Discussion:
A Lesson In Sagnac and Rotating frames, For Paul, Tom and Jerry.
Henry Wilson DSc
2010-02-11 23:33:09 UTC
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.

//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)

(the surface is drawn flat for convenience. The fibre goes right around the
planet)

Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre. The pulses are labelled according to their emission times and move at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.

Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p. At any instant, 40 million pulses are in transit
around the planet in each direction.

Tom observes that the pulses arriving simultaneously from both fibres carry
identical labels.

Using powerful rockets, the planet is sent into rotation with a period of 1
day, (86400 seconds). Its surface rotation speed 'v' wrt the nonrotating frame
is 465 m/s.

In this new situation, the pulses that arrive at C when it is at point p, were
not emitted at that point but rather from an imaginary point q, which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.

Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.

Since the distance between pulses is still 1 metre in the C's frame, there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.

"That's odd", says Tom, "now, for some reason, the pulses coming from the left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"

"This is great", says Jerry after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".

Henry Wilson...

.......provider of free physics lessons
Henry Wilson DSc
2010-02-12 21:27:47 UTC
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
(the surface is drawn flat for convenience. The fibre goes right around the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre. The pulses are labelled according to their emission times and move at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p. At any instant, 40 million pulses are in transit
around the planet in each direction.
Tom observes that the pulses arriving simultaneously from both fibres carry
identical labels.
Using powerful rockets, the planet is sent into rotation with a period of 1
day, (86400 seconds). Its surface rotation speed 'v' wrt the nonrotating frame
is 465 m/s.
In this new situation, the pulses that arrive at C when it is at point p, were
not emitted at that point but rather from an imaginary point q, which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.
Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.
Since the distance between pulses is still 1 metre in the C's frame, there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.
"That's odd", says Tom, "now, for some reason, the pulses coming from the left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"
"This is great", says Jerry after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".
Henry Wilson...
.......provider of free physics lessons

Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-12 22:35:45 UTC
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
What is q and v?
Post by Henry Wilson DSc
(the surface is drawn flat for convenience. The fibre goes right around the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre. The pulses are labelled according to their emission times and move at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p.
Doesn't matter if they are detected at p or at C
Post by Henry Wilson DSc
At any instant, 40 million pulses are in transit
around the planet in each direction.
Tom
Tom? I thought it was Paul and Jerry?
Post by Henry Wilson DSc
observes that the pulses arriving simultaneously from both fibres carry
identical labels.
Using powerful rockets, the planet is sent into rotation with a period of 1
day, (86400 seconds). Its surface rotation speed 'v' wrt the nonrotating frame
is 465 m/s.
OK .. assuming you got the math right
Post by Henry Wilson DSc
In this new situation, the pulses that arrive at C when it is at point p, were
not emitted at that point but rather from an imaginary point q,
Not an imaginary point at all. It is as real and physical point as p is.
All you are saying is that C has moved (and continues to move)
Post by Henry Wilson DSc
which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.
Angular speed I think you would mean, it is moving around the ring according
to C's rotating frame.
Post by Henry Wilson DSc
Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
So you're using an emission theory analysus, which does say that the time
taken for the two paths is the same.
Post by Henry Wilson DSc
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.
I'll take your word for it.
Post by Henry Wilson DSc
Since the distance between pulses is still 1 metre in the C's frame,
That's what emission theory would say
Post by Henry Wilson DSc
there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.
WRONG. Emmision theory does not say that. It says that the distance in C's
frame between pulses is one meter. The distance the pulses travel in C's
frame is the same. So the number of pulses in the fibre at any given time
is the same.

You can also tell it is wrong because the same number of pulses has been
emitted in each direction and you just said earlier that they take the same
time to travel around the ring. So there must be the same number of them in
there. Or does your explanation have pulse fairies that eat up the pulses?
Post by Henry Wilson DSc
"That's odd", says Tom, "now, for some reason, the pulses coming from the left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"
Emmision theory doesn't say that will happen at all. It says what you said
earlier .. the time take for them to travel is the same.

What you are now suddenly describing is what SR or an aether theory would
predict. You've changed theories in mid analysis !!!
Post by Henry Wilson DSc
"This is great", says Jerry after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".
No .. because your analysis above used SR at the end. You switched theories
mid way thru.

You started saying the time for the pulses to travel in both directions
(with the rotating tube) was the same. That is what emission theory says.
And you said the distance between the pulses (in the rotating frame) is the
same. That is what emission theory says. The consequence of that is that
there MUST be the same number of pulses in each direction in the tube at any
given time. And they MUST arrive at the detector with the same label. And
so emission theory says that a sagnac device will NOT detect rotation.

The *real* Sagnac devices have proven that THAT theory is NOT correct, and
it is NOT the same time required for the pulses (which is what SR predicts).

BAHAHAHHA
BURT
2010-02-12 22:47:36 UTC
Post by Inertial
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
What is q and v?
Post by Henry Wilson DSc
(the surface is drawn flat for convenience. The fibre goes right around the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre.  The pulses are labelled according to their emission times and move
at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p.
Doesn't matter if they are detected at p or at C
Post by Henry Wilson DSc
At any instant, 40 million pulses are in transit
around the planet in each direction.
Tom
Tom?  I thought it was Paul and Jerry?
Post by Henry Wilson DSc
observes that the pulses arriving simultaneously from both fibres carry
identical labels.
Using powerful rockets, the planet is sent into rotation with a period of 1
day, (86400 seconds). Its surface rotation speed 'v' wrt the nonrotating frame
is 465 m/s.
OK .. assuming you got the math right
Post by Henry Wilson DSc
In this new situation, the pulses that arrive at C when it is at point p, were
not emitted at that point but rather from an imaginary point q,
Not an imaginary point at all.  It is as real and physical point as p is.
All you are saying is that C has moved (and continues to move)
Post by Henry Wilson DSc
which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.
Angular speed I think you would mean, it is moving around the ring according
to C's rotating frame.
Post by Henry Wilson DSc
Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
So you're using an emission theory analysus, which does say that the time
taken for the two paths is the same.
Post by Henry Wilson DSc
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.
I'll take your word for it.
Post by Henry Wilson DSc
Since the distance between pulses is still 1 metre in the C's frame,
That's what emission theory would say
Post by Henry Wilson DSc
there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.
WRONG. Emmision theory does not say that.  It says that the distance in C's
frame between pulses is one meter.  The distance the pulses travel in C's
frame is the same.  So the number of pulses in the fibre at any given time
is the same.
You can also tell it is wrong because the same number of pulses has been
emitted in each direction and you just said earlier that they take the same
time to travel around the ring.  So there must be the same number of them in
there.  Or does your explanation have pulse fairies that eat up the pulses?
Post by Henry Wilson DSc
"That's odd", says Tom, "now, for some reason, the pulses coming from the left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"
Emmision theory doesn't say that will happen at all.  It says what you said
earlier .. the time take for them to travel is the same.
What you are now suddenly describing is what SR or an aether theory would
predict.  You've changed theories in mid analysis !!!
Post by Henry Wilson DSc
"This is great", says Jerry after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".
No .. because your analysis above used SR at the end.  You switched theories
mid way thru.
You started saying the time for the pulses to travel in both directions
(with the rotating tube) was the same.  That is what emission theory says.
And you said the distance between the pulses (in the rotating frame) is the
same.  That is what emission theory says.  The consequence of that is that
there MUST be the same number of pulses in each direction in the tube at any
given time.  And they MUST arrive at the detector with the same label.  And
so emission theory says that a sagnac device will NOT detect rotation.
The *real* Sagnac devices have proven that THAT theory is NOT correct, and
it is NOT the same time required for the pulses (which is what SR predicts).
BAHAHAHHA- Hide quoted text -
- Show quoted text -
Einstein could never reconcile the photon with the light wave. I ask
which wave is the photon in?
The magnetic or electric? Light is electric energy oscillating with
magnetism in a Dual Unfied force.

Mitch Raemsch
Henry Wilson DSc
2010-02-13 05:18:54 UTC
Post by Inertial
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
What is q and v?
Post by Henry Wilson DSc
(the surface is drawn flat for convenience. The fibre goes right around the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre. The pulses are labelled according to their emission times and move at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p.
Doesn't matter if they are detected at p or at C
C is at point p.
Post by Inertial
Post by Henry Wilson DSc
At any instant, 40 million pulses are in transit
around the planet in each direction.
Tom
Tom? I thought it was Paul and Jerry?
Tom reckon the rotating frame analysis of Sagnac proves BaTh wrong.
Post by Inertial
Post by Henry Wilson DSc
observes that the pulses arriving simultaneously from both fibres carry
identical labels.
Using powerful rockets, the planet is sent into rotation with a period of 1
day, (86400 seconds). Its surface rotation speed 'v' wrt the nonrotating frame
is 465 m/s.
OK .. assuming you got the math right
Post by Henry Wilson DSc
In this new situation, the pulses that arrive at C when it is at point p, were
not emitted at that point but rather from an imaginary point q,
Not an imaginary point at all. It is as real and physical point as p is.
All you are saying is that C has moved (and continues to move)
Hey dopey, can you see a point marked where you were 10 seconds ago?
Post by Inertial
Post by Henry Wilson DSc
which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.
Angular speed I think you would mean, it is moving around the ring according
to C's rotating frame.
v is its peripheral speed wrt the nonR frame.
Post by Inertial
Post by Henry Wilson DSc
Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
So you're using an emission theory analysus, which does say that the time
taken for the two paths is the same.
It certainly is.
Post by Inertial
Post by Henry Wilson DSc
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.
I'll take your word for it.
Post by Henry Wilson DSc
Since the distance between pulses is still 1 metre in the C's frame,
That's what emission theory would say
that's what is happening.
Post by Inertial
Post by Henry Wilson DSc
there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.
WRONG. Emmision theory does not say that. It says that the distance in C's
frame between pulses is one meter. The distance the pulses travel in C's
frame is the same.
No dopey, the travel times are the same. The path lengths are different by +/-
62m
Post by Inertial
So the number of pulses in the fibre at any given time
is the same.
that's correct. that is why the fringes on a sagnac interferometer do not move
during constant rotation.
Post by Inertial
You can also tell it is wrong because the same number of pulses has been
emitted in each direction and you just said earlier that they take the same
time to travel around the ring. So there must be the same number of them in
there.
No it doesn't mean that at all. I have explained perfectly well what happens.
It is only during a CHANGE of speed that pulses move in and out of the two
paths.
Post by Inertial
Or does your explanation have pulse fairies that eat up the pulses?
No the explanation is that you are plain dopey.
Post by Inertial
Post by Henry Wilson DSc
"That's odd", says Tom, "now, for some reason, the pulses coming from the left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"
Emmision theory doesn't say that will happen at all. It says what you said
earlier .. the time take for them to travel is the same.
Correct.
Post by Inertial
What you are now suddenly describing is what SR or an aether theory would
predict. You've changed theories in mid analysis !!!
No dopey. You are incapable of understanding what i said.
Post by Inertial
Post by Henry Wilson DSc
"This is great", says Jerry after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".
No .. because your analysis above used SR at the end. You switched theories
mid way thru.
You started saying the time for the pulses to travel in both directions
(with the rotating tube) was the same. That is what emission theory says.
And you said the distance between the pulses (in the rotating frame) is the
same. That is what emission theory says. The consequence of that is that
there MUST be the same number of pulses in each direction in the tube at any
given time.
Only when it is not rotating.
Post by Inertial
And they MUST arrive at the detector with the same label. And
so emission theory says that a sagnac device will NOT detect rotation.
No dopey , it says the sagnac effect is too hard for you to understand.
Post by Inertial
The *real* Sagnac devices have proven that THAT theory is NOT correct, and
it is NOT the same time required for the pulses (which is what SR predicts).
BAHAHAHHA
come back when you have learnt to use your brain

Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-14 11:25:20 UTC
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT
ROTATING.
Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
What is q and v?
Post by Henry Wilson DSc
(the surface is drawn flat for convenience. The fibre goes right around the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre. The pulses are labelled according to their emission times and
move
at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p.
Doesn't matter if they are detected at p or at C
C is at point p.
Yeup
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
At any instant, 40 million pulses are in transit
around the planet in each direction.
Tom
Tom? I thought it was Paul and Jerry?
Tom reckon the rotating frame analysis of Sagnac proves BaTh wrong.
You just muddled up your example by changing Jerry's name :) Guess you're
not able to admit something like that though.
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
observes that the pulses arriving simultaneously from both fibres carry
identical labels.
Using powerful rockets, the planet is sent into rotation with a period
of
1
day, (86400 seconds). Its surface rotation speed 'v' wrt the nonrotating frame
is 465 m/s.
OK .. assuming you got the math right
Post by Henry Wilson DSc
In this new situation, the pulses that arrive at C when it is at point
p,
were
not emitted at that point but rather from an imaginary point q,
Not an imaginary point at all. It is as real and physical point as p is.
All you are saying is that C has moved (and continues to move)
Hey dopey, can you see a point marked where you were 10 seconds ago?
Why do you think points need to be marked in order not to be imaginary?
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.
Angular speed I think you would mean, it is moving around the ring according
to C's rotating frame.
v is its peripheral speed wrt the nonR frame.
Fine .. I'm happy with that. Though a constant speed, it is not a constant
velocity.
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
So you're using an emission theory analysus, which does say that the time
taken for the two paths is the same.
It certainly is.
Good
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.
I'll take your word for it.
Post by Henry Wilson DSc
Since the distance between pulses is still 1 metre in the C's frame,
That's what emission theory would say
that's what is happening.
Good
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.
WRONG. Emmision theory does not say that. It says that the distance in C's
frame between pulses is one meter. The distance the pulses travel in C's
frame is the same.
No dopey, the travel times are the same. The path lengths are different by +/-
62m
In which frame?
Post by Henry Wilson DSc
Post by Inertial
So the number of pulses in the fibre at any given time
is the same.
that's correct. that is why the fringes on a sagnac interferometer do not move
during constant rotation.
I meant in each direction.
Post by Henry Wilson DSc
Post by Inertial
You can also tell it is wrong because the same number of pulses has been
emitted in each direction and you just said earlier that they take the same
time to travel around the ring. So there must be the same number of them in
there.
No it doesn't mean that at all.
Yes it does.
Post by Henry Wilson DSc
I have explained perfectly well what happens.
Post by Henry Wilson DSc
It is only during a CHANGE of speed that pulses move in and out of the two
paths.
Irrelevant
Post by Henry Wilson DSc
Post by Inertial
Or does your explanation have pulse fairies that eat up the pulses?
No the explanation is that you are plain dopey.
No .. it is that you can't put forward a logically consistent argument
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
"That's odd", says Tom, "now, for some reason, the pulses coming from
the
left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"
Emmision theory doesn't say that will happen at all. It says what you said
earlier .. the time take for them to travel is the same.
Correct.
So you cannot measure the absolute rotation according to emission theory.
Real sagnac devices operate as SR predicts, and you CAN measure it
Post by Henry Wilson DSc
Post by Inertial
What you are now suddenly describing is what SR or an aether theory would
predict. You've changed theories in mid analysis !!!
No dopey. You are incapable of understanding what i said.
I understand just fine. Which is why i know you are wrong .. yet again
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
"This is great", says Jerry after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".
No .. because your analysis above used SR at the end. You switched theories
mid way thru.
You started saying the time for the pulses to travel in both directions
(with the rotating tube) was the same. That is what emission theory says.
And you said the distance between the pulses (in the rotating frame) is the
same. That is what emission theory says. The consequence of that is that
there MUST be the same number of pulses in each direction in the tube at any
given time.
Only when it is not rotating.
Not according to emission theory. if the pulses take the same time to get
back to point C, and they travel at the same speed in C's frame, then they
must have the same time stamp on them and must have travelled the same
distance (in that frame)
Post by Henry Wilson DSc
Post by Inertial
And they MUST arrive at the detector with the same label. And
so emission theory says that a sagnac device will NOT detect rotation.
No dopey , it says the sagnac effect is too hard for you to understand.
No .. yhou've just been getting it wrong for years .. and continue to post
Post by Henry Wilson DSc
Post by Inertial
The *real* Sagnac devices have proven that THAT theory is NOT correct, and
it is NOT the same time required for the pulses (which is what SR predicts).
BAHAHAHHA
come back when you have learnt to use your brain
At least I have one that works
Henry Wilson DSc
2010-02-14 21:49:15 UTC
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT
ROTATING.
Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
What is q and v?
Post by Henry Wilson DSc
(the surface is drawn flat for convenience. The fibre goes right around the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre. The pulses are labelled according to their emission times and
move
at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p.
Doesn't matter if they are detected at p or at C
C is at point p.
Yeup
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
At any instant, 40 million pulses are in transit
around the planet in each direction.
Tom
Tom? I thought it was Paul and Jerry?
Tom reckon the rotating frame analysis of Sagnac proves BaTh wrong.
You just muddled up your example by changing Jerry's name :) Guess you're
not able to admit something like that though.
Tom has always believed that is true.
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
observes that the pulses arriving simultaneously from both fibres carry
identical labels.
Using powerful rockets, the planet is sent into rotation with a period
of
1
day, (86400 seconds). Its surface rotation speed 'v' wrt the nonrotating frame
is 465 m/s.
OK .. assuming you got the math right
Post by Henry Wilson DSc
In this new situation, the pulses that arrive at C when it is at point
p,
were
not emitted at that point but rather from an imaginary point q,
Not an imaginary point at all. It is as real and physical point as p is.
All you are saying is that C has moved (and continues to move)
Hey dopey, can you see a point marked where you were 10 seconds ago?
Why do you think points need to be marked in order not to be imaginary?
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.
Angular speed I think you would mean, it is moving around the ring according
to C's rotating frame.
v is its peripheral speed wrt the nonR frame.
Fine .. I'm happy with that. Though a constant speed, it is not a constant
velocity.
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
So you're using an emission theory analysus, which does say that the time
taken for the two paths is the same.
It certainly is.
Good
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.
I'll take your word for it.
Post by Henry Wilson DSc
Since the distance between pulses is still 1 metre in the C's frame,
That's what emission theory would say
that's what is happening.
Good
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.
WRONG. Emmision theory does not say that. It says that the distance in C's
frame between pulses is one meter. The distance the pulses travel in C's
frame is the same.
No dopey, the travel times are the same. The path lengths are different by +/-
62m
In which frame?
Post by Henry Wilson DSc
Post by Inertial
So the number of pulses in the fibre at any given time
is the same.
that's correct. that is why the fringes on a sagnac interferometer do not move
during constant rotation.
I meant in each direction.
Post by Henry Wilson DSc
Post by Inertial
You can also tell it is wrong because the same number of pulses has been
emitted in each direction and you just said earlier that they take the same
time to travel around the ring. So there must be the same number of them in
there.
No it doesn't mean that at all.
Yes it does.
Post by Henry Wilson DSc
I have explained perfectly well what happens.
Post by Henry Wilson DSc
It is only during a CHANGE of speed that pulses move in and out of the two
paths.
Irrelevant
Post by Henry Wilson DSc
Post by Inertial
Or does your explanation have pulse fairies that eat up the pulses?
No the explanation is that you are plain dopey.
No .. it is that you can't put forward a logically consistent argument
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
"That's odd", says Tom, "now, for some reason, the pulses coming from
the
left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"
Emmision theory doesn't say that will happen at all. It says what you said
earlier .. the time take for them to travel is the same.
Correct.
So you cannot measure the absolute rotation according to emission theory.
Real sagnac devices operate as SR predicts, and you CAN measure it
Post by Henry Wilson DSc
Post by Inertial
What you are now suddenly describing is what SR or an aether theory would
predict. You've changed theories in mid analysis !!!
No dopey. You are incapable of understanding what i said.
I understand just fine. Which is why i know you are wrong .. yet again
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
"This is great", says Jerry after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".
No .. because your analysis above used SR at the end. You switched theories
mid way thru.
You started saying the time for the pulses to travel in both directions
(with the rotating tube) was the same. That is what emission theory says.
And you said the distance between the pulses (in the rotating frame) is the
same. That is what emission theory says. The consequence of that is that
there MUST be the same number of pulses in each direction in the tube at any
given time.
Only when it is not rotating.
Not according to emission theory. if the pulses take the same time to get
back to point C, and they travel at the same speed in C's frame, then they
must have the same time stamp on them and must have travelled the same
distance (in that frame)
Post by Henry Wilson DSc
Post by Inertial
And they MUST arrive at the detector with the same label. And
so emission theory says that a sagnac device will NOT detect rotation.
No dopey , it says the sagnac effect is too hard for you to understand.
No .. yhou've just been getting it wrong for years .. and continue to post
Post by Henry Wilson DSc
Post by Inertial
The *real* Sagnac devices have proven that THAT theory is NOT correct, and
it is NOT the same time required for the pulses (which is what SR predicts).
BAHAHAHHA
come back when you have learnt to use your brain
At least I have one that works
I admitted my mistakes and corrected them in the next post.

the labels are the same at all speeds.

Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-14 22:44:32 UTC
"Henry Wilson DSc" <***@..> wrote in message news:***@4ax.com...
[snip]
Post by Henry Wilson DSc
I admitted my mistakes and corrected them in the next post.
If so, that is a pleasant change for you .. Where is the next post?
Henry Wilson DSc
2010-02-14 23:07:39 UTC
Post by Inertial
[snip]
Post by Henry Wilson DSc
I admitted my mistakes and corrected them in the next post.
If so, that is a pleasant change for you .. Where is the next post?
Just below the first one. posted on 14th

here's a copy.
***************
Post by Inertial
Post by Henry Wilson DSc
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
What is q and v?
Post by Henry Wilson DSc
Post by Henry Wilson DSc
(the surface is drawn flat for convenience. The fibre goes right around the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre. The pulses are labelled according to their emission times and move at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p.
Doesn't matter if they are detected at p or at C
I'll withdraw my previous reply because it was wrong. So was my initial message
because I stuffed up what I wanted to say.

p is the point where the pulses emitted at q are received.
Post by Inertial
Post by Henry Wilson DSc
Post by Henry Wilson DSc
observes that the pulses arriving simultaneously from both fibres carry
identical labels.
Using powerful rockets, the planet is sent into rotation with a period of 1
day, (86400 seconds). Its surface rotation speed 'v' wrt the nonrotating frame
is 465 m/s.
OK .. assuming you got the math right
Post by Henry Wilson DSc
Post by Henry Wilson DSc
In this new situation, the pulses that arrive at C when it is at point p, were
not emitted at that point but rather from an imaginary point q,
Not an imaginary point at all. It is as real and physical point as p is.
All you are saying is that C has moved (and continues to move)
It is imaginary in the R frame...REAL in the nonR frame.
Post by Inertial
Post by Henry Wilson DSc
Post by Henry Wilson DSc
which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.
Angular speed I think you would mean, it is moving around the ring according
to C's rotating frame.
yes
Post by Inertial
Post by Henry Wilson DSc
Post by Henry Wilson DSc
Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
So you're using an emission theory analysus, which does say that the time
taken for the two paths is the same.
Yes. The path lengths differ by 124 metres but one pulse moves at c+v and the
other at c-v, so they both arrive at p together.
Post by Inertial
Post by Henry Wilson DSc
Post by Henry Wilson DSc
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.
I'll take your word for it.
Post by Henry Wilson DSc
Post by Henry Wilson DSc
Since the distance between pulses is still 1 metre in the C's frame,
That's what emission theory would say
Post by Henry Wilson DSc
Post by Henry Wilson DSc
there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.
WRONG. Emmision theory does not say that. It says that the distance in C's
frame between pulses is one meter. The distance the pulses travel in C's
frame is the same. So the number of pulses in the fibre at any given time
is the same.
But I'm talking about the total number of pulses in transit in each path at any
instant. there are 124 more in one path than the other.
Post by Inertial
You can also tell it is wrong because the same number of pulses has been
emitted in each direction and you just said earlier that they take the same
time to travel around the ring. So there must be the same number of them in
there. Or does your explanation have pulse fairies that eat up the pulses?
Now you are being silly.

I made a mistake in my original post. This is regrettable but doesn't present a
problem for Bath.
As you rightly said, the pulses that leave together DO always arrive together
at the detector.
Post by Inertial
Post by Henry Wilson DSc
Post by Henry Wilson DSc
"That's odd", says Tom, "now, for some reason, the pulses coming from the left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"
This is wrong....and is not what I wanted to point out.
Post by Inertial
Emmision theory doesn't say that will happen at all. It says what you said
earlier .. the time take for them to travel is the same.
It is.....and the label on each arriving pulse is the same.
Post by Inertial
What you are now suddenly describing is what SR or an aether theory would
predict. You've changed theories in mid analysis !!!
Yes, basically I did. I apologise for any inconvenience.
Post by Inertial
Post by Henry Wilson DSc
Post by Henry Wilson DSc
"This is great", says Jerry after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".
No .. because your analysis above used SR at the end. You switched theories
mid way thru.
Yes I did.
Post by Inertial
You started saying the time for the pulses to travel in both directions
(with the rotating tube) was the same. That is what emission theory says.
And you said the distance between the pulses (in the rotating frame) is the
same. That is what emission theory says.
It is.
Post by Inertial
The consequence of that is that
there MUST be the same number of pulses in each direction in the tube at any
given time.
No that is YOUR mistake. The path lengths are different so there are more
pulses in one path than the other.
Post by Inertial
And they MUST arrive at the detector with the same label.
They do. I was wrong, before.
Post by Inertial
And
so emission theory says that a sagnac device will NOT detect rotation.
Now I'll tell you why that is wrong.
Post by Inertial
The *real* Sagnac devices have proven that THAT theory is NOT correct, and
it is NOT the same time required for the pulses (which is what SR predicts).
BAHAHAHHA
The true fact is that because the number of pulses is different in each path,
THE RATE at which they arrive at the detector is DIFFERENT for each path.
Between the instant of emission from point q and the detection at point p,
40000062 pulses arrive from one direction but only 39999938 from the other.
Those numbers will change only during a rotational speed change.

