On Mon, 27 Nov 2006 12:58:24 +0100, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Sun, 26 Nov 2006 21:28:53 +0100, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Thu, 23 Nov 2006 15:55:09 +0100, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Wed, 22 Nov 2006 10:45:34 +0100, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Tue, 21 Nov 2006 13:40:24 +0100, "Paul B. Andersen"
wrote:

Henri Wilson wrote: [about the H&K experiment]
You don't have to go to all this trouble.
As soon as the planes carrying the clocks are in the air, just stop the Earth
from rotating.
Somehow, relativists will claim that this action will affect the now remote
clocks. They will now read the same time when reunited.
I would like your answer the question at the end
of this posting.

Two supersonic planes are flying in opposite directions
around the Earth along equator. The ground speed is
one Earth circumference per sidereal day (ca. 1670 km/h, MACH 1.36).
Each plane is carrying a ring laser gyro detecting
the rotation around an axis parallel to the Earth axis
(the pitch axis).
The planes are starting at a point at the Earth, and
returning to the same point one sidereal day later.

Case 1:
The experiment is performed on the rotating Earth.
How many rotations are measured by the gyros in
the East- and West going plane respectively?
(The obvious answer is: 2 and 0)

Case 2:
The experiment is performed on a non-rotating Earth.
How many rotations are measured by the gyros in
the East- and West going plane respectively?
(The obvious answer is: 1 and 1)

Now the question I would like you to answer:
Do you claim that the rotation of the Earth
somehow will affect remote ring laser gyros?
THIS HAS NOUGHT TO DO WITH THE QUESTION.

We are talking about clock rates.

You people claim the two clocks will run at different rates and will read
differently when reunited.

I'm telling you that if the Earth stops rotating as soon as the clocks are in
the air, their relative rates cannot possibly be affected and they should still
read differently when reunited (according to your silly theory).

But, since their whole trip is carried out when the Earth is NOT rotating, they
obviously should NOT read differently when reunited.

Even YOU should know that observer behavior cannot affect an observed object.
So considering that observer behaviour cannot affect
an observed object, do you claim that the rotation of
the Earth somehow will affect remote ring laser gyros?
No answer, Henri?

We're talking about clocks, not ring gyros.
Doesn't matter. Same problem.
They are both instruments inside the planes.

So we could make the question more general:
Do you claim that the rotation of the Earth somehow will
affect instruments inside the plane?

How is it that the H&K clocks would miraculously know how to change rates if
the (remote) Earth suddenly stopped rotating?
How is it that the gyros in my thought experiment would miraculously
know how to change the rotation if the (remote) Earth suddenly
stopped rotating?

The answer should be obvious, Henri.
And the answer is the same whether we are talking
about clocks or gyros.

I will give you a hint:
What is it the gyros measure?
Why are the results different in case 1 and 2?
What is the relevance of the Earth?
Paul, as usual you run away from the real topic and stall for time by talking
about something entirely irrelevant.

IF THE EARTH STOPS ROTATING AS SOON AS THE PLANES ARE IN THE AIR, HOW DO THE
CLOCKS 'KNOW' HOW TO ADJUST THEIR RATES SO THEY WILL NOW READ THE SAME WHEN
REUNITED?

THIS SHOWS HOW NOINSENSICAL THE H&K REALLY IS.

WOULDN'T YOU AGREE?
IF THE EARTH STOPS ROTATING AS SOON AS THE PLANES ARE IN THE AIR, HOW DO THE
GYROS 'KNOW' HOW TO ADJUST THEIR MEASUREMENTS SO THEY WILL NOW READ THE SAME WHEN
REUNITED?

Why don't you answer?

Because the question is completely irrelevant.

Is it because you can't, or is it because you don't
like the answer you would have to give?

I will answer you.

The gyros will tell each plane when it has done a complete 360 turn.
It makes no difference whether the Earth stops spinning or not.

Of course. That is indeed the obvious answer.
The gyros and the clocks give different results in
the two scenarios, because the _planes_ move in two very
different ways in the two scenarios.
As you so correctly point out, it does not matter if the Earth
is spinning or not.

So the answers to your question:
" IF THE EARTH STOPS ROTATING AS SOON AS THE PLANES ARE IN THE AIR, HOW DO THE
CLOCKS 'KNOW' HOW TO ADJUST THEIR RATES SO THEY WILL NOW READ THE SAME WHEN
REUNITED?"
and my question:
"IF THE EARTH STOPS ROTATING AS SOON AS THE PLANES ARE IN THE AIR, HOW DO THE
GYROS 'KNOW' HOW TO ADJUST THEIR MEASUREMENTS SO THEY WILL NOW READ THE SAME WHEN
REUNITED?"
are exactly the same:

The clocks or the gyros don't have to know anything about
the rotation of the Earth because it does not affect them
in any way, it is the pathes of the planes through space and time
that determine what the instruments will show at the end of their journey.

In case 1, one plane is stationary in a non rotating frame,
while the other plane is moving twice around a circle at the speed 3340 km/h.
In case 2, both planes are moving in opposite directions once around a circle
at the same speed 1670 km/h.

"It makes no difference whether the Earth stops spinning or not."

Remember that for the future.

Case closed.

The case is indeed closed.
You have agreed that the H&K MUST give a null result.

Thank you Paul.


Paul


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

Thank christ there is one genuine physicist on the NG.

Henri Wilson wrote:
On Mon, 27 Nov 2006 12:58:24 +0100, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Sun, 26 Nov 2006 21:28:53 +0100, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Thu, 23 Nov 2006 15:55:09 +0100, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Wed, 22 Nov 2006 10:45:34 +0100, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Tue, 21 Nov 2006 13:40:24 +0100, "Paul B. Andersen"
wrote:

Henri Wilson wrote: [about the H&K experiment]
You don't have to go to all this trouble.
As soon as the planes carrying the clocks are in the air, just stop the Earth
from rotating.
Somehow, relativists will claim that this action will affect the now remote
clocks. They will now read the same time when reunited.
I would like your answer the question at the end
of this posting.

