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B Triplet paradox and a thought

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  1. May 14, 2017 #1

    Buckethead

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    First off, let me just say I love this forum and all the people in it! The world is a better place because of all of you who take the time to post here and I thank you all!

    The most popular explanation to resolve the twin paradox seems to be the asymmetry that arises because the ship undergoes acceleration. I have to admit I have never been able to find comfort in that explanation even though I know it is correct. It also led me to wonder about a triplet situation whereby a non accelerating ship (A) passes Earth, syncs its clock at that time and heads out to space where it passes a non accelerating inbound ship (B), at which time B records A’s time. B eventually does a flyby past Earth where Earth and Ship B compare times. With no acceleration involved in that scenario I then had to study more explanations still with no real comfort.

    But finally, in post #18 of the thread entitled “A Geometrical View of Time Dilation and the Twin Paradox – Comments” my non acceleration scenario was laid out perfectly along with a beautiful resolution.

    https://www.physicsforums.com/threa...me-dilation-real-or-perspective-based.911615/

    My hat is off to the_emi_guy for writing that post and I can now die in peace! I encourage anyone that is still scratching their head on the twin paradox to read it. It certainly helped me and most importantly, it’s really simple. As explained in the post it comes down the fact that although ship A sees Earth’s clock as moving more slowly than its own, the clock on ship B is moving more slowly still by virtue of the fact that the relative velocity between the two ships is greater than the relative velocity between ship A and the Earth. Hence the clock on ship B is moving slowest of all relative to ship A. When ship B arrives back at Earth, its clock will show less accumulated time than Earth and everyone is happy. Wonderful!

    My reason for this post though is because it led me to a very interesting question.

    A few years ago I asked what the clock on a twin paradox ship is actually doing while it is accelerating away from earth and before the turnaround. Was it moving faster or slower or undefined. I seem to recall the answers left me unsatisfied and I don’t think anyone could say for sure. So here is my question with some backstory to reveal why I’m asking it.

    I had a thought that there might be a definitive answer not only to my question above but to the question of what all the clocks are doing in the triplet scenario. I began by recognizing that the clocks on ship A and B, before they left Earth were fitted with identical clocks and calibrated to match the clock on Earth. So far so good. But then necessarily, ship B accelerated to begin its voyage to deep space, turned around, and headed back. Likewise ship A also accelerated in the opposite direction, turned around and headed back to do its initial Earth flyby. So both ship A and B had acceleration in their history and Earth did not.

    I thought back on all non gravitational situations involving time dilation and realized that every single one of them involved an acceleration history that resulted in a differential velocity. Even a ship orbiting a low gravity asteroid using retro-rocket would first have to accelerate (for example) off the asteroid to begin its orbit or have, at some point in its history, accelerated relative to the asteroid. In all cases the clocks that underwent acceleration had slower running clocks when they were compared to their non accelerated counterparts.

    This led me to think that in the triplet scenario, could it be possible that upon acceleration (and the resulting velocity differential) clock A runs slower than the clock on Earth not in a relative fashion but in an absolute fashion and the same for clock B. In addition, since clock A and clock B are moving at let’s say .6c relative to the Earth and since both went through identical acceleration curves, could it be both clock A and clock B are ticking at the same rate in an absolute, not a relative sense. This feels satisfying to me because it works in the triplet scenario and also in the twins scenario. In the twins paradox, what if when the ship accelerates, its clock slows, then when it decelerates and stops its clock speeds up again to match the rate of the clock on Earth. Then when it accelerates again, its clock slows again, then when it reaches Earth and comes to a stop, its clock again matches the tic rate of the clock on Earth but has accumulated less time overall.

    So my question is, is it possible that this could be true? That when a clock accelerates from a zero differential velocity to a non-zero differential velocity, its clock slows in an absolute fashion and when it decelerates back to a zero differential velocity its clock returns to match the tic rate of its counterpart?
     
