On Jan 30, 10:50*am, "Dirk Van de moortel" dirkvandemoor...@ThankS-NO-
SperM.hotmail.com wrote:
"snapdragon31" wrote in ...

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- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

Hi Dirk Vdm,

Do you know relativity? I am not interested in discussing relativity
with someone who does not even know what the length contraction and
time dilation equations are.
The formula of the length contraction can be found at
http://en.wikipedia.org/wiki/Length_contraction
The equation for the time dilation can be found at
http://en.wikipedia.org/wiki/Time_dilation

The equation "Distance = Speed * Time" was taught in grade school. I
don't expect that you do not know that. Distance is a scaler
quantity. Distance is always positive. Please don't add a negative
sign to make it negative.

"snapdragon31" wrote in message
...
Do you know relativity? I am not interested in discussing relativity
with someone who does not even know what the length contraction and
time dilation equations are.
The formula of the length contraction can be found at
http://en.wikipedia.org/wiki/Length_contraction
The equation for the time dilation can be found at
http://en.wikipedia.org/wiki/Time_dilation

The equation "Distance = Speed * Time" was taught in grade school. I
don't expect that you do not know that. Distance is a scaler
quantity. Distance is always positive. Please don't add a negative
sign to make it negative.

I think the problem was that you did not clearly define what your terms, x,
t etc meant. x is generally meant taken to mean a position in space along
the x-axis .. not a distance. using 'dx' or 's' might have been good idea
(dx meaning a difference between x values, and s a commonly used term for
displacement). Or maybe 'd' for distance (if you really do want
non-negative values).

Regardless, your equations for calculating the distances and times were
correct (to the accuracy you were using) in both cases .. but your
assumption that they should result in the same values was incorrect, as you
were calculating two quite different values (but confusingly using the same
letter 'x' in both cases).

On Jan 30, 4:55*am, "Jeckyl" wrote:
"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?- Hide quoted text -

- Show quoted text -

Both case refer to the same journey - the first leg of the journey of
M from Earth to a star 10c away.
S would come up with the following information regarding the outbound
journey of M:
|v| = 0.995c (measured)
x = 10c (measured)
t = 10.05 years (calculated)

Applying the time dilation equation to the above information, we can
calculate what twin M would measure.
|v| = 0.995c (Same as above)
t' = t / gamma = 1.005 years (Calculated by applying time dilation
eq. as M is moving w.r.t. S)
x' = |v| * t' = c (Calculated)

So long as the calculated t' = the measured t', time dilation equation
is okay. We assume that the calculated t' = the measured t' here.

Twin M did not know that he was the moving twin. He would apply the
time dilation equation to his observed data to predict what the twin S
would measure.
|v| = 0.995c (Same as above)
t = t' / gamma = 0.1005 years (Calculated by applying time
dilation eq. as S is moving w.r.t. M)
x = |v| * t = 0.1c (Calculated)

At the very beginning, we know that the value of x measured by S is
10c so the predicted value of 0.1c must be wrong. Lorentz
transformation does not work in this case.


Now consider the twin S is on the Star. S finds that M is moving
towards him at a speed of 0.995c and M also finds that S is moving
toward him at 0.995c. The condition is symmetrical between S and M.
The distance between them measured by S is 10c. That is the distance
between the Earth and the star.

After M reaches the star. The distance between Earth and the star can
be calculated
as x' = |v| * t'
If x' = c then M is the moving twin.
If x' = 10c then M and S are moving towards each other at the same
speed.
If x' = 100c then S is the moving twin.

On Jan 29, 5:14 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?

It does not have to use the GPS to point out the absolute stupidity in
the Lorentz transform as it manifests the twin's paradox through the
very combination of the time dilation and the principle of relativity.