Two supersonic planes are flying in opposite directions
around the Earth along equator. The ground speed is
one Earth circumference per sidereal day (ca. 1670 km/h, MACH 1.36).
Each plane is carrying a ring laser gyro detecting
the rotation around an axis parallel to the Earth axis
(the pitch axis).
The planes are starting at a point at the Earth, and
returning to the same point one sidereal day later.

Case 1:
The experiment is performed on the rotating Earth.
How many rotations are measured by the gyros in
the East- and West going plane respectively?
(The obvious answer is: 2 and 0)

Case 2:
The experiment is performed on a non-rotating Earth.
How many rotations are measured by the gyros in
the East- and West going plane respectively?
(The obvious answer is: 1 and 1)

Now the question I would like you to answer:
Do you claim that the rotation of the Earth
somehow will affect remote ring laser gyros?
THIS HAS NOUGHT TO DO WITH THE QUESTION.

We are talking about clock rates.

You people claim the two clocks will run at different rates and will read
differently when reunited.

I'm telling you that if the Earth stops rotating as soon as the clocks are in
the air, their relative rates cannot possibly be affected and they should still
read differently when reunited (according to your silly theory).

But, since their whole trip is carried out when the Earth is NOT rotating, they
obviously should NOT read differently when reunited.

Even YOU should know that observer behavior cannot affect an observed object.
So considering that observer behaviour cannot affect
an observed object, do you claim that the rotation of
the Earth somehow will affect remote ring laser gyros?
No answer, Henri?

We're talking about clocks, not ring gyros.
Doesn't matter. Same problem.
They are both instruments inside the planes.

So we could make the question more general:
Do you claim that the rotation of the Earth somehow will
affect instruments inside the plane?

How is it that the H&K clocks would miraculously know how to change rates if
the (remote) Earth suddenly stopped rotating?
How is it that the gyros in my thought experiment would miraculously
know how to change the rotation if the (remote) Earth suddenly
stopped rotating?

The answer should be obvious, Henri.
And the answer is the same whether we are talking
about clocks or gyros.

I will give you a hint:
What is it the gyros measure?
Why are the results different in case 1 and 2?
What is the relevance of the Earth?
Paul, as usual you run away from the real topic and stall for time by talking
about something entirely irrelevant.

IF THE EARTH STOPS ROTATING AS SOON AS THE PLANES ARE IN THE AIR, HOW DO THE
CLOCKS 'KNOW' HOW TO ADJUST THEIR RATES SO THEY WILL NOW READ THE SAME WHEN
REUNITED?

THIS SHOWS HOW NOINSENSICAL THE H&K REALLY IS.

WOULDN'T YOU AGREE?
IF THE EARTH STOPS ROTATING AS SOON AS THE PLANES ARE IN THE AIR, HOW DO THE
GYROS 'KNOW' HOW TO ADJUST THEIR MEASUREMENTS SO THEY WILL NOW READ THE SAME WHEN
REUNITED?

Why don't you answer?

Because the question is completely irrelevant.

Is it because you can't, or is it because you don't
like the answer you would have to give?

I will answer you.

The gyros will tell each plane when it has done a complete 360 turn.
It makes no difference whether the Earth stops spinning or not.

Of course. That is indeed the obvious answer.
The gyros and the clocks give different results in
the two scenarios, because the _planes_ move in two very
different ways in the two scenarios.
As you so correctly point out, it does not matter if the Earth
is spinning or not.

So the answers to your question:
" IF THE EARTH STOPS ROTATING AS SOON AS THE PLANES ARE IN THE AIR, HOW DO THE
CLOCKS 'KNOW' HOW TO ADJUST THEIR RATES SO THEY WILL NOW READ THE SAME WHEN
REUNITED?"
and my question:
"IF THE EARTH STOPS ROTATING AS SOON AS THE PLANES ARE IN THE AIR, HOW DO THE
GYROS 'KNOW' HOW TO ADJUST THEIR MEASUREMENTS SO THEY WILL NOW READ THE SAME WHEN
REUNITED?"
are exactly the same:

The clocks or the gyros don't have to know anything about
the rotation of the Earth because it does not affect them
in any way, it is the pathes of the planes through space and time
that determine what the instruments will show at the end of their journey.

In case 1, one plane is stationary in a non rotating frame,
while the other plane is moving twice around a circle at the speed 3340 km/h.
In case 2, both planes are moving in opposite directions once around a circle
at the same speed 1670 km/h.

"It makes no difference whether the Earth stops spinning or not."

Remember that for the future.

Case closed.

The case is indeed closed.
You have agreed that the H&K MUST give a null result.

There is no possible way to interpret what Paul said to mean that.

Reading for comprehension is quite difficult.


Thank you Paul.


Paul


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

Thank christ there is one genuine physicist on the NG.

Eric Gisse wrote:
Henri Wilson wrote:
On Mon, 27 Nov 2006 12:58:24 +0100, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Sun, 26 Nov 2006 21:28:53 +0100, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Thu, 23 Nov 2006 15:55:09 +0100, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Wed, 22 Nov 2006 10:45:34 +0100, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Tue, 21 Nov 2006 13:40:24 +0100, "Paul B. Andersen"
wrote:

Henri Wilson wrote: [about the H&K experiment]
You don't have to go to all this trouble.
As soon as the planes carrying the clocks are in the air, just stop the Earth
from rotating.
Somehow, relativists will claim that this action will affect the now remote
clocks. They will now read the same time when reunited.
I would like your answer the question at the end
of this posting.

Two supersonic planes are flying in opposite directions
around the Earth along equator. The ground speed is
one Earth circumference per sidereal day (ca. 1670 km/h, MACH 1.36).
Each plane is carrying a ring laser gyro detecting
the rotation around an axis parallel to the Earth axis
(the pitch axis).
The planes are starting at a point at the Earth, and
returning to the same point one sidereal day later.

Case 1:
The experiment is performed on the rotating Earth.
How many rotations are measured by the gyros in
the East- and West going plane respectively?
(The obvious answer is: 2 and 0)

Case 2:
The experiment is performed on a non-rotating Earth.
How many rotations are measured by the gyros in
the East- and West going plane respectively?
(The obvious answer is: 1 and 1)

Now the question I would like you to answer:
Do you claim that the rotation of the Earth
somehow will affect remote ring laser gyros?
THIS HAS NOUGHT TO DO WITH THE QUESTION.

