"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.

On Jan 31, 12:32 am, "Jeckyl" wrote:
"bambuu" wrote in message

...



On Jan 31, 12:05 am, "Jeckyl" wrote:
"bambuu" wrote in message

...

On Jan 30, 3:38 am, "Jeckyl" wrote:
Just as in the so-called twins paradox .. both twins watches are
correct
for
the relevant twin. One twin really is older than the other.

what is tha difference between
really older and just older?

None, of course, I was just emphasising that it is a *real* difference ..
not just some sort of optical illusion, or an error in the ticking of a
clock.
thanks, so if clock is good, he becomes
really older for no reason

And how did you conclude that nonsense from what I said? No wonder you have
problems with physics .. you can't even comprehend written language, let
alone the mathematics and logic involved.

you said clock no error, but he really gets older


In the so-called twins paradox, the travelling twin has experienced less
elapsed time because its path through space-time (which involves changes

how would you know about others experiences

in
frame of reference) is different to that of the stay-at-home twin. One can
say that more simply: the stay at home twin is older than the travelling
twin.

why, entropy goes faster staying home?

"bambuu" wrote in message
...
On Jan 31, 12:32 am, "Jeckyl" wrote:
"bambuu" wrote in message

...



On Jan 31, 12:05 am, "Jeckyl" wrote:
"bambuu" wrote in message

...

On Jan 30, 3:38 am, "Jeckyl" wrote:
Just as in the so-called twins paradox .. both twins watches are
correct
for
the relevant twin. One twin really is older than the other.

what is tha difference between
really older and just older?

None, of course, I was just emphasising that it is a *real* difference
..
not just some sort of optical illusion, or an error in the ticking of
a
clock.
thanks, so if clock is good, he becomes
really older for no reason
And how did you conclude that nonsense from what I said? No wonder you
have
problems with physics .. you can't even comprehend written language, let
alone the mathematics and logic involved.
you said clock no error, but he really gets older

Yes .. he really gets older (the twin that stays at home compared to the
twin the travels).

You really are incredibly slow understanding things

In the so-called twins paradox, the travelling twin has experienced less
elapsed time because its path through space-time (which involves changes
how would you know about others experiences

When they reunit they can have a nice little chat over some coffee .. that
is assuming the stay-at-home twin is still alive .. the travlling twin may
end up having to talk to the stay-at-home-twins great great grandchildren
instead.

in
frame of reference) is different to that of the stay-at-home twin. One
can
say that more simply: the stay at home twin is older than the travelling
twin.
why, entropy goes faster staying home?

Why are you suddenly discussing entropy.

It is very simple .. less time has elapsed for the travelling twin compared
to the stay at home twin, because the travelling twin has taken a different
path through space-time than the stay-at-home twin (from the event where
they separate, to the event where they reunite).

On Jan 30, 1:03 pm, HW@....(Dr. Henri Wilson) wrote:
On Wed, 30 Jan 2008 16:02:45 +0100, "Paul B. Andersen"

wrote:
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.

According to the 'religious jargon' the Schwarzschild metric
GPS clocks should gain 38us per day on the ground clock,
and according to continuous observations during 30+ years,
we know that GPS clocks do indeed gain 38us/day on the ground clock.

No they don't...but that's another matter.

Yes, they do. This is explained in the GPS specifications as well as
the original GPS test satellites which you have been quoted but will
never read.


Accordingly, an observer (OO) in GPS orbit would see the GC LOSING 52us per
day.

I see you are invoking your tick fairies again. :-)

When the ground clock shows X and the orbiting clock shows X+38us,
why do you think an orbiting observer would read that as X-14us and X+38us?
How can anybody disagree about what a clock shows?
What a veird idea. :-)

Well I never did see a plausible explanation or the '-7us/d speed component'.

That's because you are unwilling to understand the explanations given
to you. What you think is irrelevant.


Apparently, according to relativists, the GO is moving wrt the inertial frame
but not the OO.

Actually, according to physicists, neither are in inertial frames.


| to "Paul's tick fairies", please?
|
| Otherwise I will have to remind you about this scenario again.

Henri Wilson wrote:
| OK Paul, I will never refer to PAUL ANDERSEN'S FAMOUS TICK FAIRIES again.......

You are really something, Henry.
Is this scenario really so hard to grasp that
you have to screw it up and invoke your tick fairies
every second day?

