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More thoughts on the twin paradox. One persons view has to be an illusion?

  1. Aug 15, 2011 #1
    Everyone knows the situation. The one twin takes off in a rocket, and both twins see the others clock as running slow. Upon the return, its the one who has experienced acceleration that ends up being incorrect. So at some point, the "moving" twins view must show time on his brothers earth clock as going faster than his own. Say right from blast off, he keeps his eye on the telescope locked on a clock on earth. (Yes I know, he would have a very sore eye!) He sees that clock running slower than his own until when?? When he starts to slow down? At some point, the earths clock would HAVE to move forward very fast to its real (earths)time, because when he gets out, hes much younger than his brother. When would it be, and why isnt this usually talked about in books that present the paradox? Thanks for the help!!!!
     
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  3. Aug 15, 2011 #2

    Dale

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  4. Aug 15, 2011 #3

    Janus

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    Are you talking about what he "sees". Or what he would determine what is happening to the Earth clock. For instance, on the return trip, he would "see" the Earth Clock running fast, but after correcting for light propagation delay, would conclude that the Earth clock is running slow.
     
  5. Aug 15, 2011 #4

    ghwellsjr

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    What someone sees with their eye looking through a telescope is not time dilation although in simple cases they can be correlated.

    Here's what happens. During the outbound half of the trip, both twins see the other one's clock running slow by the same amount compared to their own.

    At the point where the moving twin stops and turns around, he immediately sees the home twin's clock speed up and it stays that way for the rest of the trip.

    But the earth twin does not see the traveling twin's clock speed up at the half way point because he has to wait for the images of the remote twin to travel to him at the speed of light which will take some time. It isn't until some time later that he finally sees the traveling twin turn around and head for home. At this time, he will see his twin's clock speed up by the same amount that the traveling twin saw the earth twin's clock speed up at the half way point.

    It's because the traveling twin sees the earth twin's clock running at two different rates for half the time and the earth twin sees the traveling twin's clock running most of the time at the slow rate, that the traveling twin's clock ends up with less time on it than the earth twin's clock does.
     
  6. Aug 15, 2011 #5

    Ben Niehoff

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    This is exactly what happens. The Earth clock appears to speed up and overtake the rocket clock during the acceleration phase in which the rocket turns around to come back home.

    The entire confusion arises from neglecting the acceleration required to turn around and come home. Typically we idealize the rocket trip with two straight line paths: one out, and one back. But this actually results in infinite acceleration at a single point where the rocket "rebounds" (like a billiard ball). So in our idealized picture, the twin on the rocket sees the earthbound twin's clock jump discontinuously at the moment of rebound. In one instant, it jumps from 12:54 PM Jul 8 1983 to 3:41 PM Nov 2 2005, or whatever.

    From that point onward, each twin sees the other twin's clock run more slowly than his own, but the damage has already been done due to the discontinuous jump. When they meet up again, the earthbound twin will be older.

    Edit: When I say "see", I mean after correcting for light propagation delay.
     
  7. Aug 15, 2011 #6

    So my original thoughts were always correct? That on the return trip, the rocket twin will see the earths clock going very fast? That should always be stated! Now what happens if on the outbound trip, the rocket twin takes pictures of himself periodically, as does the earth twin. At the midpoint, before he turns, the earth bound twin is still older? What will the picture show upon the return? Is the rocket twin aging slower even on the outbound trip? If so, an illusion must be taking place. Or taking away any acceleration. A rocket passes earth at say .75 c, and two people are born at the same time. (earths time) Both people see the other as aging slower, and they take pictures of themselves over the course of 10 earth years. After that time, the rocket turns and comes back to earth. Do the pictures show the rocket "twin" as being younger when compared to the earth "twin"?? (no acceleration or Gs have taken place in the rocket twins pictures.)
     
  8. Aug 15, 2011 #7

    Ben Niehoff

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    No, I didn't say that. Read more carefully.
     
