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I A _perfectly_ symmetric twin paradox cases

  1. Mar 13, 2017 #1
    Case 1)
    Two rockets (no Earth involved) have an exactly the same acceleration profile/flight-plan during round trip but they dispatched to opposite directions. At the start both rockets are docked to the same space station...both rockets have an identical engine operation plan during the round trip (engine vector/thrust plans are identical along the trip). After the trip both rockets docked ISS back and compare their G-record files they acquired/recorded during their trips.

    Case 2)
    one observer stays on the Earth another observer started from the Earth orbit. Let make sure that the profile of rocket engine thrust (vector and force) and astronaut orientations inside the rocket during the trip are arranged in such way that observer/astronaut in the rocket experiences all the time a vertical steady (from his perspective) 1G acceleration exactly the same 1G as observer on the Earth. After a round trip, the rocket get back to the same original position before the trip - a low earth orbit; during radio session with Huston they may compare the watches and acceleration profile along time of trip - they should be identical. This case is not a perfectly symmetric so may have some loopholes though...

    It seems that once the setup is perfectly G-force/vector symmetric the only way to avoid paradox is to have an identical record files of of G-meters readings for both observers along the trip. The alternative would be to accept that paradox indeed takes place and both observers meet with different counterpart - let call it an "Everett-3 twin solution" ;o))

    I have no difficulty to see how math works for SR and so I see no paradox at all for SR compatible setups but I failed so see the solution for symmetric cases (not compatible with SR). I'm sure there is a consensus among experts in this area about a perfectly symmetric cases and I would appreciate if someone helps me to comprehend how paradox could be solved for _perfectly_ symmetric cases.

    Thank you,
    Stephan
     
    Last edited: Mar 13, 2017
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  3. Mar 13, 2017 #2

    A.T.

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    Proper acceleration doesn't depend on perspective. And your rocket cannot take off from the surface, if its proper acceleration cannot exceed the proper acceleration of the local surface. But all that doesn't really matter, because due to the space-time curvature there is no symmetry here anyway. Gravitational time dilation comes into play.
     
  4. Mar 13, 2017 #3
    >"Proper acceleration doesn't depend on perspective"
    I'm not sure what "propper acceleration" means; from point of view of observer the only what is observable is G-force and its direction relatively observer frame and the file of G-records during round trip is the only observables to compare after trip.

    Btw, I intentional suggested two cases: one is perfectly symmetric and another is not quite symmetric; the first case is clean and simple the second is not quite clean yet you focus on second one... ;o)

    Thank you for reply,
    Stephan
     
  5. Mar 13, 2017 #4

    Dale

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    This one is symmetric so the clock readings are identical. If you swap the two rockets and rotate 180 degrees then you recover the original scenario.

    This scenario is not symmetric. If you swap the astronauts and rotate 180 degrees you do not recover the original scenario. There is no expectation by symmetry that they should be the same.
     
  6. Mar 13, 2017 #5

    Nugatory

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    @stefanbanev:
    Case A is easy: the passengers on both rockets age equally (and less than someone who stayed behind on the space station).

    However, these results do not come from considering the acceleration profiles. The two travelers and stay-at-home are following different paths through spacetime (that's spacetime! Not just space!) between the departure event and the reunion event, and in general different amounts of time pass on different paths through spacetime. We calculate these paths, and we find that the different paths followed by the travelers happen to be the same length (unsurprising, in view of the symmetry), and that that length is shorter than the path stay-at-home followed.

    Thus, the acceleration is something of a red herring. It's the path through spacetime that matters, and acceleration only comes into the picture because we couldn't set the travellers on different paths without accelerating at least one of them.

    Once we understand that, we can see that setting up the same acceleration profile in case B is also a red herring. Spaceship and Houston may experience the same acceleration profiles but they're still on very different paths through spacetime and there's no particular reason to expect both paths to have the same length.

    The "right" way to analyze case B will require doing line integrals in Schwarzschild spacetime to calculate the path lengths (fortunately, the problem is common enough that you can find these integrals already worked online). However, you will get a perfectly good and much more intuitive result for much less work if you apply the technique described in the "Doppler Shift Analysis" section of the twin paradox FAQ.
     
