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Modified Twin Paradox. (Experts required).

  1. Sep 3, 2007 #1
    I have gone through the previous posts on the subject in this forum but still have an issue.
    So please tell me what is wrong with my following (qualitative) reasoning:

    Assume the standard paradox is replaced with:

    Twin A at origin of x-axis in free space.
    Twin B accelerates away to speed near light speed (unspecified) along +ve x-axis.

    The x-axis has been priorly populated with identical clocks at each integer point and by an observer at each point using Einstein's method of clock synchroniazation such that any observation at points along the x-axis can be measured and then the results sent back to the origin. (I.e they will be 'time-stamped' such that 0 (origin) will know when they occured in his past at his clock time).

    Now according to S.R. the theory is based on relative velocities not on relative accelerations and the time-dilations observed are calculated using 'v' only. All calculations of any age difference are done this way and many books state that the actual acceleration values are not used in the calculations and can be made arbitrarily large such that Twin B can reach his velocity 'quickly'.
    Also I assume if we did factor in the acceleration, then we would only need to know the instantaneous relative velocities seen by Twin A at points along the x-axis during the accelerating phase to calculate the true time-dilation values at each point.

    To Begin:
    Twin B sets off with a number of identical clocks to Twin A. Upon reaching his target-relative-velocity his clocks appear slowed (and ARE slowed as he changed inertial frames) to observers along the x-axis.
    Now Twin B then decides to eject one of his clocks in the -ve x-axis direction at an identical acceleration value that he originally did himself such that when this ejected clock reaches relative-velocity-v w.r.t himself, the ejected clock then finds itself stationary w.r.t the x-axis. The ejected clock is now opposite one of the observers on the x-axis.

    Here's the problem:
    The ejected clock has initially been time-slowed w.r.t the x-axis; then time-slowed again w.r.t Twin B. So it has endured a compounded time slowing with no mechanism to actually set itself back to it's original 'rate' even though it ends up stationary back on the x-axis.

    Note. These time dilations must be 'real' and physical as are the Muon decay rates observed to be really time-dilated in the atmosphere of Earth.
    So from Twin B's frame,the ejected clock is experiencing a REAL time-dilation (changed inertial frames again). Yet the ejected clock is now stationary w.r.t Twin A and should tick at his rate.

    Now, I understand the formulae involved and the world-line explanation e.t.c. It just seems to me that anything that moves at various velocities w.r.t something else and comes back to rest keeps on compounding and building up time-dilations and will run slower and slower with no re-set??

    So what is the answer? Thankyou.

    P.S.
    Any measurements done by either Twin on the others frame will be symmetric and both will claim each other is the one who has been time-dilated. Yet the claim made is that the one who actually changed inertial frames (accelerated) endures the REAL net effect of the time-dilation, CONTRARY to the S.R. Theory that it is ONLY relative velocities that are used to do the calculations.
    If the whole of the x-axis including Twin A were to accelerate to match Twin B's velocity so that they then were stationary w.r.t each other, then it seems that if a different accleration value was used, this would upset any theory that accelerations affect any time-dilation value as different dilations would result.
    The only way I see a resolution is that time-dilation effects are only calculated by referral and are not real. E.g. The Muon lifetime is not a 'proper' life'time'; it's value is found by taking it's assumed path length in the Earths atmosphere and divided by it's near c value of velocity to give a referred lifetime value.
    Again, the catch-22 situation. Whatever way I look at it, it makes sense upto a point, only to contradict itself when looking at all the views.
     
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  3. Sep 3, 2007 #2

    StatusX

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    Your emphasis of the word "real" shows you're not thinking of time dilation the right way. No dilation effects are any more or less real than any others. The fact that they are observered by some observers and not others is not because some observers are right and others are deceived, but because there is no right answer: time measurements are relative, not absolute.

    This makes no sense. Go over it again, keeping in mind what I said.
     
  4. Sep 3, 2007 #3
    When I say 'real' I mean physically in that state. E.g. Muon lifetimes in the atmosphere are physically real and not 'just' observably real.
    The twin (B) said to be changing inertial frames is not just observed as being time-dilated, he is physically 'really' time-dilated; wheras Twin A is said only to be 'observably' time-dilated.