We are talking about clock rates.

You people claim the two clocks will run at different rates and will read
differently when reunited.

I'm telling you that if the Earth stops rotating as soon as the clocks are in
the air, their relative rates cannot possibly be affected and they should still
read differently when reunited (according to your silly theory).

But, since their whole trip is carried out when the Earth is NOT rotating, they
obviously should NOT read differently when reunited.

Even YOU should know that observer behavior cannot affect an observed object.
So considering that observer behaviour cannot affect
an observed object, do you claim that the rotation of
the Earth somehow will affect remote ring laser gyros?
No answer, Henri?

We're talking about clocks, not ring gyros.
Doesn't matter. Same problem.
They are both instruments inside the planes.

So we could make the question more general:
Do you claim that the rotation of the Earth somehow will
affect instruments inside the plane?

How is it that the H&K clocks would miraculously know how to change rates if
the (remote) Earth suddenly stopped rotating?
How is it that the gyros in my thought experiment would miraculously
know how to change the rotation if the (remote) Earth suddenly
stopped rotating?

The answer should be obvious, Henri.
And the answer is the same whether we are talking
about clocks or gyros.

I will give you a hint:
What is it the gyros measure?
Why are the results different in case 1 and 2?
What is the relevance of the Earth?
Paul, as usual you run away from the real topic and stall for time by talking
about something entirely irrelevant.

IF THE EARTH STOPS ROTATING AS SOON AS THE PLANES ARE IN THE AIR, HOW DO THE
CLOCKS 'KNOW' HOW TO ADJUST THEIR RATES SO THEY WILL NOW READ THE SAME WHEN
REUNITED?

THIS SHOWS HOW NOINSENSICAL THE H&K REALLY IS.

WOULDN'T YOU AGREE?
IF THE EARTH STOPS ROTATING AS SOON AS THE PLANES ARE IN THE AIR, HOW DO THE
GYROS 'KNOW' HOW TO ADJUST THEIR MEASUREMENTS SO THEY WILL NOW READ THE SAME WHEN
REUNITED?

Why don't you answer?
Because the question is completely irrelevant.

Is it because you can't, or is it because you don't
like the answer you would have to give?
I will answer you.

The gyros will tell each plane when it has done a complete 360 turn.
It makes no difference whether the Earth stops spinning or not.
Of course. That is indeed the obvious answer.
The gyros and the clocks give different results in
the two scenarios, because the _planes_ move in two very
different ways in the two scenarios.
As you so correctly point out, it does not matter if the Earth
is spinning or not.

So the answers to your question:
" IF THE EARTH STOPS ROTATING AS SOON AS THE PLANES ARE IN THE AIR, HOW DO THE
CLOCKS 'KNOW' HOW TO ADJUST THEIR RATES SO THEY WILL NOW READ THE SAME WHEN
REUNITED?"
and my question:
"IF THE EARTH STOPS ROTATING AS SOON AS THE PLANES ARE IN THE AIR, HOW DO THE
GYROS 'KNOW' HOW TO ADJUST THEIR MEASUREMENTS SO THEY WILL NOW READ THE SAME WHEN
REUNITED?"
are exactly the same:

The clocks or the gyros don't have to know anything about
the rotation of the Earth because it does not affect them
in any way, it is the pathes of the planes through space and time
that determine what the instruments will show at the end of their journey.

In case 1, one plane is stationary in a non rotating frame,
while the other plane is moving twice around a circle at the speed 3340 km/h.
In case 2, both planes are moving in opposite directions once around a circle
at the same speed 1670 km/h.

"It makes no difference whether the Earth stops spinning or not."

Remember that for the future.

Case closed.
The case is indeed closed.
You have agreed that the H&K MUST give a null result.

There is no possible way to interpret what Paul said to mean that.

Henri knows that very well.
This is only his way of diverting the attention from the fact
that he himself, by pointing out that it is the motion of
of the _planes_ and not the rotation of the Earth that determines
what instruments in the planes will show, demonstrated that his babble
about stopping the rotation of the Earth is stupid.

Paul

Paul B. Andersen wrote:
Eric Gisse wrote:
Henri Wilson wrote:
On Mon, 27 Nov 2006 12:58:24 +0100, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Sun, 26 Nov 2006 21:28:53 +0100, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Thu, 23 Nov 2006 15:55:09 +0100, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Wed, 22 Nov 2006 10:45:34 +0100, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Tue, 21 Nov 2006 13:40:24 +0100, "Paul B. Andersen"
wrote:

Henri Wilson wrote: [about the H&K experiment]
You don't have to go to all this trouble.
As soon as the planes carrying the clocks are in the air, just stop the Earth
from rotating.
Somehow, relativists will claim that this action will affect the now remote
clocks. They will now read the same time when reunited.
I would like your answer the question at the end
of this posting.

Two supersonic planes are flying in opposite directions
around the Earth along equator. The ground speed is
one Earth circumference per sidereal day (ca. 1670 km/h, MACH 1.36).
Each plane is carrying a ring laser gyro detecting
the rotation around an axis parallel to the Earth axis
(the pitch axis).
The planes are starting at a point at the Earth, and
returning to the same point one sidereal day later.

Case 1:
The experiment is performed on the rotating Earth.
How many rotations are measured by the gyros in
the East- and West going plane respectively?
(The obvious answer is: 2 and 0)

Case 2:
The experiment is performed on a non-rotating Earth.
How many rotations are measured by the gyros in
the East- and West going plane respectively?
(The obvious answer is: 1 and 1)

Now the question I would like you to answer:
Do you claim that the rotation of the Earth
somehow will affect remote ring laser gyros?
THIS HAS NOUGHT TO DO WITH THE QUESTION.

We are talking about clock rates.

You people claim the two clocks will run at different rates and will read
differently when reunited.

I'm telling you that if the Earth stops rotating as soon as the clocks are in
the air, their relative rates cannot possibly be affected and they should still
read differently when reunited (according to your silly theory).

But, since their whole trip is carried out when the Earth is NOT rotating, they
obviously should NOT read differently when reunited.