Why does the OO not see the GC running 52us'd slow, as Jeckyl so wisely
claimed?

There is no symmetry, idiot. The frames are non-inertial.


Henri Wilson. ASTC,BSc,DSc(T)

www.users.bigpond.com/hewn/index.htm

On Jan 30, 7:29 pm, "Ockham" wrote:
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.

In 1 second.

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

The other 5 come in the next second.

- Randy

In sci.physics.relativity, Jeckyl

wrote
on Thu, 31 Jan 2008 09:28:55 +1100
:
"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.

That follows from the second postulate, that says the speed of light is c ..
as the distances are the same ,the time must be the same. Why do you need
another postulate forthat?

An interesting subpoint, that; I am not certain now. However, it's
clear that three events in A's space occur, and A can only observe
two of them.

These are, of course;

(0,0)_A
(L,L/c)_A
(0,2L/c)_A

for some value L. A cannot observe the second one directly.


Indeed .. SR and Einstein agrees with that. 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.

If B is not at rest, it does raise some issue in determining the distance AB
without some definition of simultaneity.

It's not *that* difficult, though you're right; one can
get tied into semantical knots if one's not careful.
But given that A is motionless and B moves v away from A,
one can simply calculate that L, the distance the beam
travels from A to B, is such that

(0,t0)_A = (L,t0+L/c)_A = (gL-gv(t0+L/c), gL0+gL/c-gvL/c^2)_B

using the Lorentz.

Since B's origin reflects the beam, L can be calculated as follows:

gL-gv(t0+L/c) = 0
L-vt0-vL/c = 0
L(1-v/c) = vt0
L = vt0/(1-v/c)

according to SR. The total time is of course simply 2L/c, as L
was defined in A's reference frame.

But, yes, if A is at rest (ie the
light is returning to the same point in space and so travlling the same
distance) then that is sufficient for saying the time of each leg of the
trip is the same. In Einstiens paper, A and B were described as fixed
points in space.

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.




--
#191,
Useless C++ Programming Idea #23291:
void f(item *p) { if(p != 0) delete p; }

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

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

In sci.physics.relativity, Randy Poe

wrote
on Wed, 30 Jan 2008 10:42:33 -0800 (PST)
:
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.

Only in SR (though both have been observed). In Newtonian
math the frequency changes as per the Doppler, but the
wavelength is unaltered.


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


--
#191,
Windows Vista. Now in nine exciting editions. Try them all!

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

"The Ghost In The Machine" wrote in message
...
In sci.physics.relativity, Jeckyl

wrote
on Thu, 31 Jan 2008 09:28:55 +1100
:
"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.

That follows from the second postulate, that says the speed of light is c
..
as the distances are the same ,the time must be the same. Why do you
need
another postulate forthat?

An interesting subpoint, that; I am not certain now. However, it's
clear that three events in A's space occur, and A can only observe
two of them.

These are, of course;

(0,0)_A
(L,L/c)_A
(0,2L/c)_A

for some value L. A cannot observe the second one directly.

Yeup .. that's the whole point of having a definition of what it means for
clocks to be synchronised at two remote position in an intertial frame.

[snip]
If B is not at rest, it does raise some issue in determining the distance
AB
without some definition of simultaneity.
It's not *that* difficult, though you're right; one can
get tied into semantical knots if one's not careful.
But given that A is motionless and B moves v away from A,
one can simply calculate that L, the distance the beam
travels from A to B, is such that
(0,t0)_A = (L,t0+L/c)_A = (gL-gv(t0+L/c), gL0+gL/c-gvL/c^2)_B
using the Lorentz.

Yes .. but that assumes one already has Lorentz transforms derived or
postulated .. Einstein's definition for synchronisation for remote clocks in
a given frame of reference (ie that clock B shows the 1/2 the time for
A-to-B that clock A shows for A-to-B-to-A) was set up before his derivation
of Lorentz

"The Ghost In The Machine" wrote in message
...
In sci.physics.relativity, Randy Poe

wrote
on Wed, 30 Jan 2008 10:42:33 -0800 (PST)
:
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.

Only in SR (though both have been observed). In Newtonian
math the frequency changes as per the Doppler, but the
wavelength is unaltered.

It depends on who is moving relative to the medium (assuming we are talking
about doppler shifting of waves in a medium) .. the source or the observer.

In ballistic theory, there aren't really any waves in a medium (so how
*does* ballistic theory account for the wave-like behaviour of light?)