It is that difference in number that is analogous to the classical 'phase
difference' of the two light rays in a ring gyro.

Two different arrival rates produces a beat frequency at the detector.
Exactly how that gives rise to interference fringes I'm not sure at this stage
but the fact that BaTh predicts such a difference and produces the correct
equation for fringe displacement is sufficient to demonstrate that Sagnac
definitely does NOT refute BaTh.

Henry Wilson...

.......provider of free physics lessons

Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-14 23:21:23 UTC
Post by Henry Wilson DSc
Post by Inertial
[snip]
Post by Henry Wilson DSc
I admitted my mistakes and corrected them in the next post.
If so, that is a pleasant change for you .. Where is the next post?
Just below the first one. posted on 14th
here's a copy.
***************
Post by Inertial
Post by Henry Wilson DSc
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT
ROTATING.
Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
What is q and v?
Post by Henry Wilson DSc
Post by Henry Wilson DSc
(the surface is drawn flat for convenience. The fibre goes right around the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre. The pulses are labelled according to their emission times and
move
at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p.
Doesn't matter if they are detected at p or at C
I'll withdraw my previous reply because it was wrong. So was my initial message
because I stuffed up what I wanted to say.
That's fine. Everyone makes mistakes at some time or another. I am always
willing and happy to take that into account when it is admitted in that way.
Inertial
2010-02-14 23:28:50 UTC
Post by Henry Wilson DSc
Two different arrival rates produces a beat frequency at the detector.
There is no beat frequency in a Sagnac device. The two beams have the same
frequency at the detector. This is the case in both emission theory and SR.
SR, however, predicts differing arrival TIMES for pulses emitted at the same
time .. hence the shift. Emission theory says they arrive at the same time
(as you agree) .. hence there is no shift predicted (this refuting the
theory).

That your theory is even worse and somehow predicts differing arrival rates
and a beat frequency that does not occur is enough proof to refute it.
Henry Wilson DSc
2010-02-15 00:32:32 UTC
Post by Inertial
Post by Henry Wilson DSc
Two different arrival rates produces a beat frequency at the detector.
There is no beat frequency in a Sagnac device. The two beams have the same
frequency at the detector. This is the case in both emission theory and SR.
SR, however, predicts differing arrival TIMES for pulses emitted at the same
time .. hence the shift. Emission theory says they arrive at the same time
(as you agree) .. hence there is no shift predicted (this refuting the
theory).
That your theory is even worse and somehow predicts differing arrival rates
and a beat frequency that does not occur is enough proof to refute it.
You still don't get it.

The two beams DO NOT have the same frequency at the detector.
In the clock example I gave, 40000062 pulses arrive at the detector from one
direction but only 3999938 from the other DURING THE COMMON TRAVEL TIME of any
particular split pulse. Those numbers change during a speed CHANGE.

Convert that to 'wavecrests', in the case of light, and you get a beat
frequency that is somehow related to fringe displacement. Don't ask me how at
the moment but the reason will reveal quite a lot about the nature of a photon.

The fact that the frequencies in each direction are different annihilates the
Roberts, Jerry, Andersen 'rotating frame' argument as well as your own.

Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-15 00:50:45 UTC
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Two different arrival rates produces a beat frequency at the detector.
There is no beat frequency in a Sagnac device. The two beams have the same
frequency at the detector. This is the case in both emission theory and SR.
SR, however, predicts differing arrival TIMES for pulses emitted at the same
time .. hence the shift. Emission theory says they arrive at the same time
(as you agree) .. hence there is no shift predicted (this refuting the
theory).
That your theory is even worse and somehow predicts differing arrival rates
and a beat frequency that does not occur is enough proof to refute it.
You still don't get it.
I get it just fine
Post by Henry Wilson DSc
The two beams DO NOT have the same frequency at the detector.
Yes, they do in reality. That your theory predicts otherwise shows it to be
wrong.
Post by Henry Wilson DSc
In the clock example I gave, 40000062 pulses arrive at the detector from one
direction but only 3999938 from the other DURING THE COMMON TRAVEL TIME of any
particular split pulse. Those numbers change during a speed CHANGE.
And that is nonsense .. as the time taken for each pulse to make the trip is
the same (in both directions), and the time between pulses being emitted is
the same (in both directions), then in a given time, the same number of
pulses arrive (in both directions).
Post by Henry Wilson DSc
Convert that to 'wavecrests', in the case of light, and you get a beat
frequency that is somehow related to fringe displacement. Don't ask me how at
the moment but the reason will reveal quite a lot about the nature of a photon.
Because it is nonnsese. There is no beat frequency.
Post by Henry Wilson DSc
The fact
Lie
Post by Henry Wilson DSc
that the frequencies in each direction are different
They are not.
Post by Henry Wilson DSc
annihilates the
Roberts, Jerry, Andersen 'rotating frame' argument as well as your own.
Nope .. you've just proven your own argument wrong. Well done.
Henry Wilson DSc
2010-02-15 02:21:39 UTC
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Two different arrival rates produces a beat frequency at the detector.
There is no beat frequency in a Sagnac device. The two beams have the same
frequency at the detector. This is the case in both emission theory and SR.
SR, however, predicts differing arrival TIMES for pulses emitted at the same
time .. hence the shift. Emission theory says they arrive at the same time
(as you agree) .. hence there is no shift predicted (this refuting the
theory).
That your theory is even worse and somehow predicts differing arrival rates
and a beat frequency that does not occur is enough proof to refute it.
You still don't get it.
I get it just fine
Post by Henry Wilson DSc
The two beams DO NOT have the same frequency at the detector.
Yes, they do in reality. That your theory predicts otherwise shows it to be
wrong.
Post by Henry Wilson DSc
In the clock example I gave, 40000062 pulses arrive at the detector from one
direction but only 3999938 from the other DURING THE COMMON TRAVEL TIME of any
particular split pulse. Those numbers change during a speed CHANGE.
And that is nonsense .. as the time taken for each pulse to make the trip is
the same (in both directions), and the time between pulses being emitted is
the same (in both directions), then in a given time, the same number of
pulses arrive (in both directions).
Try thinking for once. I know this is difficult...but you should be able to
understand.
The path lengths are different. The distance between pulses is 1 metre in all
frames. The number of pulses in each path differs by 124. When a pulse is split
into two and sent down both fibres, it takes the same time to get to the
detector in both directions. Both halves arrive together.
So 124 more pulses have arrived from one direction than the other.
IN OTHER WORDS, the pulse arrival rate is different from both paths.
Post by Inertial
Post by Henry Wilson DSc
Convert that to 'wavecrests', in the case of light, and you get a beat
frequency that is somehow related to fringe displacement. Don't ask me how at
the moment but the reason will reveal quite a lot about the nature of a photon.
Because it is nonnsese. There is no beat frequency.
Post by Henry Wilson DSc
The fact
Lie
Post by Henry Wilson DSc
that the frequencies in each direction are different
They are not.
Post by Henry Wilson DSc
annihilates the
Roberts, Jerry, Andersen 'rotating frame' argument as well as your own.
Nope .. you've just proven your own argument wrong. Well done.

Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-15 02:32:23 UTC
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Two different arrival rates produces a beat frequency at the detector.
There is no beat frequency in a Sagnac device. The two beams have the same
frequency at the detector. This is the case in both emission theory and SR.
SR, however, predicts differing arrival TIMES for pulses emitted at the same
time .. hence the shift. Emission theory says they arrive at the same time
(as you agree) .. hence there is no shift predicted (this refuting the
theory).
That your theory is even worse and somehow predicts differing arrival rates
and a beat frequency that does not occur is enough proof to refute it.
You still don't get it.
I get it just fine
Post by Henry Wilson DSc
The two beams DO NOT have the same frequency at the detector.
Yes, they do in reality. That your theory predicts otherwise shows it to be
wrong.
Post by Henry Wilson DSc
In the clock example I gave, 40000062 pulses arrive at the detector from one
direction but only 3999938 from the other DURING THE COMMON TRAVEL TIME
of
any
particular split pulse. Those numbers change during a speed CHANGE.
And that is nonsense .. as the time taken for each pulse to make the trip is
the same (in both directions), and the time between pulses being emitted is
the same (in both directions), then in a given time, the same number of
pulses arrive (in both directions).
Try thinking for once. I know this is difficult...but you should be able to
understand.
The path lengths are different. The distance between pulses is 1 metre in all
frames. The number of pulses in each path differs by 124. When a pulse is split
into two and sent down both fibres, it takes the same time to get to the
detector in both directions. Both halves arrive together.
So 124 more pulses have arrived from one direction than the other.
IN OTHER WORDS, the pulse arrival rate is different from both paths.
Post by Inertial
Post by Henry Wilson DSc
Convert that to 'wavecrests', in the case of light, and you get a beat
frequency that is somehow related to fringe displacement. Don't ask me
how
at
the moment but the reason will reveal quite a lot about the nature of a photon.
Because it is nonnsese. There is no beat frequency.
Post by Henry Wilson DSc
The fact
Lie
Post by Henry Wilson DSc
that the frequencies in each direction are different
They are not.
Post by Henry Wilson DSc
annihilates the
Roberts, Jerry, Andersen 'rotating frame' argument as well as your own.
Nope .. you've just proven your own argument wrong. Well done.
I did .. and you just refuted your own theory. Saying 'it predicts
something that doesn't happen and somehow instead what is observed does
happen anyway and my theory cannot explain why' means your theory is
refuted.
Henry Wilson DSc
2010-02-15 02:45:18 UTC
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
annihilates the
Roberts, Jerry, Andersen 'rotating frame' argument as well as your own.
Nope .. you've just proven your own argument wrong. Well done.
I did .. and you just refuted your own theory. Saying 'it predicts
something that doesn't happen and somehow instead what is observed does
happen anyway and my theory cannot explain why' means your theory is
refuted.
Don't kid yourself.

By counting the pulses arriving from the two directions, I can calculate the
earth's rotation speed. I don't need an interferometer. A clock will do just as
well.

Sagnac fully supports BaTh.

Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-15 02:55:08 UTC
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
annihilates the
Roberts, Jerry, Andersen 'rotating frame' argument as well as your own.
Nope .. you've just proven your own argument wrong. Well done.
I did .. and you just refuted your own theory. Saying 'it predicts
something that doesn't happen and somehow instead what is observed does
happen anyway and my theory cannot explain why' means your theory is
refuted.
Don't kid yourself.
I'm not
Post by Henry Wilson DSc
By counting the pulses arriving from the two directions, I can calculate the
earth's rotation speed.
Nope. as the tube has a fixed length, the same number must arrive per
second as are sent per second. Otherwise either pulses are disappearing
along the way, or being created from nowhere. You theory relies on pulse
fairies.
Post by Henry Wilson DSc
I don't need an interferometer. A clock will do just as
well.
Sagnac fully supports BaTh.
Nope .. your supposed 'BaTh' is pure impossible nonsense.
Inertial
2010-02-15 02:52:59 UTC
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Two different arrival rates produces a beat frequency at the detector.
There is no beat frequency in a Sagnac device. The two beams have the same
frequency at the detector. This is the case in both emission theory and SR.
SR, however, predicts differing arrival TIMES for pulses emitted at the same
time .. hence the shift. Emission theory says they arrive at the same time
(as you agree) .. hence there is no shift predicted (this refuting the
theory).
That your theory is even worse and somehow predicts differing arrival rates
and a beat frequency that does not occur is enough proof to refute it.
You still don't get it.
I get it just fine
Post by Henry Wilson DSc
The two beams DO NOT have the same frequency at the detector.
Yes, they do in reality. That your theory predicts otherwise shows it to be
wrong.
Post by Henry Wilson DSc
In the clock example I gave, 40000062 pulses arrive at the detector from one
direction but only 3999938 from the other DURING THE COMMON TRAVEL TIME
of
any
particular split pulse. Those numbers change during a speed CHANGE.
And that is nonsense .. as the time taken for each pulse to make the trip is
the same (in both directions), and the time between pulses being emitted is
the same (in both directions), then in a given time, the same number of
pulses arrive (in both directions).
Try thinking for once. I know this is difficult...
It is for you
Post by Henry Wilson DSc
but you should be able to
understand.
I do
Post by Henry Wilson DSc
The path lengths are different.
The distance between pulses is 1 metre in all
frames. The number of pulses in each path differs by 124. When a pulse is split
into two and sent down both fibres, it takes the same time to get to the
detector in both directions. Both halves arrive together.
So 124 more pulses have arrived from one direction than the other.
IN OTHER WORDS, the pulse arrival rate is different from both paths.
So the following is equivalent to what you have here:

You have two paths from (say) A to B.

You have n objects leaving A over a period of 1 second down one path and
taking time T to get to B. The first object arrives at time T, the last at
time T+1.

You have another n objects leaving A over a period of 1 second down the
other path and taking the same time T to get to B. The first object arrives
at time T, the last at time T+1.