Even YOU should know that observer behavior cannot affect an observed object.
So considering that observer behaviour cannot affect
an observed object, do you claim that the rotation of
the Earth somehow will affect remote ring laser gyros?
No answer, Henri?

We're talking about clocks, not ring gyros.
Doesn't matter. Same problem.
They are both instruments inside the planes.

So we could make the question more general:
Do you claim that the rotation of the Earth somehow will
affect instruments inside the plane?

How is it that the H&K clocks would miraculously know how to change rates if
the (remote) Earth suddenly stopped rotating?
How is it that the gyros in my thought experiment would miraculously
know how to change the rotation if the (remote) Earth suddenly
stopped rotating?

The answer should be obvious, Henri.
And the answer is the same whether we are talking
about clocks or gyros.

I will give you a hint:
What is it the gyros measure?
Why are the results different in case 1 and 2?
What is the relevance of the Earth?
Paul, as usual you run away from the real topic and stall for time by talking
about something entirely irrelevant.

IF THE EARTH STOPS ROTATING AS SOON AS THE PLANES ARE IN THE AIR, HOW DO THE
CLOCKS 'KNOW' HOW TO ADJUST THEIR RATES SO THEY WILL NOW READ THE SAME WHEN
REUNITED?

THIS SHOWS HOW NOINSENSICAL THE H&K REALLY IS.

WOULDN'T YOU AGREE?
IF THE EARTH STOPS ROTATING AS SOON AS THE PLANES ARE IN THE AIR, HOW DO THE
GYROS 'KNOW' HOW TO ADJUST THEIR MEASUREMENTS SO THEY WILL NOW READ THE SAME WHEN
REUNITED?

Why don't you answer?
Because the question is completely irrelevant.

Is it because you can't, or is it because you don't
like the answer you would have to give?
I will answer you.

The gyros will tell each plane when it has done a complete 360 turn.
It makes no difference whether the Earth stops spinning or not.
Of course. That is indeed the obvious answer.
The gyros and the clocks give different results in
the two scenarios, because the _planes_ move in two very
different ways in the two scenarios.
As you so correctly point out, it does not matter if the Earth
is spinning or not.

So the answers to your question:
" IF THE EARTH STOPS ROTATING AS SOON AS THE PLANES ARE IN THE AIR, HOW DO THE
CLOCKS 'KNOW' HOW TO ADJUST THEIR RATES SO THEY WILL NOW READ THE SAME WHEN
REUNITED?"
and my question:
"IF THE EARTH STOPS ROTATING AS SOON AS THE PLANES ARE IN THE AIR, HOW DO THE
GYROS 'KNOW' HOW TO ADJUST THEIR MEASUREMENTS SO THEY WILL NOW READ THE SAME WHEN
REUNITED?"
are exactly the same:

The clocks or the gyros don't have to know anything about
the rotation of the Earth because it does not affect them
in any way, it is the pathes of the planes through space and time
that determine what the instruments will show at the end of their journey.

In case 1, one plane is stationary in a non rotating frame,
while the other plane is moving twice around a circle at the speed 3340 km/h.
In case 2, both planes are moving in opposite directions once around a circle
at the same speed 1670 km/h.

"It makes no difference whether the Earth stops spinning or not."

Remember that for the future.

Case closed.
The case is indeed closed.
You have agreed that the H&K MUST give a null result.

There is no possible way to interpret what Paul said to mean that.

Henri knows that very well.
This is only his way of diverting the attention from the fact
that he himself, by pointing out that it is the motion of
of the _planes_ and not the rotation of the Earth that determines
what instruments in the planes will show, demonstrated that his babble
about stopping the rotation of the Earth is stupid.

Then we get to hear about his confusion about why there is an
asymmetry, then someone will try to explain the clocks in the planes
were compared to the clocks on the ground, and then Henri will
misunderstand in a new way...etc


Paul

On Wed, 29 Nov 2006 10:47:37 +0100, "Paul B. Andersen"
wrote:

Eric Gisse wrote:
Henri Wilson wrote:
On Mon, 27 Nov 2006 12:58:24 +0100, "Paul B. Andersen"
wrote:


IF THE EARTH STOPS ROTATING AS SOON AS THE PLANES ARE IN THE AIR, HOW DO THE
CLOCKS 'KNOW' HOW TO ADJUST THEIR RATES SO THEY WILL NOW READ THE SAME WHEN
REUNITED?

THIS SHOWS HOW NOINSENSICAL THE H&K REALLY IS.

WOULDN'T YOU AGREE?
IF THE EARTH STOPS ROTATING AS SOON AS THE PLANES ARE IN THE AIR, HOW DO THE
GYROS 'KNOW' HOW TO ADJUST THEIR MEASUREMENTS SO THEY WILL NOW READ THE SAME WHEN
REUNITED?

Why don't you answer?
Because the question is completely irrelevant.

Is it because you can't, or is it because you don't
like the answer you would have to give?
I will answer you.

"It makes no difference whether the Earth stops spinning or not."

Remember that for the future.

Case closed.
The case is indeed closed.
You have agreed that the H&K MUST give a null result.

There is no possible way to interpret what Paul said to mean that.

Henri knows that very well.
This is only his way of diverting the attention from the fact
that he himself, by pointing out that it is the motion of
of the _planes_ and not the rotation of the Earth that determines
what instruments in the planes will show, demonstrated that his babble
about stopping the rotation of the Earth is stupid.

Paul

I think you should have a closer look at the H&K...and its stupidity.

The clocks are presumed to run at different rates RELATIVE TO EACH OTHER.



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

Thank christ there is one genuine physicist on the NG.

Henri Wilson wrote:
On Wed, 29 Nov 2006 10:47:37 +0100, "Paul B. Andersen"
wrote:

Eric Gisse wrote:
Henri Wilson wrote:
On Mon, 27 Nov 2006 12:58:24 +0100, "Paul B. Andersen"
wrote:


IF THE EARTH STOPS ROTATING AS SOON AS THE PLANES ARE IN THE AIR, HOW DO THE
CLOCKS 'KNOW' HOW TO ADJUST THEIR RATES SO THEY WILL NOW READ THE SAME WHEN
REUNITED?

