Instantaneous Age Changes in the Twin "paradox"

Discussion in 'Physics & Math' started by Mike_Fontenot, Sep 2, 2018.

  1. Janus58 Valued Senior Member

    You can actually work the problem out using a 1g acceleration if you want.
    Let's assume that the relative velocity between our observer and train is 0.6c
    At the moment a clock on the train that reads 0 passes our observer, he begins to accelerate at 1g (proper acceleration) to 0.6 c and matches the train's velocity (Mind you, in order for him to be still next to any part of the train after matching velocity with it, you are going to need a long train. )
    It will take our observer ~ 245.59 days by his own clock to reach 0.6c, during which time he will travel ~88.58 light days as measured by his original rest frame, and the clock he started next to will be 88.58 light days ahead of him (as measured in the train frame.)
    Assuming that all clock on the train are synchronized in the train frame, then the clock the observer ends up next to once he matches speed will read 265.73 days, as will the clock he started next to (according to our observer and train)

    Now just before he starts his acceleration, the clock he will eventually end up next to will, at that moment, read 53.15 days before zero. Thus according to our observer, the clock he started next to advances from 0 to 265.73 days during his acceleration, while the clock he ends up next to advances from -53.15 days to 265.73 days or by 318.88 days. The clock he left will, on average, run slower than the clock he arrives at during the acceleration.

    This is not limited to clocks on the train either. If we assume that our observer is on his own train, with its own clocks that are synchronized to each other prior to the acceleration, our observer will find that these clocks will go out of sync with each other once his starts his acceleration. Clocks in the direction of his acceleration will run fast ( the further away, the faster) and clocks in the opposite direction will run slow (the further away, the slower)
    Mike_Fontenot and Neddy Bate like this.
  2. Google AdSense Guest Advertisement

    to hide all adverts.
  3. phyti Registered Senior Member

    In fig.1, A is the stationary twin and B is the wandering twin. B moves at .6c relative to A outbound and inbound, with an instantaneous reversal* at B4. The light signals (blue) are sent at unit intervals (tick marks) from A. B receives 2 signals in 4t outbound and 8 signals in 4t inbound. All A events are observed by B in a continuous sequence with no gaps, and all B events (shown only as tick marks for clarity) are observed by A. This must be the case for signals emitted from one boundary of a closed path to the opposite boundary.

    The interpretation of a time jump for A events observed by B is based on the axis of simultaneity (aos) (red). The aos is defined by the SR clock synch convention. If B pings A for a clock reading, (events B1, A2, B4), the clock reading is assigned to half the round trip transit time, B 2.5. If B pings A for a 2nd clock reading, (events B4, A8, B7), the clock reading is assigned to half the round trip transit time, B5.5.

    Since B cannot assign a time to the A clock readings until after they are detected, the aos is irrelevant to B observations of A.

    The 'time jump' is the result of a poorly rendered drawing.

    Fig.2 shows a more realistic reversal which rotates the aos clockwise between the two positions shown in fig.1.

    Each observes clock rates as doppler effects while diverging or converging, since clocks are frequencies. Aging is the accumulation of time, and determined by a comparison of clocks at reunion. Mutual observations are irrelevant.

    This is a case of making a simple problem unnecessarily complicated. Even as a 1st approximation, it still serves the purpose of illustrating time dilation and aging.

    In the 3rd drawing, using the ct scale of A, draw an arc at 5 centered on 0 that meets the x coordinate of the reversal point. That point projected horizontally indicates ct=4, the same value resulting from the red hyberbolic calibration curve (as labeled by Max Born). It's simpler to draw an arc! Rotate the drawing 180 and apply the same method from ct=10.

    The net result B loses 2t and ages 8t.

    * Since the duration of the change is zero, there are no forces.

    Attached Files:

  4. Google AdSense Guest Advertisement

    to hide all adverts.
  5. Neddy Bate Valued Senior Member

    Applying the brakes in your car very gently, you come to a stop over a relatively long period of time, and you hardly feel any force.

    Applying the brakes in your car very harshly, you come to a stop over a relatively short period of time, and you feel a lot of force.

    So in the limit as the turn-around time tends toward zero, your force should be going up toward infinity, not toward zero.
  6. Google AdSense Guest Advertisement

    to hide all adverts.
  7. phyti Registered Senior Member

    If there is in fact an acceleration, there is a radius of curvature for each increment of the curve. The force would be inversely proportional to the radius and last for the duration of the curve. In the idealized, fictional, instantaneous change in direction, the radius and the duration are zero. If the duration of any process is zero, there is no change!
  8. Neddy Bate Valued Senior Member

    If the force is inversely proportional to the radius, then as the radius approaches zero, the force approaches infinity. The idealized instantaneous acceleration is pure fiction, so you don't have to worry about its duration actually being zero. You can think of it as an arbitrarily small duration, and an arbitrarily small radius. The forces would not be pleasant to experience!
  9. phyti Registered Senior Member

    When you get 'infinity' in calculations, it's a flag that the calculations aren't working. Infinity is not quantifiable and therefore meaningless in a math sense. Obviously that doesn't deter people from debating the nonsensical (an ongoing thread).
  10. Neddy Bate Valued Senior Member

    I said "approaches infinity" which does not imply that any calculations are not working.

