Re: GPS CLOCK PARADOX

On Jan 30, 11:27 am, "Ockham" <m...@xxxxxxxxxxxxxxxxxx> wrote:
"The Ghost In The Machine" <ew...@xxxxxxxxxxxxxxxxxxxxxxx> wrote in messagenews:l2d675-gfp.ln1@xxxxxxxxxxxxxxxxxxxxxxxxxx
| In sci.physics.relativity, Jeckyl
| <no...@xxxxxxxxxxx>
| wrote
| on Wed, 30 Jan 2008 21:28:28 +1100
| <13q0kaft9d71...@xxxxxxxxxxxxxxxxxx>:
| > "Ockham" <m...@xxxxxxxxxxxxxxxxxx> wrote in message
| >news:SAXnj.14454$3m6.7071@xxxxxxxxxxxxxxxxxxxxxxxxxxxx
| >>
| >> "snapdragon31" <snapdrago...@xxxxxxxxx> wrote in message
| >>news:dc482e20-b4df-43ff-a84e-70909a0b93b2@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
| >> On Jan 29, 8:54 pm, Randy Poe <poespam-t...@xxxxxxxxx> wrote:
| >>> On Jan 29, 8:14 pm, HW@....(Dr. Henri Wilson) wrote:
| >>>
| >>> > According to relativists, GPS clocks GAIN 38us per day on the ground
| >>> > clock.
| >>> > That is due to two components, 45us for gravity and -7us for
relative
| >>> > speed.
| >>>
| >>> > Accordingly, an observer (OO) in GPS orbit would see the GC LOSING
52us
| >>> > per
| >>> > day.
| >>>
| >>> > After one year, the OO would calculate that the OC was about 19ms
ahead
| >>> > of the
| >>> > GC.
| >>> > However, the GO would calculate that his GC was only 13ms behind.
| >>>
| >>> > What happens when the clocks are reunited?
| >>> > Who is right?
| >>>
| >>> Two people drive different routes from city A to
| >>> city B. When they are reunited, one odometer reads
| >>> 220 km and the other reads 230 km. Which one is
| >>> right?
| >>>
| >>> - Randy
| >>
| >> | According to relativity, both odometer readings are wrong. They do
| >> | not represent the true distance of the routes travelled because of
the
| >> | length contraction effect.
| >> | According to Newton's law, both odometer readings are right.
| >>
| >> | The GPS clock paradox is a variation of the twin paradox, so no valid
| >> | solution.
| >>
| >> The paradox resides in the third postulate.
| >
| > Androcles .. we've told you .. there is no third postulate
|
| Yes there is; it's not usually expressed as a postulate, but
| it is a simple one:
|
| - If a TWLS be conducted between a source and a moving mirror,
| then the time taken (as observed by the source) of the
| light beam from source to mirror and back to source is
| exactly twice that of the time taken from source to
| mirror. In other words, t_AB = t_BA.

Not true, the reflected beam will be doppler shifted.

Yes, both wavelength and frequency experience a doppler shift.

That's how doppler radar works.
Since c1 = lamba1 * f outbound and c2 = lambda2 * f inbound
it follows that c1 <> c2.

How does that follow without a statement about
how both lambda and f shift?

As it turns out, if lambda2 = p*lambda1, then
f inbound = f outbound/p. As a result,
lambda2 * f_inbound = (lambda1*p)*(f_outbound/p)
= lambda1 * f_outbound.

- Randy
.

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