As will be shown, the Twin Paradox is not a paradox because it can be fully explained using The Theory of Velji Relativity. However, one cannot properly explain the Twin Paradox using only the Theory of Einstein Relativity. Thus, using only that system, it does remain a paradox! When most people argue that the Twin Paradox is really not a paradox, they use this explanation: There is a time dilation effect because one of the twins is travelling at a velocity while the other is at rest. Then, when one considers the Relativity Postulate, it is asked why one twin is travelling at a velocity while the other isn't? That is, if measuring velocities are relative, then who is actually in motion? When asked that, these people argue that one twin is in motion because she experienced an acceleration, and the other isn't because she did not experience an acceleration. Within this page we will demonstrate a situation in which one can be "in motion" while not having to undergo an "acceleration". Although this idea is simple in mind, it requires a lot of "machinery" to be properly explained. And so, the fact that time dilates, which has been shown to happen in various physical experiments, cannot be because of velocity in a regular Lorentz factor! This does not discount the fact that general Lorentz factors can explain time dilation. This is because general Lorentz factors are unique, whereas regular Lorentz factors are relative. Recall: We define a "regular Lorentz factor", γ, as:> γ = 1/(1-v²/c²)^½ • where v is some velocity As Einstein's Relativity dictates, time dilates proportionally to regular Lorentz factors. Now, the following is a thought-experiment that employs Captain Picard and Guinan of the Starship Enterprise (from the TV show Star Trek: The Next Generation). In it, we will construct a situation where the one "in motion" has not undergone an "acceleration".Captain Picard has been hearing Guinan mouth off for years on how she can confound him with a pair of science experiments. So Picard, having nothing to do one day, exhausting the library of the holodeck, gives Guinan the keys to the Enterprise, to do two experiments. Being resourceful, Guinan makes a few calls and rents two identical space stations - Station A and Station B. They are identical, of the same build and model, same mass M, etc. Picard is to stay on Station A while Commander Riker must stay on Station B. On Station A: there exist 2 identical twins, Twin A and Twin B, and a space ship in the docking bay, none other than the USS Enterprise. (Twin A and Twin B have been together in the same frame of reference for their entire lives.) The mission of these experiments is to complete The Twin Paradox experiment, in two different ways. First, Guinan instructs both space stations to enter a region of space of pure blackness - called the "black-expanse". In that space, nothing but the two stations are visible to each other. There are no stars, no galaxies, etc., visible from within that space. Setup to The Twin Paradox Experiment #1: To start, Space Stations A and B are at rest together. Now, before the experiment begins both stations synchronize their clocks, and they decide to set the following quantities appropriately: a force F _{1} and two durations of time T_{1}, and T_{2}. Their itinerary is planned: Starting in unison, at time T _{0}, they disembark from each other by exerting a force F_{1} for a time T_{1}, and so quickly travel away from each other at the same speed for a fixed time T_{2}. Then they decelerate, that is, undo the acceleration by exerting a force F_{1} for a time T_{1} in the *opposite* direction as before. By both stations doing this in *unison* we can be assured that after completion they will be at rest with each other, but a certain distance apart. They are at rest at time T_{3}. Now, the crew on both stations can determine the distance D between them by:
The Twin Paradox Experiment #1: The crew on Station A will now commence the Twin Paradox experiment, at time T _{4}. For that experiment Twin B is to be sent away towards Station B while Twin A is left on Station A. The experiment will be over when Twin A reaches Station B, at time T_{5}. The period of time between T _{4} and T_{5} will be called the "coasting period". So now, Twin B boards the Enterprise and exits the docking bay of Station A. Once out, she accelerates very quickly to a velocity v in the direction of Station B, leaving her twin sister, Twin A, behind on Station A. Exactly when Twin B ignites her engine, the crew on Space Station A starts a Clock C at zero. Now, the Enterprise is at all times sending telemetry, a pulse of light, to Station A. Twin B will set the frequency of this light at f _{s}, and the crew on Station A will be privy to this quantity as they have discussed it prior to the mission. The crew on Station A will measure the frequency of this light as f_{o}. Now, the crew on Station A can determine *exactly* when Twin B reaches Station B. First, using relativistic doppler shift:
(Guinan secretly records this velocity v for later use. Wink-wink.) So, when Twin B reaches Station B, Twin A will know *exactly* this instant because her measured time on the Clock C will match the derived value of t right then. Their times have been synchronized! The Twin Paradox Experiment #1 - Diagram:_{5}----------------------• X@ • X o • X o • X o • X o • X o • X o • X o • X o • X o • X o • X o • X o • X o • X o • X o • X o T _{4}----------------------• X o • o • o • o • o T _{3}----------------------• o • o • o T _{2}------------------------• o • o • o T _{1}----------------------------------• o • o • o T _{0}---------------------------------------•oStation A is represented by • Station B is represented by o The USS Enterprise is represent by X the x-axis: we have the axis of movement of both Stations the y-axis: we have time So we can check to see which twin is older. "How?" Picard and Riker asked Guinan this very question before the experiment? She