    Start with some chosen finite radius, and calculate the force. Divide your chosen radius by 2 and the force doubles. Divide your chosen radius by 3 and the force triples. Divide your chosen radius by 4 and the force quadruples. Note that as the radius is "approaching zero" the force is "approaching infinity".

    It is a matter of limits. From wikipedia:
  11. phyti Registered Senior Member

    You have been indoctrinated into the mathmagic school of thought.
    The examples you gave are for finite values.
    If infinity has no end, how can you approach it? Can you approach the horizon?
    Can the horse approach the carrot on a stick?
  12. Neddy Bate Valued Senior Member

    Oh brother.

    Yes, let's talk about finite values. As the finite value of the radius decreases, the finite value of the force increases. The idealized instantaneous acceleration could be approximated by a really, really small radius. And then it could be approximated better by dividing that small radius by 2. And it could be approximated even better by dividing that small radius by 10 or 20, etc. The force keeps going up the better your approximation. What does that tell you?
  13. marco$bossco Registered Member

    Why is everybody still trying to explain away the twin paradox, while claiming that its not really a paradox at all? There are more excuses that try to dismiss the errors of special Relativity than there are flies on a piece of meat. Has no one thought that maybe its a paradox because the theory is flawed? What happened to critical thinking in Physics? Question everything. start with all those old so called heroes of physics beginning at Maxwell, and don't skip out the Geniuses. You may find that all those paradoxes, and there are many, are all correctly pointing out that Relativity is just wrong. period.
  14. DaveC426913 Valued Senior Member

    Because it's not. It only appears to be a paradox when people try to apply Newtonian Physics to relativistic scenarios.
    The key to unravelling the apparent paradox is to discard the Newtonian assumptions - and that happens in the details, which is why one needs to go through it step by step.

    Many people who don't understand it will give up and draw that simplistic conclusion. That does not make it wrong - the universe is not sympathetic to your inability to grasp it.
    The rest of us grasp it just fine.
  15. phyti Registered Senior Member

    The critical point is: you can't divide by zero!
  16. Neddy Bate Valued Senior Member

    I never said you could divide by zero. But you can divide by a smaller and smaller radius, and this tells you that as the radius gets smaller, the force gets bigger. So you should conclude that the force would be quite large for a very quick turn-around. You had said the force would be zero, which is what I replied to.
  17. DaveC426913 Valued Senior Member

    By using limits.

    1 divided by 0.5 = 2
    1 divided by 0.05 = 20
    1 divided by 0.0005 = 2,000
    In other words: as the denominator approaches zero, the result approaches infinity.

    While it is true that you can't divide by zero, that is a red herring, since relativity never requires you to.

    At .9c, the time dilation factor is 2.29.
    At .99c, the time dilation factor is 7.0.
    At .9999c, the time dilation factor is 70.0.
    In other words: as the fraction of c approaches one, the dilation factor approaches infinity.
  18. phyti Registered Senior Member

    ‘Approach infinity’ is a contradiction of terms. Infinite and infinity is a state or property of not being measurable, or not quantifiable. You can’t approach something that is always distant. It was a poor choice of words in the history of mathematical definitions.
    A more sensible phrase would be ‘as x increases without limit’, which is what the definition implies, ‘with no end’, ‘without bounds’.

    If a group of irish people are merged with a 2nd group of irish people, the larger group is not more irish then either of the original two. The point is ‘irish’ is independent of the size of any group, just as ‘infinite’ is independent of the size.

    In the case of instantaneous change of direction, when the radius equals 0, the circumference equals x=0, and the expression t=x/v becomes t=0/v =0 for all speeds. It’s not complicated.

    If you fly at a constant altitude, the distance from you to the horizon is constant, i.e. it moves ahead of you at the same speed. You never get closer to the horizon.
  19. DaveC426913 Valued Senior Member

    Well, that's why I invoked "limits", as the more accurate term.

    If that phrase sits better with you, I'm all for it. That's just a semantic issue.
    As long as you acknowledge that the math works out just fine.
  20. Confused2 Registered Senior Member

    I have a thought experiment to clarify my difficulty with some of this.
    Anyone unhappy with impulses please sit this one out.

    Alice stays at station A.
    Bob travels to station B at velocity v.

    Bob's train stops at station B.