Yet by your logic, you would claim that more would arrive per second at B
from one path than the other.

That is just pure impossible nonsense.

You also claim that even though in any given time you emit the same number
of pulses in both directions, and that they are all in the tube for the same
time, that there are more pulses in the tube going in one direction than
their are in the other. Where do the extra pulses come from?

That is also just pure impossible nonsense.

At any given time the tube has the same length in both directions .. the
length is the circumference of the tube, it does not change. And as you say
that the pulses are the same distance apart, there must be the same number
of them in the tube at a time. But you claim that at any given time there
are more pulses in the tube going in one direction than that other.

That is also just pure impossible nonsense.

You really have just totally refuted your 'thoery'. It is all based on pure
impossible nonsense.
Henry Wilson DSc
2010-02-15 05:44:11 UTC
Post by Inertial
Post by Henry Wilson DSc
The path lengths are different.
The distance between pulses is 1 metre in all
frames. The number of pulses in each path differs by 124. When a pulse is split
into two and sent down both fibres, it takes the same time to get to the
detector in both directions. Both halves arrive together.
So 124 more pulses have arrived from one direction than the other.
IN OTHER WORDS, the pulse arrival rate is different from both paths.
You have two paths from (say) A to B.
You have n objects leaving A over a period of 1 second down one path and
taking time T to get to B. The first object arrives at time T, the last at
time T+1.
You have another n objects leaving A over a period of 1 second down the
other path and taking the same time T to get to B. The first object arrives
at time T, the last at time T+1.
THIS IS WRONG. HOW LONG DOES IT TAKE FOR SOMETHING TO SINK INTO YOUR BRAIN.

You have n+m objects in one path and n-m objects in the other.

One path moves faster than the other so the nth object in each path reunites
with its counterpart simultaneously at the detector. While it is traveling, n+m
objects reach the detector one way and n-m the other way.
Post by Inertial
Yet by your logic, you would claim that more would arrive per second at B
from one path than the other.
Do you know what a bicycle chain looks like? The faster you move it the more
Post by Inertial
That is just pure impossible nonsense.
What you are saying is pure nonsense , yes.
Post by Inertial
You also claim that even though in any given time you emit the same number
of pulses in both directions, and that they are all in the tube for the same
time, that there are more pulses in the tube going in one direction than
their are in the other. Where do the extra pulses come from?
They enter or leave each path during a speed change...That is when the fringes
are seen to MOVE.
Post by Inertial
That is also just pure impossible nonsense.
What you are saying is pure nonsense, yes.
Post by Inertial
At any given time the tube has the same length in both directions .. the
length is the circumference of the tube, it does not change. And as you say
that the pulses are the same distance apart, there must be the same number
of them in the tube at a time.
Only when it is not rotating. and p = q.

You have to consider the distance C moves while a pulse is in transit. That is
the reason for the path length difference.

I might remind you that your own SR says exactly that too.
Post by Inertial
But you claim that at any given time there
are more pulses in the tube going in one direction than that other.
That is also just pure impossible nonsense.
You really have just totally refuted your 'thoery'. It is all based on pure
impossible nonsense.
You will probably feel very embarrased when and if the penny drops but I doubt
if you have the intelligence to make that happen.

Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-15 06:12:31 UTC
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
The path lengths are different.
The distance between pulses is 1 metre in all
frames. The number of pulses in each path differs by 124. When a pulse
is
split
into two and sent down both fibres, it takes the same time to get to the
detector in both directions. Both halves arrive together.
So 124 more pulses have arrived from one direction than the other.
IN OTHER WORDS, the pulse arrival rate is different from both paths.
You have two paths from (say) A to B.
You have n objects leaving A over a period of 1 second down one path and
taking time T to get to B. The first object arrives at time T, the last at
time T+1.
You have another n objects leaving A over a period of 1 second down the
other path and taking the same time T to get to B. The first object arrives
at time T, the last at time T+1.
THIS IS WRONG. HOW LONG DOES IT TAKE FOR SOMETHING TO SINK INTO YOUR BRAIN.
No .. its not wrong.
Post by Henry Wilson DSc
You have n+m objects in one path and n-m objects in the other.
Where did the m extra objects come from? The same number go down each path
and the same number arrive at the other end. So each pair that leave, they
arrive together. Where are these extra/missing pulses?
Post by Henry Wilson DSc
One path moves faster than the other so the nth object in each path reunites
with its counterpart simultaneously at the detector.
So that means there must be the same number of objects in the path.
Post by Henry Wilson DSc
While it is traveling, n+m
objects reach the detector one way and n-m the other way.
Nope. Impossible. For each pulse entering path 1, a pulse enters path 2.
For each pulse exiting path 1, a pulse exits path 2. When do the m extra
pulses enter one path, and when do m fewer enter the other?
Post by Henry Wilson DSc
Post by Inertial
Yet by your logic, you would claim that more would arrive per second at B
from one path than the other.
Do you know what a bicycle chain looks like? The faster you move it the more
Irrelevant
Post by Henry Wilson DSc
Post by Inertial
That is just pure impossible nonsense.
What you are saying is pure nonsense , yes.
I'm only repeating what you are claiming
Post by Henry Wilson DSc
Post by Inertial
You also claim that even though in any given time you emit the same number
of pulses in both directions, and that they are all in the tube for the same
time, that there are more pulses in the tube going in one direction than
their are in the other. Where do the extra pulses come from?
They enter or leave each path during a speed change...
There is no speed change. The rotation is a constant rate.
Post by Henry Wilson DSc
That is when the fringes
are seen to MOVE.
Post by Inertial
That is also just pure impossible nonsense.
What you are saying is pure nonsense, yes.
I'm only repeating what you are claiming
Post by Henry Wilson DSc
Post by Inertial
At any given time the tube has the same length in both directions .. the
length is the circumference of the tube, it does not change. And as you say
that the pulses are the same distance apart, there must be the same number
of them in the tube at a time.
Only when it is not rotating. and p = q.
Wrong
Post by Henry Wilson DSc
You have to consider the distance C moves while a pulse is in transit. That is
the reason for the path length difference.
Path length is different .. BUT length of tube is unchanged. The distance
between pulses is the same .. so number of pulses in each direction in the
tube must be the same.
Post by Henry Wilson DSc
I might remind you that your own SR says exactly that too.
SR's correctness is not relevant in a discussion of your nonsense.
Post by Henry Wilson DSc
Post by Inertial
But you claim that at any given time there
are more pulses in the tube going in one direction than that other.
That is also just pure impossible nonsense.
You really have just totally refuted your 'thoery'. It is all based on pure
impossible nonsense.
You will probably feel very embarrased when and if the penny drops but I doubt
if you have the intelligence to make that happen.
You won't ever admit your mistakes .. you're too stupid a moron to realise
and too dishonest to admit it if you did.
Henry Wilson DSc
2010-02-15 10:33:10 UTC
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
You have two paths from (say) A to B.
You have n objects leaving A over a period of 1 second down one path and
taking time T to get to B. The first object arrives at time T, the last at
time T+1.
You have another n objects leaving A over a period of 1 second down the
other path and taking the same time T to get to B. The first object arrives
at time T, the last at time T+1.
THIS IS WRONG. HOW LONG DOES IT TAKE FOR SOMETHING TO SINK INTO YOUR BRAIN.
No .. its not wrong.
Post by Henry Wilson DSc
You have n+m objects in one path and n-m objects in the other.
Where did the m extra objects come from? The same number go down each path
and the same number arrive at the other end. So each pair that leave, they
arrive together. Where are these extra/missing pulses?
Look, that is the case when it is not rotating.

What you and many others cannot get into your heads is that when it IS
rotating, the source/detector moves a distance vt during the travel time of a
pulse.

This is what makes the path lengths unequal...and results in more pulses being
present in one path than the other.

I know this is difficult but you should at least try to understand it.
It is the basis of the SR sagnac explanation as well as BaTh's.
Post by Inertial
Post by Henry Wilson DSc
One path moves faster than the other so the nth object in each path reunites
with its counterpart simultaneously at the detector.
So that means there must be the same number of objects in the path.
You are ignoring the distance vt. In both SR and BaTh, the path lengths are
2piR + vt and 2piR - vt.
Post by Inertial
Post by Henry Wilson DSc
While it is traveling, n+m
objects reach the detector one way and n-m the other way.
Nope. Impossible. For each pulse entering path 1, a pulse enters path 2.
For each pulse exiting path 1, a pulse exits path 2. When do the m extra
pulses enter one path, and when do m fewer enter the other?
I have told you already...during an acceleration.
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Yet by your logic, you would claim that more would arrive per second at B
from one path than the other.
Do you know what a bicycle chain looks like? The faster you move it the more
Irrelevant
no it isn't . You can obviously only understand very simple language,
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
time, that there are more pulses in the tube going in one direction than
their are in the other. Where do the extra pulses come from?
They enter or leave each path during a speed change...
There is no speed change. The rotation is a constant rate.
At constant rate, vt is constant and there is fringe displacement but no fringe
movement.
Post by Inertial
Post by Henry Wilson DSc
That is when the fringes
are seen to MOVE.
Post by Inertial
That is also just pure impossible nonsense.
What you are saying is pure nonsense, yes.
I'm only repeating what you are claiming
Post by Henry Wilson DSc
Post by Inertial
At any given time the tube has the same length in both directions .. the
length is the circumference of the tube, it does not change. And as you say
that the pulses are the same distance apart, there must be the same number
of them in the tube at a time.
Only when it is not rotating. and p = q.
Wrong
Post by Henry Wilson DSc
You have to consider the distance C moves while a pulse is in transit. That is
the reason for the path length difference.
Path length is different .. BUT length of tube is unchanged. The distance
between pulses is the same .. so number of pulses in each direction in the
tube must be the same.
The length of the fibre is NOT the path length. You are ignoring the distance
vt.
You are also ignoring the fact that every rotating frame has an associated
intertial frame.
Post by Inertial
Post by Henry Wilson DSc
I might remind you that your own SR says exactly that too.
SR's correctness is not relevant in a discussion of your nonsense.
I was merely emphasising your inability to understand plain physics.
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
But you claim that at any given time there
are more pulses in the tube going in one direction than that other.
That is also just pure impossible nonsense.
You really have just totally refuted your 'thoery'. It is all based on pure
impossible nonsense.
You will probably feel very embarrased when and if the penny drops but I doubt
if you have the intelligence to make that happen.
You won't ever admit your mistakes .. you're too stupid a moron to realise
and too dishonest to admit it if you did.
You are as useless as Einstein...zero spatial and logical ability.

Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-15 10:59:11 UTC
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
You have two paths from (say) A to B.
You have n objects leaving A over a period of 1 second down one path and
taking time T to get to B. The first object arrives at time T, the last at
time T+1.
You have another n objects leaving A over a period of 1 second down the
other path and taking the same time T to get to B. The first object arrives
at time T, the last at time T+1.
THIS IS WRONG. HOW LONG DOES IT TAKE FOR SOMETHING TO SINK INTO YOUR BRAIN.
No .. its not wrong.
Post by Henry Wilson DSc
You have n+m objects in one path and n-m objects in the other.
Where did the m extra objects come from? The same number go down each path
and the same number arrive at the other end. So each pair that leave, they
arrive together. Where are these extra/missing pulses?
Look, that is the case when it is not rotating.
Rotation doesn't change it.
Post by Henry Wilson DSc
What you and many others cannot get into your heads is that when it IS
rotating, the source/detector moves a distance vt during the travel time of a
pulse.
Of course they do
Post by Henry Wilson DSc
This is what makes the path lengths unequal...
Of course they are
Post by Henry Wilson DSc
and results in more pulses being
present in one path than the other.
The number of pulses in the tube at any given time going one way is the same
as the number going the other.
Post by Henry Wilson DSc
I know this is difficult but you should at least try to understand it.
It is the basis of the SR sagnac explanation as well as BaTh's.
Nope. Your lies are not a basis for anything
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
One path moves faster than the other so the nth object in each path reunites
with its counterpart simultaneously at the detector.
So that means there must be the same number of objects in the path.
You are ignoring the distance vt.
Not ignoring it. It is compensated by the pulses moving faster along the
longer path.
Post by Henry Wilson DSc
In both SR and BaTh, the path lengths are
2piR + vt and 2piR - vt.
Nope. SR and Emission theory have different pairs of values for the two
path lengths for the two path lengths.
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
While it is traveling, n+m
objects reach the detector one way and n-m the other way.
Nope. Impossible. For each pulse entering path 1, a pulse enters path 2.
For each pulse exiting path 1, a pulse exits path 2. When do the m extra
pulses enter one path, and when do m fewer enter the other?
I have told you already...during an acceleration.
There is no acceleration of the tube rotation rate. The tube rotates at a
constant speed
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Yet by your logic, you would claim that more would arrive per second at B
from one path than the other.
Do you know what a bicycle chain looks like? The faster you move it the more
Irrelevant
no it isn't .
It isn't relevant.
Post by Henry Wilson DSc
You can obviously only understand very simple language,
I can understand things that you have no hope of.
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
time, that there are more pulses in the tube going in one direction than
their are in the other. Where do the extra pulses come from?
They enter or leave each path during a speed change...
There is no speed change. The rotation is a constant rate.
At constant rate, vt is constant and there is fringe displacement but no fringe
movement.
Your theory does not predict that. You say that there is a frequency beat
which means a constantly changing fringe displacement.
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
That is when the fringes
are seen to MOVE.
Post by Inertial
That is also just pure impossible nonsense.
What you are saying is pure nonsense, yes.
I'm only repeating what you are claiming
Post by Henry Wilson DSc
Post by Inertial
At any given time the tube has the same length in both directions .. the
length is the circumference of the tube, it does not change. And as you say
that the pulses are the same distance apart, there must be the same number
of them in the tube at a time.
Only when it is not rotating. and p = q.
Wrong
Post by Henry Wilson DSc
You have to consider the distance C moves while a pulse is in transit. That is
the reason for the path length difference.
Path length is different .. BUT length of tube is unchanged. The distance
between pulses is the same .. so number of pulses in each direction in the
tube must be the same.
The length of the fibre is NOT the path length.
I didn't say it was
Post by Henry Wilson DSc
You are ignoring the distance
vt.
No .. it is simply not relevant to the number of pulses in the tube at a
given time. Only the tube length is relevant.
Post by Henry Wilson DSc
You are also ignoring the fact that every rotating frame has an associated
intertial frame.
Irrelevant.
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
I might remind you that your own SR says exactly that too.
SR's correctness is not relevant in a discussion of your nonsense.
I was merely emphasising your inability to understand plain physics.
I understand it just fine. And I can see your nonsense very clearly. Its
hilarious
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
But you claim that at any given time there
are more pulses in the tube going in one direction than that other.
That is also just pure impossible nonsense.
You really have just totally refuted your 'thoery'. It is all based on pure
impossible nonsense.
You will probably feel very embarrased when and if the penny drops but I doubt
if you have the intelligence to make that happen.
You won't ever admit your mistakes .. you're too stupid a moron to realise
and too dishonest to admit it if you did.
You are as useless as Einstein...zero spatial and logical ability.
You're the one who has pulse fairies creating pulses and fringe fairies
stoping the beat frequency changing the fringe displacement. Your theory is
just a work of fiction and unrelated to reality.
Henry Wilson DSc
2010-02-15 22:17:32 UTC
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
You have two paths from (say) A to B.
You have n objects leaving A over a period of 1 second down one path and
taking time T to get to B. The first object arrives at time T, the last at
time T+1.
You have another n objects leaving A over a period of 1 second down the
other path and taking the same time T to get to B. The first object arrives
at time T, the last at time T+1.
THIS IS WRONG. HOW LONG DOES IT TAKE FOR SOMETHING TO SINK INTO YOUR BRAIN.
No .. its not wrong.
Post by Henry Wilson DSc
You have n+m objects in one path and n-m objects in the other.
Where did the m extra objects come from? The same number go down each path
and the same number arrive at the other end. So each pair that leave, they
arrive together. Where are these extra/missing pulses?
Look, that is the case when it is not rotating.
Rotation doesn't change it.
It changes the path lengths. Are you not aware that light takes time to travel
and during its passage around the Earth, the Earth moves a distance vt.
Gawd, just look up any SR analysis of sagnac and you will see this.
Post by Inertial
Post by Henry Wilson DSc
What you and many others cannot get into your heads is that when it IS
rotating, the source/detector moves a distance vt during the travel time of a
pulse.
Of course they do
Post by Henry Wilson DSc
This is what makes the path lengths unequal...
Of course they are
Post by Henry Wilson DSc
and results in more pulses being
present in one path than the other.
The number of pulses in the tube at any given time going one way is the same
as the number going the other.
It fucking is NOT.
The numbers arriving from each direction between the emission of pulse X and
its arrival at the detector differ by 124 in my example.
If you cannot understand this I will waste no further time on you.
Post by Inertial
Post by Henry Wilson DSc
I know this is difficult but you should at least try to understand it.
It is the basis of the SR sagnac explanation as well as BaTh's.
Nope. Your lies are not a basis for anything
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
One path moves faster than the other so the nth object in each path reunites
with its counterpart simultaneously at the detector.
So that means there must be the same number of objects in the path.
You are ignoring the distance vt.
Not ignoring it. It is compensated by the pulses moving faster along the
longer path.
They DO move faster. That's why more arrive at the detector in the same time.
Post by Inertial
Post by Henry Wilson DSc
In both SR and BaTh, the path lengths are
2piR + vt and 2piR - vt.
Nope. SR and Emission theory have different pairs of values for the two
path lengths for the two path lengths.
They do not. Both theories asy the path lengths are 2piR+/vt

I think it's time you did some reading and used your brain...if you have one.
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
While it is traveling, n+m
objects reach the detector one way and n-m the other way.
Nope. Impossible. For each pulse entering path 1, a pulse enters path 2.
For each pulse exiting path 1, a pulse exits path 2. When do the m extra
pulses enter one path, and when do m fewer enter the other?
I have told you already...during an acceleration.
There is no acceleration of the tube rotation rate. The tube rotates at a
constant speed
gawd! what an idiot.

go away you are not capable of discussing physics...

Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-15 22:25:09 UTC
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
You have two paths from (say) A to B.
You have n objects leaving A over a period of 1 second down one path and
taking time T to get to B. The first object arrives at time T, the
last
at
time T+1.
You have another n objects leaving A over a period of 1 second down the
other path and taking the same time T to get to B. The first object arrives
at time T, the last at time T+1.
THIS IS WRONG. HOW LONG DOES IT TAKE FOR SOMETHING TO SINK INTO YOUR BRAIN.
No .. its not wrong.
Post by Henry Wilson DSc
You have n+m objects in one path and n-m objects in the other.
Where did the m extra objects come from? The same number go down each path
and the same number arrive at the other end. So each pair that leave, they
arrive together. Where are these extra/missing pulses?
Look, that is the case when it is not rotating.
Rotation doesn't change it.
It changes the path lengths.
It doesn't change the length of the tube, or the number of pulses in transit
at any time nor the arrival rate.
Post by Henry Wilson DSc
Are you not aware that light takes time to travel
and during its passage around the Earth, the Earth moves a distance vt.
Gawd, just look up any SR analysis of sagnac and you will see this.
You don't understand the SR analysis .. nor SR itself.
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
What you and many others cannot get into your heads is that when it IS
rotating, the source/detector moves a distance vt during the travel time of a
pulse.
Of course they do
Post by Henry Wilson DSc
This is what makes the path lengths unequal...
Of course they are
Post by Henry Wilson DSc
and results in more pulses being
present in one path than the other.
The number of pulses in the tube at any given time going one way is the same
as the number going the other.
It fucking is NOT.
Of course it is .. if you put the same number of pulses into a fixed length
tube and they are in the tube for the same time, its the same number of
pulses
Post by Henry Wilson DSc
The numbers arriving from each direction between the emission of pulse X and
its arrival at the detector differ by 124 in my example.
Nopwe .. you're a blatant liar or a fool. My bet is both.
Post by Henry Wilson DSc
If you cannot understand this I will waste no further time on you.
So you run away when you are proven wrong. Typical cowardly lying crackpot
Henry.
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
I know this is difficult but you should at least try to understand it.
It is the basis of the SR sagnac explanation as well as BaTh's.
Nope. Your lies are not a basis for anything
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
One path moves faster than the other so the nth object in each path reunites
with its counterpart simultaneously at the detector.
So that means there must be the same number of objects in the path.
You are ignoring the distance vt.
Not ignoring it. It is compensated by the pulses moving faster along the
longer path.
They DO move faster. That's why more arrive at the detector in the same time.
No .. they don't. The same number of pulses are emitted per second, so the
same number must be detected per second
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
In both SR and BaTh, the path lengths are
2piR + vt and 2piR - vt.
Nope. SR and Emission theory have different pairs of values for the two
path lengths for the two path lengths.
They do not.
Yes .. the ydo have different values
Post by Henry Wilson DSc
Both theories asy the path lengths are 2piR+/vt
The t values are different for the two pulses in SR, not in emission theory

You just have no idea.
Post by Henry Wilson DSc
I think it's time you did some reading and used your brain...if you have one.
I use mine .. your only use of your brain is to lie and cheat. What a
waste.
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
While it is traveling, n+m
objects reach the detector one way and n-m the other way.
Nope. Impossible. For each pulse entering path 1, a pulse enters path 2.
For each pulse exiting path 1, a pulse exits path 2. When do the m extra
pulses enter one path, and when do m fewer enter the other?
I have told you already...during an acceleration.
There is no acceleration of the tube rotation rate. The tube rotates at a
constant speed
gawd! what an idiot.
Indeed you are
Post by Henry Wilson DSc
go away you are not capable of discussing physics...
A dsicssion of physics with a moron like you is pretty pointless. But it is
worth having your lies exposed for the benefit of others.

And they have been .. your assertions are soundly refuted.
Inertial
2010-02-15 22:31:25 UTC
Post by Henry Wilson DSc
The numbers arriving from each direction between the emission of pulse X and
its arrival at the detector differ by 124 in my example.
So, according to you in some fixed period of time, there are 62 more pulses
arriving in one direction than were emitted in that same period of time. So
if you leave it running for a while, you end up with more pulses coming out
of the device than have gone into it. You've created free energy.
Congratulations.
Henry Wilson DSc
2010-02-15 23:04:25 UTC
Post by Inertial
Post by Henry Wilson DSc
The numbers arriving from each direction between the emission of pulse X and
its arrival at the detector differ by 124 in my example.
So, according to you in some fixed period of time, there are 62 more pulses
arriving in one direction than were emitted in that same period of time.
Yes. It's called doppler shift.
Post by Inertial
So
if you leave it running for a while, you end up with more pulses coming out
of the device than have gone into it. You've created free energy.
Congratulations.
moron. The increased number in one path is matched by a decrease in the other.
Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-15 23:11:37 UTC
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
The numbers arriving from each direction between the emission of pulse X and
its arrival at the detector differ by 124 in my example.
So, according to you in some fixed period of time, there are 62 more pulses
arriving in one direction than were emitted in that same period of time.
Yes. It's called doppler shift.
There is no Doppler shift, you moron.
Post by Henry Wilson DSc
Post by Inertial
So
if you leave it running for a while, you end up with more pulses coming out
of the device than have gone into it. You've created free energy.
Congratulations.
moron. The increased number in one path is matched by a decrease in the other.
So .. you only emit down the path with more pulses coming out than going in
and you create free energy.

You really are an idiot if you continue this pretense that what you said was
right.

If two pulses go into a fixed length circular tube (in opposite directions)
and travel for the same time, then they MUST arrive at the detector at the
same time and so the arrival rate of successive pulses MUST be the same.
Your claim to the contrary is ridiculous.
Henry Wilson DSc
2010-02-16 02:58:03 UTC
Post by Inertial
Post by Henry Wilson DSc
moron. The increased number in one path is matched by a decrease in the other.
So .. you only emit down the path with more pulses coming out than going in
and you create free energy.
The extra pulses are already in there.
Post by Inertial
You really are an idiot if you continue this pretense that what you said was
right.
If two pulses go into a fixed length circular tube (in opposite directions)
and travel for the same time, then they MUST arrive at the detector at the
same time and so the arrival rate of successive pulses MUST be the same.
Your claim to the contrary is ridiculous.
This is just far too hard for you I'm afraid.

You cannot get it into your head that, when it is rotating, the path length
DOES NOT EQUAL the tube length.

Here it is, SR style: http://en.wikipedia.org/wiki/Sagnac_effect

Go down to 'theory'. SEE THE DIAGRAM WITH THE OMEGA IN THE MIDDLE?

Henry Wilson...

.......provider of free physics lessons
eric gisse
2010-02-16 03:34:13 UTC
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
moron. The increased number in one path is matched by a decrease in the other.
So .. you only emit down the path with more pulses coming out than going
in and you create free energy.
The extra pulses are already in there.
...and where do they come from?