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  3. May 14, 2017 #2

    Ibix

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    No. Build two clocks in rockets at rest with respect to each other but separated by some distance d. Ascertain that they tick at the same rate by each watching the other through a telescope. Now accelerate the rockets towards each other. If the clocks slow in some absolute sense then they should tick at the same rate, whereas relativity would say they both measure the other ticking slow.

    You've introduced an absolute rest frame, basically. That gross an error would show up all over the place, and we've never seen it.
     
  4. May 14, 2017 #3

    PeroK

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    There are a number of problems with your approach. First, you are working in a non-mathematical mode that is fine for simple scenarios, but the more complicated you make your scenarios, the more you get lost in your forest of words and descriptions and explanations.

    Put simply, an accelerating clock is continuously changing its velocity with respect to any inertial reference system. The same rules of time dilation apply, except they apply only instantaneously for each velocity. But, this is really no different from any other continuously changing quantity in physics.

    To get the time dilation at any instant, you simply look at the velocity at that instant. And, to get the "proper" time on an accelerating clock over an interval, you integrate this quantity over the interval.

    There is really nothing mysterious about this, once you adopt mathematics as your language.
     
  5. May 14, 2017 #4

    Buckethead

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    Yes, they would see each other's clock moving more slowly as per SR, but when they crash into each other and compare times, their times would match, so their clocks were ticking at the same rate.
     
  6. May 14, 2017 #5

    PeroK

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    Their clocks were "ticking at the same rate" in a reference frame in which they were moving at the same speed, but not in any other reference frame. The key to this paradox lies, as usual, in the relativity of simultaneity.

    Another problem with your approach is you try to explain everything through time dilation alone. The relativity of simultaneity is absent from your analysis and that is why you are leading yourself astray.
     
  7. May 14, 2017 #6

    Ibix

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    Why, if they are both slow? Once they accounted for the distance change they would measure the same tick rate.

    No. That they show the same time means that they ticked the same number of times since they were synchronised. It doesn't mean they were always ticking at the same rate (see @PeroK's last comment about the relativity of simultaneity).
     
  8. May 14, 2017 #7

    Buckethead

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    Yes, I apologize for this.

    Yes I understand this, and of course at any time any of the 3 observers will observe the other clocks as moving more slowly, but still, the accelerating body makes for an unsymmetrical situation as is brought to life in the twin paradox, so someones clock is acting physically differently. Even though at each point in the acceleration it is simply a matter of relative velocity, in the end, it does not result in a symmetrical situation (all clocks reading the same when they meet up) and slowing one clock down physically seems to resolve this as if giving energy or something to an object (accelerating it to a higher relative velocity) would appears to slow the clock down in an absolute sense.
     
  9. May 14, 2017 #8

    Buckethead

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    Yes, they would measure the same tic rate. No argument there. They can both tic more slowly and this would still be the case.

    I see. Yes this is interesting. It's true it does not mean that they always clicked the same rate, but in the end, when they recombine, their clocks would be the same so what you have is a black box in the middle. Same time when they start, then they both accelerate and show the same time when they crash but how can one say if in the middle whether or not they clicked at the same rate all the time or if instead one speeded up then slowed down so that they matched in the end. It seems to me because of the symmetry of the situation, having the same rate at all times is the more Occum's razor like.
     
  10. May 14, 2017 #9

    Buckethead

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    I see. But let's be an observer on one ship. To start the two ships are moving at a 0 differential speed and they both measure the others clock to be ticing at the same rate. Then they accelerate toward each other and the observer sees the other's clock ticking more slowly. When they crash and compare, the times compare, meaning in absolute terms, both clocks were ticing at the same rate.
     
  11. May 14, 2017 #10

    PeroK

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    Saying the clocks did or did not tick at the same rate is meaningless without specifying the reference frame. In the Earth frame, because of the symmetry, they do at all times. In the frame of one of the rockets, the situation is more complicated. At any instant time in one frame is dilated when measured in the other. You, therefore, need a more thorough analysis of how time is measured.

