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.

|
| 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.

That you think the data should be changed when
it doesn't match your theory is yours.

- Randy

Jeckyl skrev:
"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)

So what do these equations mean?

x' = x * sqrt(1 - v^2/c^2)
---------------------------
x = gamma*(x' + vt')
when t'=0, x = gamma*x' (which is your equation)
That is, it tells us that the event with coordinates (x',0)
in the primed frame has the coordinates (gamma*x',?) in the unprimed frame.
To find the temporal coordinate, you have to use:
t = gamma(t'+vx'/c^2) = gamma*(vx'/c^2)

So: primed frame (x',0) = unprimed frame (gamma*x',gamma*(vx'/c^2))
================================================== =================

t' = t * sqrt(1 - v^2/c^2)
--------------------------
t = gamma*(t' + x'v/c^2)
when x' = 0, t = gamma*t' (which is your equation)
That is, it tells us that the event with primed frame coordinates (0,t')
has the unprimed frame coordinates (?,gamma*t')
To find the spatial coordinate, we must use:
x = gamma*(x' + vt') = gamma*vt'

So: primed frame (0,t') = unprimed frame (gamma*vt',gamma*t')
================================================== =============

Conclusion:
the (x',t'), and the (x,t) in the equations:
x' = x * sqrt(1 - v^2/c^2)
t' = t * sqrt(1 - v^2/c^2)
cannot be the coordinates of an event in the respective frames
unless the event has coordinates (0,0) in both frames.

What did you want to find?
Wasn't it the primed frame coordinates of the event with
unprimed frame coordinates (x,t), where x = vt?

x' = gamma(x - vt) = gamma(vt - vt) = 0
t' = gamma(t - vx/c^2) = gamma(t(1-v^2/c^2)) = t/gamma

So: unprimed frame (vt,t) = primed frame (0,t/gamma)
================================================== =============


Can't you see that this is incompatible with the equations:
x' = x * sqrt(1 - v^2/c^2)
t' = t * sqrt(1 - v^2/c^2)
which yields:
unprimed frame (vt,t) = primed frame (vt/gamma,t/gamma)
which is wrong for all t but zero. And then x' = x = t = 0 as well.


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

Right.
So your "double check" showed that all was not fine,
quite the contrary, the equations you checked proved to be wrong.

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 !!

But the equations you double checked do not work for values other than 0.

Did you possibly miss what Dirk said was not fine?

--
Paul

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

On Wed, 30 Jan 2008 21:24:38 +1100, "Jeckyl" wrote:

"Dr. Henri Wilson" HW@.... wrote in message
.. .
On Wed, 30 Jan 2008 19:29:15 +1100, "Jeckyl" wrote:

"Dr. Henri Wilson" HW@.... wrote in message
...
On Wed, 30 Jan 2008 17:05:59 +1100, "Jeckyl" wrote:

"Koobee Wublee" wrote in message
...
On Jan 29, 7:54 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?

Is Mr. Poe really as blind as Androcles has claimed you to be?

This is a classical case of the twin's paradox if you have not
realized it finally. It is absolutely impossible to resolve because
of the mathematics of the Lorentz transform.

It doesn't need to be 'resolved' as there is no real paradox involved ..
and
it is *due* to the Lorentz transforms .. they predict it. Get it right,
dumbo.

So what's the answer?

There is no question. Other than "can one explain the so-called twin
paradox result using relativity", and one certainly can (and via a number
of
approaches to the problem)

When the clocks are reunited, do they differ by 19ms or 13ms?

In the twins paradox? Once one has the parameters of the experiment, one
can calculate the difference in elapsed time for the two paths. I've seen
it done many times. Haven't you?

In the case of a GPS satellite returning to earth, the calculations are
quite complex and it would not be possible to say without a *lot* of
information what the difference would be .. I wouldn't like to attempt it
myself. That there would be some difference, we already know from other
experiments/observations.

You rally have no idea what you are talking about, do you.

Yes .. I do

The twins paradox is a thought experiment.

Yes .. it is

GPS clocks are real.

Yes .. they are

If the OC runs 19ms/y faster than the GC and the GC runs 13 ms/y slower
than
the OC, what is the difference in clock readings when the OC is brought
back to
Earth?

Way too complex to work out with the limited information you have given, if
you are talking about a difference in total elapsed time. But when it
returns to earth and is placed next to the ground clock, it would still be
running at the same rate as the ground clock (assuming the OC survived
re-entry ok) .. because both clocks have been keeping time corrrectly and
are unchanged, so there is no reason to expect otherwise.

You should enlist the services of the tick fairies again.




Henri Wilson. ASTC,BSc,DSc(T)

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

On Wed, 30 Jan 2008 13:06:38 +0100, "Dirk Van de moortel"
wrote:


"Dr. Henri Wilson" HW@.... wrote in message ...
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.

You should inform Jeckyl then if you know so much about it.

Dirk Vdm



Henri Wilson. ASTC,BSc,DSc(T)

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

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.

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'.

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




| 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?



Henri Wilson. ASTC,BSc,DSc(T)

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

"Dr. Henri Wilson" HW@.... wrote in message ...
On Wed, 30 Jan 2008 13:06:38 +0100, "Dirk Van de moortel"
wrote:


"Dr. Henri Wilson" HW@.... wrote in message ...
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.

You should inform Jeckyl then if you know so much about it.

Unlike you, Jeckyl is learning.
But you obviously are not familiar with that concept, so never mind.

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.

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?

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. 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.

"Ockham" wrote in message
. uk...

"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.

yes .. the frequency and wavelength will be changed .. We know that that
happens experimentally.

That's how doppler radar works.
Since c1 = lamba1 * f outbound and c2 = lambda2 * f inbound
it follows that c1 c2.

No .. it doesn't. Are you really that bad at math?

| There's no elegant method by which to verify this postulate
| experimentally,
Doppler radar is very elegant.

Yes

It falsifies the postulate which is why
there is no elegant way to verify it.

No .. it doesn't .. it just shows there is doppler shift in frequency (as SR
predicts) .. it say nothing about the speed of light.

| 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.

No .. the light goes at c in both directions relative to both observers ..
there is only a change in frequency.

| 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

And surely then the Doppler radar (which sees a difference in frequency)
refutes that position

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

You failed to demonstrate a contradiction.

[snip]
| 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.

No .. doppler does not refute it at all

Bigotry is not elegant.

Who said it was?

"Ockham" wrote in message
. uk...
"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.

But that has been shows experimentally, has it not?

| 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.

That you do not understand physics or mat is yours (and it becomes ours when
we have to deal with cretins like you)

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

"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,

Sorry .. what I mean was that we have when t=0, t'=0 and when x=0 x'=0 .. my
apologies for not expressing that correct, and thanks for spotting my
mistake in expressing that

But the rest of the analysis should be correct.