[...]
Inertial
2010-02-16 03:36:22 UTC
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
moron. The increased number in one path is matched by a decrease in the other.
So .. you only emit down the path with more pulses coming out than going in
and you create free energy.
The extra pulses are already in there.
Post by Inertial
You really are an idiot if you continue this pretense that what you said was
right.
If two pulses go into a fixed length circular tube (in opposite directions)
and travel for the same time, then they MUST arrive at the detector at the
same time and so the arrival rate of successive pulses MUST be the same.
Your claim to the contrary is ridiculous.
This is just far too hard for you I'm afraid.
No .. you're just wrong. And I (and everyone else here but you) sees that
clearly.
Post by Henry Wilson DSc
You cannot get it into your head that, when it is rotating, the path length
DOES NOT EQUAL the tube length.
Of course it is different .. I didn't say it was the same. That doesn't
make any difference to the complete fuckup you made here. You claim of
different arrival rates for the pulses is contradicted by your own claim
that pairs of pulses that are emitted at the same time arrive at the same
time.
Post by Henry Wilson DSc
Here it is, SR style: http://en.wikipedia.org/wiki/Sagnac_effect
Go down to 'theory'. SEE THE DIAGRAM WITH THE OMEGA IN THE MIDDLE?
No need. I'm familiar with Sagnac, and unlike you, I understand it (and SR,
which you never could understand because when you tried to it was just
incoherent to you. Fortunately, those of us with a bit more brain power and
understanding of physics find it quite easy to follow).
Henry Wilson DSc
2010-02-16 06:03:39 UTC
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
moron. The increased number in one path is matched by a decrease in the other.
So .. you only emit down the path with more pulses coming out than going in
and you create free energy.
The extra pulses are already in there.
Post by Inertial
You really are an idiot if you continue this pretense that what you said was
right.
If two pulses go into a fixed length circular tube (in opposite directions)
and travel for the same time, then they MUST arrive at the detector at the
same time and so the arrival rate of successive pulses MUST be the same.
Your claim to the contrary is ridiculous.
This is just far too hard for you I'm afraid.
No .. you're just wrong. And I (and everyone else here but you) sees that
clearly.
Post by Henry Wilson DSc
You cannot get it into your head that, when it is rotating, the path length
DOES NOT EQUAL the tube length.
Of course it is different .. I didn't say it was the same. That doesn't
make any difference to the complete fuckup you made here. You claim of
different arrival rates for the pulses is contradicted by your own claim
that pairs of pulses that are emitted at the same time arrive at the same
time.
Post by Henry Wilson DSc
Here it is, SR style: http://en.wikipedia.org/wiki/Sagnac_effect
Go down to 'theory'. SEE THE DIAGRAM WITH THE OMEGA IN THE MIDDLE?
No need. I'm familiar with Sagnac, and unlike you, I understand it (and SR,
which you never could understand because when you tried to it was just
incoherent to you. Fortunately, those of us with a bit more brain power and
understanding of physics find it quite easy to follow).
you don't have a clue about any branch of physics...
you are a moron.

<plonk>

Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-16 06:32:53 UTC
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
moron. The increased number in one path is matched by a decrease in
the
other.
So .. you only emit down the path with more pulses coming out than going in
and you create free energy.
The extra pulses are already in there.
Post by Inertial
You really are an idiot if you continue this pretense that what you said was
right.
If two pulses go into a fixed length circular tube (in opposite directions)
and travel for the same time, then they MUST arrive at the detector at the
same time and so the arrival rate of successive pulses MUST be the same.
Your claim to the contrary is ridiculous.
This is just far too hard for you I'm afraid.
No .. you're just wrong. And I (and everyone else here but you) sees that
clearly.
Post by Henry Wilson DSc
You cannot get it into your head that, when it is rotating, the path length
DOES NOT EQUAL the tube length.
Of course it is different .. I didn't say it was the same. That doesn't
make any difference to the complete fuckup you made here. You claim of
different arrival rates for the pulses is contradicted by your own claim
that pairs of pulses that are emitted at the same time arrive at the same
time.
Post by Henry Wilson DSc
Here it is, SR style: http://en.wikipedia.org/wiki/Sagnac_effect
Go down to 'theory'. SEE THE DIAGRAM WITH THE OMEGA IN THE MIDDLE?
No need. I'm familiar with Sagnac, and unlike you, I understand it (and SR,
which you never could understand because when you tried to it was just
incoherent to you. Fortunately, those of us with a bit more brain power and
understanding of physics find it quite easy to follow).
you don't have a clue about any branch of physics...
you are a moron.
<plonk>
So .. you cannot refute any of my claims and proofs that you are WRONG.

If the pulses leave in pairs (as you imply), and take the same time to
travel all the way around their path and arrive back again (as emission
theory says), then they MUST arrive back in the same pairs they were
emitted, and so the SAME NUMBER OF PULSES arrive back from each fibre path
in any given time interval.

You cannot determine the rotation of the device (according to emission
theory) by looking at the arrival times and rates of the pulses. And the
fact that your CAN in reality (with a real sagnac device) is resounding and
refutation of emission theory.

Why do you keep lying, Henry?
eric gisse
2010-02-16 07:01:08 UTC
***@..(Henry Wilson DSc) wrote:

[...]
Post by Henry Wilson DSc
you don't have a clue about any branch of physics...
you are a moron.
<plonk>
Henry Wilson...
.......provider of free physics lessons
Looks like the 'provider of free physics lessons' couldn't find a way out.
eric gisse
2010-02-15 00:57:14 UTC
***@..(Henry Wilson DSc) wrote:

[...]
Post by Henry Wilson DSc
The fact that the frequencies in each direction are different annihilates
the Roberts, Jerry, Andersen 'rotating frame' argument as well as your
own.
...and your evidence that the frequencies are different is based on what?
Post by Henry Wilson DSc
Henry Wilson...
.......provider of free physics lessons
Henry Wilson DSc
2010-02-13 21:53:21 UTC
Post by Inertial
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
What is q and v?
Post by Henry Wilson DSc
(the surface is drawn flat for convenience. The fibre goes right around the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre. The pulses are labelled according to their emission times and move at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p.
Doesn't matter if they are detected at p or at C
I'll withdraw my previous reply because it was wrong. So was my initial message
because I stuffed up what I wanted to say.

p is the point where the pulses emitted at q are received.
Post by Inertial
Post by Henry Wilson DSc
observes that the pulses arriving simultaneously from both fibres carry
identical labels.
Using powerful rockets, the planet is sent into rotation with a period of 1
day, (86400 seconds). Its surface rotation speed 'v' wrt the nonrotating frame
is 465 m/s.
OK .. assuming you got the math right
Post by Henry Wilson DSc
In this new situation, the pulses that arrive at C when it is at point p, were
not emitted at that point but rather from an imaginary point q,
Not an imaginary point at all. It is as real and physical point as p is.
All you are saying is that C has moved (and continues to move)
It is imaginary in the R frame...REAL in the nonR frame.
Post by Inertial
Post by Henry Wilson DSc
which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.
Angular speed I think you would mean, it is moving around the ring according
to C's rotating frame.
yes
Post by Inertial
Post by Henry Wilson DSc
Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
So you're using an emission theory analysus, which does say that the time
taken for the two paths is the same.
Yes. The path lengths differ by 124 metres but one pulse moves at c+v and the
other at c-v, so they both arrive at p together.
Post by Inertial
Post by Henry Wilson DSc
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.
I'll take your word for it.
Post by Henry Wilson DSc
Since the distance between pulses is still 1 metre in the C's frame,
That's what emission theory would say
Post by Henry Wilson DSc
there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.
WRONG. Emmision theory does not say that. It says that the distance in C's
frame between pulses is one meter. The distance the pulses travel in C's
frame is the same. So the number of pulses in the fibre at any given time
is the same.
But I'm talking about the total number of pulses in transit in each path at any
instant. there are 124 more in one path than the other.
Post by Inertial
You can also tell it is wrong because the same number of pulses has been
emitted in each direction and you just said earlier that they take the same
time to travel around the ring. So there must be the same number of them in
there. Or does your explanation have pulse fairies that eat up the pulses?
Now you are being silly.

I made a mistake in my original post. This is regrettable but doesn't present a
problem for Bath.
As you rightly said, the pulses that leave together DO always arrive together
at the detector.
Post by Inertial
Post by Henry Wilson DSc
"That's odd", says Tom, "now, for some reason, the pulses coming from the left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"
This is wrong....and is not what I wanted to point out.
Post by Inertial
Emmision theory doesn't say that will happen at all. It says what you said
earlier .. the time take for them to travel is the same.
It is.....and the label on each arriving pulse is the same.
Post by Inertial
What you are now suddenly describing is what SR or an aether theory would
predict. You've changed theories in mid analysis !!!
Yes, basically I did. I apologise for any inconvenience.
Post by Inertial
Post by Henry Wilson DSc
"This is great", says Jerry after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".
No .. because your analysis above used SR at the end. You switched theories
mid way thru.
Yes I did.
Post by Inertial
You started saying the time for the pulses to travel in both directions
(with the rotating tube) was the same. That is what emission theory says.
And you said the distance between the pulses (in the rotating frame) is the
same. That is what emission theory says.
It is.
Post by Inertial
The consequence of that is that
there MUST be the same number of pulses in each direction in the tube at any
given time.
No that is YOUR mistake. The path lengths are different so there are more
pulses in one path than the other.
Post by Inertial
And they MUST arrive at the detector with the same label.
They do. I was wrong, before.
Post by Inertial
And
so emission theory says that a sagnac device will NOT detect rotation.
Now I'll tell you why that is wrong.
Post by Inertial
The *real* Sagnac devices have proven that THAT theory is NOT correct, and
it is NOT the same time required for the pulses (which is what SR predicts).
BAHAHAHHA
The true fact is that because the number of pulses is different in each path,
THE RATE at which they arrive at the detector is DIFFERENT for each path.
Between the instant of emission from point q and the detection at point p,
40000062 pulses arrive from one direction but only 39999938 from the other.
Those numbers will change only during a rotational speed change.

It is that difference in number that is analogous to the classical 'phase
difference' of the two light rays in a ring gyro.

Two different arrival rates produces a beat frequency at the detector.
Exactly how that gives rise to interference fringes I'm not sure at this stage
but the fact that BaTh predicts such a difference and produces the correct
equation for fringe displacement is sufficient to demonstrate that Sagnac
definitely does NOT refute BaTh.

Henry Wilson...

