"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.

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

You still didn't get it right. What valid (and handy) coordinate system do
you think that observer (OO) would use?

Harald

"snapdragon31" wrote in message ...

[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.

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

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

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

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

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.


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 29, 10:40 pm, "Jeckyl" wrote:
"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?
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.

It was an anlogy only .. derr .. to illustrate that taking different paths
in space gives you different elapsed distances .. and that similarly
different paths in space time can give you different elapsed times. And
there is no such thing as 'true distance' in any case.

According to Newton's law, both odometer readings are right.

Just as in SR, both clocks are right in the so-called twins paradox. They
are simply measuring different quantities.

The GPS clock paradox is a variation of the twin paradox, so no valid
solution.

Why not .. the so-called twins paradox is well explained by relativity by
a
number of methods (all giving the same results) .. why do you think there
is
no 'solution'? Why do you even think there is something there that needs
solving?- Hide quoted text -

- Show quoted text -

Yes, there are tons of solutions to the twin paradox

There is nothing to solve

but none of them is a valid solution.

They all predict the same result. Why do you think they are all invalid ..
is it because they all give the same reults which is different to what you'd
like it to be?

Let me show you why it is a logical problem that has no solution.

It isn't a problem .. its a statement of what would happen.

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

What distances and times are you measuring?

The information we have is:
1. v - velocity of moving twin.
2. x = v * t
3. x' = v * t'

OK .. so we area assuming here that x and t are the distance and duration
the stationary twin S sees the moving twin M travel on the first leg of the
journey, and that corresponds to a distance of x' and t' that the moving
twin M seems the stationary twin S travel.

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)

I hate equations.

I wonder why?

Let me convert them into numbers.
Let v = 0.995c and x = 10c both v and x can be measured accurately
1/gamma = sqrt(1 - 0.995^2) = 0.1

close enough .. and gamma = 10.013

From the point of view of twin S.
Eq 2. t = x / v = 10c / 0.995c = 10.05 years
Eq 4. x' = x * sqrt(1 - 0.995^2) = x * 0.1 = c
Eq 5. t' = t * 0.1 = 1.005 years (Calculated)
x' = c and
v * t' = 0.995c * 10.05 = c
Eq 3. x' = v * t' = c

Sounds fine so far

Using Lorentz transformation the calculated time twin M used is 1.005
years.
So far so good if the time measured by twin M is 1.005 years for the
whole journey.

The first leg of it .. ie the trip away from S .. yes

So I assume now you are going to look at it all in reverse

1'. v - velocity of moving twin (S).
2'. x' = v * t'
3'. x = v * t
4'. x = x' * sqrt(1 - v^2/c^2)
5'. t = t' * sqrt(1 - v^2/c^2)
Assuming that the measured time is the same as the calculated time =
1.005 year.

What is the difference between 'measured' and 'caclulated' as you see it?

From twin M's point of view:
v = 0.995c Velocity of twin S and it can be measured by accurately
t' = 1.005 years (measured)
Eq 2'. x' = 0.995c * 1.005 = c
Eq 4'. x = x' * 0.1 = c * 0.1 = 0.1c (Calculated)
Eq 5'. t = t1 * 0.1 = 0.1005 year
x = 0.1c and v * t = 0.995c * 0.1005 = 0.1c
Eq 3'. x = v * t = 0.1c

So far there is still no problem as long as the calculated x and the
measured x are the same. Unfortunately, the calculated x = 0.1c and
the measured x = 10c.
As a conclusion, Lorentz transformation is not valid at least in one
situation.

As explained to you before .. you have used the same name x here for two
different values. You are confusing yourself as a result.

In the first case x is how far M travels in time t
In the second case x is how far M travels in time t/gamma

How could you expect them to be equal?

"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

Dirk Vdm

"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.

Dirk Vdm

"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 !!

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

"snapdragon31" wrote in message
...

[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

No .. x = vt is valid for all times t. It is the position of object M as
measured from S at time t

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.

Yes .. as I think I noted, he got some signs wrong .. not a fatal error
though in this case .. we can pretend he is talking about speeds

4. x' = x * sqrt(1 - v^2/c^2)
Only valid for events satifying
t' = 0
Exercise: Why?

No .. it is valid for all times t .. at time t, S sees M at x=vt, then at
the corresponding time t' for M, M sees S at x' = -x * sqrt(1 - v^2/c^2)

5. t' = t * sqrt(1 - v^2/c^2)
Only valid for events satisfying
x' = 0
Exercise: Why?


No .. by the same logic as above

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?

No .. it is not

On Jan 30, 12:28 am, snapdragon31 wrote:
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?

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.

They both represent the distance traveled according
to both Newton and GR.

- Randy

On Jan 30, 12:40 am, Koobee Wublee wrote:
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.

Well, "impossible to resolve" except for the simple
resolution: That the two twins follow different world lines,
and thus the total elapsed time along those lines
is different.

Exactly analogous to my two odometers.

- Randy

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.

There's no elegant method by which to verify this postulate
experimentally, as the source cannot directly observe the
light returning without the light returning, which takes
the rest of the time, and the moving mirror cannot observe
the source emission at all, until t_AB is already past.
Therefore, this is an assumption, albeit a very reasonable
one.

Besides, as Ockham should well know by now, if the light
goes c+v in one direction and c-v in the other, 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).

Also, various other measurable effects are well-documented.
For example, SR postulates changes in wavelength and frequency;
Newton merely postulates changes in frequency.

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


'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. 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.

--
#191,
Useless C/C++ Programming Idea #11823822:
signal(SIGKILL, catchkill);

--
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