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.

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. :-)

Yesterday I wrote:
| I bet you will forget that you yet again have admitted
| that no tick fairies are required, and invoke them yet again.
| Tomorrow maybe?
| You have a short memory.

Wow. I can predict the future!

I see you have forgotten this as well:
It is the very same scenario as yours above.
(only per orbit in stead of per day.
One orbit = half a sidereal day.)

See this conversation from October 2004:

Paul B. Andersen wrote:
| Consider the following scenario.
| We have two identical clocks on the ground, both running
| at the frequency 1MHz. Let's call each cycle a "tick".
| One of the clocks is launched into GPS orbit.
| Let N be the number of ticks counted by the ground
| clock during one orbit of the orbiting clock, as observed
| from the ground.
|
| It is then experimentally verified that the orbiting clock
| will tick out N+19 ticks during one orbit.
|
| - The ground clock can observe that the orbiting clock
| ticks out N+19 ticks while itself is ticking out N ticks.
| - The orbiting clock can observe that the ground clock
| ticks out N ticks while itself is ticking out N+19 ticks.
| - The ground clock will measure one orbit to last N us.
| - The orbiting clock will measure one orbit to last N+19 us.
| - The ground clock will measure the average frequency
| of the orbiting clock to be (N+19)/N MHz.
| - The orbiting clock will measure the average frequency
| of the ground clock to be N/(N+19) MHz.
|
| Can you please point out what your 'tick fairies' are
| supposed to do in this scenario?
|
| Are there any missing ticks?
| Are there any ticks to many?
|
| I expect another miserable failure by Henri Wilson.

Henri Wilson wrote:
| No Paul you are actually learning and have eliminated the fairies....

Paul B. Andersen wrote:
| This is the very same scenario as always.
| I never saw any fairies in it.
| YOU were the one that asserted their presence.
|
| Am I now to understand that you are retracting that claim?
| Are you admitting that there were no need for any fairies?
| So when you are referring to my "tick fairies", that was
| actually YOUR tick fairies which you now have killed?
|
| Fine. It's settled then. Henri Wilson have given up.
| There never was any need for 'tick fairies.'

Henri Wilson wrote:
| I never DID believe in fairies.

Paul B. Andersen wrote:
| The point is that you have claimed that the scenario above
| is impossible without those fairies.
|
| You have referred to "Paul's tick faires" enumerable times, Henri.
| The statement that made me respond was:
| Henri Wilson wrote:
| Jesus christ, I do believe Tom has just discovered
| Paul Andersen's famous tick fairies.
|
| But now you have admitted that the scenario above is
| perfectly possible without any fairies.
|
| So will you remember this for the future, and stop referring
| 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?

--
Paul

http://home.c2i.net/pb_andersen/

"Jeckyl" wrote in message ...
"Dirk Van de moortel" wrote
in message ...

"Jeckyl" wrote in message
...
"snapdragon31" wrote in message
...

[snip]

4. x' = x * sqrt(1 - v^2/c^2)
5. t' = t * sqrt(1 - v^2/c^2)

Lets just double check that .. LT tells us that (assuming we have that at
t=t'=0 x=x'=0, the corresponding event for (x,t) for S is (x',t') for M,
where
x' = gamma(x - vt)
t' = gamma(t - xv/c^2)
where gamma = 1/sqrt(1-v^2/c^2)
So when M (as seen from S) has gone a distance x = vt, M will be at
(x1',t1') in its own frame of reference
x1' = gamma(vt - vt)
x1' = 0 (as expected .. M has not moved relative to itself)
t1' = gamma(t - vtv/c^2)
t1' = gamma(t(1 - v^2/c^2))
t1' = gamma(t(1 - v^2/c^2))
t1' = t/gamma
and at that time, M will see S as being at (x2',t2')
x2' = -vt1'
x2' = -vt/gamma
x2' = -x/gamma
So that's all fine (other than getting the sign correct there)

All fine?
Yes all fine, provided x' = t' = x = t = 0

No .. all fine for all values where x = vt (ie for all spacetime events
along the path of M). Lorentz transforms do work for values other than 0
you know !!