.......provider of free physics lessons
Androcles
2010-02-13 22:37:33 UTC
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT
ROTATING.
Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
What is q and v?
Post by Henry Wilson DSc
(the surface is drawn flat for convenience. The fibre goes right around the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre. The pulses are labelled according to their emission times and
move
at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p.
Doesn't matter if they are detected at p or at C
I'll withdraw my previous reply because it was wrong. So was my initial message
because I stuffed up what I wanted to say.
p is the point where the pulses emitted at q are received.
Post by Inertial
Post by Henry Wilson DSc
observes that the pulses arriving simultaneously from both fibres carry
identical labels.
Using powerful rockets, the planet is sent into rotation with a period
of
1
day, (86400 seconds). Its surface rotation speed 'v' wrt the nonrotating frame
is 465 m/s.
OK .. assuming you got the math right
Post by Henry Wilson DSc
In this new situation, the pulses that arrive at C when it is at point
p,
were
not emitted at that point but rather from an imaginary point q,
Not an imaginary point at all. It is as real and physical point as p is.
All you are saying is that C has moved (and continues to move)
It is imaginary in the R frame...REAL in the nonR frame.
Post by Inertial
Post by Henry Wilson DSc
which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.
Angular speed I think you would mean, it is moving around the ring according
to C's rotating frame.
yes
Post by Inertial
Post by Henry Wilson DSc
Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
So you're using an emission theory analysus, which does say that the time
taken for the two paths is the same.
Yes. The path lengths differ by 124 metres but one pulse moves at c+v and the
other at c-v, so they both arrive at p together.
Post by Inertial
Post by Henry Wilson DSc
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.
I'll take your word for it.
Post by Henry Wilson DSc
Since the distance between pulses is still 1 metre in the C's frame,
That's what emission theory would say
Post by Henry Wilson DSc
there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.
WRONG. Emmision theory does not say that. It says that the distance in C's
frame between pulses is one meter. The distance the pulses travel in C's
frame is the same. So the number of pulses in the fibre at any given time
is the same.
But I'm talking about the total number of pulses in transit in each path at any
instant. there are 124 more in one path than the other.
His "Emmision" theory has nothing to do with emission theory.
Post by Henry Wilson DSc
Post by Inertial
You can also tell it is wrong because the same number of pulses has been
emitted in each direction and you just said earlier that they take the same
time to travel around the ring. So there must be the same number of them in
there. Or does your explanation have pulse fairies that eat up the pulses?
Now you are being silly.
I made a mistake in my original post. This is regrettable but doesn't present a
problem for Bath.
As you rightly said, the pulses that leave together DO always arrive together
at the detector.
Post by Inertial
Post by Henry Wilson DSc
"That's odd", says Tom, "now, for some reason, the pulses coming from
the
left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"
This is wrong....and is not what I wanted to point out.
Post by Inertial
Emmision theory doesn't say that will happen at all. It says what you said
earlier .. the time take for them to travel is the same.
It is.....and the label on each arriving pulse is the same.
Post by Inertial
What you are now suddenly describing is what SR or an aether theory would
predict. You've changed theories in mid analysis !!!
Yes, basically I did. I apologise for any inconvenience.
Post by Inertial
Post by Henry Wilson DSc
"This is great", says Jerry after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".
No .. because your analysis above used SR at the end. You switched theories
mid way thru.
Yes I did.
Post by Inertial
You started saying the time for the pulses to travel in both directions
(with the rotating tube) was the same. That is what emission theory says.
And you said the distance between the pulses (in the rotating frame) is the
same. That is what emission theory says.
It is.
Post by Inertial
The consequence of that is that
there MUST be the same number of pulses in each direction in the tube at any
given time.
No that is YOUR mistake. The path lengths are different so there are more
pulses in one path than the other.
Post by Inertial
And they MUST arrive at the detector with the same label.
They do. I was wrong, before.
Post by Inertial
And
so emission theory says that a sagnac device will NOT detect rotation.
Now I'll tell you why that is wrong.
Post by Inertial
The *real* Sagnac devices have proven that THAT theory is NOT correct, and
it is NOT the same time required for the pulses (which is what SR predicts).
BAHAHAHHA
The true fact is that because the number of pulses is different in each path,
THE RATE at which they arrive at the detector is DIFFERENT for each path.
Between the instant of emission from point q and the detection at point p,
40000062 pulses arrive from one direction but only 39999938 from the other.
Those numbers will change only during a rotational speed change.
It is that difference in number that is analogous to the classical 'phase
difference' of the two light rays in a ring gyro.
Two different arrival rates produces a beat frequency at the detector.
Exactly how that gives rise to interference fringes I'm not sure at this stage
It's simple enough.
There are two beat frequencies, the sum and the difference.
http://en.wikipedia.org/wiki/Circle_of_fifths
So if you have 4 Hz and 5 Hz you'll also get 1 Hz and 9 Hz, but it won't
be very musical.
Post by Henry Wilson DSc
but the fact that BaTh predicts such a difference and produces the correct
equation for fringe displacement is sufficient to demonstrate that Sagnac
definitely does NOT refute BaTh.
Henry Wilson...
.......provider of free physics lessons
Henry Wilson DSc
2010-02-13 23:43:15 UTC
Post by Androcles
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT
ROTATING.
Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
What is q and v?
Post by Henry Wilson DSc
(the surface is drawn flat for convenience. The fibre goes right around the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre. The pulses are labelled according to their emission times and
move
at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p.
Doesn't matter if they are detected at p or at C
I'll withdraw my previous reply because it was wrong. So was my initial message
because I stuffed up what I wanted to say.
p is the point where the pulses emitted at q are received.
Post by Inertial
Post by Henry Wilson DSc
observes that the pulses arriving simultaneously from both fibres carry
identical labels.
Using powerful rockets, the planet is sent into rotation with a period
of
1
day, (86400 seconds). Its surface rotation speed 'v' wrt the nonrotating frame
is 465 m/s.
OK .. assuming you got the math right
Post by Henry Wilson DSc
In this new situation, the pulses that arrive at C when it is at point
p,
were
not emitted at that point but rather from an imaginary point q,
Not an imaginary point at all. It is as real and physical point as p is.
All you are saying is that C has moved (and continues to move)
It is imaginary in the R frame...REAL in the nonR frame.
Post by Inertial
Post by Henry Wilson DSc
which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.
Angular speed I think you would mean, it is moving around the ring according
to C's rotating frame.
yes
Post by Inertial
Post by Henry Wilson DSc
Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
So you're using an emission theory analysus, which does say that the time
taken for the two paths is the same.
Yes. The path lengths differ by 124 metres but one pulse moves at c+v and the
other at c-v, so they both arrive at p together.
Post by Inertial
Post by Henry Wilson DSc
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.
I'll take your word for it.
Post by Henry Wilson DSc
Since the distance between pulses is still 1 metre in the C's frame,
That's what emission theory would say
Post by Henry Wilson DSc
there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.
WRONG. Emmision theory does not say that. It says that the distance in C's
frame between pulses is one meter. The distance the pulses travel in C's
frame is the same. So the number of pulses in the fibre at any given time
is the same.
But I'm talking about the total number of pulses in transit in each path at any
instant. there are 124 more in one path than the other.
His "Emmision" theory has nothing to do with emission theory.
Post by Androcles
Post by Henry Wilson DSc
Now you are being silly.
I made a mistake in my original post. This is regrettable but doesn't present a
problem for Bath.
As you rightly said, the pulses that leave together DO always arrive together
at the detector.
Post by Inertial
Post by Henry Wilson DSc
"That's odd", says Tom, "now, for some reason, the pulses coming from
the
left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"
This is wrong....and is not what I wanted to point out.
Post by Inertial
Emmision theory doesn't say that will happen at all. It says what you said
earlier .. the time take for them to travel is the same.
It is.....and the label on each arriving pulse is the same.
Post by Inertial
What you are now suddenly describing is what SR or an aether theory would
predict. You've changed theories in mid analysis !!!
Yes, basically I did. I apologise for any inconvenience.
Post by Inertial
Post by Henry Wilson DSc
"This is great", says Jerry after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".
No .. because your analysis above used SR at the end. You switched theories
mid way thru.
Yes I did.
Post by Inertial
You started saying the time for the pulses to travel in both directions
(with the rotating tube) was the same. That is what emission theory says.
And you said the distance between the pulses (in the rotating frame) is the
same. That is what emission theory says.
It is.
Post by Inertial
The consequence of that is that
there MUST be the same number of pulses in each direction in the tube at any
given time.
No that is YOUR mistake. The path lengths are different so there are more
pulses in one path than the other.
Post by Inertial
And they MUST arrive at the detector with the same label.
They do. I was wrong, before.
Post by Inertial
And
so emission theory says that a sagnac device will NOT detect rotation.
Now I'll tell you why that is wrong.
Post by Inertial
The *real* Sagnac devices have proven that THAT theory is NOT correct, and
it is NOT the same time required for the pulses (which is what SR predicts).
BAHAHAHHA
The true fact is that because the number of pulses is different in each path,
THE RATE at which they arrive at the detector is DIFFERENT for each path.
Between the instant of emission from point q and the detection at point p,
40000062 pulses arrive from one direction but only 39999938 from the other.
Those numbers will change only during a rotational speed change.
It is that difference in number that is analogous to the classical 'phase
difference' of the two light rays in a ring gyro.
Two different arrival rates produces a beat frequency at the detector.
Exactly how that gives rise to interference fringes I'm not sure at this stage
It's simple enough.
There are two beat frequencies, the sum and the difference.
http://en.wikipedia.org/wiki/Circle_of_fifths
So if you have 4 Hz and 5 Hz you'll also get 1 Hz and 9 Hz, but it won't
be very musical.
http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/MechSagnac.gif
In the case of light it will be the difference that creates the pattern.
....and fringe DISPLACEMENT will be proportional to beat frequency.

Henry Wilson...

.......provider of free physics lessons
Androcles
2010-02-14 01:20:44 UTC
On Sat, 13 Feb 2010 22:37:33 -0000, "Androcles"
Post by Androcles
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT
ROTATING.
Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
What is q and v?
Post by Henry Wilson DSc
(the surface is drawn flat for convenience. The fibre goes right
around
the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both
ends
of the
fibre. The pulses are labelled according to their emission times and
move
at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were
emitted
and
detected by Jerry at point p.
Doesn't matter if they are detected at p or at C
I'll withdraw my previous reply because it was wrong. So was my initial message
because I stuffed up what I wanted to say.
p is the point where the pulses emitted at q are received.
Post by Inertial
Post by Henry Wilson DSc
observes that the pulses arriving simultaneously from both fibres carry
identical labels.
Using powerful rockets, the planet is sent into rotation with a period
of
1
day, (86400 seconds). Its surface rotation speed 'v' wrt the
nonrotating
frame
is 465 m/s.
OK .. assuming you got the math right
Post by Henry Wilson DSc
In this new situation, the pulses that arrive at C when it is at point
p,
were
not emitted at that point but rather from an imaginary point q,
Not an imaginary point at all. It is as real and physical point as p is.
All you are saying is that C has moved (and continues to move)
It is imaginary in the R frame...REAL in the nonR frame.
Post by Inertial
Post by Henry Wilson DSc
which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.
Angular speed I think you would mean, it is moving around the ring according
to C's rotating frame.
yes
Post by Inertial
Post by Henry Wilson DSc
Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
So you're using an emission theory analysus, which does say that the time
taken for the two paths is the same.
Yes. The path lengths differ by 124 metres but one pulse moves at c+v
and
the
other at c-v, so they both arrive at p together.
Post by Inertial
Post by Henry Wilson DSc
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.
I'll take your word for it.
Post by Henry Wilson DSc
Since the distance between pulses is still 1 metre in the C's frame,
That's what emission theory would say
Post by Henry Wilson DSc
there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.
WRONG. Emmision theory does not say that. It says that the distance in C's
frame between pulses is one meter. The distance the pulses travel in C's
frame is the same. So the number of pulses in the fibre at any given time
is the same.
But I'm talking about the total number of pulses in transit in each path at any
instant. there are 124 more in one path than the other.
His "Emmision" theory has nothing to do with emission theory.
Post by Androcles
Post by Henry Wilson DSc
Now you are being silly.
I made a mistake in my original post. This is regrettable but doesn't present a
problem for Bath.
As you rightly said, the pulses that leave together DO always arrive together
at the detector.
Post by Inertial
Post by Henry Wilson DSc
"That's odd", says Tom, "now, for some reason, the pulses coming from
the
left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"
This is wrong....and is not what I wanted to point out.
Post by Inertial
Emmision theory doesn't say that will happen at all. It says what you said
earlier .. the time take for them to travel is the same.
It is.....and the label on each arriving pulse is the same.
Post by Inertial
What you are now suddenly describing is what SR or an aether theory would
predict. You've changed theories in mid analysis !!!
Yes, basically I did. I apologise for any inconvenience.
Post by Inertial
Post by Henry Wilson DSc
"This is great", says Jerry after some deliberation, "now we can build
a
ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".
No .. because your analysis above used SR at the end. You switched theories
mid way thru.
Yes I did.
Post by Inertial
You started saying the time for the pulses to travel in both directions
(with the rotating tube) was the same. That is what emission theory says.
And you said the distance between the pulses (in the rotating frame) is the
same. That is what emission theory says.
It is.
Post by Inertial
The consequence of that is that
there MUST be the same number of pulses in each direction in the tube at any
given time.
No that is YOUR mistake. The path lengths are different so there are more
pulses in one path than the other.
Post by Inertial
And they MUST arrive at the detector with the same label.
They do. I was wrong, before.
Post by Inertial
And
so emission theory says that a sagnac device will NOT detect rotation.
Now I'll tell you why that is wrong.
Post by Inertial
The *real* Sagnac devices have proven that THAT theory is NOT correct, and
it is NOT the same time required for the pulses (which is what SR predicts).
BAHAHAHHA
The true fact is that because the number of pulses is different in each path,
THE RATE at which they arrive at the detector is DIFFERENT for each path.
Between the instant of emission from point q and the detection at point p,
40000062 pulses arrive from one direction but only 39999938 from the other.
Those numbers will change only during a rotational speed change.
It is that difference in number that is analogous to the classical 'phase
difference' of the two light rays in a ring gyro.
Two different arrival rates produces a beat frequency at the detector.
Exactly how that gives rise to interference fringes I'm not sure at this stage
It's simple enough.
There are two beat frequencies, the sum and the difference.
http://en.wikipedia.org/wiki/Circle_of_fifths
So if you have 4 Hz and 5 Hz you'll also get 1 Hz and 9 Hz, but it won't
be very musical.
http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/MechSagnac.gif
In the case of light it will be the difference that creates the pattern.
Yeah,
....and fringe DISPLACEMENT will be proportional to beat frequency.
Bullshit. You've never worked with an oscilloscope.
How come you never produce a web page to back up your stupid claims?
Henry Wilson DSc
2010-02-14 06:27:48 UTC
Post by Androcles
On Sat, 13 Feb 2010 22:37:33 -0000, "Androcles"
Post by Androcles
It's simple enough.
There are two beat frequencies, the sum and the difference.
http://en.wikipedia.org/wiki/Circle_of_fifths
So if you have 4 Hz and 5 Hz you'll also get 1 Hz and 9 Hz, but it won't
be very musical.
http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/MechSagnac.gif
In the case of light it will be the difference that creates the pattern.
Yeah,
http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/Sagnac2.JPG
Yes. That shows the difference.
Post by Androcles
....and fringe DISPLACEMENT will be proportional to beat frequency.
Bullshit. You've never worked with an oscilloscope.
How come you never produce a web page to back up your stupid claims?
...you're starting to sound more like little eric every day.....

Henry Wilson...

