Compiled from various web sources by Dr. K. S. Park

Especially for Christian Fundies

"There is no such thing as 'universal time'!"

Genesis 1:5And God called the light Day, and the darkness he called Night.
And the evening and the morning were the first day.

Dali: Clock ExplosionThe so-called Twin Paradox derives from comments made by Albert Einstein (1879~1955) in his 1905 paper On the Electrodynamics of Moving Bodies, the paper which introduced his version of relativity. At the end of Section I.4 of that paper, he discusses a "peculiar consequence" which has two clocks (at the equator & at one of the poles) experiencing different time rates because of the motion of one of the clocks.

Einstein's theory of relativity assumes 2 things:

  • The laws of physics are the same in all inertial(≈non-accelerating) reference frames.
  • The speed of light is always the same regardless of reference frame.
    • Twin (Non-)Paradox Simulation

    A Long Trip

    'Aboard' and 'Based' are twins. 'A' takes a trip on the train to a distant station (distance L), at uniform velocity v. Upon reaching the distant station, 'A' takes a return train at the same speed, returning to meet 'B' at the original station. Upon arriving at the original station, 'A' and 'B' discover that 'B' has aged considerably, yet 'A' hasn't aged much at all.
    The "paradox": Special relativity says that physics is the same in reference frames that move at a uniform velocity relative to one another. Observers in any two frames moving relative to one another should not be able to make any observation that indicates which one is "actually" in motion. So, how is it that one of the twins ('A') is younger than the other? Does this mean one of the twins was "actually" in motion while the other was not?

    Like all such paradoxes the paradox is only apparent. In this case, 'A' has actually taken a trip. That is, 'A' has left the Frame of Reference of 'B' (feeling an initial acceleration at the beginning of the trip), then, after some time, 'A' again feels an acceleration as 'A' slows to a halt and turns around (boards the return train), etc. 'A' has felt lots of accelerations; 'B' has felt no accelerations. Therefore, 'A' did actually do something "out of the ordinary" — it might not be unreasonable to find that 'A' and 'B' have had different experiences.

    However, it might be objected that although the two twins have had different experiences, it is not at all clear that one should be YOUNGER than the other! In the explanation that follows, lengths and times as measured by 'A' will have a " ' ", or apostrophe attached to them to distinguish these measurements from those of 'B'. The bottom line: both 'B' and 'A' will agree that 'A' has aged less than 'B'.

    Let's ignore the duration of time during the accelerations undergone by 'A'. (Let's say these acceleration occur almost instantly, so we don't need to keep track of how long they take.) As you will see, the accelerations therefore have NO BEARING on the quantitative difference in duration of the separation experienced by 'A' and 'B'.

    At the start, both A's and B's wrist watches read the same time. Let's call that time t = t' = 0. In order to have some numbers to play with, let's say L = 1 light-year (a very distant train station!) and v = 0.8c, in other words, v is 80% of the speed of light (c = 300,000 km/s). The trains will be very fast.

    In B's Frame of Reference

    B's Observations: At the Start

    'B' observes his own watch to read t = 0 as 'A' is racing off at velocity v. 'B' also sees that A's watch reads t' = 0.

    B's Observations: When 'A' Reaches the Distant Station

    'B' is not at the distant station, but there is a clock on the station platform, and it is synchronized with the B's wrist watch (by an obliging alien). That clock reads t = L/v = 1cy/0.8c = 1.25 years, since that is how long it would take a train traveling at speed v = 0.8c to travel a distance L = 1 light-year. A photo taken of the platform clock at the time of the arrival of the train is transmitted to 'B' (at the speed of light), so 'B' can check it for himself.

    When 'A' momentarily stops and is about to change trains for the return trip, a photo is also taken of A's watch; this photo is also transmitted to 'B' (at the speed of light) so 'B' can take a look at it. The photo shows that A's watch reads t' = 0.75 years. 'B' understands this result since 'B' observed that A's wrist watch was running slowly while 'A' was moving (at the start of the trip). So, of course less time would have elapsed on A's watch. 'B' even knows why:

    B's Observations: At the Finish

    When 'A' arrives back at the original station, 'B' notes that his own watch reads t = 2 × 1.25 years = 2.5 years, and 'B' realizes this makes perfect sense: if it took the train 1.25 years to travel the 1 light-year distance to the other station, then it would take the second train, traveling at the same speed, another 1.25 years to make the return trip. The total trip would take t = 2.5 years.