    I would state that a blue coloured ball at rest is physically 'blue', wheras if it is travelling away from me at substantial velocity, it may be Doppler shifted so that it is observably 'red'. If it comes back to my frame and stays 'red' then thats the problem.
     
  5. Sep 3, 2007 #4

    Janus

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    While there is a real difference in the Twin's ages when they re-uinite, there isn't only one "real" reason why this is so. Twin A says that twin B is younger because he aged slower during the whole trip. Twin B says that he is younger because even though Twin A aged more slowly during the outbound and return trips, Twin A's age jumped ahead greatly when he(Twin B) turned around and headed back. (Due to the Relativity of Simultaneity) The standard Yime dilation formulas are only designed for use in non-accelerated frames. From an accelerated frame such as B's when he turns around there is a different rule: Clocks in the direction of your acceleration will run fast by a factor dependant on your acceleration and their distance from you. The greater the acceleration or distance, the faster they will run. (If the change of velocity is considered instantaneous, then the clocks instantly jump forward by an amount determined by their distance, again due to the Relativity of Simultaneity.)

    Now, the next question on your mind will be, then how come we can ignore the value of the acceleration when we do the calculation for the final time difference? Well, for finite values of acceleration, we can't. This is because from A's view, B's velocity does not remain constant and changes during the launch and turn around points, meaning that the time dilation rate changes over these periods. (for example, if you want to know the total time dilation of Twin B as seen by Twin A, as Twin B accelerates at acceleration "a" you use the formula:
    [tex]t_A = \frac{c}{a} \sinh \left ( \frac{at_B}{c} \right )[/tex]
    For Twin B it becomes even more messy, because you not only have to take A's changing velocity wrt B into account, but also, the changing distance between A and B for the Relativity of Simultaneity to be taken into account.

    This is why for simplicity sake, we usually assume instantaneous velocity changes.

    But even if we take acceleration into account, It won't change the fact that both Twins still agree as to the age difference when they return. All the changes that A sees in B clocks adds up to the same time difference between the two as all the changes that B see's in A;s clock do.
    See above. The only difference would be due to the fact that the velocities would not be constant throughout.
    The resolution comes from the fact that time is Relative and not absolute. Time dilation being "real" does not mean that every frame of reference must agree as to who is undergoing what time dilation when. In all frames everyone will agree as to the age difference between A and B when they come back together, but they will disagree as to how it came about. And each frame's view of the sequence of events are just as real as any others. Such is the nature of "time".
     
    Last edited: Sep 3, 2007
  6. Sep 3, 2007 #5

    JesseM

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    You have no obligation to say that the muon's lifespan has "really" been dilated--in the muon's rest frame, the lifespan is not dilated at all, and the explanation for why it reaches the surface of the Earth is length contraction which shrinks the distance from the upper atmosphere to the surface, so the muon takes less time to cross that distance.
    His total time is dilated, but there is no truth about whether his clock was ticking slower or faster than the Earth's clock during any given section of the trip. For example, you could analyze things from the perspective of the frame where twin B was at rest during the outbound phase of the trip while the Earth was moving, and in this frame the Earth's clock was ticking slower than B's clock during this phase, then after twin B turned around his speed was even greater than the Earth's in this frame so his clock speed was even slower. The end result is that this frame will make exactly the same prediction as any other frame about the total elapsed time on B's clock at the moment that B catches up with Earth.
     
  7. Sep 3, 2007 #6

    pervect

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    Note that you can ignore accelerations only when you adopt an inertial frame of reference. So for clarity, we can say that you can ignore accelerations as long as you yourself are in an inertial frame. Dealing with accelerated frames is possible, but should be avoided until you have a firm grasp on the non-accelerated case.

    No, you seem to be missing the point. From B's point of view, his clocks are not slow. From B's point of view, his clocks are running just fine, at 1 second per second. So what B sees is that his clock, which is running at the same rate as all his other clocks (i.e. at one second per second) slows down to match A's clocks when he ejects it.

