most intuitive to work in a reference frame in which the point at which the twins start their journeys is the same as where they finish - Earth, if they both start there and meet up there at the end to compare their ages.
  4. Dec 9, 2012 #3


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    One example would be having the "stay at home" twin in orbit about his planet, while the "traveling" twin starts off flying upward at just short of escape velocity, so that he flies out for a long way, gradually slows to a stop, then falls back inward and comes back to the stay-at-home twin at exactly the same point in his orbit. In this example, the "stay at home" twin is the one who will age less; the "traveling" twin will be older when they meet up again. But both of them are in free fall (except for the traveling twin starting and stopping, but that can be idealized away just as it is in the standard twin paradox), so the difference in aging can't be accounted for by acceleration; it can only be understood by recognizing that in curved spacetime, there can be multiple geodesic paths between the same two events, and they can have different lengths.
  5. Dec 9, 2012 #4
    Something where the traveling twin sends a signal to another spaceship traveling with a constant velocity that happens to pass the twin.

    The problem starts out the same. However, instead of the traveling twin turning on his rockets to go home, he sees a spaceship from Alpha Centauri headed toward earth. The aliens are friendly, so they are glad to pass on mail. The twin sends an email to the Alpha Centauri trip stating his biological age at the time of passing. The alien ship continues until it gets to earth. While passing the earth, the aliens email a copy of the letter to the twin on earth as well as documentation for the time they spent traveling to earth since they met the twin.

    The brother on earth adds the age of his traveling brother when the alien ship passed him plus the extra time experienced by the aliens while continuing to earth. The sum is less than the amount of aging experience by the brother when since the other brother left earth.

    This has been presented to me as a counter example of a hypothesis to the idea that there is a literal acceleration that causes the asymmetry. Note that the role of forces in this problem are hidden within the concept of "signalling".

    Although there is no literal acceleration in this problem, there is an interaction that breaks the symmetry. The traveling brother signals the alien space ship while they are passing. The aliens receive the signal but with a large Doppler shift caused by the large relative velocity of the two ships.

    The interaction that breaks the symmetry is the large Doppler shifted email transmission. I think this interaction falls under what another poster called transmission of information. The physical means of sending information from one ship to another breaks the symmetry. The twin on earth doesn't receive a similar transmission of information. Therefore, his situation is not quite the same as his traveling twin.

    The composite system of traveling twin and alien space craft can be considered a type of observer. Although each of them are in different inertial frames, their joint path through space time is not an inertial frame.

    My point wasn't that one needs a literal acceleration to break the symmetry. I was trying to say that there is always some interactions needed to define the path of the observer through space time. If there are two observers passing through the same spatial point twice, the observer who is closest to traveling a geodesic is going to experience a longer round trip. The physical events that cause this observer to deviate from a geodesic are the symmetry breaking interactions.

    The famous time-dilation experiment involving muons and cosmic rays were like this. There wasn't a literal observer making a round trip. In a sense, the lifetimes were describing information stored in the atoms. However, there were some very violent symmetry breaking mechanisms. When a cosmic ray hit an atom, the formation of the muon was quite physical. Being violently brought into existence can arguably be called an acceleration !-)

    I handle problems where information is exchanged by constructing a hypothetical observer who is carrying around the information. If that hypothetical observer requires a hypothetical force to turn around, then I deduce that this observer is not in an inertial frame.

    I will recant a little bit. The symmetry breaking mechanism doesn't have to be an obvious external force. One can reformulate this in terms of information, measurements, or whatever other jargon is appropriate to the dynamics one is studying. However, there have to be some type of two body interaction necessary for the observer to switch between inertial frames.

    The symmetry breaking mechanism is whatever events make the observers different. External force, birth, death, and so forth are events needed to break the symmetry. The trajectories through space-time are punctuated by events that cause deviations from geodesic trajectories.
  6. Dec 9, 2012 #5
    In terms of special relativity, the traveling twin is accelerating. His turning around is accelerating. In terms of special relativity, gravity is just like any other force. Gravitational acceleration is just like rocket thrust. The external force breaks the symmetry by acting on the traveling twin while he is a large distance from the earth twin.

    I don't know general relativity enough to be sure about how it relates to this scentario. However, I will try to conjecture. In terms of general relativity, gravitational potential of the traveling twin is changing. I think this means that he is not on a geodesic.

    The traveling twin could monitor his motion by measuring the "tidal force" on his body. If there is a gravitational gradient, it will stretch his body. The closer that he gets to the earth, the greater the stress from the gradient of the gravitational potential.

    The earth twin feels a constant "tidal force" while the traveling twin feels a "time varying tidal force". So I would say that it is the "tidal force" that is breaking the asymmetry. I don't have the mathematics to back this up, though.

    In both cases, we see that the gravitational potential breaks the symmetry. The local stress caused by gravity can be used to distinguish between trajectories.
  7. Dec 9, 2012 #6


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    In that case, perhaps you should have stayed quiet because the previous poster does know about GR
    In the scenario described by PeterDonis the twin in orbit is in free fall all the time and the other twin is in free fall except during the initial outward impulse. But one must take into account gravitational time dilation in this setup.
    Last edited: Dec 9, 2012
  8. Dec 9, 2012 #7


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    First of all, special relativity can't handle gravity at all. It's *general* relativity that deals with gravity.

    Second, gravity in relativity is *not* like any other force; that's the whole point. What is just like rocket thrust is *sitting at rest* in a gravitational field--in other words, what is just like rocket thrust is *not* accelerating the way the local gravity field would accelerate you, for example, by standing on the surface of the Earth. This is one way of stating what is called the principle of equivalence.