    'B' now looks at A's watch. It reads t' = 2 × 0.75 years = 1.5 years. Again, 'B' understands this result perfectly. 'B' saw the photo of A's watch taken at the distant station; it showed t' = 0.75 years. So, now, at the end of the total trip it should read t' = 1.5 years. 'B' understands exactly why t' is not the same as t: 'B' chalks it all up to Time Dilation.

    'B' summarizes it all:

    In A's Frame of Reference

    A's Observations: At the Start

    'A' observes his watch to read t' = 0, and 'A' sees that B's watch also reads t = 0. Motion is relative. Once the instantaneous acceleration is over, 'A' observes and feels nothing to suggest that 'A' is moving; instead 'A' observes 'B' racing away at the speed v = 0.8c. 'A' notices that B's watch is running slowly. Finally, 'A' also observes the Earth's surface racing under the train at the speed v = 0.8c.

    A's Observations: When 'A' Reaches the Distant Station

    Eventually, the distant station arrives at A's location (this is how 'A' observes things to happen). 'A' notices that his own watch reads t' = 0.75 years. 'A' understands this perfectly. He knows, from the map he took on the train, that the distance between the two stations is 1 light-year. 'A' knows that this distance was measured by people holding rulers end-to-end (for example) and standing still with respect to the stations (just as 'B' is standing still at the original station). But that is not the distance that 'A' would measure between the two moving stations! 'A' knows (and could verify by elaborate measurements, if he wanted) that the distance between the two stations along the MOVING Earth's surface is L/γ = 0.6 light-years. 'A' knows that this result is something called Lorentz Contraction. 'A' says,
  • Relativistic Length Contraction

    'A' also observes the clock at the station reading t = 1.25 years. He is momentarily baffled by this clock's reading, but then remembers his Special Relativity. The Time Dilation and Lorentz Contraction results do not sum up all there is to Special Relativity. For example, there is also the "Relativity of Simultaneity" — events that are simultaneous in one observer's frame of reference will not be simultaneous in another observer's frame of reference (if the two observers are moving relative to each other, and if the two events have some separation in space along the direction of the relative motion).

    So what is A's explanation of the t = 1.25 years versus the t' = 0.75 years? According to 'B', his watch is synchronized with the clock at the distant station. But according to 'A', not only are B's watch and the clock on the distant station running slowly, but they are NOT synchronized. More specifically, according to Special Relativity, the clock at the distant station is ahead of B's watch, according to 'A'! 'A' could verify this by enlisting an elaborate system of cohorts, each with a clock, spread out along the very long train track. But 'A' also knows how to calculate this effect and the result is that the clock at the station is ahead of B's watch by an amount vL/c2 (= 0.8 years). That clock STARTED ahead of B's watch by this amount (when 'A' noticed that 'B' was racing off at v = 0.8c), so of course it reads a later time than 0.75 years when the distant station arrives at A's location. 'A' calculates the time that will be displayed on the station clock as

    Estimated Time of Arrival.

    The first term L/γ/v/γ (= 0.45 years) in this result is equal to the distance the stations travel L/γ as observed by 'A' divided by the speed of their travel v, all as observed by 'A', but then reduced by the Lorentz factor γ to take account of the slow rate of the station clock (the Time Dilation effect). The second term is the initial amount by which the station clock was ahead at the beginning of the motion. So, 'A' says,
    Top figure:
    In the inertial frame of the wagon, the lamps are switched on simultaneously and the two light impulses reach the girl at the same time.
    Space-time Diagram

    The vertical line is the worldline of the twin that stays at home, while the traveling twin has the curved worldline (outbound slope=1/β, inbound slope= -1/β). The slanted lines between the world lines are lines of simultaneity at various times for the traveling twin. During the period of deceleration and acceleration near the distant star, the line of simulaneity of the traveling twin rotates (or jumps), such that the twin staying at home ages rapidly in the reference frame of the traveler.

    Bottom figure:
    In the inertial frame of the observer outside the wagon, it seems that the left lamp is switched on first, although for the girl in the wagon the lamps are switched on simultaneously.

  • Simultaneous — but to whom?

    A's Observations: At the Finish

    When the original station arrives at A's location, 'A' notices that his watch reads t' = 1.5 years, just as he expects, since it took 0.75 years for the first half of the trip.

    'A' now looks at B's watch and sees that it reads t = 2.5 years, and 'A' is not surprised by that since he observed that the station clock read t = 1.25 years at the halfway point of the experiment, and since 'A' understands why (as we just explained above).

