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Two Observers with clocks in space moving at vrel relative to each other

  1. Oct 15, 2012 #1
    As the title says, imagine two observers with clocks A and B in space which move at vrel relative to each other.

    Just when clock A passes by clock B, the clocks counter shows zero. Observer A will conclude clock B to tick slower, while observer B will conclude clock A to move slower (both using data composed by an army of observers with clocks being at rest in their frame).


    Observer A decides to accelerate until he is at rest in the frame B is at rest in.
    When he accelerates (non-instantaneous), clocks that are towards B will time shift (tick faster than before the acceleration) depending on the distance, vrel and acceleration. Or the change of the inertial reference system he is at rest in if you want when at a distance to B.


    Now the question. Is it possible for Observer A to chose the appropriate acceleration depending on the distance and vrel, which allows him to make up for the slower ticking clock of B caused by moving at vrel to B, in such a way that he would conclude the clock of B to be ticking at the same pace as his clock until he is at rest in B's frame?
    He would have to accelerate stronger at close distance to B, and less at high distance as i see it. (might see it wrong)


    Of course, if that was possible, then observer A (now at rest in the frame B is at rest in too, but at a distance) could do the similar in respect to a clock in his current position when wanting to travel back to B (locally), and yet the clocks would be in sync still, even thought observer A traveled and was accelerating at all times with a non-constant acceleration chosen according to distance and vrel.


    Is this possible or did my mind trick me? I could try to draw this in a minkowski diagram, but it would take quite a bit.
     
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  3. Oct 15, 2012 #2

    Simon Bridge

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    Hmmm ... whichever twin accelerates to join the other one is the one that end up younger. The other twin's clock must go faster during the accelerating phase for this to happen. This what you are thinking of?

    By accelerating you would pass through a speed where the clocks are at the same rate - stopping accelerating at that point would have you in the same reference frame...

    In the usual treatment, the journeying twin accelerates to catch-up with the other twin, so they can compare notes, not just join him at the same speed.

    http://www.physicsguy.com/ftl/html/FTL_part2.html
     
  4. Oct 15, 2012 #3
    I know about the standard case. Here i am discussing the case where the "twin" starts accelerating towards the other twin right after they are synced/pass by each other with both their clocks showing 0s at this event.

    One of the twins accelerates back to the frame the other twin is at rest in right after the sync event at a non-constant acceleration. Starting at a quite high acceleration which drops continuously the more distance is between the two, in a way that the shift of the counter of the non-accelerating twin just makes up exactly for the slower ticking clock, so both clocks tick at the same pace at all times until the acceleration comes to a halt and both find themselves in the same frame at rest.

    If the non-accelerating twin placed a line of synced clocks all along the path the accelerating twin went towards, it is only logical to assume that if the accelerating twin's clock is in sync with the non-accelerating twin's clock, it will also be in sync with the clock at a distance when he becomes at rest in that frame.


    So accordingly, he can repeat the process in the other direction now using the synced clock he is at locally as a reference until he reaches the non-accelerating twin. Again, if he manages to be in sync with one of the clocks in the non-accelerating twin's rest frame at all times until he comes to a rest in the non-accelerating twin's frame, then he will be in sync with all clocks the non-accelerating twin placed.


    If above is correct, then it is not only the acceleration at a distance which matters, but also the type of acceleration at a distance which matters in order to get an age difference.

    Furthermore, if above is correct (might not be), then a very low acceleration becoming even weaker towards the end, would have the accelerating twin aging more actually.
     
  5. Oct 15, 2012 #4

    ghwellsjr

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    In Observer B's rest frame, his own clock has never been time dilated while the clock of A has been time dilated ever since they passed each other. There is nothing that can be done to get clock A to gain time no matter how it accelerates. Time dilation can only slow clocks down, never speed them up faster than that of the coordinate time (which is the same as that of clock B).

    Of course it is trivially easy to get clock A to run at the same pace as clock B but not to be in sync with clock B (unless clock B accelerates but I don't think that is what you have in mind).
     
  6. Oct 15, 2012 #5

    Simon Bridge

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    Yes, that is what I understood you to be talking about ... in the standard example, the twin who accelerates ends up younger because the acceleration speeds the other twins clock up in his frame (as in the example I linked to).

    I stand by the answer:
    You cannot recover lost time by accelerating to be stationary in another's frame.
    You won't be able to have the same tick-rate without being the same speed.

    Of course you are welcome to do the math and prove me wrong.
     
  7. Oct 15, 2012 #6
    Yes, but observer's B clock has been dilated seen from A's rest frame. BOTH conclude (contrary to see if we include Doppler effects) using an army of observer, the other twin's clock to be ticking slower.


    definition: red clocks are clocks at rest in A's frame, which were placed and synced all along the axis B is traveling on.