You wrote:
| "...assuming we have that at
| t=t'=0 x=x'=0, the corresponding event for (x,t) for S is (x',t') for M,
| where
| x' = gamma(x - v t)
| t' = gamma(t - x v / c^2)"

This is equivalent with
x = gamma(x' + v t')
t = gamma(t' + x v'/c^2)
so, the minute you write
x' = x * sqrt(1 - v^2/c^2)
in this context, you also write
t' = 0
and the minute you write
t' = t * sqrt(1 - v^2/c^2)
in this context, you also write
x' = 0 .

So the combined equations
x' = x * sqrt(1 - v^2/c^2)
t' = t * sqrt(1 - v^2/c^2)
are only valid if
x' = t' = x = t = 0,
in other words for only one event.

Dirk Vdm

"Dirk Van de moortel" wrote in message ...

[snip]

This is equivalent with
x = gamma(x' + v t')
t = gamma(t' + x v'/c^2)

typo:
t = gamma(t' + x' v/c^2)

Dirk Vdm

"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.

|
| 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.


|
| 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.

| 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.

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

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


| 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



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

Assertion carries no weight.



|
|
| '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.

On Jan 30, 4:40*am, "Dirk Van de moortel" dirkvandemoor...@ThankS-NO-
SperM.hotmail.com wrote:
"snapdragon31" wrote in ...

[snip]

Yes, there are tons of solutions to the twin paradox but none of them
is a valid solution.
Let me show you why it is a logical problem that has no solution.
Assuming that Lorentz transformation can predict the time and distance
of the other frame.
Let v = velocity of the moving twin M
x = distance measured by stationary twin S
t = time measured by twin S
x' = distance measured by twin M
t' = time measured by twin M

The information we have is:
1. v - velocity of moving twin.
2. x = v * t

This equation is valid for events satisfying
* * * * x' = 0,
expressing the fact that is the velocity of the origin of
the primed system is v w.r.t. the unprimed system

3. x' = v * t'

Expressing the (erroneous) fact that is the velocity of
the origin of the unprimed system is *also* v w.r.t.
the primed system. That is wrong. The velocity is -v
This is your first error.

Please note that x' is defined as a distance measured by twin M not
displacement.
Both x and x' are not vectors.
A more correct equation should be
x' = |v| * t'
distance = speed * time


4. x' = x * sqrt(1 - v^2/c^2)

Only valid for events satifying
* * * * t' = 0
Exercise: Why?

x' = x * sqrt(1 - v^2/c^2) Length contraction


5. t' = t * sqrt(1 - v^2/c^2)

Only valid for events satisfying
* * * * x' = 0
Exercise: Why?


t' = t * sqrt(1 - v^2/c^2) Time dilation

So if you combine equations 4 and 5, you are *talking about
events that satisfy
* * * * x' = 0
* * * * t' = 0
and therefore also
* * * * x = 0
* * * * t = 0
Exercise: Why?

Congratulations.


Congratulation, you figure out that both the Length Contraction and
Time Dilation equations can be true only when x' = 0, x = 0, t' = 0
and t = 0. If they are not valid equations then twin paradox is not a
paradox any more.



I hate equations.

Of course you hate equations.
You understand nothing about these equations.
Before you write an equation, you should understand what
the variables mean.
Physics is not an exercise in algebra.
As I told you before, *that* is your problem.

Dirk Vdm

"snapdragon31" wrote in message ...
On Jan 30, 4:40 am, "Dirk Van de moortel" dirkvandemoor...@ThankS-NO-
SperM.hotmail.com wrote:
"snapdragon31" wrote in ...