THIS SHOWS HOW NOINSENSICAL THE H&K REALLY IS.

WOULDN'T YOU AGREE?
IF THE EARTH STOPS ROTATING AS SOON AS THE PLANES ARE IN THE AIR, HOW DO THE
GYROS 'KNOW' HOW TO ADJUST THEIR MEASUREMENTS SO THEY WILL NOW READ THE SAME WHEN
REUNITED?

Why don't you answer?
Because the question is completely irrelevant.

Is it because you can't, or is it because you don't
like the answer you would have to give?
I will answer you.

"It makes no difference whether the Earth stops spinning or not."

Remember that for the future.

Case closed.
The case is indeed closed.
You have agreed that the H&K MUST give a null result.
There is no possible way to interpret what Paul said to mean that.
Henri knows that very well.
This is only his way of diverting the attention from the fact
that he himself, by pointing out that it is the motion of
of the _planes_ and not the rotation of the Earth that determines
what instruments in the planes will show, demonstrated that his babble
about stopping the rotation of the Earth is stupid.

Paul

I think you should have a closer look at the H&K...and its stupidity.

The clocks are presumed to run at different rates RELATIVE TO EACH OTHER.

The clocks in the planes will measure different proper times between
the same two events because their paths through space and time are different,
and the gyros in the planes will measure different rotations of the very
same reason.

This is what GR predicts for a K&H like experiment:

Let's start with the Schwarzschild Solution:

ds^2 = (1 - 2m/r)dt^2 - 1/(1 - 2m/r)dr^2 - r2 (dp^2 + (sin p)^2 dq^2)

where t is the temporal coordinate,
r is the radial coordinate,
p is latitude, and q is longitude.

m = G*M/c2 where G is the gravitational constant
and M is the mass of the Earth.

If the speed in the Earth centred non rotating frame
(Schwarzschild frame, hereafter called ECI-frame) is v,
we have v*dt = r*dp.
Constant height (dr=0) around equator (dq=0)
and the equation above becomes:

ds^2 = (1 - 2*m/r - v^2/c^2)*dt^2 (v in conventional units - thus c^2)
or
ds = sqrt(1 -2*m/r - v^2/c^2)*dt
or a first order approximation (m/r 1, v c):
T' = (1 - m/r - 0.5*v^2/c^2)*T
where T' is the proper time of the clock, and T is the travelling
time in the ECI-frame.
Since we are interested in finding the difference between two
clocks at different height and speed (Earth clock and plane clock),
we get:
(T1 - T2)/T = -(m/r1) + (m/r2) + (0.5*v2^2/c^2 - 0.5*v1^2/c^2)

Inserting m = G*M/c^2 and G*M/r1^2 = g, acceleration at Earth's surface,
we get:
(T1 - T2)/T = -(g/c^2)*r1*(1-r1/r2) + (0.5*v2^2/c^2 - 0.5*v1^2/c^2)
if we set: r2 = r1 + h, and approximate (r1+h)/r1 = ca. 1 since h r1,
we can write this:
(T1 - T2)/T = -g*h/c^2 + (0.5*v2^2/c^2 - 0.5*v1^2/c^2)

Let's use the equation above on our idealized case with
two planes flying in opposite direction around equator, same ground speed
and height. We will use "reasonable values" for commercial planes.
(They will have to fly non stop, though.)

Let's suppose the ground speeds of the aircrafts are 232.5 m/s,
and that they are flying at a height 9000m. The aircrafts
will use two sideral days on the journey.
The speeds of the clocks in the ECI-frame will be:
Earth clock A: va = 465 m/s
West bound B: vb = va - 232.5 m/s = 232.5 m/s
East bound C: vc = va + 232.5 m/s = 697.5 m/s

West bound:
TA - TB = 2*sideral_day*(-g*h/c^2 + 0.5*vb^2/c^2 - 0.5*va^2/c^2) sec
TA - TB = 2*86160*(-1 + 0.3 - 1.2)*10^-12 sec = - 327 ns
West bound clock gains 327 ns (H&K 273 ns)

East bound:
TA - TC = 2*sideral_day*(-g*h/c^2 + 0.5*vc^2/c^2 - 0.5*va^2/c^2) sec
TA - TC = 2*86160*(-1 + 2.7 - 1.2)*10-12 sec = + 86 ns
East bound clock loses 86 ns (H&K 59 ns)

A gyro in the west going plane would measure one rotation,
a gyro in the east going plane would measure 3 rotations,
and a gyro on the ground would measure 2 rotations.

Now tell me, Henry:
Where in the above do you see the rotation speed of the Earth?
The only speeds that appear are the speeds of the CLOCKS.
Since these are different, the three clocks and gyros will all
measure different values because their paths through space and
time are different.

GR's predictions for what clocks will show are tested in
a number of experiments, and have in all cases proven to
be correct within the precision of the measurements.
H&K is the least precise of these experiments.

You know this, Henri, and you have accepted the results
of these experiments (except the H&K) before.

You will of course keep insisting that the predictions
of GR are stupid even if they invariably are "accidentally"
correct. :-)

Paul

On Thu, 30 Nov 2006 12:57:21 +0100, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Wed, 29 Nov 2006 10:47:37 +0100, "Paul B. Andersen"

The clocks in the planes will measure different proper times between
the same two events because their paths through space and time are different,
and the gyros in the planes will measure different rotations of the very
same reason.

Hahaha! This is becoming quite funny.

Tell me how the clocks will somehow be affected by the fact that the Earth
stops rotating midway through the experiment?

This is what GR predicts for a K&H like experiment:

Let's do a similar exeriment with satellites.


Let's start with the Schwarzschild Solution:

ds^2 = (1 - 2m/r)dt^2 - 1/(1 - 2m/r)dr^2 - r2 (dp^2 + (sin p)^2 dq^2)

where t is the temporal coordinate,
r is the radial coordinate,
p is latitude, and q is longitude.

m = G*M/c2 where G is the gravitational constant
and M is the mass of the Earth.