    Last edited: Mar 13, 2017
  7. Mar 13, 2017 #6

    Nugatory

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    That's proper acceleration, the acceleration that would be recorded by an accelerometer being carried along by the accelerated object. You are making a good start by focusing on proper acceleration (as opposed to "coordinate acceleration").
     
  8. Mar 14, 2017 #7

    vanhees71

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    I'm still puzzled, how complicated you can still make discussions about the twin paradox, which is not more paradoxical than the fact that you can travel different distances when going from a point A to another point B, depending on the path you take for your travel. The only slightly more abstract point here is that we talk about Minkowski distances in spacetime rather than usual Euclidean distances in space.
     
  9. Mar 14, 2017 #8
    To "Nugatory" :

    Thank you for the prompt replies...

    Thus, if I understand you correctly, the asymmetry for case 2 is the rotation of observer frame to mountain a steady vertical 1G force (to comply with case 2). Yet, from point of view of G-force meter observer holds on its knee, it seems (*) (may be I'm wrong) possible to arrange the rocket engine thrust profile during the round trip that G-meter record may have a flat steady 1G record directed from observer head to its feet exactly the same as Earth' observer has.

    The reason I focus on G-meter is that from point view of observer the _only_ available observables is the record of G-force value and its direction recorded during the round trip. There is no other objective measurements to compare after trip to establish the symmetry or asymmetry. Apparently this record can not depend on the apriory knowledge about rocket engine operation profile. The weak point in this logic is assumption (*) though... I apologize for puzzling some experts by my stupidity ;o)

    Thanks,
    Stephan
     
  10. Mar 14, 2017 #9

    PeterDonis

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    It's not. The rocket can get from orbit to ground under 1G acceleration (though the direction of the acceleration will have to change during the trip), but he can't get from ground back up to orbit under 1G acceleration, as has already been pointed out, because 1G acceleration pointed straight up is just enough to keep him "hovering" at the Earth's surface--he can't get any higher.
     
  11. Mar 14, 2017 #10

    PeterDonis

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    This is not true. The observer's elapsed clock time is also an observable. What's more, this observable directly reflects the length of the observer's path through spacetime, so it is the direct comparison that is used to show symmetry/asymmetry.
     
  12. Mar 14, 2017 #11

    Dale

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    So when we are talking about symmetry in physics this is not what we mean. In physics we talk about the symmetry of the laws of physics and the symmetry of the boundary conditions.

    If the laws of physics obey a certain symmetry AND the boundary conditions of a particular problem also obey that same symmetry then the solutions will have that symmetry. In this case the boundary conditions for 2) are not symmetric so there is no expectation of symmmetry in the solutions.

    The astronauts ignorance of other measurements is not a justification of symmetry. It is always possible to reduce the available information to the point that you cannot get a correct answer.
     
  13. Mar 14, 2017 #12

    Jonathan Scott

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    Note that proper time rates are unrelated to acceleration. They are related to velocity (for SR) or potential (for gravity). When those change there is obviously acceleration involved, but that only affects the result by changing the velocity or potential.

    You could for example have two observers sitting on different-sized planets where the surface gravity ##Gm/r^2## is the same but the potential ##-Gm/r## and the corresponding time dilation fraction ##-Gm/rc^2## are very different. For example, if you had a planet with a mass four times that of Earth but with a radius twice that of Earth (so half the average density), then the surface gravity would be the same as that of Earth but the fractional time dilation at that surface would be twice that at the surface of the Earth (compared with outer space).
     
  14. Mar 14, 2017 #13

    Boing3000

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    May I suggest a cheaper setup with truly symmetrical G-meter reading. Put two clock in centrifuge at 2G. But one centrifuge two time bigger than the other. It should move slower than twice the speed because the relation ##a= \frac{v^2}{r}##

    As per SR, from the ground, the outer clock should record more time, because it travel less space.

    From any of the two clock frame (if we imagine they only see each other), they would see themselves moving like planet around some hidden sun. Thus the exterior clock should tick faster due to being above in the "fake" gravitational field (as per GR).

    All this is based on wild guesses.
     
  15. Mar 14, 2017 #14

    pervect

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    I will mention my usual alternate approach to the general topic of the "Twin Paradox".