    I don't think so. Acceleration changes the frame observer B is at rest in. Following the two postulates of SR you conclude that events happening at x,t in the frame B is at rest in before accelerating (the second, part of the turn around acceleration which gets him back into the rest frame of A), will happen time and space shifted at x' t' in the new frame he is at rest in.
    This time and space shift of events results in red clocks in front of the observer B to display a higher count after the acceleration, while red clocks behind him, to display a lower count after the acceleration.

    (if former posts are correct,then it also depends on the value and direction of the acceleration when we have two objects moving at vrel at a distance)


    Time-interval expansion(dilation) does indeed result in the moving clocks to tick slower. I never objected that. Changing the inertial frame of reference however, also has an effect on clocks which overlaps.


    Maybe you should elaborate on that, because i do not know what YOU have in mind.
     
    Last edited: Oct 15, 2012
  8. Oct 15, 2012 #7
    The red clocks (the synced clocks which are at rest in A's frame) that travel towards observer B will ALWAYS be ahead (higher count) of observer B's clock when he passes by them.
    So no matter how he accelerates back into A's frame, he will never be able to have his clock display anything higher than any of the red clocks along his path (locally x=x' t=t'). In case of an instantaneous acceleration, he will end up with what the red clock local to him will show, which is always higher than what his clock shows. Since all red clocks are in sync with A, then this is what A's clock shows too.

    When B accelerates back into A's frame, almost instantaneously, he will conclude A's clock at a distance to run much faster than his. If he accelerates less, he could certainly get himself to conclude that A's clock is running at his own clock's pace, but then he would not get back into A's rest frame and would have to continue the process.

    So the answer to this will most likely be, that B can get his clock to run at the same pace as A's clock, but he would have to lower the acceleration the greater the distance, to keep the same pace, which would probably result in never managing to accelerate back into the frame A is at rest.

    I am not going to burn my brain-cells trying to proof this mathematically however.
     
    Last edited: Oct 15, 2012
  9. Oct 15, 2012 #8

    Simon Bridge

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    You start out by asking if the situation you describe is possible, then don't believe it when you are told it is not?

    So do the math and prove us wrong.
     
  10. Oct 16, 2012 #9
    Well, i was not happy with the answers as i did not feel they addressed the issue properly. So i kept thinking what could be the case, and found the solution myself in the end or what i believe to be the solution.

    I claim that it is possible to have A's clock run at the same pace at all times as B's clock, seen from B's point of view, moving at vrel to A. He would have to pick his acceleration according to the distance to A, resulting in having to pick a lower acceleration the further away.

    The total acceleration would not suffice to bring B at a halt relative to A (into A's rest frame). It would be an endless process.

    But i do not see where your problem lies with this description, because i assume that you would have to agree, given B meets any of the red clocks locally (at rest in A's rest frame) which are in sync with A's clock (in A's rest frame of course), those clocks will always have a higher count than A's clocks seen from within B's system which simply follows from the two postulates of SR.

    It also follows (SR postulates) that any one of the red clocks moving towards B at vrel seen from B's point of view, when local next to B, will show a higher count that B's clock.

    If B saw a red clock next to him(locally) with a count of let's say 10 seconds, given he accelerated instantaneous back into A's rest frame, locally x=x' t=t'. This follows directly of SR's postulates as well. B's clock which let's say shows 9s at this point, would of course also remain at 9s after the instantaneous acceleration because of x=x' t=t'.
    Logically, without having to calculate, once B is inside A's rest frame again, all red clocks are synced still, and so A's clock would have to display the same count, 10 seconds.


    In order for this to happen, while B was accelerating back, A's clock had to do a jump up, since it's count was lower before the acceleration, seen(actually concluded) from B's point of view. It was lower than both, the local red clock to B and B's clock, concluded by observer B, having an army of observers with clocks at rest in his frame, sending him data.
    Yet after the instantaneous acceleration, it becomes even with the red clock and ahead of B's clock. In case of a non-instantaneous high acceleration this jump would be seen(concluded) as A's clock moving much much faster.
    Picking a lower acceleration this effect would be less. It also overlaps with the effect of clocks ticking slower because of vrel.
    Picking the right acceleration, you would be able to see this effect cancel out the slower tick rate of the clock caused by vrel.


    Even if i did this mathematically, i would have to go by the description above, which you say you do not agree on. So what would be the point?
     
    Last edited: Oct 16, 2012
  11. Oct 16, 2012 #10

    Simon Bridge

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    If you did it mathematically, you'd have to start with the mathematical description of space-time in Einstein's relativity - then you'd be able to show whether the description you give is consistent with that model.

    I'd have to either concede or find something wrong with the math... or reveal myself a hypocrite. It stops being a matter of opinion.
     
  12. Oct 17, 2012 #11

    ghwellsjr

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    I can analyze any scenario (not involving gravity) using any single Inertial Reference Frame (IRF) I want. I choose one in which observer B is at rest, simply because he never accelerates and that makes it easier to conceptualize what will happen. If you want to make things more challenging and difficult for yourself by choosing a different IRF or a non-inertial frame, go ahead, but you can never claim that it will produce any different results than my choice of frame.