    'A' summarizes things:

    Conclusion

    Thus both observers agree, based on their own observations, that B's watch reads 2.5 years for the duration of their separation, and A's watch reads 1.5 years. Both observers also understand why there is such a difference. 'A' is now younger than his twin!

    Note, on the return trip, because of the Relativity of Simultaneity effect, 'A' observes B's watch to be (0.8 years) ahead of the distant station's clock!! That, with Time Dilation, explains the t = 1.25 years reading for the return trip on B's watch. If you are baffled by how, as observed by 'A', the distant station clock was at one point ahead of B's watch, and then later behind B's watch, the reason lies not in a some mistake in the synchronization of the two clocks by 'B', but instead is the fact that 'A' observes those clocks from DIFFERENT frames of reference on the way out and on the way back.

    Remember, way back at the beginning, BEFORE any acceleration by the train, A's and B's watches were synchronized (at t = t' = 0), and the distant station's clock was synchronized with B's watch and A's watch. But then 'A' changed his frame of reference: he accelerated (momentarily) into a new uniform velocity frame. Back in B's frame (A's original frame), 'B' still noted that all clocks in his frame were synchronized. It is 'A' that saw the clocks in B's frame to be out of synchronization. Since 'A' was then moving toward the distant clock, he observed it to be ahead.

    Once 'A' stopped at the distant station, he rejoined B's frame of reference: 'A' saw all the clocks in that frame to be synchronized again (but reading 1.25 years, while his own watch said 0.75 years). Then 'A' started up again, now traveling in the reverse direction: 'A' then had to observe B's watch to be ahead of the distant station's clock, and by just the right amount so that when 'A' arrives at the original station, B's watch will display 2.5 years.

    If you think this whole affair is difficult to understand, then you're not alone. But it has been observed (in particle experiments), and can be explained (as, for example, in the above story), through Special Relativity.

    Twin Paradox Without Accelerations

    In the above argument it is stated that 'A' undergoes accelerations whereas 'B' does not, therefore, it is perhaps no surprise that they experience different things (in particular, a different duration for A's trip). Just in case the reader might decide that it must be the accelerations that cause the difference in duration for the twins, it is possible to explain the paradox entirely without accelerations.

    Instead of having 'A' start at rest, next to 'B', imagine the following demonstration.

    First Passby at the Start

    'B' sits still at the original station. According to 'B', 'A' is moving along (and always has been) at the speed v, and passes 'B' at the moment that B's and A's watches read t = t' = 0.

    Second Passby at the Distant Station

    'A' continues on, ultimately passing the distant station. At the moment he passes the distant station, A's watch reads t' = 0.75 years (for reasons he understands, and were explained above). At that exact moment, 'C' (another person), passes the same distant station in another train heading back toward the original station at speed v. Both 'A' and 'C' notice that the station clock reads t = 1.25 years. They also notice that both their watches display the time t' = t" = 0.75 years (C's watch is just coincidently equal to 0.75 years). According to 'B' (or his cohorts spread all over the place) 'C' has always been moving this way.

    Final Passby at the Finish

    Eventually 'C' passes the original station. As he passes, 'C' notices that his own watch reads 1.5 years, and that the watch on B's wrist reads 2.5 years. 'B' notices the two watches also. B's explanation is     Time Dilation (just as in the above explanation). C's explanation is that B's clocks were running slowly and out of synchronization (again, just as in the above explanation).

    Same Conclusion

    The resulting difference between the reading on B's watch and that on C's watch is not the result of any accelerations experienced by anyone (nobody experienced any accelerations). But notice that the full duration measured by 'B' was, of course, measured in one reference frame; the duration for the full experiment recorded by 'A' and 'C' required the combined results acquired in two different reference frames. That is the source of the asymmetry in the results. This explanation of the twin paradox (without accelerations) shows that it takes TWO different reference frames to keep track of the time duration experienced by the twin who actually takes the trip, while it takes only one frame to keep track of the duration for the twin who stays at home. Their situations are fundamentally different, and the different time durations they experience are the result.

    The time dilation effect is taken into account every day to help keep the atomic clocks on the 24 Global Positioning System (GPS) satellites that encircle the Earth in sync with Earth-based atomic clocks.

    Since the satellites are moving relative to the ground atomic clocks, they have dilated time intervals. The GPS requires that the satellite clocks be coordinated to a very high accuracy. This is absolutely necessary to predict where the location is of people using the GPS receivers on the Earth's surface. Even though the special relativity effect is very very small, the atomic clocks are accurate enough to be affected by the time dilation effect.
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