replies in jest: "By counting gray hairs!" So, *you* tell me now, who is older, Twin A or Twin B? And why? (Notice, there is no turnaround acceleration!) What Picard and Riker would believe is the way Relativity explains the experiment: Twin B will be younger than Twin A because she is the one to have travelled at a velocity and also underwent an acceleration. Let's suppose for now that this is exactly what happens and redo the experiment with a little twist. Setup to The Twin Paradox Experiment #2: So now, Guinan has one last experiment to befuddle Picard. She sets this experiment similar to the previous one. Guinan finds two other twins, Twin A2 and Twin B2, who are willing to undergo this experiment and they are put appropriately on Station A. (Twin A2 and Twin B2 have been together in the same frame of reference for their entire lives.) To start, Space Stations A and B are at rest together. Now, before the experiment begins both stations synchronize their clocks, and they again choose the following quantities they used previously: a force F _{1} and two durations of time T_{1}, and T_{2}. Their itinerary is the same: Starting in unison, they disembark from each other by exerting a force F _{1} for a time T_{1}, and so quickly travel away from each other at fixed velocity for fixed time T_{2}. Then they decelerate, that is, undo the acceleration by exerting a force F_{1} for a time T_{1} in the *opposite* direction as before. By both stations doing this in *unison* we can be assured that after completion they will be at rest with each other, but a certain distance apart. They are at rest at time T_{3}. Again, the crew on both stations can determine the distance D between them by:
Then, at time T _{3}, they already have special instructions by Guinan to *immediately* accelerate in a common direction for a time t: Station A is to experience a force F_{3} for a time t in a direction away from Station B; while Station B is to experience the very same force for the very same time, that is, a force F_{3} for a time t in the direction towards Station A. The direction that both stations accelerate is the same, and the times that both stations accelerate is the same; and so, the distance between the two do not change during this period. To keep the amount of time in acceleration a minimum, she sets t to be small, like 0.1 seconds. Using the mass M of each individual Station, and the velocity v she secretly recorded above (wink-wink), she calculates F _{3} as such:
After this, at time T _{4}, the Stations will again be at rest relative to each other, but now traveling at a velocity in unison, that velocity being v.The Twin Paradox Experiment #2: The crew on Station A will now commence the Twin Paradox experiment, at time T _{4}. For that experiment Twin B2 is to be sent away towards Station B while Twin A2 is left on Station A. The experiment will be over when Twin B2 reaches Station B, at time T_{5}. So now, Twin B2 boards the Enterprise and exits the docking bay of Station A. Once out, towards Station B she accelerates - NO! - she DECELERATES! Remember, Station A and Station B are travelling in unison at a velocity v. So, she does not *accelerate* to a velocity *v*, instead, she *decelerates* to a velocity of *zero*! So now at first glance it may seem that Twin B2 will not be able reach Station B, and in a way this is right. Twin B2 won't travel to Station B, rather, Station B will travel to Twin B2! Remember, Station B is travelling at a velocity v towards Station A, and Twin B2 is at rest on the Enterprise where it had initially disembarked Station A and decelerated. So, until Station B reaches Twin B2, her sister Twin A2 remains on Station A, travelling in the abyss at velocity 0 m/s for the entire coasting period. Now, exactly as before when Twin B2 had ignited her engines, the crew on Space Station A had started a Clock C at zero. Again, as in the previous experiment, the Enterprise is at all times sending telemetry, a pulse of light, to Station A. Twin D will set the frequency of this light at f _{s}, and the crew on Station A will be privy to this quantity as they have discussed it prior to the mission. The crew on Station A will measure the frequency of this light as f_{o}. Now, the crew on Station A can determine *exactly* when Twin B2 reaches Station B. First, using relativistic doppler shift:
So, when Station B reaches Twin B2, Twin A2 will know *exactly* this instant because her measured time on the Clock C will match the derived value of t right then. Their times have been synchronized! The Twin Paradox Experiment #2 - Diagram:T5------------------@o X o X o X o X o X o X o X o • X o • X o • X o • X o • X o • X o • X o • X o • X o T4----------------•-X o • o • o • o T3----------------------• o • o • o • o T2------------------------• o • o • o T1----------------------------------• o • o • o T0---------------------------------------•o Station A is represented by • Station B is represented by o The Enterprise is represent by X the x-axis: we have the axis of movement of both Stations the y-axis: we have time So it's time to count gray hairs again! So, *you* tell me no, who is older, Twin A2 or Twin B2? And why? (Notice again, there is no turnaround acceleration!) Examine the following table:
So, it would seem sound to say that (assuming that other parts of Einstein's Relativity to be correct) Twin A is younger and Twin B is older. The paradox that remains: Who is younger, Twin A2 or Twin B2? If we believe that Twin A2 is younger, then she must have experienced time dilation because she was in motion. If we believe that Twin B2 is younger, then she must have experienced time dilaiton because she experienced some form of acceleration/deceleration. Thus, the standard explanation of the Twin Paradox cannot be relied upon because the twin in motion (Twin A2) is *not* the twin to have experienced some form of acceleration/deceleration (Twin B2). |

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