If the speed in the Earth centred non rotating frame
(Schwarzschild frame, hereafter called ECI-frame) is v,
we have v*dt = r*dp.
Constant height (dr=0) around equator (dq=0)
and the equation above becomes:

ds^2 = (1 - 2*m/r - v^2/c^2)*dt^2 (v in conventional units - thus c^2)
or
ds = sqrt(1 -2*m/r - v^2/c^2)*dt
or a first order approximation (m/r 1, v c):
T' = (1 - m/r - 0.5*v^2/c^2)*T
where T' is the proper time of the clock, and T is the travelling
time in the ECI-frame.
Since we are interested in finding the difference between two
clocks at different height and speed (Earth clock and plane clock),
we get:
(T1 - T2)/T = -(m/r1) + (m/r2) + (0.5*v2^2/c^2 - 0.5*v1^2/c^2)

So where is the proof that this is true.

Inserting m = G*M/c^2 and G*M/r1^2 = g, acceleration at Earth's surface,
we get:
(T1 - T2)/T = -(g/c^2)*r1*(1-r1/r2) + (0.5*v2^2/c^2 - 0.5*v1^2/c^2)
if we set: r2 = r1 + h, and approximate (r1+h)/r1 = ca. 1 since h r1,
we can write this:
(T1 - T2)/T = -g*h/c^2 + (0.5*v2^2/c^2 - 0.5*v1^2/c^2)

Let's use the equation above on our idealized case with
two planes flying in opposite direction around equator, same ground speed
and height. We will use "reasonable values" for commercial planes.
(They will have to fly non stop, though.)

Let's suppose the ground speeds of the aircrafts are 232.5 m/s,
and that they are flying at a height 9000m. The aircrafts
will use two sideral days on the journey.
The speeds of the clocks in the ECI-frame will be:
Earth clock A: va = 465 m/s
West bound B: vb = va - 232.5 m/s = 232.5 m/s
East bound C: vc = va + 232.5 m/s = 697.5 m/s

Observer speed makes no difference to time flow anywhere else..

West bound:
TA - TB = 2*sideral_day*(-g*h/c^2 + 0.5*vb^2/c^2 - 0.5*va^2/c^2) sec
TA - TB = 2*86160*(-1 + 0.3 - 1.2)*10^-12 sec = - 327 ns
West bound clock gains 327 ns (H&K 273 ns)

According to your stupid theory.

East bound:
TA - TC = 2*sideral_day*(-g*h/c^2 + 0.5*vc^2/c^2 - 0.5*va^2/c^2) sec
TA - TC = 2*86160*(-1 + 2.7 - 1.2)*10-12 sec = + 86 ns
East bound clock loses 86 ns (H&K 59 ns)

According to your stupid theory.

A gyro in the west going plane would measure one rotation,
a gyro in the east going plane would measure 3 rotations,
and a gyro on the ground would measure 2 rotations.

Now tell me, Henry:
Where in the above do you see the rotation speed of the Earth?
The only speeds that appear are the speeds of the CLOCKS.
Since these are different, the three clocks and gyros will all
measure different values because their paths through space and
time are different.

Their 'paths through time' are identical.


GR's predictions for what clocks will show are tested in
a number of experiments, and have in all cases proven to
be correct within the precision of the measurements.
H&K is the least precise of these experiments.

You know this, Henri, and you have accepted the results
of these experiments (except the H&K) before.

The results have been officially withdrawn.

You will of course keep insisting that the predictions
of GR are stupid even if they invariably are "accidentally"
correct. :-)

There is absolutely no evidence that any of Einstein's version of relativity is
correct.
There is insurmounable evidence that it is nonsense from start to finish.
..


Paul


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

Thank christ there is one genuine physicist on the NG.

"Henri Wilson" HW@.. wrote in message
...
On Thu, 30 Nov 2006 12:57:21 +0100, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Wed, 29 Nov 2006 10:47:37 +0100, "Paul B. Andersen"

The clocks in the planes will measure different proper times between
the same two events because their paths through space and time are
different,
and the gyros in the planes will measure different rotations of the very
same reason.

Hahaha! This is becoming quite funny.

Tell me how the clocks will somehow be affected by the fact that the Earth
stops rotating midway through the experiment?

This is what GR predicts for a K&H like experiment:

Let's do a similar exeriment with satellites.


Let's start with the Schwarzschild Solution:

ds^2 = (1 - 2m/r)dt^2 - 1/(1 - 2m/r)dr^2 - r2 (dp^2 + (sin p)^2 dq^2)

where t is the temporal coordinate,
r is the radial coordinate,
p is latitude, and q is longitude.

m = G*M/c2 where G is the gravitational constant
and M is the mass of the Earth.

If the speed in the Earth centred non rotating frame
(Schwarzschild frame, hereafter called ECI-frame) is v,
we have v*dt = r*dp.
Constant height (dr=0) around equator (dq=0)
and the equation above becomes:

ds^2 = (1 - 2*m/r - v^2/c^2)*dt^2 (v in conventional units - thus c^2)
or
ds = sqrt(1 -2*m/r - v^2/c^2)*dt
or a first order approximation (m/r 1, v c):
T' = (1 - m/r - 0.5*v^2/c^2)*T
where T' is the proper time of the clock, and T is the travelling
time in the ECI-frame.
Since we are interested in finding the difference between two
clocks at different height and speed (Earth clock and plane clock),
we get:
(T1 - T2)/T = -(m/r1) + (m/r2) + (0.5*v2^2/c^2 - 0.5*v1^2/c^2)

So where is the proof that this is true.

Inserting m = G*M/c^2 and G*M/r1^2 = g, acceleration at Earth's surface,
we get:
(T1 - T2)/T = -(g/c^2)*r1*(1-r1/r2) + (0.5*v2^2/c^2 - 0.5*v1^2/c^2)
if we set: r2 = r1 + h, and approximate (r1+h)/r1 = ca. 1 since h r1,
we can write this:
(T1 - T2)/T = -g*h/c^2 + (0.5*v2^2/c^2 - 0.5*v1^2/c^2)

Let's use the equation above on our idealized case with
two planes flying in opposite direction around equator, same ground speed
and height. We will use "reasonable values" for commercial planes.
(They will have to fly non stop, though.)

