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New Twin Paradox

  1. Feb 26, 2013 #1
    Suppose you have two people who are in identical orbits around a large star. The only difference between them is the direction they are traveling. At a certain point where they meet ( there are two such points ), they sync clocks.

    After a few orbits, they meet again and again compare clocks. From "A" perspective, they don't move but instead are in perpetual free fall, while "B" moves rapidly around the star. Ditto but reversed for B. Both experience the exact same acceleration during our experiment, so who's clock is right when they compare them again? Both would expect the other's clock to run slower since, from their own perspective, the other is moving while they stand still.

    How is this paradox actually resolved in relativity?

    Also, if the universe is closed ( spatially ), then one could create the same experiment but without any acceleration at all. The twins would compare notes after passing each other again once they've lapped all the way around the universe. How does relativity resolve that paradox?
     
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  3. Feb 26, 2013 #2

    Dale

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    Regardless of how you specify the details of what you mean by "their frames" there are some rather weird gravitational effects. The net result is that they expect the other guys clock to read the same as theirs.
     
  4. Feb 26, 2013 #3

    PAllen

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    From context, I assume you are looking for a special relativity answer. There really isn't a valid one because of gravity. To correctly treat gravity, you need general relativity. If you don't mind, I will translate to avoid any gravity issues:

    Two rockets, using thrusters, maintain perfectly symmetrical opposite direction circular trajectories around a space buoy that has not force or thrust acting on it.

    Then, SR (special relativity) predicts that both rockets have the same clock readings on each meeting; and further that if each rocket sends clock readings to the buoy on each meeting, the buoy would find the difference between these sent readings to be less than difference in times on the buoy's clock between receipts of these messages.

    There have been many threads on this recently here. You should check them out. Special relativity does not say that 'all motion is relative', it says (among other things) that inertial motion (in a closed 'lab') cannot be distinguished from rest by any measurement inside the 'lab', however large the 'lab' is. Note that non-inertial motion of a 'lab' is easily distinguished from inertial motion. Specifically, the familiar formula for time dilation depending only on velocity is true only in an inertial frame. You can derive a different formula for specific coordinates in which a non-inertial observer is at coordinate rest. This will be a more complex formula that will involve position and time as well as speed. The result would be that both non-inertial rockets would compute that the other rocket ages the same between meetings as they do.

    For a closed universe (which is not part of conventional SR), the answer is different. Please note that if we are staying in this 'generalize SR', the closure must be such as to not introduce curvature. This means, among other things, it cannot be spherically symmetric, spatially. This means the closure must introduce preferred directions (imagine a cylinder). As a result you get:

    - multiple inertial paths connecting the same events (meetings)
    - the different inertial paths may have different different clock readings depending on their
    orientation to the preferred direction introduced by the closure.

    If you set up Minkowski type coordinates for different inertial observers in a closed universe, the form of the closure will be differ for different observers; this introduces objective differences; the principle of relativity loses global validity in such a universe.

    [Edit: FYI, the GR analysis of the opposite orbit situation would produce the same result as the SR analysis in this case (because distance from star is constant). However, the explanatory model would be different. Let me know if you really meant or want a GR explanation. ]
     
    Last edited: Feb 26, 2013
  5. Feb 26, 2013 #4

    russ_watters

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    How can "A" do that without actually falling and hitting the planet? And conversely how can "B" be moving at twice orbital velocity and not fly off into space?
     
  6. Feb 26, 2013 #5
    There are situations where two inertial observer with no relative motion between them can nonetheless have differing clock rates wrt each other. Also there are situations where a non-inertial observer and an inertial observer with a constant distance between them will have the same clock rate wrt each other.

    Place observer A inside a uniform massive hollow sphere, such that for A there is no local gravitational acceleration. Place observer B, at a constant distance from A, on the surface of the sphere. Place observer C, at a constant distance from A and B, in a position far removed from this mass.
    1) All observers will agree that B's clock is slower than C's clock, though not necessarily by how much, due to gravitational time dilation.
    2) A and B will agree that their clocks share a common relative rate, even though A is an inertial observer and B is not.
    3) A's clock will be slowed relative to C's clock even though they maintain a constant relative distance and both are inertial observers.

    In cases involving gravity the distinction between gravitational depth and curvature is often overlooked. Gravitational time dilation is the consequence of gravitational depth, not curvature. What time dilation is caused by curvature is a product of SR, and the result of gravitational acceleration imparting a relative velocity.

