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Resolve to the Twins Paradox – Frame of Origin

  1. Sep 6, 2010 #1
    I have read many arguments concerning the issue of the famed “Twin Clock Paradox”. Very many arguments go through considerably complex explanations so as to defend Special Relativity or dispute it. But the explanation is actually much simpler.

    In every proposed twins paradox, there is a Frame of Origin. The frame of origin is where the twins were at rest with respect to each other and to where they return. In the original scenario, that was “the Earth” and is referred to as “the inertial frame”.

    It is called “the inertial frame” because it is the one frame declared “inertial; not movable”. But the issue comes to mind as to how either frame can be considered the one that isn’t moving. SR declares that such a notion is irrelevant. But if it is ignored, an apparent paradox arises.

    So as to clear up the picture a bit, let’s presume that neither twin stands perfectly still, but rather one twin (Twin A) just very minutely drifts away from the Frame of Origin while the other twin (Twin B) takes off in his ultra fast rocket. This is merely to provide distinction and reveal that there are always 3 frames involved.

    And further, so as to get away from the complexities of gravitation issues, lets put them out in space originally standing on a small floating space station far away from any gravity field.

    The time dilation factor is always applied to whichever one leaves the Frame of Origin.

    Twin B gets in his rocket and rockets away. Twin A gets into his rocket and sticks ores out his windows and begins to row hoping to catch a few gas particles drifting by from Twin B’s exhaust.

    If we apply the time dilation to the one doing all of the serious moving, Twin B’s clock will run slower and he will age less. We cannot simply reverse the reasoning and say that it is all relative because Twin A is not moving away from the Frame of Origin.

    That is pretty much the end of the story, but…

    Some scenarios remove any sign of the Frame of Origin so as to make their point clear. The problem is that in every case, there must always be a Frame of Origin where their clocks were staying in sync whether anything is standing as a sign of it or not. If two items are moving away from each other, there must be a frame indicating the prior situation before they began moving. The issue becomes how to find out where that frame would be.

    If the ships merely see nothing but each other separating, they might not know which is accelerating from the Frame of Origin, but that is only a matter of their awareness. If they later find that one has greater time dilation than the other, it can be deduced as to which was accelerating more from the Frame of Origin.

    If they see no difference in their clocks, then they can deduce that they were both leaving the Frame of Origin merely in different directions.

    Without determining the Frame of Origin, the Twins cannot know which of them will age more until they meet again.



    Make sense?
     
    Last edited: Sep 7, 2010
  2. jcsd
  3. Sep 7, 2010 #2

    JesseM

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    No--the twin paradox still works fine if you don't assume the two observers actually started out at rest relative to each other, you could imagine one astronaut who's spent his whole life on one spaceship and another who's spent his whole life on a different spaceship, with the two ships having been in motion relative to another since the astronauts were born. Then if the two ships pass right next to each other at some moment in time, they can quickly compare ages at the moment they are right next to each other, then both ships travel on inertially at the same relative speed. Then after they have moved apart for a while, either one can fire its rockets in the direction of the other ship, so that the distance between them decreases until they pass one another again and compare their ages again. Then whichever ship fired its rockets, the astronaut on that ship will have aged less in total than the ship that traveled inertially between the two meetings.
     
  4. Sep 7, 2010 #3
    So its the acceleration that 'decides' who ages slower?
    Is there any formula involving acceleration that gives a math solution to the paradox?
     
  5. Sep 7, 2010 #4
    Well other than being more complicated, I don't see how that is any different than what I said. But consider this;

    Twin B takes off for a few years and returns. As he passes by at constant velocity (no rockets), Twin A accelerates to catch up to him. Which has aged more?

    Obviously (I hope) Twin B will still be the more aged. Yet it was Twin A that fired his rockets.

    But on the other hand, upon returning, Twin B could have fired his rockets so as to slow and meet Twin A. Still Twin B is the more aged.

    Who fired their rockets after leaving the Frame of Origin, doesn't resolve who is going to age more.
     
  6. Sep 7, 2010 #5

    JesseM

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    The fact that one twin accelerated between meetings and the other didn't means that the twin that accelerated is guaranteed to have aged less, regardless of the actual magnitude and duration of the acceleration. This is closely analogous to the fact in ordinary 2D Euclidean geometry that if you have a straight line between two points (constant slope in any Cartesian coordinate system) and a non-straight path between the same two points (change in slope somewhere along the path, analogous to change in velocity) then the non-straight path is always guaranteed to have a greater total length, not because it accumulated all the extra length on the section of the path with a changing slope (the non-straight path might consist of two straight segments with different slopes joined by a very short curved section, like a bendy straw), but rather because in Euclidean geometry a straight line is the shortest distance between points. I fleshed out the mathematical details of this analogy in post #8 here if you're interested.
     