[snip]

Yes, there are tons of solutions to the twin paradox but none of them
is a valid solution.
Let me show you why it is a logical problem that has no solution.
Assuming that Lorentz transformation can predict the time and distance
of the other frame.
Let v = velocity of the moving twin M
x = distance measured by stationary twin S
t = time measured by twin S
x' = distance measured by twin M
t' = time measured by twin M

The information we have is:
1. v - velocity of moving twin.
2. x = v * t

This equation is valid for events satisfying
x' = 0,
expressing the fact that is the velocity of the origin of
the primed system is v w.r.t. the unprimed system

3. x' = v * t'

Expressing the (erroneous) fact that is the velocity of
the origin of the unprimed system is *also* v w.r.t.
the primed system. That is wrong. The velocity is -v
This is your first error.

Please note that x' is defined as a distance measured by twin M not
displacement.
Both x and x' are not vectors.
A more correct equation should be
x' = |v| * t'
distance = speed * time

then the other equation should be
x = - |v| * t
so you induce a new error.



4. x' = x * sqrt(1 - v^2/c^2)

Only valid for events satifying
t' = 0
Exercise: Why?

x' = x * sqrt(1 - v^2/c^2) Length contraction

Only valid for measurements of two sides of an object
when measured simultaneously in the primed frame,
expressed by
t' = 0
Check the definition of length contraction.



5. t' = t * sqrt(1 - v^2/c^2)

Only valid for events satisfying
x' = 0
Exercise: Why?


t' = t * sqrt(1 - v^2/c^2) Time dilation

Only valid for measurements of two ticks of a clock
at rest in the primed frame, expressed by
x' = 0
Check the definition of time dilation.


So if you combine equations 4 and 5, you are talking about
events that satisfy
x' = 0
t' = 0
and therefore also
x = 0
t = 0
Exercise: Why?

Congratulations.


Congratulation, you figure out that both the Length Contraction and
Time Dilation equations can be true only when x' = 0, x = 0, t' = 0
and t = 0. If they are not valid equations then twin paradox is not a
paradox any more.

The two are *together* valid if and only if everything
is zero, so when you write
x' = x * sqrt(1 - v^2/c^2) Length contraction
t' = t * sqrt(1 - v^2/c^2) Time dilation,
you actually write
0 = 0
0 = 0 ,
and that is something we all know, thank you.




I hate equations.

Of course you hate equations.
You understand nothing about these equations.
Before you write an equation, you should understand what
the variables mean.
Physics is not an exercise in algebra.
As I told you before, *that* is your problem.

Of course you hate equations.
You understand nothing about these equations.
Before you write an equation, you should understand what
the variables mean.
Physics is not an exercise in algebra.
As I told you before, *that* is your problem.

Dirk Vdm

On Jan 30, 4:06 am, "Dirk Van de moortel" dirkvandemoor...@ThankS-NO-
SperM.hotmail.com wrote:
"Dr. Henri Wilson" HW@.... wrote in messagenews:h9jvp3pll90m0a8q92vum1s2j23sd255go@4ax .com...

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.

If you would understand where this -7 comes from,
you could answer the question you should ask next.



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

No no no, here you should *ask*:
"Accordingly, does an observer (OO) in GPS orbit would
see the GC LOSING 52us per day?
and if you would understand where the -7 comes from,
you could answer it yourself.
It's very simple but wouldn't understand where the -7
comes from.

Dirk Vdm



Don't you EVER tell him. Let him die an idiot (he lives as an idiot
anyways)

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.

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

"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.
|
| 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 is your problem.

"Dono" wrote in message ...
On Jan 30, 4:06 am, "Dirk Van de moortel" dirkvandemoor...@ThankS-NO-
SperM.hotmail.com wrote:
"Dr. Henri Wilson" HW@.... wrote in messagenews:h9jvp3pll90m0a8q92vum1s2j23sd255go@4ax .com...

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.

If you would understand where this -7 comes from,
you could answer the question you should ask next.



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

No no no, here you should *ask*:
"Accordingly, does an observer (OO) in GPS orbit would
see the GC LOSING 52us per day?
and if you would understand where the -7 comes from,
you could answer it yourself.
It's very simple but wouldn't understand where the -7
comes from.

Dirk Vdm



Don't you EVER tell him. Let him die an idiot (he lives as an idiot
anyways)

No worries :-)

Dirk Vdm