    Actually, from the standpoint of relativity, you *are* accelerated when you are standing on the surface of the Earth, because you feel weight. That's the relativistic definition of acceleration: feeling weight. (When absolute clarity is needed, this is called "proper acceleration".) This definition also works for the standard twin paradox scenario, but it doesn't really help with the variants we've been discussing. See below.

    Yes, because the traveling twin is accelerated, in the above sense. In order to turn around, in the standard scenario, the traveling twin has to feel weight, even if only for an instant, while his rockets are firing (or some other external force is acting on him).

    However, in the variant scenario I proposed, the "traveling twin", who flies upward and then slows to a stop and falls back down, because of the gravity of the massive object, is in free fall the whole time; he never feels any weight. That means he's not accelerated in the correct relativistic sense.

    You can change your gravitational potential and still be on a geodesic. Throw a ball upward and it will change its gravitational potential, but it is freely falling and so is moving on a geodesic.

    In the standard twin paradox, where the traveling twin has to fire rockets to turn around, one way of analyzing the scenario uses the principle of equivalence to view the acceleration the traveling twin feels as being due to his rocket holding him at rest in a "gravitational field", while free-falling objects fall "downward" past him. On this view, the stay at home twin is much higher up in the field and so is at a much higher gravitational potential, and this is why the stay at home twin ages faster. The Usenet Physics FAQ gives a good treatment of this way of analyzing the scenario, along with its limitations:


    In the standard scenario, there is no tidal force on the traveling twin; his rocket acts on him all at once, with no variation in thrust across his body. More generally, spacetime in the standard scenario is flat, which means that tidal gravity is zero, because tidal gravity is equivalent to spacetime curvature.

    In my variant scenario, there is indeed tidal gravity present; in fact, according to general relativity, that's how you can tell a "true" gravitational field, like the one due to the Earth, from the "pseudo" field created by acceleration due to a rocket. However, the tidal gravity does not come into play in calculating the aging of the traveling twin as compared to the stay at home twin in my variant scenario; it's not what "breaks the symmetry". See below.

    Tidal gravity is not the same as gravitational potential. For one thing, tidal gravity is much more general; it can always be defined, even in non-static scenarios where there is no useful definition of "gravitational potential", such as the expansion of the universe. For another, "gravitational potential" can be present in flat spacetime, where tidal gravity is zero; the standard twin paradox scenario, where the traveling twin has to fire rockets to turn around, is an example of this.

    In the case of the standard twin paradox scenario, where the traveling twin has to fire rockets, what breaks the symmetry is that he feels acceleration, even if only for an instant, while the stay at home twin is in free fall the whole way. That's all there is to it.

    In the case of my variant scenario, both twins are in free fall the whole time, so whatever it is that breaks the symmetry, it can't possibly be acceleration. So the two cases are different, and have to be analyzed differently. In this case, what breaks the symmetry is simply that the two twins are traveling on different worldlines in a curved spacetime, and those worldlines can have different lengths even though they are both geodesics.
  9. Dec 10, 2012 #8


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  10. May 7, 2014 #9
    Contradicting yourself here. From a general relativity point of view, sitting on a thrusting rocket puts you (and the rest of the universe) in a homogeneous gravitational field (just like you said) and, hence, is indistinguishable from standing on earth.

    On an unrelated note: The scenario of a time signal sent from the outbound ship to a ship flying towards earth is not acceleration-less for the following reason:

    The time signal consists of photons (e.g. radio waves). When the outbound ship is sending the signal, it loses a very small mass and the inbound ship gains this very mass when it is receiving the signal. It is this mass that was accelerated toward earth and it is not possible to transmit any information without this type of acceleration.
  11. May 8, 2014 #10


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    It's indistinguishable from *feeling weight* because you're standing on Earth. But it's certainly *not* indistinguishable from *freely falling* in the Earth's gravitational field; if you're freely falling, you're weightless, which is easily distinguishable from feeling weight. That was my point.

    Photons don't experience proper acceleration when they are sent and received as radio waves; and the (negligibly small) impulse felt by each ship as the signal is sent and received can be arranged to be perpendicular to the direction of their relative motion, so that there is no acceleration in that direction and therefore no acceleration that matters for the problem under discussion.
  12. May 8, 2014 #11

    Filip Larsen

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    I also like the "three-event" way of modelling the twin-paradox, namely that the issue is modeled by three non-accelerating observers that, in pairs, are involving in three point events where they pass each other.

    However ...
    I can't see why Doppler shifts should matter at all. Surely the exchange between the "out-going twin" and the "home-going twin" (the alien ship in Darwins post) can be modeled as a point event that, in the context of classical physics, can be arbitrarily approximated in a "real" experiment. The important thing is that all three observers are able to make arbitrarily precise measurement of their own clock time at the instant of the two point events where they pass another observer.

    The fact that the three observers, when all three passing are done, at leisure can transmit the time interval (i.e difference in clock time between the two events they participate in) digitally to each of the other observers such that all can arrive the conclusion that the staying-home twin has seen a longer clock time than the sum of the out-going and home-going twin, does not add anything extra in order to resolve the paradox.

    To me the resolution of the paradox (the breaking of symmetry) with this model comes right when you introduce the event where the out-going and home-going twins pass.
  13. May 8, 2014 #12
    But that is not what you wrote.

    The small mass that was coming from earth and due to the transmission of the signal returns to earth is the only thing relevant for the twin paradox (everything else either has not been on earth or does not return to earth). This mass had a velocity directed away from earth and after the transmission towards earth, no matter if the transmission occurs before, during ("perpendicular" transmission), or after the rendezvous and the delta v is exactly the delta v of the ships.
    Has this mass not been accelerated?
  14. May 8, 2014 #13


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    This thread is two years old. Please don't necropost, particularly if the sole purpose is to nitpick (particularly baseless nitpicks like this).
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