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Twin paradox

  1. May 31, 2005 #1
    I'm a little confused on the twin paradox, and perhaps someone can clear my understanding up. You have a pair of twins that are born on some planet. One takes off in a rocket and travels at nearly the speed of light for 4 years relative to himself in the spacecraft. When he returns he is merely 4 years older however his twin has aged some far greater number, lets say 30. WHY DOES THE GUY IN THE SPACESHIP AGE SLOWER? I know of time dilation and that equation would solve the problem. But, what if out inertial oberserver was in the frame of the space craft. It would appear that the spacecraft is stationary and that the planet is moving at close to the speed of light. Should't the twin on the planet now age only 4 years and the person in the spacecraft age 30? Basically what this comes down to is that I don't like the notion that scientists state that there is no such thing as a completely stationary object is space and that all velocities are relative. I think space is like a big giant grid with a bunch of "sinks" and "sources" that move through it and warp it. I dont really know can someone clear my thinking up about the twin thing though
     
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  3. May 31, 2005 #2
    The answer is that time dilation does happen from both frames of reference. The twin on Earth is justified in saying that the twin in the spaceship's clock is moving slower, and, at the same time, the twin in the spaceship is justified in saying the twin on the Earth's clock is moving slower. This is where the "paradox" arises; the solution to the paradox is the relativity of simultaneity. The twin on the spaceship will claim the clock on Earth actually started a long time before his clock, while the twin on the Earth will claim the clocks were synchronized. For an example of the relativity of simultaneity see here: https://www.physicsforums.com/showthread.php?t=75943#post572557
     
  4. Jun 1, 2005 #3
    There are thousands of articles and many books by notable authors that treat the twin paradox - all agree that in fact there is no paradox - but they do not agree upon why there is no paradox - some of the explanations are equivalent - they lead to the same conclusions from different viewpoints - but others (even by well know authors such as Scima and Born) are based upon General Relativity and do not comport with the explanations based upon SR.
     
  5. Jun 1, 2005 #4
    You can say that the SR is applicable only if you're in the inertial reference frame. By inertial reference frame I mean the frame of reference where the Newton's first law is hold. The spaceship cannot be the inertial reference frame since we know there is a force (combustion of the gas molecules off the engines..)acting up on it. So it is not clear to me, what would the observer from the spaceship will have to say about the age of the observer on earth. I hope this will be resolved once one understands GR??
     
  6. Jun 2, 2005 #5

    pervect

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    Try reading the sci.physics FAQ on the twin paradox

    http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

    My favorite explanation is the geometrical one. In Euclidean geometry, the shortest distance between two points is a straight line, and any path that takes an observer in a path other than a straight line will be longer.

    The observer who travels is like the observer who takes two sides of the triiangle to reach his destination rather than walking in a straight line. The important difference is that while distance is shorter, for the twin paradox the elapsed time is _longest_ for the observer who follows the straight path.

    My next favorite explanation is the "doppler shift" explanation, but I won't go into it. It's not quite as powerful as the geometric explanation, but it may be easier to come to terms with.
     
  7. Jun 2, 2005 #6
    This is correct
     
  8. Jun 3, 2005 #7
    Yep...pretty sure it is.
     
  9. Jun 4, 2005 #8
    The attached picture visualizes this geometric approach. Note that the picture indeed uses Euclidean geometry (as standard for Euclidean relativity). The picture used in John Baez' explanation uses Minkowsky geometry.

    The difference between the two approaches is that in Euclidean geometry the twins end up at different space-time events (separated in the timedimension) but are still able to shake hands in space. In the Minkowski diagram this gap does not show in the drawing because the 4D displacement [itex]ds[/itex] of the traveling twin is drawn "stretched" (the easiest way to see that is by considering that for a photon [itex]ds=0[/itex] while its worldline in the Minkowski diagram has positive length).

    In Euclidean relativity the twins can still shake hands due to the universal velocity [itex]c[/itex] in 4D space-time that causes the time dimension to fully contract in any rest frame (similar to spatial length contraction for an object that moves at velocity [itex]c[/itex]).
     

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  10. Jun 4, 2005 #9

    robphy

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    Can you clarify?
    Do you mean that the "reunion event" appears as two distinct points in your diagram? In the Minkowski diagram for Special Relativity, the reunion event is drawn as one point in the diagram.


    Huh? For a photon [itex]ds_{Minkowski}^2=0[/itex] (but not [itex]d\vec s_{Minkowski}=0[/itex], "the displacement is nonzero"). On a Minkowski-spacetime diagram, the photon worldline has zero Minkowski-length, although it appears to have positive length to someone who uses a Euclidean metric on the diagram.


    Concerning your alternate interpretation of Einstein's Relativity, can you clarify something about your attachment?

    As you may know, there are alternative "diagrams" for Einstein's Relativity, which differ in how the physical situation is drawn for different inertial reference frames. In the Minkowski diagram, under a sequence of Lorentz transformations, events slide along a concentric set of hyperbolas. (This is of course analogous to points in the ordinary Euclidean plane sliding along a set of concentric circles under a sequence of rotations.)

    With your "Euclidean relativity", can you draw a sequence of diagrams illustrating the transformation from one inertial reference frame to another?
    Please include a set of light-rays in your diagrams. Ideally, I would like to see how you render this Minkowski diagram http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_vase.html#doppler under a sequence of transformations with your "Euclidean relativity".

    Maybe this clarification belongs in a different thread.
     
    Last edited: Jun 4, 2005
  11. Jun 5, 2005 #10
    Yes.
    You are of course correct. That was an incorrect use of [itex]ds[/itex]. Please ignore it.
    I wouldn't call it "my" Euclidean relativity. Perhaps that credit belongs to Hans Montanus, a Dutch mathematician. Although the topic interests me a lot, I am by no means the best expert on the terrain (I am still learning it myself and am still struggling with some of its concepts). As to your concrete questions, I have collected a series of references that may explain things:
    - Jose Almeida, K-calculus in 4-dimensional optics gives on pages 5 and 6 the Minkowski and Euclidean representation of a series of radar pulses sent between two objects that move relative to each other.
    - Hans Montanus, Proper time physics (Hadronic Journal 22, 625-673, 1999) gives a couple of pictures that show the relation between Minkowski and Euclidean diagrams. Since this article is not available online I have scanned them in the attachment. Note that Fig. 2 is the top-down view of Fig. 1. Fig. 3 is the front view of Fig. 1.
    - Alexander Gersten, Euclidean special relativity gives a short overview of the topic.
    - One of my own pages gives a possible way to derive the Lorentz transformation equations in Euclidean space-time.

    I agree with you that this should best be moved to another thread. Perhaps to "Theory development".
     

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