    Because B is an inertial observer, the acceleration does not matter - as we pointed out earlier, as long as you are in an inertial frame (and B is in an inertial frame), it's OK to ignore accelerations.

    The point is that A's clocks run slow in B's frame, and B's clocks run slow in A's frame. This is possible without logical contradiction only because of the relativity of simultaneity.

    I was going to talk more about the relativity of simultaneity, but that would be "too much at once". So let's get this much straight, first.
     
  8. Sep 3, 2007 #7
    Thanks for the replies.

    I think I need to mull this over a bit before I reply much more at the moment.
    But I'd just like to add this bit to see if you agree or not:

    'O.K So far I have the conclusion that a measurement is only relevent to the observers point of view. So my analogy of a blue ball seeming red when receeding at a great velocity due to the Doppler effect means that it is only the red view that is physicaly relevent to my frame. (The fact that it's blue in it's rest frame is only relevent to that rest frame.
    In the same way, a Muon decay in the atmosphere is only relevent to the distance it has travelled in the atmosphere - which is the only important determinent in any physical observation result. The fact that it shows different times and distances in it's own frame is irrelevent to our (Earth) frame.'
     
  9. Sep 4, 2007 #8

    JesseM

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    In physicist-speak the only quantities or facts which can be considered genuinely "physical" are the ones that don't depend on your choice of coordinate system. So, the color of the ball in its own rest frame (which depends on its composition) or the fact that a muon created in the upper atmosphere will reach the Earth's surface are both "physical", but the wavelength of the light emitted by the ball or the question of the muon's decay time or the distance it must travel from the upper atmosphere to the surface are coordinate-dependent and so not really "physical" in this sense. Certainly the "physical observation result" that the muon does make it to the surface can be calculated in any frame, there is nothing about the situation that requires you to use the Earth's frame.
     
  10. Sep 5, 2007 #9
    A few thoughts/questions:


    This is very helpfull. I was indeed missing this point.

    Are these different (simplified) scenario's correct? (after B has reached rel v):
    1) Instead of turning around B decelerates and stops w.r.t A's frame.
    Result: B is still younger.
    2) As above , but Twin A accelerates to catch upto B's spatial position and decelerates to a stop.
    Result: A now same age as B.
    3) Twin B is still at rel v, but Twin A then decides to catch upto him. This means he has to accelerate to a greater velocity than B to make up ground. Upon meeting (and decellerating) then:
    Result A younger than B.


    Is it correct to say that you have to think of the combination: 'space-time' as a unity and that any non-inertial movement from an initial space-time reference point involves a 'trade-off' between the space component and the time component?
    So someone in an inertial frame is purely 'moving' through time and not moving through space (in their own frame), whereas someone accelerating and then moving at rel v w.r.t their start point is moving through time and space with the time part being somewhat 'sacrificed' for the space part?
    Is this space-time unity the space-time metric? I.e, the part that doesn't change.


    So Kinetic-Energy, mass, frequency/wavelength of light, velocity, magnetic fields e.t.c are not genuinely physical?

    Surely, the only things that should be considered as physical are the 'measurements' and 'results' obtained in any particular experiment?

    Are there any definitions of 'Physicallity'?
     