    Another approach is to analyse light from one clock in the frame of the other and that would give you the reading on the other clock throughout and should eventually end up with synchronisation.

    In any case, you cannot analyse the problem by time dilation alone.
     
  12. May 14, 2017 #11

    PeroK

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    What the "observer sees" and how time in the two reference frames is related are very different things. See my above post for the two approaches. If you take the reference frame approach, you analyse the relation between reference frames. And, if you take the "observer sees" approach, you analyse the path of light from one of the clocks to the other ship.

    You need to tighten up your language seriously. My first suggestion is the phrase "two clocks ticking at the same rate" is abolished from your vocabulary as it is meaningless.
     
  13. May 14, 2017 #12

    Ibix

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    But that's not what relativity says will happen. And if the world did work like that, police speed radar wouldn't work as designed because the base frequency would vary through the day.

    A situation that is symmetrical in one frame isn't necessarily symmetrical in any other.

    You cannot think about this without getting into the maths. You are being mislead by your incomplete mental models.
     
  14. May 14, 2017 #13

    Buckethead

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    Why?
     
  15. May 14, 2017 #14

    Ibix

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    Why wouldn't speed radar work? Because a point on the surface of the Earth is moving in a circle (in fact, it circles daily, moves in an ellipse annually, plus a monthly circuit around the Earth-Moon barycentre). So, it is accelerating or decelerating all the time, so a clock (such as the emitter in a radar set) would its vary rate all the time. You'd need to correct for the changing frequency of the emitted radar pulse in order to get a speed from a Doppler radar. We don't have to.

    There are far reaching implications to throwing out the principle of relativity.
     
  16. May 14, 2017 #15

    Buckethead

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    I understand why you are saying this. Because clocks can tic at different rates depending on the observer as per SR. When I talk about two clocks ticking at the same rate or at different rates, I am talking about what we can say about two clocks after conducting specific experiments where the clocks start at 0 relative velocity and end at 0 relative velocity.

    An example of this would be a ship taking off from Earth, going through a black box for 1 month Earth time and returning. When the clocks are compared we find that only 1 day has elapsed on the ships clock. Therefore we can say that regardless of what happened in the black box, the ships clock ticked at a rate of 1/30 the Earths clock on average over the course of the month and that is absolute. No observer can disagree with that, not even the ships pilot.

    So, it may not be possible to confirm what a clock may be doing in absolute terms at any given time when the clock is off on a joy ride. And I'm fine with that. But what if I suggest that this be used as a tool, or a visual aid to describe the outcome of certain experiments. For example what if I say that a clock does not actually tic slower in absolute terms but instead say, that if we imagine a clock ticking more slowly (or more quickly) in absolute terms as a result of any historical accelerations as mentioned in my opening post, then it seems the outcomes of certain experiments can be predicted. The twin and triplet paradoxes are offered as examples of this as is a ship taking off from an asteroid, circling the asteroid under retro-rocket fire, then returning to the asteroid. In other words, can you think of any experiment (barring gravitational experiments) that can be performed where the prediction of this tool would vary from the prediction made using a more traditional method?
     
  17. May 14, 2017 #16

    Buckethead

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    I don't see this as being a problem. The fact that the Earth moves in a circle is outside of the effect I suggested. Although the Earth is accelerating, it is going at a constant velocity in it's orbit. The effect I was suggesting is only applicable to objects that change relative velocities such as a ship accelerating away from Earth. However you did mention non circles as well and this is not a problem either since not only the radar but the object of the radar are co-moving. When the frequency of the radar changes, so does the filter that reads the radar change. In other words, their absolute tic rates change together. (Please excuse my use of tic rates again as I explained in my post above).
     
  18. May 14, 2017 #17

    Ibix

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    The speed of a point on the surface of the Earth varies throughout the day unless you are working in an Earth-centred frame. Why would your absolute rest frame be an Earth-centred frame?

    I think that depends on what the reflected frequency is, which depends on the details of your model.

