Re: The Twin Paradox explained from the moving twin ?
- From: The Ghost In The Machine <ewill@xxxxxxxxxxxxxxxxxxxxxxx>
- Date: Sun, 17 Sep 2006 22:00:09 GMT
In sci.physics.relativity, Sorcerer
<Headmaster@xxxxxxxxxxxxxxxxxx>
wrote
on Sun, 17 Sep 2006 18:41:38 GMT
<CFgPg.5510$2Q4.2286@xxxxxxxxxxxxxxxxxxxxxxxxx>:
"The Ghost In The Machine" <ewill@xxxxxxxxxxxxxxxxxxxxxxx> wrote in message
news:c8d0u3-cjl.ln1@xxxxxxxxxxxxxxxxxxxxxxxxxx
| In sci.physics.relativity, G. L. Bradford
| <glbrad01@xxxxxxxxxxxxx>
| wrote
| on Sun, 17 Sep 2006 08:05:06 -0400
| <HM-dndgFQdDTp5DYnZ2dnUVZ_u2dnZ2d@xxxxxxxxxxxxx>:
| > (Sigh)
| >
| > A traveler is traveling between two stars measured at each end to be
ten
| > light years apart. Any observer on an Earth type planet at star A will
| > observe star B to be zero minus ten years relative to his own atomic
clock
| > time, thus ten light years distant from him. Any observer on an Earth
type
| > planet at star B will observe star A to be zero minus ten years relative
to
| > his own atomic clock time, thus ten light years distant from him.
| >
| > The traveler stands beside the observer on the planet at star A and
| > observes with him star B. They both observe star B to be ten light years
| > away or zero minus ten years in time relative to star A. The traveler
boards
| > his space supership and takes off for star B.
| >
| > It takes the traveler two years to get one light year from star A.
Star A,
|
| [snip rest of very long-winded diatribe for brevity by the Ghost In
| The Machine]
|
| I'm assuming this means v = 0.5 c; this gives a transit time of
| about 40 years objective but only 34.64 subjective,
Not for tachyon that does 27 km in 1/11245 seconds.
OK, lemme put it to you this way.
A proton's mass is 1.673 * 10^-27 kg. At Newtonian lightspeed this
becomes 469 MeV. This means that a 7.00 TeV proton has
15,000 times the amount of energy required to achieve lightspeed.
Since KE = 1/2 m_p v^2, and 469 MeV = 1/2 m_p c^2, it follows that
1/2 m_p v^2 = 15000 * (1/2 m_p c^2)
v^2 = 15000 * c^2
v = 122.5 * c, or about 3.674 * 10^10 m/s.
The circumference is 27 km, give or take. Therefore, the frequency --
1 over the lap time, if you will -- should be about 0.735 MHz.
But it's not specified at 0.735 MHz. Why is that?
[rest snipped by The Ghost In The Machine]
--
#191, ewill3@xxxxxxxxxxxxx
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