  11. Sep 5, 2007 #10

    JesseM

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    In A's frame, but not in all frames--since they are at different positions in space, the question of which is younger at a "single moment" depends on your definition of simultaneity. In the frame where B was at rest before "decelerating" to the same velocity as A (actually an increase in speed in this frame...physicists usually don't use the word 'decelerate' at all, and just refer to all change in velocity as acceleration), A will be younger because A was moving at relativistic velocity the whole time B was at rest, then after B changes velocity they are both moving at the same velocity and so age at the same rate afterwards.
    Assuming that A accelerates to the same velocity B was formerly moving at, then their worldlines will be symmetrical so they'll be the same age when they meet.
    Yes, assuming B moved at constant velocity throughout the trip, then since A's worldline is non-inertial while B's is inertial, we can be sure that when they reunite A has aged less.
    You could think of it this way--the faster you move through space the slower you "move through time", i.e. the slower your clock ticks--but it only makes sense relative to a particular frame, there can be no absolute sense in which one observer is moving through space more quickly or in which one observer is "moving through time" more quickly. Also, you say "any non-inertial movement ... involves a 'trade-off'", but it isn't really relevant whether the movement is inertial or non-inertial, relative to a given frame any clock that's moving faster through space is ticking slower, regardless of whether the clock is moving inertially or non-inertially.
    Not sure what you mean by this question, the metric is a more complicated idea than the trade-off you describe. But the metric does tell you the "distance" between points spacetime, and one feature of the metric of flat spacetime that resembles this trade-off is the fact that, for two different points in spacetime A and B, if the spatial distance between these points is dx and the time between the events is dt in your inertial coordinate coordinate system, the spacetime interval ds^2 = c^2*dt^2 - dx^2 is an invariant quantity which will always come out the same even if you use the dx and dt from different inertial coordinate systems.
    Actually, googling a bit I don't think this definition is universal, it's just something I've picked from other discussions (others do you use 'physical' this way, see here for example)...ultimately it's just a matter of terminology, although note that your broad definition would also allow things like x-coordinate to be defined as "physical" quantities. Anyway, it would be more clear say that some quantities and facts are "invariant" while others are not, and physicists are typically most interested in these invariant facts, with the coordinate-dependent quantities just used as tools to calculate predictions about invariant quantities and events.
     
    Last edited: Sep 5, 2007
  12. Sep 12, 2007 #11
    Another question.

    I have another question:

    In the scenario I described originally, If Twin B is travelling along an extended baseline of Twin A which are populated by clocks that were previously synchronized using Einsteins method; would Twin B infact be seeing two types of time info:

    type1) He see's A's original clock slowed as he receives A's signals further and further away from A's start point. (Doppler effect?).

    type2) He see's clocks in A's extended frame at points outside his window which actually show snapshots of time in A's frame indicating A's clocks are running faster.

    Say if each clock in A's frame showed the time/day/month/year, then it would be B that determines he is the one who is actually time-slowed without the need to turn around or slow to A's frame?

    Is this correct?
     
  13. Sep 12, 2007 #12

    JesseM

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    Are you asking what B would see using light, or what would be true of the clocks in B's frame? In B's frame the clocks are all running slow by [tex]\sqrt{1 - v^2/c^2}[/tex], but if they are synchronized in A's frame and the distance between each clock and its neighbor is x, then in B's frame each clock is out-of-sync with its nearest neighbor by a factor of [tex]vx/c^2[/tex] (this is the relativity of simultaneity), with each successive clock that passes B being ahead of the previous one by this amount rather than behind it. For example, if B is traveling at 0.8c relative to A and A's clocks are 10 light-seconds apart in their own rest frame, in B's frame they are each ticking at 0.6 the rate of his own clock, then are only 6 light-seconds apart, and each one is ahead of the previous one by 8 seconds. So if he passes one and it reads 100 seconds, the next one reads 108 seconds at the same moment in B's frame, and it takes the next one 6 l.s./0.8c = 7.5 seconds to reach B's location, during which time it advances by 7.5*0.6 = 4.5 seconds, so the time on this next clock as it passes B is 108 + 4.5 = 112.5. So, the difference between the time on the first clock as B passed it and the time on the second clock as B passed it is 12.5 seconds, while only 7.5 seconds passed on B's clock, and 12.5*0.6=7.5 so this result makes sense in A's frame too. But again, in B's frame this result is still consistent with the idea that all of A's clocks are running slower than his, but are also out-of-sync with one another.

    I gave some visual illustrations of a scenario much like this, except with two different rows of clocks moving alongside one another and each synchronized in their own rest frame, on this thread a while ago.

    On the other hand, if you're interested in what B sees rather than what's happening in his frame, you have to take into account the Doppler effect, which will mean that the clocks moving towards him look like they're ticking faster than his own, while the clocks moving away from him look like they're ticking even slower than would be predicted by the time dilation equation. Equations for the relativistic Doppler effect can be found here. Also, since the light from each clock takes 6 seconds longer to reach him than the light from the closer clock in his frame, instead of seeing them as out-of-sync by 8 seconds, he sees the time on each clock moving away from him differing from the next one by 6+8=14 seconds at any given moment, while he sees the time on each clock moving towards him differing from the next one by 8-6=2 seconds at any given moment. edit: the figures in this last sentence are wrong, I forgot to take into account that the clocks are moving in this frame, so the light that's reaching him now won't have been emitted from a distance of 6 light-seconds away in spite of the fact that the spacing between the clocks is 6 light-seconds.
     