    Imagine that we take the block universe interpretation of physics. Note that I'm not claiming this is the only way to understanding time, just that it is "a way", one that is recognized as working with special relativity.

    By visualizing time as if it were another dimension, we can see that the reading of a clock is a lot like an odometer, except the clock reads the time dimension. But we're considering the time dimension as if it were "another dimension" in our block universe interpretation.

    Given this, we can construct an analog of the twin paradox, the "odometer paradox". Two cars drive from city A to city B, each measuring the mileage. One car takes a straight line, the other doesn't. The car that takes a straight line has the lowest reading on their odometer.

    I'm not aware of anyone who fines this paradoxical. We don't even expect the two odometers to read the same.

    The situation with time is similar, there is no reason to require that the time readings on the two clocks (which are measuring this extra dimension in the block universe) should read the same if the clocks take different routes through space-time.

    Due to a belief in what is commonly called "absolute time", some people have an intuition that the clocks (unlike the odometers) should have the same reading even if they take different routes. The point is that this may be an intuitive expectation, but it's not a logical requirement for a theory to be self-consistent, and that violating this intuitive expectation is in and of itself not paradoxical.

    There are some important differences between time and space that impact the power and scope of this analogy. The main thing to note is that the odometer will read the shortest distance for a straight line path, while the clocks will read the longest elapsed time for the equivalent "shortest path". I could say more on this, but it would detract from the main point of the thread, I think, so I'll mention it in passing but not go into details at this point.

    We can apply this analogy usefully to the current problem. If two routes are exactly symmetrical, the odometer readings will match. If the odometer readings don't match, the routes cannot be exactly symmetrical. The later case seems to be what's under discussion, there is a problem where we have clocks (odometers in the analogy) that aren't matching. This is enough basis for us to reject the claims that the routes are "completely symmetrical". Having two routes through space that were "totally symmetrical", but had different lengths, would be logically incosistent.
     
  16. Mar 15, 2017 #15

    Dale

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    This also is not symmetrical. If you swap the clocks and rotate 180 degrees you do not recover the original scenario.

    Furthermore, the difference in angular velocity is detectable by standard 6 degree of freedom accelerometers. It would only be a 1 degree of freedom accelerometer that would be tricked here. See my above comments about ignorance.
     
  17. Mar 15, 2017 #16

    Boing3000

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    I don't get this part.

    That's fine, now I am less ignorant. I was simply trying to make a simpler "more" symmetrical setup for the OP.
    I suppose what you describe is that the local curvature (in the sense of Einsten equivalence principle) is different even for the same G, just like Jonathan Scott describe on post #12. I didn't know infinitely small acceleratormeter would be able to distinguish the difference (aren't they poking at tidal forces ?)

    Now, what about two clocks a different heights in a steadily accelerating rocket ? Wouldn't the situation be truly symmetrical ?
    OK no, if the rocket goes the other way, the clocks will diverge the other way too. But even inside one experiment, a perfect G-Meter records comparison would still no be enough IMO.
     
  18. Mar 15, 2017 #17

    Nugatory

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    Your example is using centrifuges - that is a flat-space special relativity situation, no curvature involved.
     
  19. Mar 15, 2017 #18

    A.T.

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    Jonathan was talking about the gravitational potential and gravitational time dilation, not the local curvature related to tidal forces.
     
  20. Mar 15, 2017 #19

    Boing3000

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    Indeed, but I was just trying to fix the OP second scenario, with "truly"(not really as Dale explained to me) similar "G-Meter profile" . Cannot we use centrifuge for that ? Aren't any accelerated frame of references equivalent (in principle) ? And don't both centrifuges would create two different "kind" of "identical" acceleration ?

    But as Nugatory says, this problem is probably better treated from SR perspective, although the OP intent is not to use an external frame of reference, or so I think.

    Am I wrong in saying that even a stronger symmetrical "G-Profile" would still not justify the clock to be synchronized (as the classic accelerating ship gravitational time dilation example shows) ?
     
  21. Mar 15, 2017 #20

    PeterDonis

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    No. A centrifuge and a rocket accelerating in a straight line are not equivalent, except over a small enough patch of spacetime. The centrifuge is rotating and the rocket is not, and that will have observable effects (which in Newtonian physics go by the name of "Coriolis forces").
     
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