  7. Sep 7, 2010 #6
    No. It is not the acceleration deciding which by calculation (not the amount of acceleration, but merely which accelerated away). The acceleration merely identifies who moved from the Frame of Origin.

    You could assume instantaneous acceleration without affecting the results. The question is only who left the Frame of Origin where the clocks were in sync. If they both left, both will age with comparative dilations.
     
  8. Sep 7, 2010 #7

    JesseM

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    It's different because the two twins don't have a common "frame of origin", they started in different frames. I thought your argument was saying it was important to identify the "frame of origin"? Which astronaut's frame is the "frame of origin" in my scenario where they just pass each other at constant velocity, having lived their whole lives on ships moving inertially with different rest frames?
    Does "takes off for a few years" mean A and B were moving apart for a while, so the distance between them was increasing? If so, how can Twin B "return" to "pass by" A unless he fires his rockets so that he is traveling back towards A for a while?
    Not if Twin B has to fire his rockets at the midpoint of the journey while A only fires his at the end, as suggested above. If that's not what you're imagining, please explain the scenario in more detail.
    If what I'm saying above is right, they both fired their rockets, at different times. The fact that Twin B had to fire them at the point where the distance between the twins was greatest, while Twin A only fired them at the very end when they were already very close, helps explain why twin B has aged less. Consider the geometric analogy above I mentioned in my post to Sakha. Suppose you have two points A and B on a 2D plane, and one path between them is mostly straight (and almost parallel to a perfect straight-line path between A and B) and only has a bend right before it meets B. But the other path takes off in a straight line at a totally different angle from the perfect straight-line path, then has a bend in the middle followed by another straight section at a different angle which goes to B (like a V shape with a slightly rounded bottom and the two tips of the V ending at A and B). These paths both contain sections of changing slope (analogous to changing velocity), but the second path with a bend in the middle is obviously going to be longer than the first path that is much closer to the straight-line path and only has a bend at the very end.
    In a scenario where one of them moves totally inertially between their two meetings while the other fires his rockets at some point between the two meetings, yes it doesn, the one that accelerated will always have aged less.
     
  9. Sep 7, 2010 #8
    So I deduce acceleration is not relative, i.e only one of them feels the force.
    This takes me back to my high school physics teacher - "That tree got in my way, not my fault!" (Relatives velocities week), so with SR its really his fault, he 'accelerated' towards the tree and not viceversa.
     
  10. Sep 7, 2010 #9

    JesseM

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    Right, in SR you know if you're accelerating since you can feel G-forces, as measured by an accelerometer.
     
  11. Sep 7, 2010 #10
    Ok, you cleared a lot of confusing thought from my head. Now I'll sleep very well.
    I don't know why I assumed that acceleration could be relative.

    Nighties. It's your fault if I dream with spaceships and elder astronauts!
     
  12. Sep 7, 2010 #11
    Jesse, in your version, they have to remember who has fired rockets and how much. In mine, they have to remember who left the Frame of Origin wherein their clocks were in sync. If they have no memories, they are pretty much screwed until they meet gain. Knowing who fired rockets last doesn't really help.
     
  13. Sep 7, 2010 #12

    Dale

    Staff: Mentor

    No, the age (proper time) is a Lorentz invariant, so it can be calculated in any inertial frame and all inertial frames will agree on its value. It is certainly not restricted to the frame of origin.
     
  14. Sep 7, 2010 #13

    JesseM

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    Well, you didn't address the question of which is supposed to be the "frame of origin" in my scenario where they pass each other inertially--are you saying that your approach just can't tell us who'll be older in this scenario?

    And consider another scenario: suppose we two twins that grew up on a small space station, then at some point one twin gets on a shuttle that's docked to the station and accelerates to build up a large velocity relative to the station, then coasts away inertially for a while. Would you say that the station's frame is the "frame of origin" in this case? If so, what do you predict would happen if, after the two twins have been moving apart inertially for a while, the station fires its own rockets to accelerate in the direction of the shuttle, reaching a high enough velocity that it is eventually able to catch up with the shuttle? Would you say that since the shuttle was the one that initially left the frame of origin, the twin on the shuttle should be younger when the two twins reunite?
     
