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.


|
| 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.



|
| 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, as one can readily
work out using Galilean relativity; remember that the beam
reflects off the car at c-v, as the car is moving away at
velocity v. The reflected beam is moving at speed v-c (i.e.,
negative, away from the car); one has to add v back to get
back into the radar guns' reference frame, so the beam
should be coming back at speed 2v-c.

Except that it doesn't anyway.


| the average
| speed thereby is less than c because of a variant of the
| "headwind/tailwind" effect; the MMX was designed to measure
| that effect (and failed to show any variance).

Non sequitur, GPS doesn't use average speed.

Ah yes. For once you are entirely correct; the original context
was a GPS problem.


| Also, various other measurable effects are well-documented.

Non sequitur, Ptolemy's epicycles are well-documented (and wrong).

Also correct, though I'm not sure where these come into the picture.



| For example, SR postulates changes in wavelength and frequency;
| Newton merely postulates changes in frequency.

Nope. Newton postulates NO change in frequency.
c1 = lambda1 * f
c2 = lambda2 * f

x' = x-vt
t' = t

Car moves at v; in A's frame, beam moves at c.

(0,t)_A = (ct1,t+t1)_A = (ct1-v(t+t1), t+t1)_B

Since the beam reflects off the car ct1-v(t+t1) = 0,
or t = (c-v)t1/v, or t1 = vt/(c-v). t+t1 = ct/(c-v).

At this point, the beam reflects, incoming at velocity c-v,
outgoing at velocity v-c, back towards the gun. Therefo

(0, ct/(c-v))_B = ((v-c)t2, ct/(c-v)+t2)_B
= ((v-c)t2+v(ct/(c-v)+t2), ct/(c-v)+t2)_A

Since we want to see when the beam comes back, we equate
(v-c)t2+v(ct/(c-v)+t2) = 0
(v-c)^2t2+v(-ct+t2(v-c)) = 0
(v-c)^2t2+t2(v-c)-cvt = 0
((v-c)^2+(v-c))t2 = cvt
t2 = cvt/((v-c)^2+(v-c))
= cvt/((c-v)^2-(c-v))
= cvt/( (c-v)(c-v-1) )

Since cv/( (c-v)(c-v-1) ) != 1, there is a frequency change
in Galilean mathematics. What is invariant is the *wavelength*,
or lambda1 = lambda2.

(In SR, of course, both change.)




| It is all paradoxical, to be sure -- but there's no real contradiction.

Assertion carries no weight.


Correct. Of course, you do have evidence of an invariant wavelength
from a moving source, right?



|
|
| 'the "time" required by light to travel from A to B equals
| the "time" it requires to travel from B to A' -- Albert Einstein
|
| The time for a signal to get from the satellite to the receiver
| does not equal the time for an uplink because the satellite has
| moved, obviously.
|
| Indeed .. SR and Einstein agrees with that.

Nope. In SR the uplink time is the same as the signal time.


| Time from A to B for light is
| only the same as the time from B to A when A and B both at rest in some
| frame of reference (ie they are not moving relative to each other)
|
|
| B does not have to be at rest. Of course the actual time
| at which the ray of light impacts B (and the position
| of B at the point of impact) might be a little hard to
| specify unless one has an alternate "infinite speed"
| particle, which is currently (and probably forever will
| be) impossible.

You have elegant way of verifying it, but there are many
elegant ways to falsify it, of which doppler is the easiest.
Bigotry is not elegant.


--
#191,
GNU and improved.

--
Posted via a free Usenet account from http://www.teranews.com