    Bob syncs his clock with station_clock_B which is synced to Alice and station_clock_A.

    Now it starts...

    Overview - impulse start, short interval to observe, impulse stop.
    The impulse stop is irrelevant except to clarify the point that the train has moved no distance (or distance->0 )relative to the station.

    Bob uses a telescope to take a picture of Alice's Station_clock_A.

    Bob's train then does an impulse acceleration to velocity -v (v heading back towards Alice).
    Bob uses his telescope to take a picture of Alice's Station_clock_A.
    Bob also takes a picture of station_clock_B to avoid later confusion.

    The favoured result seems to be that the image Bob sees of Station_clock_A changes instantaneously - I'm kinda's with that but...

    What Bob's telescope is picking up is EM radiation (or photons if you like). After the impulse to -v his view of Alice's clock changes BUT he is in the same space as before the impulse so we seem to have different photons in the same space. After the stop impulse he's back to the same photons as were passing through Station B.

    Have I misunderstood?
  21. Janus58 Valued Senior Member

    Yes, you have misunderstood.
    Bob will not visually see any change in the time on Station_clock_A( though he will see a Doppler shift). What does change is what he concludes the time on Station_clock_A is at that moment.

    So, let's say that he accelerates to 0.6 c and the stations are 1 light hr apart (as measured by the stations).
    While he is at rest with respect to Station B he sees that Station_clock_B reads 12:00 and Station_clock_A reads 11:00. He knows that the light left 1 hr ago and that it is actually 12:00 at Station_A

    He accelerates up to 0.6c. He still sees the same times on both clocks, but he sees the light from Station_1 as being blue-shifted by a factor of 2.
    Due to length contraction, Station_A is 0.8 light hrs away by his measurement. At 0.6c this take 1 1/3 hrs to cross by his clock, and 1:20 shows on his clock when he meets up with Station_A.
    During which time, he will visually see Station_A_clock advance 2 2/3 hrs, meaning it will read 01:40 when he arrives. (Station_B_clock would be seen as red-shifted and he will see it tick off just 40 min and reading 12:40 when he arrives at Station_B.

    Since both Station clocks would have been time dilated during the trip they ticked at a rate of 0.8 of that of Bob's clock, despite what he visually saw, ticking off 64 minutes during his trip. If Station_clock_A reads 1:40 when he arrived, and had ticked off 64 minutes, then it was 12:36 at Station_A when he left Station_B (even though he visually saw it reading 11:00).

    Also, if Station_B_clock started at 12:00 and ticked off 1:04 during the trip, it reads 1:04 upon Bob's arrival to Station_A ( while Bob visually sees it as reading 12:40)
    If Bob now comes to a rest with respect to both stations, he still measures 1:40 on Station_A_Clock and still visually sees Station_clock_B as reading 12:40. But now, being at rest with respect to the stations, he measures them as being 1 light hr apart and knows that it is 1:40 at Station_B.

    Thus he started with a clock that read 12:00 and ended with a clock reading 1:20, while both station clocks started at 12:00 and ended at 1:40 ( after he come to rest with respect to them.)
    Confused2 likes this.
  22. Confused2 Registered Senior Member

    Very clear - thank you.
    So, seen by Bob -
    Alice's clock never changes instantly. The Doppler shift does change 'instantly' with instantaneous acceleration and (given a known source frequency) contains all the information Bob needs to know to predict Alice's clock time when he arrives after a length contracted distance of (say) x' in his frame and x in Alice's frame. This is in no way contradicts what Neddy Bates has been writing - just that what he has written and what I have read might not be the same thing.
    Hopefully all good now. Thanks again.
  23. Neddy Bate Valued Senior Member

    Thank you. It is important to note that the 'time-jumps' which we have been discussing do not show up directly in what the traveler visually sees.

    So the 'time jump' manifests itself as Bob concluding that Station_clock_A was reading 12:00 (not visually) just before he accelerated, and that Station_clock_A was reading 12:36 (not visually) just after he accelerated. So Bob's first acceleration involves a 36 minute 'time jump' on Station_clock_A, and this allows Bob to justify how that clock only ticked off 64 minutes during the trip, yet ended up reading 1:40 at the end of the trip.

    Likewise, Bob knows that Station_clock_B was reading 12:00 (even visually) just before he accelerated, and that it only ticked off 64 minutes during the trip. So Station_clock_B must have been displaying 1:04 (not visually) just before Bob decelerated. And yet, we know that Station_clock_B ended up reading 1:40 at the end of the trip. So Bob's deceleration at the end must involve a similar 36 minute 'time jump' for the time on Station_clock_B (not visually).

    In this scenario, both 'time jumps' were in the forward direction of time. But with careful construction, there can also be scenarios where there are 'time jumps' which are in the backward direction of time.
    Last edited: Sep 23, 2018

Share This Page