Let's suppose the ground speeds of the aircrafts are 232.5 m/s,
and that they are flying at a height 9000m. The aircrafts
will use two sideral days on the journey.
The speeds of the clocks in the ECI-frame will be:
Earth clock A: va = 465 m/s
West bound B: vb = va - 232.5 m/s = 232.5 m/s
East bound C: vc = va + 232.5 m/s = 697.5 m/s

Observer speed makes no difference to time flow anywhere else..

West bound:
TA - TB = 2*sideral_day*(-g*h/c^2 + 0.5*vb^2/c^2 - 0.5*va^2/c^2) sec
TA - TB = 2*86160*(-1 + 0.3 - 1.2)*10^-12 sec = - 327 ns
West bound clock gains 327 ns (H&K 273 ns)

According to your stupid theory.

East bound:
TA - TC = 2*sideral_day*(-g*h/c^2 + 0.5*vc^2/c^2 - 0.5*va^2/c^2) sec
TA - TC = 2*86160*(-1 + 2.7 - 1.2)*10-12 sec = + 86 ns
East bound clock loses 86 ns (H&K 59 ns)

According to your stupid theory.

A gyro in the west going plane would measure one rotation,
a gyro in the east going plane would measure 3 rotations,
and a gyro on the ground would measure 2 rotations.

Now tell me, Henry:
Where in the above do you see the rotation speed of the Earth?
The only speeds that appear are the speeds of the CLOCKS.
Since these are different, the three clocks and gyros will all
measure different values because their paths through space and
time are different.

Their 'paths through time' are identical.


GR's predictions for what clocks will show are tested in
a number of experiments, and have in all cases proven to
be correct within the precision of the measurements.
H&K is the least precise of these experiments.

You know this, Henri, and you have accepted the results
of these experiments (except the H&K) before.

The results have been officially withdrawn.

You will of course keep insisting that the predictions
of GR are stupid even if they invariably are "accidentally"
correct. :-)

There is absolutely no evidence that any of Einstein's version of
relativity is
correct.
There is insurmounable evidence that it is nonsense from start to finish.

If Henri accepts this, he wouldn't have anything to rave about.

On Fri, 1 Dec 2006 12:38:07 +0200, "Norman Bates" wrote:


"Henri Wilson" HW@.. wrote in message
.. .
On Thu, 30 Nov 2006 12:57:21 +0100, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Wed, 29 Nov 2006 10:47:37 +0100, "Paul B. Andersen"

The clocks in the planes will measure different proper times between
the same two events because their paths through space and time are
different,
and the gyros in the planes will measure different rotations of the very
same reason.

Hahaha! This is becoming quite funny.

Tell me how the clocks will somehow be affected by the fact that the Earth
stops rotating midway through the experiment?

This is what GR predicts for a K&H like experiment:

Let's do a similar exeriment with satellites.


Let's start with the Schwarzschild Solution:

ds^2 = (1 - 2m/r)dt^2 - 1/(1 - 2m/r)dr^2 - r2 (dp^2 + (sin p)^2 dq^2)

where t is the temporal coordinate,
r is the radial coordinate,
p is latitude, and q is longitude.

m = G*M/c2 where G is the gravitational constant
and M is the mass of the Earth.

If the speed in the Earth centred non rotating frame
(Schwarzschild frame, hereafter called ECI-frame) is v,
we have v*dt = r*dp.
Constant height (dr=0) around equator (dq=0)
and the equation above becomes:

ds^2 = (1 - 2*m/r - v^2/c^2)*dt^2 (v in conventional units - thus c^2)
or
ds = sqrt(1 -2*m/r - v^2/c^2)*dt
or a first order approximation (m/r 1, v c):
T' = (1 - m/r - 0.5*v^2/c^2)*T
where T' is the proper time of the clock, and T is the travelling
time in the ECI-frame.
Since we are interested in finding the difference between two
clocks at different height and speed (Earth clock and plane clock),
we get:
(T1 - T2)/T = -(m/r1) + (m/r2) + (0.5*v2^2/c^2 - 0.5*v1^2/c^2)

So where is the proof that this is true.

Inserting m = G*M/c^2 and G*M/r1^2 = g, acceleration at Earth's surface,
we get:
(T1 - T2)/T = -(g/c^2)*r1*(1-r1/r2) + (0.5*v2^2/c^2 - 0.5*v1^2/c^2)
if we set: r2 = r1 + h, and approximate (r1+h)/r1 = ca. 1 since h r1,
we can write this:
(T1 - T2)/T = -g*h/c^2 + (0.5*v2^2/c^2 - 0.5*v1^2/c^2)

Let's use the equation above on our idealized case with
two planes flying in opposite direction around equator, same ground speed
and height. We will use "reasonable values" for commercial planes.
(They will have to fly non stop, though.)

Let's suppose the ground speeds of the aircrafts are 232.5 m/s,
and that they are flying at a height 9000m. The aircrafts
will use two sideral days on the journey.
The speeds of the clocks in the ECI-frame will be:
Earth clock A: va = 465 m/s
West bound B: vb = va - 232.5 m/s = 232.5 m/s
East bound C: vc = va + 232.5 m/s = 697.5 m/s

Observer speed makes no difference to time flow anywhere else..

West bound:
TA - TB = 2*sideral_day*(-g*h/c^2 + 0.5*vb^2/c^2 - 0.5*va^2/c^2) sec
TA - TB = 2*86160*(-1 + 0.3 - 1.2)*10^-12 sec = - 327 ns
West bound clock gains 327 ns (H&K 273 ns)

According to your stupid theory.

East bound:
TA - TC = 2*sideral_day*(-g*h/c^2 + 0.5*vc^2/c^2 - 0.5*va^2/c^2) sec
TA - TC = 2*86160*(-1 + 2.7 - 1.2)*10-12 sec = + 86 ns
East bound clock loses 86 ns (H&K 59 ns)

According to your stupid theory.

A gyro in the west going plane would measure one rotation,
a gyro in the east going plane would measure 3 rotations,
and a gyro on the ground would measure 2 rotations.

Now tell me, Henry:
Where in the above do you see the rotation speed of the Earth?
The only speeds that appear are the speeds of the CLOCKS.
Since these are different, the three clocks and gyros will all
measure different values because their paths through space and
time are different.

Their 'paths through time' are identical.