    When it is said that acceleration breaks the symmetry in the so called clock paradox its character differs somewhat between gravitational and inertial boost cases (SR). The key is that the total energy of the system, wrt any observer, must be conserved, and the the lowest sum of the potential energy states correspond to the slowest relative sum of clock rates. The clock paradox is constructed from the fact that each observer assigns all kinetic energy the the other, i.e., disagrees on the localization of the energy of the system. This localization problem has lead some to claim energy is not conserved in GR, but this is no more true than in the SR case.

    In the GR case acceleration between two observers can occur even when both observers effectively remain inertial. Hence, unlike the SR cases, tidal forces (g forces) are not strictly required in order for acceleration to occur. Yet this can only happen if a gravitational field exist, regardless of how different various observers may define it. Changing your location wrt a gravitational field involves changing the ratio of kinetic to potential energy wrt some observer, which is a more general definition of what a boost consist of in SR. Hence the symmetries involved with an acceleration can either involve changing an observers boost in spacetime, or changing the gravitational depth of spacetime itself. The symmetries work exactly the same, even if the manner in which an observer defines the situation differs significantly.

    It seems to me that quiet often people attempt to conserve an observer definition of the situation. Technically possible, but worthless in general. However, if you limit the definition of acceleration to strictly non-inertial observers, such that two inertial observers cannot accelerate wrt each other, GR provides very real situations under which any observer definition of the situation can be falsified. Most directly in cases where two inertial observers accelerate wrt each other.
     
  7. Feb 27, 2013 #6

    PAllen

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    This and your following explanation are primarily about GR. My suspicion is the OP was thinking about SR, but that is not clear. More to the point, none of your scenarios resemble the OP, so I am afraid all of this will be a confusing distraction from their question.

    I will assume by 'clock rate between observers' you mean something not some arbitrary coordinate dependent quantity, but something invariant like the result of sending uniformly timed signals to each other.
    This confuses things. If you mean the invariant comparison I described, it is independent of all coordinates or choice of extra observer. If you really mean to allow any choice of observer, using their own observation of B and C clock rate (rather than computing what B and C mutually observe exchanging signals), then this is false. An observer rapidly approaching B from C will observe B to be going faster than C.
    This is not true in any sense. B on the outside of the shell will be at higher potential than A inside the shell (unless the shell is massless, in which case the whole scenario is vacuous).
    This is certainly true in the invariant sense I described.
    Overlooked by who? It is extremely well known that for static GR solutions you can define a potential that predicts clock comparisons.
    It is worth emphasizing that the ability to even define 'gravitational time dilation' and gravitational potential (better word than depth) requires a static GR solution. For general GR solutions it is impossible to define these things. Except as a computational convenience for static solutions, there is really a single form of clock comparison in GR that covers all solutions, all cases: parallel transport 4-velocity of world line 1 tangent along null geodesic to world line 2; express transported 4-velocity and null geodesic propagation vector in local frame of reception event on world line 2; apply SR doppler formula. This gives observed clock comparison (and Doppler) for any circumstance at all: rapidly rotating binaries; cosmological expansion; any 4-velocity of emitter and receiver.
    This is quite wrong. As I mentioned above, in dynamic GR solutions you cannot even define potential. Conservation of energy applies only globally - at spatial infinity, and only if the the global solution is asymptotically flat (or some slight generalization thereof). Where there is no spatial infinity, you cannot define ADM energy, and there is no global conservation.
    Here you must be careful about coordinate dependent versus invariant quantities. All inertial observers have zero proper acceleration in GR, by definition. Coordinate acceleration between them can basically be made almost whatever you want. You could give this statement some meaning by defining some observation, e.g. : it is true that two inertial observers can see time varying Doppler between each other.
    This is very misleading terminology. In GR (and Newtonian gravity too!), tidal forces refer to the tendency of free falling bodies initially at mutual rest to converge or diverge. Meanwhile, GR universally defines 'g forces' if you will to be proper acceleration, which is neither a question of potential (where it can be defined), nor tidal forces. It is simply a measure of deviation from free fall.
    More confusion from mixing coordinate quantities (acceleration without any definition; best I can see is coordinate acceleration) with invariant quantities (proper acceleration, which is certainly zero for any inertial observer in GR).

    All in all, your long post did not address the OP scenario at all, and added much confusion and many erroneous statements.
     