    Last edited: Sep 12, 2007
  14. Sep 12, 2007 #13

    pervect

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    A somewhat related point that might help the OP.

    Suppose you have a 1-d array of clocks that are synchronized in the lab frame.

    You then take a picture of them. Do they all appear to read the same time on the picture?

    i.e. the clocks are arranged like this

    x---x---x---x---x---x---x---x---x

    you are at some point on this linear array of clocks, and you take a photo in which the clocks are all visible.

    Well, to give away the answer, the clocks do not all show the same time in the photo. The further away the clock is, the earlier time it reads, because light travels at a finite speed, and the light has to reach your camera from the clock in order to show up on the photo.

    So you see something like

    (-5)---(-4)---(-3)----(-2)---(-1)---(0)---(-1)---(-2)---(-3)---(-4)---(-5)

    Now, if you move in the same direction as the array of clocks, and take a photo, you'll find that that what you see on the photo doesn't depend on your speed. But your interpretation of the photo changes.

    In order for the clocks to all be synchronized properly, they must read (-c/distance), where (c/distance) is how far away they are from you.

    But when you move at relativistic velocities, while the photo stays the same, the distance between the clocks does not stay the same due to Lorentz length contraction. The only way this can be explained is to realize that while the clocks appear to be properly synchronized in the lab frame, they do not appear to be properly synchronized in the moving frame.
     
  15. Sep 12, 2007 #14
    It's hard for me to convey in writing what Iam trying to resolve, so I will try to be more precise:

    In the post from Janus , he states the resultant difference in the Twins ages upon re-uniting is from the fact that B accelerates by turning around. There are 3 accelerations. The initial outward bound one to gain v rel. Then a -ve acceleration to turn back and a final +ve one again to slow down to A's spatial point.
    Now from my understanding, the accelerations are only a means used to change inertial frames and although they cause the resultant velocity differences, the actual effect of time-dilation is the fact that someone (B) has changed inertial space-time frames.
    The whole of A's axis extending to the nearest star is the same inertial frame and every spatial point that is not moving w.r.t A is in that same s-t frame.
    As B changes from this initial A frame the only thing that is measurably invariant is the metric interval ds where [ (ds)²= (dt)² - (dx)² ],so (dx) and (dt) are 'traded' to make ds a constant. (1D case).

    Now, a clock is just a physical process that is cyclic and regular such as a pendulum/ spring-mass/ Earth orbit/ quartz crystal oscillator / biological process etc. Some are more accurate than others. Time is just the comparison of one cyclic physical process to another. (My definition).

    What Iam trying to deduce is that with the added advantage of there being an extended 1D axis running away from A which stretches to say the nearest star; and that B is following along this axis that is priorly populated at equal intervals with synchronized clocks in A's frame............is there anyway to determine that B is in fact time-dilated (biologically say) w.r.t someone on A's frame just due to the first acceleration?
    It seems to me that B who follows the axis of A's frame see's this axis contracted (i.e. If A's distance to the nearest star is 10 light years and B has attained a gamma of 10, then B see's A's axis as 1 light year in length). He also see's A's clocks as time-slowed.
    BUT , if A's clocks show the day/month and year of everyone in A's frame (all previously synchronized by Einsteins light signal method) then as B passes a newly 'local' clock along the axis of A's frame he is momentarily able to read that clock at the same spatial position and determine that it shows A's time in advance of his own by a factor of 10 time intervals since he left A's point? He still sees the clocks rates as slowed though, and he will see the added Doppler effects as the clocks advance towards him and recede away.