GR's predictions for what clocks will show are tested in
a number of experiments, and have in all cases proven to
be correct within the precision of the measurements.
H&K is the least precise of these experiments.

You know this, Henri, and you have accepted the results
of these experiments (except the H&K) before.

The results have been officially withdrawn.

You will of course keep insisting that the predictions
of GR are stupid even if they invariably are "accidentally"
correct. :-)

There is absolutely no evidence that any of Einstein's version of
relativity is
correct.
There is insurmounable evidence that it is nonsense from start to finish.

If Henri accepts this, he wouldn't have anything to rave about.

When the clocks are initially in the air, they are supposed to run at differet
rates, according to you. If the Earth stops spinning, both clocks experience
exactly the same acceleration for the same time. They have identical forces on
them.
From a purely physical point of view, there is no reason why this should cause
one to change more than the other.


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

Thank christ there is one genuine physicist on the NG.

"Henri Wilson" HW@.. wrote in message
...
On Fri, 1 Dec 2006 12:38:07 +0200, "Norman Bates"
wrote:


"Henri Wilson" HW@.. wrote in message
. ..
On Thu, 30 Nov 2006 12:57:21 +0100, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Wed, 29 Nov 2006 10:47:37 +0100, "Paul B. Andersen"

The clocks in the planes will measure different proper times between
the same two events because their paths through space and time are
different,
and the gyros in the planes will measure different rotations of the very
same reason.

Hahaha! This is becoming quite funny.

Tell me how the clocks will somehow be affected by the fact that the
Earth
stops rotating midway through the experiment?

This is what GR predicts for a K&H like experiment:

Let's do a similar exeriment with satellites.


Let's start with the Schwarzschild Solution:

ds^2 = (1 - 2m/r)dt^2 - 1/(1 - 2m/r)dr^2 - r2 (dp^2 + (sin p)^2 dq^2)

where t is the temporal coordinate,
r is the radial coordinate,
p is latitude, and q is longitude.

m = G*M/c2 where G is the gravitational constant
and M is the mass of the Earth.

If the speed in the Earth centred non rotating frame
(Schwarzschild frame, hereafter called ECI-frame) is v,
we have v*dt = r*dp.
Constant height (dr=0) around equator (dq=0)
and the equation above becomes:

ds^2 = (1 - 2*m/r - v^2/c^2)*dt^2 (v in conventional units - thus c^2)
or
ds = sqrt(1 -2*m/r - v^2/c^2)*dt
or a first order approximation (m/r 1, v c):
T' = (1 - m/r - 0.5*v^2/c^2)*T
where T' is the proper time of the clock, and T is the travelling
time in the ECI-frame.
Since we are interested in finding the difference between two
clocks at different height and speed (Earth clock and plane clock),
we get:
(T1 - T2)/T = -(m/r1) + (m/r2) + (0.5*v2^2/c^2 - 0.5*v1^2/c^2)

So where is the proof that this is true.

Inserting m = G*M/c^2 and G*M/r1^2 = g, acceleration at Earth's surface,
we get:
(T1 - T2)/T = -(g/c^2)*r1*(1-r1/r2) + (0.5*v2^2/c^2 - 0.5*v1^2/c^2)
if we set: r2 = r1 + h, and approximate (r1+h)/r1 = ca. 1 since h r1,
we can write this:
(T1 - T2)/T = -g*h/c^2 + (0.5*v2^2/c^2 - 0.5*v1^2/c^2)

Let's use the equation above on our idealized case with
two planes flying in opposite direction around equator, same ground
speed
and height. We will use "reasonable values" for commercial planes.
(They will have to fly non stop, though.)

Let's suppose the ground speeds of the aircrafts are 232.5 m/s,
and that they are flying at a height 9000m. The aircrafts
will use two sideral days on the journey.
The speeds of the clocks in the ECI-frame will be:
Earth clock A: va = 465 m/s
West bound B: vb = va - 232.5 m/s = 232.5 m/s
East bound C: vc = va + 232.5 m/s = 697.5 m/s

Observer speed makes no difference to time flow anywhere else..

West bound:
TA - TB = 2*sideral_day*(-g*h/c^2 + 0.5*vb^2/c^2 - 0.5*va^2/c^2) sec
TA - TB = 2*86160*(-1 + 0.3 - 1.2)*10^-12 sec = - 327 ns
West bound clock gains 327 ns (H&K 273 ns)

According to your stupid theory.

East bound:
TA - TC = 2*sideral_day*(-g*h/c^2 + 0.5*vc^2/c^2 - 0.5*va^2/c^2) sec
TA - TC = 2*86160*(-1 + 2.7 - 1.2)*10-12 sec = + 86 ns
East bound clock loses 86 ns (H&K 59 ns)

According to your stupid theory.

A gyro in the west going plane would measure one rotation,
a gyro in the east going plane would measure 3 rotations,
and a gyro on the ground would measure 2 rotations.

Now tell me, Henry:
Where in the above do you see the rotation speed of the Earth?
The only speeds that appear are the speeds of the CLOCKS.
Since these are different, the three clocks and gyros will all
measure different values because their paths through space and
time are different.

Their 'paths through time' are identical.


GR's predictions for what clocks will show are tested in
a number of experiments, and have in all cases proven to
be correct within the precision of the measurements.
H&K is the least precise of these experiments.

You know this, Henri, and you have accepted the results
of these experiments (except the H&K) before.

The results have been officially withdrawn.

You will of course keep insisting that the predictions
of GR are stupid even if they invariably are "accidentally"
correct. :-)

There is absolutely no evidence that any of Einstein's version of
relativity is
correct.
There is insurmounable evidence that it is nonsense from start to
finish.

If Henri accepts this, he wouldn't have anything to rave about.

When the clocks are initially in the air, they are supposed to run at
differet
rates, according to you. If the Earth stops spinning, both clocks
experience
exactly the same acceleration for the same time. They have identical
forces on
them.
From a purely physical point of view, there is no reason why this should
cause
one to change more than the other.


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

Thank christ there is one genuine physicist on the NG.

Henri, I'm not in the mood for this discussion anymore, take it to the nobel
science committee, see what they think, maybe you are right then you can go
buy a suit.