  8. Feb 27, 2013 #7
    Identical orbits means symmetry; thus per turn they will expect to age the same. The expectation of mutual time dilation is for reference systems in uniform motion.
    That's an interesting one as I never understood that concept; however it appears that you again propose a symmetrical problem, which necessarily has a symmetrical solution.
     
  9. Feb 27, 2013 #8

    A.T.

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    Classically, in A's rest frame there are inertial forces that keep A in place. In GR you would model those inertial forces as space-time geometry, so the static Schwarzshild metric is modified.

    I wonder if anyone here knows a treatment like this: A simple circular orbit in the rotating frame where the mass and testbody are both at rest. The metric should be static. The modified gravitational time dilation along the radial line should have a minimum at the orbit's radius. Could this modified gravitational time dilation in the rotating frame be computed by simply multiplying:
    - gravitational time dialtion from the inertial frame
    - tangential movement time dialtion from the inertial frame
    ?
     
  10. Feb 27, 2013 #9

    PAllen

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    You could derive this from the Schwarzschild metric by a coordinate transform where one of the angular coordinates is a linear function of time: [itex]\theta[/itex]' = [itex]\theta[/itex] - [itex]\omega[/itex]t. The metric components would still be time independent, but there would be a new theta',t cross term and as well as an additional expression for the t,t term. For some radius, this would make a circular orbit have constant theta'.
     
  11. Feb 27, 2013 #10

    PAllen

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    This is wrong, as I explained above. Any closure that preserves the Minkowski metric locally must introduce a preferred direction (or directions).
     
  12. Feb 27, 2013 #11
    I'm not going over every detail of your criticism, unless you request specific clarifications. By clock rates I mean what the observers measure wrt each other, irrespective of what interpretation you put on those measurements.

    That's debatable, since you can't analyze orbits involving time dilation under strictly SR conditions.

    Don't you mean here to say: unless the mass is distributed on a shell in the limit as the thickness of the shell approaches 0? If you want to avoid such a limit, with no margin of error, the statement can be also made arbitrarily close to true simply by placing B on a separate sphere of differing mass at sufficient distance from the first in otherwise flat spacetime.

    In fact I'm not really seeing any criticism that's not merely splitting hairs, and if I give you the benefit of the doubt about what you were thinking wrt a massless shell, for only being strictly valid in the some limit, regardless of how little practical difference that makes, or conflating the phenomenology of what observers measure with a claim about what formalism that entails, or similar.

    Even in the case where you are correct that an observer between B and C rapidly approaching B will momentarily invalidate my statement, given a period of time much larger than required for this observer to get from C to B then this no longer holds.

    Perhaps you are correct that it is a distraction, but the criticism is not substantive, or simply wrong if you intended the massless shell statement the way it came out.
     
    Last edited: Feb 27, 2013
  13. Feb 27, 2013 #12

    A.T.

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    You mean like done here?:
    http://en.wikipedia.org/wiki/Geodetic_effect#Formulae
     
  14. Feb 27, 2013 #13

    PAllen

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    Good, that's what I proposed at the beginning of my post as a way to give the comparison invariant significance.
    Agreed, which is why I translated the OP description to a pure SR case matching what I guessed was their intent. It is just a guess.
    It may be splitting hairs, but there is no way to treat a zero thickness shell of nozero mass in GR. However, I certainly agree that you can make the potential difference as small as you want.

    My biggest issue is that the whole post created a bunch of new scenarios completely unrelated to the OP, and explained them with confusing terminology and some substantive errors (e.g. the whole bit about conservation of energy in GR).
     
  15. Feb 27, 2013 #14

    PAllen

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  16. Feb 27, 2013 #15
    This they would indeed note this as they passed each other. However since neither is in an inertial reference frame, they wouldn't be able to build a coherent story of how the other's clock was running at other times during the orbits. Per recent thread here, if A were sitting in one place - and thus in an inertial reference frame - and B were making circles, passing A regularly, then A & B would agree that B's clock was running more slowly on average.

    https://www.physicsforums.com/showthread.php?t=668580&page=5
     
  17. Feb 27, 2013 #16

    Dale

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    Hi Everyone,

    There is some serious confusion in this thread as to whether the OP is interested in SR or GR. I am closing the thread. The OP is welcome to make a new thread on the same topic, but should clearly specify if he/she is considering SR or GR.
     
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