    You mean, as part of teaching about a theory named "relativity" precisely because it does not have an absolute rest frame, we introduce an absolute rest frame? Almost all the problems people have with learning relativity stem from a refusal to let go of the notion of an absolute simultaneity. Encouraging them to hang on to it is the last thing I would recommend.

    You mean, apart from the one I offered in #2?
     
  19. May 15, 2017 #18

    vanhees71

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    It has been shown in many experiments that an (ideal) clock shows its proper time. That's all what's behind the socalled "twin paradoxes". It's not very surprising, if one takes the very close mathematical analogy of distances in usual Euclidean three-space: The distance traveling between two points depends on the curve along which you travel. To make the distance between the two points unique, one defines a "proper distance" as the length of the geodesic between them, i.e., the distance when traveling along a straight line. The same holds true for times in relativity: It's dependent on the worldline of the clock in 4D relativistic spacetime and defined by
    $$c \tau=\int_{\lambda_1}^{\lambda_2} \mathrm{d} \lambda \sqrt{\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \lambda}\frac{\mathrm{d} x^{\nu}}{\mathrm{d} \lambda} g_{\mu \nu}}.$$
    Here, ##\lambda## is an arbritrary parameter for the time-like worldline. Obviously the so defined proper time is independent (a) under coordinate transformations and (b) under changes of the parametrization, i.e., a geometrical invariant for a given time-like trajectory in spacetime.
     
  20. May 15, 2017 #19
    We can examine your scenario with specific numbers, looking at what one of the ships will actually observe.

    Let's have 2 relatively stationary ships in inertial frames separated by 6 light seconds. Both ships clocks are synchronized, which is to say, each ship will observe the other's clock being precisely 6 seconds behind its own (due to the travel time of the light between them). Both ships begin moving toward each other at 3/5 of c relative to their original frame at the same time. From this original frame, they should meet in 5 seconds. However, each ship should only experience 4 seconds before they meet, as their own proper velocities relative to the initial frame are 0.75c. But, when a ship begins moving, it still won't have seen the other ship start its own movement yet, since it is 6 seconds behind. But now it's gaining on the signals it's observing, so they should appear to be actually faster.

    We can use the Doppler formula to figure out how much faster the other ship's clock will appear to be ticking based on our velocity, which is ##\sqrt {\frac {1 + \beta} {1 - \beta}} ## where ##\beta = v/c##. Now, since they're approaching each other, beta will be negative, so it basically flips the signs of these values. With beta being -3/5, this gives us a value of 0.5. Since the other ship's clock appears to be 6 seconds behind the departure time when our ship starts moving, it will take 6 * 0.5 = 3 seconds of our time before we observe the other ship's clock reach the departure time and start heading towards us.

    Now, the other ship is moving toward us at 3/5 the speed of light relative to the original frame, which, with the velocity addition formula, will mean we observe it closing with us at a velocity of about 0.882c. So now, with the Doppler formula, we get a value of 0.25. That means in the remaining 1 second of our time before we meet, we'll observe the other ship's clock pass 1/0.25 = 4 seconds, which is exactly how much of our own time passed since we started moving.

    From an instantaneous perspective, when we start moving towards the other ship, our perspective of "now" will change due to the relativity of simultaneity, and our instantaneous view of the other ship will have it already well on its way to us, about 2.12 seconds into it's trip already. It's time dilated clock will take 4 of our seconds to tick the remaining 1.88 seconds of its own before we collide. The instantaneous perspective is weird because it has the other ship jumping from synchronized time to already over half way to us and its clock ahead of us as we accelerate. But the instantaneous perspective is only an academic one, and not what we could actually observe.
     
  21. May 15, 2017 #20

    Dale

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    I am not fine with that. Do you intend "absolute" to have any meaning or is it simply a label for an arbitrary reference frame. If it is simply a label then please choose one less connotation. If it is not intended as simply a meaningless label then it is contradicted by experiment.
     
    Last edited: May 15, 2017
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