Say two piles of identical radio-active materials each with a half-
life of 500uSec orbit each other at 300m/Sec of orbiting speed. After
1 billion years according to each pile, the time dilation as observed
by any observer comoving with each pile of this toxic material becomes
500uSec. All of a sudden, each pile goes through the same but low
acceleration and brings each other to rest. How much of the original
radio-active material is left according to each pile?

CHECKMATE by Dr. Wilson

Ahahaha... The Lorentz transform is dead. SR and LET as independent
interpretations to the Lorentz transform should also be tossed into
the trashcan. Would the Einstein Dingleberries drop their 'burden'
then?

On Jan 30, 10:37 pm, snapdragon31 wrote:

[...]

The simple Lorentz transform makes the explicit assumption that the
frame being transformed to is traveling at constant velocity v.

Now, what do you call a change of velocity from v --- -v ? I call it
acceleration.

The acceleration breaks the symmetry that you are expecting.

On Wed, 30 Jan 2008 23:12:55 +0100, "Dirk Van de moortel"
wrote:


"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.

Well you should also inform Paul Andersen since he seems to agree with Jeckyl.

The GO sees the OC running fast by 12ms/y. The OO sees the GC running SLOW by
19ms/y.

When the clocks are reunited after 1 year, what is the difference between
their readings?


Dirk Vdm



Henri Wilson. ASTC,BSc,DSc(T)

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

On Wed, 30 Jan 2008 19:36:23 -0800 (PST), Eric Gisse wrote:

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.

Hahahahahahhahaha!

the ultimate SRian escape route. "IT ISN'T INERTIAL''

hahahahahaha!




Henri Wilson. ASTC,BSc,DSc(T)

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

"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

Correct.


| (though both have been observed).

That is actually funny since you cannot say by whom, when
and under what circumstances.


| In Newtonian
| math the frequency changes as per the Doppler, but the
| wavelength is unaltered.

Incorrect.

|
|
| 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, 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.

Assertion carries no weight.

| | 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.

Then you'll have to measure the wavelength as well.
It still will not change the fact that the time for a satellite signal
to reach a ground receiver is different to the time for the phone
signal to answer the satellite because the satellite moved.

|
| |
| | 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,

Ah, now you are being sensible by reversing c.

| as one can readily
| work out using Galilean relativity; remember that the beam
| reflects off the car at c-v,

Wrong again...
The car's velocity is still v, not -v, relative to the light source.
You really should get these very basic ideas straight in your mind.

On Thu, 31 Jan 2008 00:35:23 -0800 (PST), Koobee Wublee
wrote:

On Jan 29, 5:14 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?

It does not have to use the GPS to point out the absolute stupidity in
the Lorentz transform as it manifests the twin's paradox through the
very combination of the time dilation and the principle of relativity.

Say two piles of identical radio-active materials each with a half-
life of 500uSec orbit each other at 300m/Sec of orbiting speed. After
1 billion years according to each pile, the time dilation as observed
by any observer comoving with each pile of this toxic material becomes
500uSec. All of a sudden, each pile goes through the same but low
acceleration and brings each other to rest. How much of the original
radio-active material is left according to each pile?

CHECKMATE by Dr. Wilson

Ahahaha... The Lorentz transform is dead. SR and LET as independent
interpretations to the Lorentz transform should also be tossed into
the trashcan. Would the Einstein Dingleberries drop their 'burden'
then?

I think the problem lies in the fact that students are taught Einstein's
relativity by someone they respect and trust and feel naturally obliged to
accept it. Like all religions, some of the arguments seem plausible to
innocent eyes and are hard to reject, particularly when one knows one will
probably fail one's exams if one does.

Some of the brighter students like you and I, smell a rat right from the
beginning.... and when finally given the freedom to investigate further,
discover in no time at all that the whole theory is baseless, totally unproven
and indeed absolute crap from start to finish..

Now another sacred relativist cow has been slaughtered....the fictitious GPS
'GR correction'.



Henri Wilson. ASTC,BSc,DSc(T)

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