  15. Sep 7, 2010 #14
    No matter what frame you are in, you have to know in what frame their clocks would be in sync else you can't even use Lorentz. That is the "Frame of Origin".

    If they do not know at what frame their clocks would be in sync, they cannot predict who has aged more other than to just look and see.

    The station in that is not the Frame of Origin. The station was "at the frame of origin" until it accelerated. And the end result would be that they simply aged the same because they both moved away and established a knew frame of sync when they met up again.
     
  16. Sep 7, 2010 #15

    Dale

    Staff: Mentor

    The clocks never need to be in sync. Age (proper time) is a frame invariant quantity that is not dependent on any frame variant synchronization convention. In fact, the proper time is even invariant under transformations to non-inertial coordinate systems or in curved spacetime, both of which may have arbitrary synchronization conventions.

    The frame of origin is certainly a very convenient frame to use, but it is not at all necessary.
     
  17. Sep 7, 2010 #16

    JesseM

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    Yes they can, they can just use SR to predict how much each one ages. If you know each twin's velocity as a function of time v(t) in the coordinates of any inertial frame, you can predict how much each ages between the time coordinate t0 when they depart from one another and the coordinate time t1 when they reunite, by doing the integral [tex]\int_{t_0}^{t_1} \sqrt{1 - v(t)^2 /c^2 } \, dt[/tex]. This will give the same answer regardless of what frame you use. Also, if we assume the acceleration times are extremely brief, then the elapsed time can be approximated by just knowing the time in the inertial frame of each constant-velocity phase of the trip: if one twin travels at speed v1 for time t1 for the first part, then quickly turns around and travels at speed v2 for time t2 for the second part until reuniting with the other twin, then that twin's total aging will be [tex]t_1 * \sqrt{1 - v_1^2 /c^2 } + t_2 * \sqrt{1 - v_2^2 /c^2 }[/tex]. Again, you can use any frame you want to define the coordinate times of different phases of the trip and the coordinate velocities in each phase, you'll always get the same answer for the total aging. I gave a numerical illustration in post #63 of this thread:
    Well, your prediction that they aged the same is incorrect. Even if the magnitude and duration of their accelerations relative to the inertial frame where the station was originally at rest was exactly the same, in this scenario the twin on the station would have aged less when they reunited. I can give a numerical derivation if you want, but the reason is not hard to understand intuitively in terms of the geometric analogy, where a path that's mostly straight with a small curve near one endpoint will be shorter than a path with two straight segments and a curve in the middle:
     
  18. Sep 7, 2010 #17
    This doesn't seem to approach the paradox question. "Proper time" has to get determined. This issue is about who gets to make that determination between two separating bodies.

    You can't know that t0 without knowing the Frame of Origin. From "any" arbitrarily chosen frame, you can determine a change in their aging, but you can't know which frame to choose. If you choose a frame that is inertial to Twin B, then you will see yourself as the one moving away and not aging.

    The Frame of Origin gives "proper time" to your choice.


    No actually, I didn't give it much thought and considered the idea that you would come back with critique. After a little thought, I can tell that because one left earlier and stayed at high velocity longer, he would have aged a little less until the other over accelerated to catch up. At that point, it is an issue of the exact details.

    And I am a little curious now exactly how that would pan out. I'm not certain the resolution would be consistent.

    If Twin B shot off at c then stopped and later Twin A did the same, they would be the same age due to the symmetry of their situation.

    If Twin B shot off at c and didn't stop, then obviously Twin A would never catch him and they would not age any more, leaving Twin A still older.

    Anywhere between those gets a little vague, but implies that Twin A would be merely less older than he would have been if he had stayed on the station. So you are very probably right even without the math.
     
  19. Sep 7, 2010 #18
    The equation you're asking about DOES exist. Here's how I described it in a previous post:
    ________________________________________________________

    In the simplest version of the traveling twin example (with a single instantaneous speed change at the midpoint), it IS possible to INFER the change in the home twin's age (according to the traveler), during the instantaneous turnaround: you know that both twins obviously must agree about their respective ages when they are reunited, and you know how much the traveler says his twin ages during the two inertial portions of his trip. So the home twin's ageing during the turnaround (according to the traveler) must be just enough to make the totals agree at the reunion.