    In my chosen IRF, B's clock is always at rest, therefore it runs at the same rate as all the clocks in your army of observers (which represent coordinate time at different distances from B) and is never time dilated. A's clock, starts out at some speed and therefore is time dilated which means its pace is slower than that of B and his army's clocks.
    Ok, so A's army of clocks are all painted red. That doesn't change the fact that they are all time dilated according to my chosen IRF.
    No, I don't want to complicate things by considering more than one IRF and I'm sticking to the one in which observer B remains at rest in. I don't know why you think that when observer A accelerates, it requires that observer B must change frames. Nobody accelerating anywhere changes anything about any frame.
    It looks like you have forgotten your own scenario. In your first post, you said that it was observer A that accelerates and observer B remains at rest. Now it appears that it is B that accelerates while A remains at rest. Otherwise, I can't tell what you are talking about. Maybe you should start over.
    I'm convinced that you forgot your original scenario so I have given up trying to understand what you are describing. However, it appears that you are promoting the idea that in the Twin Paradox, you have figured out a way that the accelerating twin can end up at the same age as the inertial twin, which is wrong and very easily disproved by using the IRF in which the inertial twin remains at rest. The other twin's clock gets behind and can never recover. Do you doubt this simple analysis based on the inertial twin's IRF?
     
  13. Oct 17, 2012 #12
    Sure you can, but that is only because someone else already did the groundwork of working the formulas of SR according to the two postulates.
    If you had no clue about the formulas and only knew the two postulates, you would not be able to get the proper results by just considering a single frame of reference.


    This is something we agree on. I have really bad memory and it tricked me when i was replying. A became B and B became A in my replies.


    Other than that, the whole point was to look at it from the twin's point of view who keeps changing which inertial reference system he is at rest with.
    I was puzzled how he would be able to consider the twin's clock which is in the same IRF at rest at all times to be ticking at his own clock's pace at all times and yet to be the one who aged less. It is possible, because it is an endless process, him having to accelerate less the further away in order to conclude the same pace.
     
    Last edited: Oct 17, 2012
  14. Oct 17, 2012 #13

    ghwellsjr

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    That's for sure, along with everyone else in the world except Einstein, oh yes, and you.
    If you agree with me, why didn't you start over? Aren't you going to present your scenario once more in a consistent way?
    Are you still puzzled?
    Are you claiming that if the traveling twin accelerates in some particular "endless process" way, then when he returns to the inertial twin, they will have aged the same amount?

    If so, then why don't you start over and present your argument in a consistent way, fill in the details, provide some numbers and give us a specific example?
     
  15. Oct 17, 2012 #14
    The point is, that if he keeps accelerating less, the further away he is, in a way that he will always conclude A's clock to be ticking at his own clock's pace, then he will never be able to become at rest in A's rest frame. There are an infinite amount of frames he can become at rest in before becoming at rest in the frame A is at rest in.

    I will when i find time draw some diagrams for this, but it won't be easy, because i will have to draw many diagrams, for each frame he becomes at rest in. And it won't be a "smooth" acceleration, but instead i will "cheat" by doing it in many little instantaneous acceleration steps.

    I am sure some of the more advanced scientists among us would have an easy time drawing this, but for me it is always a loooong process.
     
  16. Oct 17, 2012 #15

    pervect

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    I don't think I've seen Jeronimus' analysys as to why he believes this. I wouldn't be too surprised if a fuller exposition revealed some flaws in his analysis.

    Using the usual notion of simultaneity, a situation like this is possible only if the direction of the twin is "above" the accelerating twin, I believe.
     
  17. Oct 17, 2012 #16

    ghwellsjr

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    This question can be answered in one word:

    Yes.

    or

    Nope

    Can you please answer the question-no elaboration.
     
  18. Oct 17, 2012 #17
    Nope.
     
  19. Oct 17, 2012 #18

    ghwellsjr

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    Are you claiming that if the traveling twin accelerates in some particular "endless process" way, then when he comes to rest with respect to the inertial twin and his army of observers, the traveling twin's clock will read the same time as the clock of the army observer he stops next to?
     
  20. Oct 18, 2012 #19

    Simon Bridge

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    OK - anyone can do the math using any number of frames they like - the math is simplest if they choose at least one reference frame that is inertial and work the math from there while also considering the other frames (some of which may not be useful).

    In your example there are two frames under direct consideration - those of observer A and B. Observer B is the only one that is inertial. One could choose some other inertial frame, such as the frame in which observer A starts out at rest in.

    Of course - I could have misread and both A and B are accelerated.
    [edit] removed extraneous remark - confused questioner and questionee.

    Perhaps we should now revisit the claim?
     
    Last edited: Oct 18, 2012
  21. Oct 18, 2012 #20

    ghwellsjr

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    Jeronimus got mixed up. He started off with A accelerated and B inertial. But in post #6, he accidentally switched the twos. There's no point in trying to understand what Jeronimus is proposing until and unless he decided to repost his scenario.
     
    Last edited: Oct 18, 2012
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