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Re: Inescapable (symmetric) twins paradox

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On 9/7/15 9/7/15   12:01 AM, RichD wrote:
>   On September 2, tjrob137 wrote:
>>> A thought experiment: consider a closed universe,
>>> as per GR, smaller than c/H, where H is Hubble's
>>> constant.  There, everything recedes at speed less than c.
>>
>> Not all closed universes have a Hubble constant. So let's just
>> assume a traveler moving sufficiently fast can circumnavigate
>> the universe.
>>
>>> Now, if the traveler is moving at constant speed close
>>> to c, in a straight line away from earth, he will
>>> forever remain a single inertial frame.
>>
>> Not really, because in such a universe there ARE no inertial
>> frames. Remember "inertial frame" assumes it is of infinite
>> extent and covers the universe, but:
>>   a) a closed universe admits no infinite frames, because
>> the universe itself is of finite extent
>>   b) such a closed universe cannot be flat, and thus cannot
>> be covered by a set of inertial coordinates.
>>
>>> Eventually, he 'circles the universe', and returns to earth.
>>> What does his clock read?
>>
>> Your description is woefully inadequate to determine the answer.
>> So let me discuss a situation that is similar to what you are
>> apparently thinking of, for which an answer is known:
>> Consider a flat universe with spatial topology SxR^2 --
>> that is a cylinder with 3 dimensions. Spacetime has topology
>> SxR^2xR, and just like the usual 2-d
>> cylinder SxR, it admits a flat metric.
>>
>> Consider two twins, one who remains at home, and one who travels
>> around the universe inertially, returning home. In a space-time
>> diagram omitting the two R spatial dimensions, this universe is
>> a cylinder, and we can select coordinates with the time axis
>> running along the axis of the cylinder and the spatial
>> coordinate going around the cylinder. In these coordinates,
>> the stay-at-home twin's trajectory is parallel to the time axis
>> along the cylinder, while the traveling twin's trajectory wraps
>> once around the cylinder ("diagonally", along a helix), leaving
>> and then rejoining the stay-at-home twin's trajectory. For
>> this case, the traveling twin's elapsed proper time between
>> meetings is less than that of the stay-at-home twin.
>
> Interesting.  But if the traveler never experiences any acceleration, he remains in a single inertial frame
> throughout, how do we explain the result?

It is a topological difference.


> Also, in the simpler example I offered, a similar question:
> if the traveler never experiences acceleration, he's in
> free fall the whole time, why isn't he an inertial frame?

Because in your example there ARE no inertial frames.


Tom Roberts

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Inescapable (symmetric) twins paradox Jon Price <jonelwoodprice@gmail.com> - 2015-08-10 13:07 -0700
  Re: Inescapable (symmetric) twins paradox Maciej Woźniak <mlwozniak@wp.pl> - 2015-08-10 22:10 +0200
    Re: Inescapable (symmetric) twins paradox Jon Price <jonelwoodprice@gmail.com> - 2015-08-10 13:36 -0700
      Re: Inescapable (symmetric) twins paradox Maciej Woźniak <mlwozniak@wp.pl> - 2015-08-10 22:54 +0200
  Re: Inescapable (symmetric) twins paradox Tom Roberts <tjroberts137@sbcglobal.net> - 2015-08-10 16:53 -0500
    Re: Inescapable (symmetric) twins paradox Jon Price <jonelwoodprice@gmail.com> - 2015-08-10 18:24 -0700
    Re: Inescapable (symmetric) twins paradox Koobee Wublee <koobee.wublee@gmail.com> - 2015-08-10 23:06 -0700
    Re: Inescapable (symmetric) twins paradox kenseto <setoken@att.net> - 2015-08-11 05:43 -0700
    Re: Inescapable (symmetric) twins paradox RichD <r_delaney2001@yahoo.com> - 2015-09-01 09:32 -0700
      Re: Inescapable (symmetric) twins paradox Tom Roberts <tjroberts137@sbcglobal.net> - 2015-09-02 10:32 -0500
        Re: Inescapable (symmetric) twins paradox RichD <r_delaney2001@yahoo.com> - 2015-09-06 22:01 -0700
          Re: Inescapable (symmetric) twins paradox Tom Roberts <tjroberts137@sbcglobal.net> - 2015-09-07 09:52 -0500
  Re: Inescapable (symmetric) twins paradox Koobee Wublee <koobee.wublee@gmail.com> - 2015-08-10 21:34 -0700
    Re: Inescapable (symmetric) twins paradox rotchm <rotchm@gmail.com> - 2015-08-11 07:39 -0700
      Re: Inescapable (symmetric) twins paradox kenseto <setoken@att.net> - 2015-08-11 08:06 -0700
        Re: Inescapable (symmetric) twins paradox rotchm <rotchm@gmail.com> - 2015-08-11 08:34 -0700
          Re: Inescapable (symmetric) twins paradox kenseto <setoken@att.net> - 2015-08-11 08:47 -0700
            Re: Inescapable (symmetric) twins paradox Odd Bodkin <bodkinodd@gmail.com> - 2015-08-11 11:02 -0500
              Re: Inescapable (symmetric) twins paradox kenseto <setoken@att.net> - 2015-08-11 14:49 -0700
                Re: Inescapable (symmetric) twins paradox Gary Harnagel <hitlong@yahoo.com> - 2015-08-11 15:11 -0700
                Re: Inescapable (symmetric) twins paradox Odd Bodkin <bodkinodd@gmail.com> - 2015-08-11 17:16 -0500
                Re: Inescapable (symmetric) twins paradox kenseto <setoken@att.net> - 2015-08-12 07:11 -0700
                Re: Inescapable (symmetric) twins paradox Odd Bodkin <bodkinodd@gmail.com> - 2015-08-12 09:53 -0500
                Re: Inescapable (symmetric) twins paradox kenseto <setoken@att.net> - 2015-08-12 09:09 -0700
                Re: Inescapable (symmetric) twins paradox Odd Bodkin <bodkinodd@gmail.com> - 2015-08-12 11:29 -0500
                Re: Inescapable (symmetric) twins paradox kenseto <setoken@att.net> - 2015-08-12 13:09 -0700
                Re: Inescapable (symmetric) twins paradox Odd Bodkin <bodkinodd@gmail.com> - 2015-08-12 15:39 -0500
                Re: Inescapable (symmetric) twins paradox kenseto <setoken@att.net> - 2015-08-12 14:47 -0700
                Re: Inescapable (symmetric) twins paradox Odd Bodkin <bodkinodd@gmail.com> - 2015-08-12 16:52 -0500

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