    But in only slightly more complicated examples (e.g., with multiple instantaneous velocity changes), it becomes important to KNOW how to directly calculate the amount of the home twin's ageing during each of the velocity changes. And in the case of FINITE accelerations, being able to directly calculate the home twin's ageing during each of the traveler's segments of finite acceleration is indispensable.

    It turns out to be easy to do that. For the cases of instantaneous velocity changes, the required calculations are almost trivial to carry out. For piecewise-constant accelerations, the calculations are a bit more complex, but they can still be done, if necessary, with a good calculator. And with a computer program, they are very easy.

    Both of the above types of problems can be handled with a simple equation that I derived many years ago, which I call "the CADO equation". The CADO equation follows directly from the Lorentz equations ... the CADO equation really just automates what you can deduce from the geometry of the Minkowski diagram.

    I also, many years ago, implemented the CADO equation in a computer program I call "the CADO program".

    It's also possible to use the CADO equation to do completely general acceleration profiles, but you (usually) can't do it in a closed-form way...it requires some numerical integrations. Fortunately, piecewise-constant accelerations are usually all you really need to be able to handle.

    Here is a description of the CADO equation that I've posted previously, in other threads:
    __________________________________________________ __________


    Years ago, I derived a simple equation that relates the current ages of the twins, ACCORDING TO EACH TWIN. Over the years, I have found it to be very useful. To save writing, I write "the current age of a distant object", where the "distant object" is the stay-at-home twin, as the "CADO". The CADO has a value for each age t of the traveling twin, written CADO(t). The traveler and the stay-at-home twin come to DIFFERENT conclusions about CADO(t), at any given age t of the traveler. Denote the traveler's conclusion as CADO_T(t), and the stay-at-home twin's conclusion as CADO_H(t). (And in both cases, remember that CADO(t) is the age of the home twin, and t is the age of the traveler).

    My simple equation says that

    CADO_T(t) = CADO_H(t) - L*v/(c*c),

    where

    L is their current distance apart, in lightyears,
    according to the home twin,

    and

    v is their current relative speed, in lightyears/year,
    according to the home twin. v is positive
    when the twins are moving apart.

    (Although the dependence is not shown explicitly in the above equation, the quantities L and v are themselves functions of t, the age of the traveler).

    The factor (c*c) has value 1 for these units, and is needed only to make the dimensionality correct. For simplicity, you can generally just ignore the c*c factor when using the equation.

    The equation explicitly shows how the home twin's age will change abruptly (according to the traveler, not the home twin), whenever the relative speed changes abruptly.

    For example, suppose the home twin believes that she is 40 when the traveler is 20, immediately before he turns around. So CADO_H(20-) = 40. (Denote his age immediately before the turnaround as t = 20-, and immediately after the turnaround as t = 20+.)

    Suppose they are 30 ly apart (according to the home twin), and that their relative speed is +0.9 ly/y (i.e., 0.9c), when the traveler's age is 20-. Then the traveler will conclude that the home twin is

    CADO_T(20-) = 40 - 0.9*30 = 13

    years old immediately before his turnaround. Immediately after his turnaround (assumed here to occur in zero time), their relative speed is -0.9 ly/y. The home twin concludes that their distance apart doesn't change during the turnaround: it's still 30 ly. And the home twin concludes that neither of them ages during the turnaround, so that CADO_H(20+) is still 40.

    But according to the traveler,

    CADO_T(20+) = 40 - (-0.9)*30 = 67,

    so he concludes that his twin ages 54 years during his instantaneous turnaround.

    The equation works for arbitrary accelerations, not just the idealized instantaneous speed change assumed above. I've got an example with +-1g accelerations on my web page:

    http://home.comcast.net/~mlfasf [Broken]

    The derivation of the equation is given in my paper

    "Accelerated Observers in Special Relativity",
    PHYSICS ESSAYS, December 1999, p629.

    Mike Fontenot
     
    Last edited by a moderator: May 4, 2017
  20. Sep 7, 2010 #19

    Dale

    Staff: Mentor

    Anyone in any spacetime using any reference frame and any synchronization convention may make that determination. That is the whole point of using frame-invariant quantities.
     
    Last edited: Sep 7, 2010
  21. Sep 7, 2010 #20
    The point to the paradox confusion is that they do not have a "synchronization convention" so as to make the decision. The Frame of Origin gives that convention for them so as to resolve the question.
     
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