    [Note, iam assuming that ALL clocks along A's frame have been synchronized like this:
    Say a clock is at equal intervals along A such that they are 1 light-day apart. Then the synchronization process would be such that clock 0 is at A, clock 1 is at 1 light-day away along A's axis, clock2 at 2 light-days away etc.
    The clocks are set by an initial light signal from A where A's clock shows say Jan 1st 2000. But clock 1 is pre-set at Jan 2nd 2000, clock 2 at Jan 3rd 2000 etc.
    So when each clock receives the signal travelling from A, they all end up with synchronized clocks and hence can 'time-stamp' an event and send the information to any other clock to say that event occured at 'that time in your past'].

    So in conclusion Iam trying to find out that just by doing the first acceleration of B to rel v, is this a method of determining that B has biologically altered his age?
     
  16. Sep 12, 2007 #15

    StatusX

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    It's a good point you make, and these initial and final accelerations are often simply ignored. One reason is that you could modify the experiment so that the twins are never at rest relative to each other (except maybe instantaneously at the turning point), but just compare their clocks as the spaceship passes by earth.

    But if you want to include these accelerations, which is perfectly valid, here's how you could deal with them. According to Einstein's equivalence principle, an accelerated observer is observationally the same as one at rest in a gravitational field. Now you've probably heard that clocks run slower in gravity wells. More precisely, the farther down a clock is in the gravity well (relative to you of course), the slower it runs, where as the higher up it is, the faster it runs.

    Now when the space twin turns around, he essentially observes a uniform gravitational field throughout all of space. If you think about the way he's pulled during this maneuver, you can see that he'll observe the earthbound twin to be much "higher up" in the well, so that he ages extremely fast during the turning. However, the initial and final accelerations occur very close to the earthbound twin, so the difference in the rates of the two clocks is much smaller.

    (This is really circular though, as the fact that clocks run differently in gravitational fields is usually motivated by showing how they compare to an accelerated observer, by keeping careful track of the light rays the observer is receiving from the inertial clock. You could approach the twin problem this way too if you like.)
     
  17. Sep 12, 2007 #16

    JesseM

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    The initial and final acceleration aren't really important, the analysis of the twin paradox would be completely unchanged if you assumed B had been travelling at constant velocity forever and just passed next to A on his journey, then later accelerated to turn around and travelled at constant velocity to pass A again in the other direction, comparing clocks on both passes without changing velocities.
    Be careful--the only situation in which time dilation can be considered an objective frame-invariant fact is when two observers compare clocks at some common point in spacetime, then later compare clocks again at another point in spacetime. If B just accelerates away from Earth and never returns, there's no frame-invariant sense in which B's clock is dilated compared to A's--you're free to consider things from the point of view of an inertial observer who initially sees the Earth moving at some large velocity with A and B on it, then sees B change velocities so he's now at rest in the observer's frame while the Earth and A continue off at the same large velocity as before. In this frame B's clock speeds up after he accelerates rather than slowing down, and there is no reason to see this inertial frame's perspective as any worse than A's frame.
    The axis is at rest in A's frame, but you're free to analyze its behavior from the perspective of some other inertial frame, you'll get the same predictions about local physical events like the readings on each clock on that axis at the moment B passes them.
    Measurements of time and distance change in different frames, but it's important to keep in mind that predictions about local events like the clock-passings must also be invariant from one frame to another.
    Sure, as long as you agree that you can only compare different processes to another in an objective frame-independent way by bringing them together to the same point in spacetime.
    It's easy to determine that B is time-dilated in A's frame by just comparing how much time elapses on B's clock between the moments he passes each clock on the axis with the readings of those clocks on the axis. If B passes clock #1 and it reads 100 seconds while B's own clock reads 30 seconds, and then B passes clock #2 and it reads 112.5 seconds while B's own clock reads 37.5 seconds, then since clock #1 and clock#2 were synchronized in A's frame, this must mean B's clock only advanced forward by 7.5 seconds while both of those synchronized clocks advanced forward by 12.5 seconds. But this doesn't show that B's clock was running slower than those clocks in any frame-independent sense, since other frames will disagree that clock #1 and clock #2 were properly synchronized to begin with.
    If you're talking about what B sees rather than what is true in B's current inertial rest frame, then any of A's clocks that are moving towards B will be seen to be speeded up, not slowed, thanks to the Doppler effect. In my example above where B was moving at 0.8c relative to A, those of A's clocks moving towards him will be seen to be ticking at [tex]\sqrt{\frac{1 + 0.8}{1 - 0.8}}[/tex] using the equation here, which works out to a factor of 3 faster than his own. Meanwhile he'll see the next furthest clock reading 90 seconds at that moment, since in his frame it's really reading 108 seconds at that moment and is 6-light seconds away and moving at 0.8c, so 30 seconds ago it would have read 108 - 30*0.6 = 90 seconds due to its ticking slower by a factor of 0.6, and it would have been at a distance of 30*0.8 + 6 = 30 light seconds away, so the light from the clock at that point would just be reaching him now 30 seconds later. Then since it takes B 7.5 seconds to pass from the first clock to the second by his own watch, if he sees the approaching clock ticking at 3 times the rate of his own due to the Doppler effect he'll see it advance forward by 7.5*3=22.5 seconds, so if he initially saw it reading 90 seconds when he passed the first clock he'll now see it reading 90+22.5=112.5 seconds by the time it reaches him.
    He can see that the difference in time between each successive clock he passes is greater than the time elapsed on his own clock, so he knows that if these clocks represent the time in A's frame, his clock must be dilated in A's frame. But in his own frame, he doesn't explain it that way--instead, he concludes that each of these clocks was initially out-of-sync and was always ticking slower than his own clock.
    Yes, this is just a version of the Einstein sychronization convention, based on A assuming light moves at the same speed c in all directions in A's own frame. But just keep in mind that in B's rest frame this method will cause them to be out-of-sync, since he observes clocks on one side of A moving away from the point where A sent out the signal while clocks on the other side are moving towards it, so the even if two clocks on opposite sides are the same distance from A, in B's frame it takes light a different amount of time to reach them (based on the assumption that light moves at the same speed c in all directions in B's frame), and A's method implies they'll both be set to the same time at the moment the light hits them.
    It only shows that B's clock is slowed down if you use the definition of synchronization in A's rest frame, but in other frames the different clocks which A synchronized this way will actually be out-of-sync and it may be that it's A that's aging slower than B.
     
