"Randy Poe" wrote in message
...
| On Jan 30, 1:54 pm, "Ockham" wrote:
| "Randy Poe" wrote in message
|
|
...
| | On Jan 30, 11:27 am, "Ockham" wrote:
| | "The Ghost In The Machine" wrote in
| ...
| | | In sci.physics.relativity, Jeckyl
| | |
| | | wrote
| | | on Wed, 30 Jan 2008 21:28:28 +1100
| | | :
| | | "Ockham" wrote in message
| | | . uk...
| | |
| | | "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.
| |
| | Yes, both wavelength and frequency experience a doppler shift.
|
| Nope.
|
| Yep. Measurably.

After replacing the head on your broom 5 times because it broke off,
now the handle's broken off.
Assertion carries no weight.

|
| |
| | 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?
|
| It was explained to you using a standing wave. That you don't
| understand Galilean relativity is your problem.
|

Quite simply if a source emits 10 cycles in a second at 10 fps and
10 cycles are received in two seconds at 5 fps that doesn't change
the count, it only changes the relative velocity.

Henri's way:- change frequency
Transmit: wavelength = 1 foot, frequency 10 Hz for one second,
speed relative to source = 10 fps.
Receive: wavelength = 1 foot, frequency 5 Hz for two seconds
speed = 5 fps, 10 cycles received.

The right way:- change wavelength
Transmit: wavelength = 1 foot, frequency 10 Hz for one second,
speed relative to source = 10 fps.
Receive: wavelength = 6", 10 cycles for two seconds
speed = 5 fps, 10 cycles received.

Your way:- change both wavelength and frequency
Transmit: wavelength = 1 foot, frequency 10 Hz for one second,
speed relative to source = 10 fps.
Receive: wavelength = 2 foot, frequency 5 Hz for one second,
speed relative to receiver = 10 fps, 5 cycles received.

That you think the count should be changed when it doesn't match
your theory is your problem.
Henri calls your problem "tick fairies".

That you don't understand Galilean relativity even at velocities
as low as 10 feet per second is your problem, you'll always
get the same relative velocity for both source and receiver.