  9. Aug 15, 2011 #8

    ghwellsjr

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    Your definition of "see" is not what he asked about.

    There are an infinite number of ways to correct for light propagation delay and they can all produce different interpretations. You have select one that involves using two different inertial Frames of Reference or a single non-inertial Frame of Reference. If you use just one inertial FoR, there will be no discontinuous jump at the point of acceleration.

    For example, you can analyze this from a FoR in which the home twin remains stationary. Then, only the traveling twin's clock experiences time dilation and ends up with less time on it than the earth twin's clock. In this single inertial frame you can demonstrate what each twin actually sees through their telescopes as the original poster asked about and it will be as I said earlier.

    Or you can analyze this from an inertial FoR in which the traveling twin is stationary after then initial acceleration. There will again be no jump in anybody's clocks but you will arrive at the same conclusion and you can demonstrate what both twin's see through their telescopes.

    Or you can analyze this from an inertial FoR in which the traveling twin is stationary after the turn-around acceleration. There will again be no jump in anybody's clocks but you will arrive at the same conclusion and you can demonstrate what both twin's see through their telescopes.
     
  10. Aug 15, 2011 #9

    ghwellsjr

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    Yes, you are correct, as I stated earlier.
    When the twins are separated in space, eg, at the place you are calling the midpoint, you cannot say that both twins take a picture at the same time. Time is running at different rates for both of them and their clocks are out of sync. There is no meaning to the question of which one is older.
    If each twin takes a picture of himself each year according to their own clocks, the pictures would look the same when they bring them back to compare. However, traveling twin would end up with fewer pictures because he is younger.
     
  11. Aug 15, 2011 #10

    D H

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    Emphasis mine:
    The title of the thread ("One person's view has to be an illusion") is incorrect. The traveling twin will indeed end up with fewer pictures because he truly is younger on return than is the Earth-bound twin. No illusion. Fact.
     
  12. Aug 15, 2011 #11
    Maybe I'm just missing something, but if the earth twin has to wait for the images of the remote twin to travel to him at the speed of light, why doesn't the moving twin have to wait for the images of the earth twin to travel to him at the speed of light?
     
  13. Aug 15, 2011 #12

    ghwellsjr

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    They both have to wait for the images to reach them from the other twin. If the earth twin took off in a rocket halfway through the scenario, the traveling twin wouldn't see this happen for a long time.
     
  14. Aug 15, 2011 #13

    Dale

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  15. Aug 15, 2011 #14
    This basically makes sense to me, however, why or how could he (rocket twin) see the earths clock moving faster upon return if he would measure the speed of light the same on the outgoing trip as on the return trip? Common sense tells me the clock would appear to be moving faster because he is moving into the light compared to moving away from it, but relativity says light speed is the same in both directions. Very confusing!
     
  16. Aug 15, 2011 #15

    Dale

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    The Doppler shift (frequency) doesn't imply a variable speed of light. As long as the wavelength decreases by the same factor that the frequency increases the speed remains the same.
     
  17. Aug 15, 2011 #16

    ghwellsjr

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    Remember two things:

    1) The traveling twin has to be inertial (non-accelerating) while he is making a measurement of the speed of light. This means that he has to make two separate measurements during both parts of the trip.

    2) The twin cannot tell how fast light is traveling just by looking at an image of some distant object because that is a one-way trip for the light. What he has to do is place a mirror some measured distance further away from earth (in front of him during the outbound portion of the trip and behind him during the inbound portion of the trip). Then when he sees a particular time on the earth clock, he starts his stopwatch and then he waits for the image to propagate to the mirror and reflect back to him at which point he stops his stopwatch. If he repeats this experiment during both halves of the trip, he will measure the same length of time.
     
  18. Aug 15, 2011 #17

    Janus

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    The same way that you hear the pitch of a train whistle differently if it is coming towards you rather than going away. The speed at which the sound travels relative to you doesn't change, but the pitch will.
     
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