    Last edited: Sep 13, 2007
  18. Sep 13, 2007 #17
    More questions!

    In reference to StausX's reply:

    Firstly, I think I have read that Einstein considered that the act of accelerating under say a rocket type propulsive force makes no time-dilation effect?
    I also read a thread between 'MeJennifer' and 'Pervect' (I think?) where they were debating the equivalence principle and one was saying that a rocket type acceleration would produce an equivalent gravitational field and a space-time curvature with a time-dilation effect....and the other dis-agreeing.

    So does any type of acceleration caused by other means than gravity cause a local s-t curvature?

    I was under the impression that the turn-around in the Twins case was just required to sort out the mathematics involved neatly so that the signals each sees could be resolved fully at the final meet-up point.

    In reference to JesseM's reply:

    Iam assuming all the clocks in A's frame are synchronized before B leaves. Hence he has no simultaneity argument over the synchronization pulse. Also every clock he passes outside his window, he sees at his own space-time point and each snapshot he sees at that point he can tell the time (and date) that A is seeing---compared to what he is seeing on his own clock.
    What enables this extra information to B is the fact that A has an extended frame. In the usual case, this information is unknown as A and B are just 'points'.

    So from B's point of view, he doesn't 'turn around', he just does a single acceleration towards A and as the situation is completely symmetrical upto this point, then it just this single acceleration that determines the objective-resultant time-dilation when they meet (on passing)?
     
  19. Sep 13, 2007 #18

    JesseM

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    See the general relativity explanation of the twin paradox from the twin paradox page, it talks about how an accelerating observer in flat spacetime sees a uniform "gravitational" field, but the word is in quotes because it's a kind of pseudo-field, distinguished from "real" gravitational fields which are associated with spacetime curvature.
    Yes, but in his new inertial rest frame after accelerating, he will disagree about whether the clocks are synchronized. For example, if he set off a flash when he was at the midpoint of two of A's clocks, since one clock is moving towards the flash and the other is moving away from it in his frame, if he assumes light travels at the same speed in both directions in his frame he'll have to conclude the light from the flash reached the two clocks at different times, and yet both clocks will read the same time at the moment the light hits them (which makes sense in their own rest frame, since neither is moving in that frame and light is assumed to move at the same speed in both directions from the flash halfway between them in that frame).
    But again, B will disagree that A's clocks are synchronized in his current rest frame, so he can still view A's clocks as running slower than his.

