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.
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  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.

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  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.
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  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!

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