"The Ghost In The Machine" wrote in message
...
| In sci.physics.relativity, Ockham
|
| wrote
| on Wed, 30 Jan 2008 16:27:43 GMT
| :
|
| "The Ghost In The Machine" wrote in
message
| ...
| | In sci.physics.relativity, Jeckyl
| |
| | wrote
| | on Wed, 30 Jan 2008 21:28:28 +1100
| | :
| | "Ockham" wrote in message
| | k...
| |
| | "snapdragon31" wrote in message
| |
|
...
| | On Jan 29, 8:54 pm, Randy Poe 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.
| That's how doppler radar works.
| Since c1 = lamba1 * f outbound and c2 = lambda2 * f inbound
| it follows that c1 c2.
|
| Except that lambda1 lambda2 as well.

Assertion carries no weight.

| | There's no elegant method by which to verify this postulate
| | experimentally,
|
| Doppler radar is very elegant. It falsifies the postulate which is why
| there is no elegant way to verify it.
|
| It falsifies nothing, as it doesn't measure the wavelength,
| merely the frequency, though a heterodyning circuit.

Then you'll have to measure the wavelength as well.
It still will not change the fact that the time for a satellite signal
to reach a ground receiver is different to the time for the phone
signal to answer the satellite because the satellite moved.

|
| |
| | Besides, as Ockham should well know by now, if the light
| | goes c+v in one direction and c-v in the other,
|
| That's just plain silly, the car doesn't change direction.
| the radar goes at 0+c leaving the gun and returns at v-c.
|
| Actually, the beam returns at speed 2v-c,

Ah, now you are being sensible by reversing c.

| as one can readily
| work out using Galilean relativity; remember that the beam
| reflects off the car at c-v,

Wrong again...
The car's velocity is still v, not -v, relative to the light source.
You really should get these very basic ideas straight in your mind.