    It might help if you imagined that B was sitting at the center of his own measuring-rod which also had a series of clocks along it, with these clocks synchronized in B's rest frame once he starts moving inertially (it might be easier just to assume B has been moving inertially for all time). In this case the situation is perfectly symmetrical--each time B passes one of A's clocks, the time on that clock is greater than the time on the previous clock that B passed by 12.5 seconds (using the figures from my example) while B's clock has only elapsed 7.5 seconds, but it's also true that each time A passes one of B's clocks, the time on that clock is greater than the time on the previous clock A passed by 12.5 seconds, while A's clock has only elapsed 7.5 seconds. So how can this example tell you anything about whose time is "really" dilated?
    Well, that single acceleration is what I meant by "turning around", since he was moving away from A before the acceleration and afterwards he is moving towards A. But yes, this single acceleration is what determines the fact that B's clock will show less elapsed time between the two crossings of their paths than A's clock between those same two events.
     
    Last edited: Sep 13, 2007
  20. Sep 13, 2007 #19

    StatusX

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    The curvature of spacetime is not really what directly causes the force of gravity we experience on earth. This force is simply a result of treating a non-inertial frame as if it were inertial. Specifically, the true inertial frames are the freely falling ones. Standing on the ground is not an inertial frame: we are constantly accelerating, being pushed up by the ground, and since we don't appear to move, we create a "ficticious" balancing gravitational force. It is exactly the same as the centrifugal force: not a "real" force, only an artifact of a non-inertial frame.

    The role of curvature is to explain why distant inertial observers can appear to be accelerating (and why we can appear to continually accelerate without really "getting anywhere"). This would never happen in flat space, but in the curved space around earth, a falling body on one side of the planet would observe a falling body on the other to be accelerating with 2g's, even though both are legitimate inertial observers. The local manifestation of this is the tidal force: because of small deviations in what it means to be inertial as you move from place to place, there appear to be forces distorting the shapes of systems of inertial objects.
     
  21. May 7, 2011 #20
    I have a question concerning this old thread and the answer "YES" of JesseM to the scenario #3 of RelConfuse:
    "3) Twin B is still at rel v, but Twin A then decides to catch upto him. This means he has to accelerate to a greater velocity than B to make up ground. Upon meeting (and decellerating) then:
    Result A younger than B."

    I would like to understand where is the error in my own reasoning (I would have answered "A" OLDER than "B" to this scenario).
    Reasoning:
    "A" and "B" are initialy in the same inertial frame (and same position). Let say that they are both 30 years old.
    Then "B" accelerates very quickly close to the speed of light and he keeps this speed for 4 years (in "A" reference frame).
    Let say that in the new "B" inertial frame, this same period last 2 years.
    After that period, "A" accelerates the exact same way that "B" did 4 years ago until he reach the same speed than "B".
    After that brief acceleration, "A" and "B" are back again in a common inertial frame.
    They now both agree that they are 2 light years apart.
    "A" will say that he is 34 years old (same as he was a moment ago just before his acceleration).
    "B" will say that he is 32 years old (the time it took for him before "A" start accelerating after their separation).
    They could also exchange this age information by light signal if they are willing to wait 2 years for the answer.
    But instead, let say that, soon after, "A" start accelerating again quickly close to the speed of light in the new common referential and he keeps this speed for 2 years (in "B" referential).
    Let say that in the new "A" inertial frame, this same period last 1 year.
    After that period, "A" decelerates very quickly to meet with B back in the same "B" inertial frame again.
    At that time, "A" is 35 years old and "B" is 34 years old.

    To me, this scenario looks closely equivalent to the scenario proposed by RelConfuse.
    Where is the error in my reasoning? Or where is the difference between the 2 scenarios?
     
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