.......provider of free physics lessons
eric gisse
2010-02-14 07:19:17 UTC
***@..(Henry Wilson DSc) wrote:
[...]
Post by Henry Wilson DSc
Post by Androcles
Post by Henry Wilson DSc
....and fringe DISPLACEMENT will be proportional to beat frequency.
Bullshit. You've never worked with an oscilloscope.
How come you never produce a web page to back up your stupid claims?
...you're starting to sound more like little eric every day.....
Which makes me wonder why you post here at all. Nobody here likes you, you
don't like anyone here, and you have your own webspace. Why not just post to
it and be smug in the knowledge that you are right and the world is wrong?
Post by Henry Wilson DSc
Henry Wilson...
.......provider of free physics lessons
Androcles
2010-02-14 07:31:01 UTC
On Sun, 14 Feb 2010 01:20:44 -0000, "Androcles"
Post by Androcles
On Sat, 13 Feb 2010 22:37:33 -0000, "Androcles"
Post by Androcles
It's simple enough.
There are two beat frequencies, the sum and the difference.
http://en.wikipedia.org/wiki/Circle_of_fifths
So if you have 4 Hz and 5 Hz you'll also get 1 Hz and 9 Hz, but it won't
be very musical.
http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/MechSagnac.gif
In the case of light it will be the difference that creates the pattern.
Yeah,
http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/Sagnac2.JPG
Yes. That shows the difference.
Post by Androcles
....and fringe DISPLACEMENT will be proportional to beat frequency.
Bullshit. You've never worked with an oscilloscope.
How come you never produce a web page to back up your stupid claims?
...you're starting to sound more like little eric every day.....
...you're starting to sound more like little Einstein every day.....
--
Androcles
.......provider of expensive physics lessons Awilson can't afford.
bert
2010-02-13 18:48:44 UTC
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
(the surface is drawn flat for convenience. The fibre goes right around the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre.  The pulses are labelled according to their emission times and move at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p. At any instant, 40 million pulses are in transit
around the planet in each direction.
Tom observes that the pulses arriving simultaneously from both fibres carry
identical labels.
Using powerful rockets, the planet is sent into rotation with a period of 1
day, (86400 seconds). Its surface rotation speed 'v' wrt the nonrotating frame
is 465 m/s.
In this new situation, the pulses that arrive at C when it is at point p, were
not emitted at that point but rather from an imaginary point q, which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.
Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.
Since the distance between pulses is still 1 metre in the C's frame, there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.
"That's odd", says Tom, "now, for some reason, the pulses coming from the left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"
"This is great", says Jerry after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".
Henry Wilson...
.......provider of free physics lessons
My "Spin is in Theory" Tells it all. The universe is #1 energy,and
rotational spin is in reality what all there is is all about TreBert
BURT
2010-02-13 19:11:32 UTC
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
(the surface is drawn flat for convenience. The fibre goes right around the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre.  The pulses are labelled according to their emission times and move at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p. At any instant, 40 million pulses are in transit
around the planet in each direction.
Tom observes that the pulses arriving simultaneously from both fibres carry
identical labels.
Using powerful rockets, the planet is sent into rotation with a period of 1
day, (86400 seconds). Its surface rotation speed 'v' wrt the nonrotating frame
is 465 m/s.
In this new situation, the pulses that arrive at C when it is at point p, were
not emitted at that point but rather from an imaginary point q, which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.
Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.
Since the distance between pulses is still 1 metre in the C's frame, there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.
"That's odd", says Tom, "now, for some reason, the pulses coming from the left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"
"This is great", says Jerry after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".
Henry Wilson...
.......provider of free physics lessons
My "Spin is in Theory"  Tells it all.  The universe is #1 energy,and
rotational spin is in reality what all there is is all about  TreBert- Hide quoted text -
- Show quoted text -
Spin is misused for Rotation. A rotation speed flow that changes sizes
is what spin is.

Mitch Raemsch
bert
2010-02-13 19:44:33 UTC
Post by BURT
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
(the surface is drawn flat for convenience. The fibre goes right around the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre.  The pulses are labelled according to their emission times and move at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p. At any instant, 40 million pulses are in transit
around the planet in each direction.
Tom observes that the pulses arriving simultaneously from both fibres carry
identical labels.
Using powerful rockets, the planet is sent into rotation with a period of 1
day, (86400 seconds). Its surface rotation speed 'v' wrt the nonrotating frame
is 465 m/s.
In this new situation, the pulses that arrive at C when it is at point p, were
not emitted at that point but rather from an imaginary point q, which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.
Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.
Since the distance between pulses is still 1 metre in the C's frame, there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.
"That's odd", says Tom, "now, for some reason, the pulses coming from the left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"
"This is great", says Jerry after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".
Henry Wilson...
.......provider of free physics lessons
My "Spin is in Theory"  Tells it all.  The universe is #1 energy,and
rotational spin is in reality what all there is is all about  TreBert- Hide quoted text -
- Show quoted text -
Spin is misused for Rotation. A rotation speed flow that changes sizes
is what spin is.
Mitch Raemsch- Hide quoted text -
- Show quoted text -
Burt lectron Speed rotation is the heart of my c speed of c TreBert
Paul B. Andersen
2010-02-15 13:13:02 UTC
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
(the surface is drawn flat for convenience. The fibre goes right around the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre. The pulses are labelled according to their emission times and move at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p. At any instant, 40 million pulses are in transit
around the planet in each direction.
Tom observes that the pulses arriving simultaneously from both fibres carry
identical labels.
Using powerful rockets, the planet is sent into rotation with a period of 1
day, (86400 seconds). Its surface rotation speed 'v' wrt the nonrotating frame
is 465 m/s.
In this new situation, the pulses that arrive at C when it is at point p, were
not emitted at that point but rather from an imaginary point q, which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.
Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.
Since the distance between pulses is still 1 metre in the C's frame, there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.
"That's odd", says Tom, "now, for some reason, the pulses coming from the left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"
"This is great", says Jerry after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".
Good grief, Ralph. :-)

Yet again you have made a giant fool of yourself.
I wonder how long it will take this time before you
realize that you have made a giant, unbelievable stupid blunder.

I bet it will take a very long time, and if it eventually should
dawn to you, you will never admit it. What will you do then?
Start a new thread with a modified 'experiment'? :-)

I won't bother to tell you what your blunder is.

I will tell you a story, though.
I have a circular tube. It is not rotating.
At one point in that tube, there is a source/target device (STD)
which is emitting marbles at a constant speed 1 m/s relative
to the STD. When a marble has moved around the circuit
and gets back to the STD, it is re-emitted by the STD.
The circumference of the circle is 10m. So there are at any time
10 marbles in the tube, and it takes 10 seconds for a marble to
go around the circle.

Now we let the circle rotate with one rotation in 10 seconds,
in the same direction as the marbles are moving, which means that
the STD is moving at 1 m/s in the non rotating frame.

In this new situation, the marble that arrive at the STD when it is at point p,
were not emitted at that point but rather from an imaginary point q, which is at
rest in the nonR frame and moving away from the STD at speed -1 m/s in the STD R frame.

Each marble still takes 10 seconds to pass through the tube.
Therefore the distance p-q = (1 m/s * 1 second) metres.....or 1 metre.

Since the distance between marbles is still 1 metre in the STD's frame,
there are now 11 marbles moving through the tube.

"This is great", says Ralph after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Ralph Rabbidge's
marble creating fairies.
--
Paul

http://home.c2i.net/pb_andersen/
Henry Wilson DSc
2010-02-15 22:24:16 UTC
On Mon, 15 Feb 2010 14:13:02 +0100, "Paul B. Andersen"
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
I recognized my mistake and immediately corrected it.

The pulses that leave together arrive together.

If you want to comment, use my later version in which the pulse arrival rates
are different from both directions. The theory should be obvious to anyone with
a higher IQ than inertial.

Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-15 22:35:15 UTC
Post by Henry Wilson DSc
On Mon, 15 Feb 2010 14:13:02 +0100, "Paul B. Andersen"
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
I recognized my mistake and immediately corrected it.
The pulses that leave together arrive together.
So the SAME NUMBER of pulses arrive in both directions per second. So your
original claim was wrong and you cannot measure absolute rotation with the
device.
Post by Henry Wilson DSc
If you want to comment, use my later version in which the pulse arrival rates
are different from both directions.
But you just said "The pulses that leave together arrive together."

So that means they arrive with the same arrival rate. Otherwise they can't
arrive in pairs. Unless you have pulse fairies creating extra pulses from
nowhere.
Post by Henry Wilson DSc
The theory should be obvious to anyone with
a higher IQ than inertial.
Your blatant error should be obvious to even you. Yet either you're a
bigger moron than I thought, or as big a liar.
Henry Wilson DSc
2010-02-15 23:05:43 UTC
Post by Inertial
Post by Henry Wilson DSc
On Mon, 15 Feb 2010 14:13:02 +0100, "Paul B. Andersen"
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
I recognized my mistake and immediately corrected it.
The pulses that leave together arrive together.
So the SAME NUMBER of pulses arrive in both directions per second. So your
original claim was wrong and you cannot measure absolute rotation with the
device.
Post by Henry Wilson DSc
If you want to comment, use my later version in which the pulse arrival rates
are different from both directions.
But you just said "The pulses that leave together arrive together."
So that means they arrive with the same arrival rate. Otherwise they can't
arrive in pairs. Unless you have pulse fairies creating extra pulses from
nowhere.
Post by Henry Wilson DSc
The theory should be obvious to anyone with
a higher IQ than inertial.
Your blatant error should be obvious to even you. Yet either you're a
bigger moron than I thought, or as big a liar.
When are you going to say something intelligent?

Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-15 23:22:09 UTC
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
On Mon, 15 Feb 2010 14:13:02 +0100, "Paul B. Andersen"
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
I recognized my mistake and immediately corrected it.
The pulses that leave together arrive together.
So the SAME NUMBER of pulses arrive in both directions per second. So your
original claim was wrong and you cannot measure absolute rotation with the
device.
Post by Henry Wilson DSc
If you want to comment, use my later version in which the pulse arrival rates
are different from both directions.
But you just said "The pulses that leave together arrive together."
So that means they arrive with the same arrival rate. Otherwise they can't
arrive in pairs. Unless you have pulse fairies creating extra pulses from
nowhere.
Post by Henry Wilson DSc
The theory should be obvious to anyone with
a higher IQ than inertial.
Your blatant error should be obvious to even you. Yet either you're a
bigger moron than I thought, or as big a liar.
When are you going to say something intelligent?
This is Henry's typical results when he knows that he is beaten.

Of course, everything I say is intelligent. But you are just too big of a
moron to understand it.

And here's something else intelligent for you:

If two pulses go into a fixed length circular tube (in opposite directions)
at the same time, and travel for the same duration, then they MUST arrive at
the detector at the same time and so the arrival rate of successive pairs of
pulses MUST be the same. Your claim to the contrary is ridiculous.

At least you've retracted your claim that there are differing numbers of
pulses in the tube going one way to the other. Are you ready to retract
your ridiculous claim that the arrival rates are different when you admit
that every pair of pulses arrives at the same time?

Note that even with what you have admitted now, there can be no phase shift
because the pairs of pulses emitted together arrive together at the same
time. Phase shift occurs when pulses arrive at different times (as, for
example, SR predicts). Emission theory is once again soundly refuted, using
Henry Wilson DSc
2010-02-16 10:34:02 UTC
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
On Mon, 15 Feb 2010 14:13:02 +0100, "Paul B. Andersen"
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
I recognized my mistake and immediately corrected it.
The pulses that leave together arrive together.
So the SAME NUMBER of pulses arrive in both directions per second. So your
original claim was wrong and you cannot measure absolute rotation with the
device.
Post by Henry Wilson DSc
If you want to comment, use my later version in which the pulse arrival rates
are different from both directions.
But you just said "The pulses that leave together arrive together."
So that means they arrive with the same arrival rate. Otherwise they can't
arrive in pairs. Unless you have pulse fairies creating extra pulses from
nowhere.
Post by Henry Wilson DSc
The theory should be obvious to anyone with
a higher IQ than inertial.
Your blatant error should be obvious to even you. Yet either you're a
bigger moron than I thought, or as big a liar.
When are you going to say something intelligent?
This is Henry's typical results when he knows that he is beaten.
Of course, everything I say is intelligent. But you are just too big of a
moron to understand it.
If two pulses go into a fixed length circular tube (in opposite directions)
at the same time, and travel for the same duration, then they MUST arrive at
the detector at the same time and so the arrival rate of successive pairs of
pulses MUST be the same. Your claim to the contrary is ridiculous.
At least you've retracted your claim that there are differing numbers of
pulses in the tube going one way to the other. Are you ready to retract
your ridiculous claim that the arrival rates are different when you admit
that every pair of pulses arrives at the same time?
Note that even with what you have admitted now, there can be no phase shift
because the pairs of pulses emitted together arrive together at the same
time. Phase shift occurs when pulses arrive at different times (as, for
example, SR predicts). Emission theory is once again soundly refuted, using
I see I'll have to unplonk you for your own sake. You need help.

1st question: Do you understand how and why the path lengths are different in
the nonR frame?

Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-16 11:39:14 UTC
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
On Mon, 15 Feb 2010 14:13:02 +0100, "Paul B. Andersen"
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
I recognized my mistake and immediately corrected it.
The pulses that leave together arrive together.
So the SAME NUMBER of pulses arrive in both directions per second. So your
original claim was wrong and you cannot measure absolute rotation with the
device.
Post by Henry Wilson DSc
If you want to comment, use my later version in which the pulse
arrival
rates
are different from both directions.
But you just said "The pulses that leave together arrive together."
So that means they arrive with the same arrival rate. Otherwise they can't
arrive in pairs. Unless you have pulse fairies creating extra pulses from
nowhere.
Post by Henry Wilson DSc
The theory should be obvious to anyone with
a higher IQ than inertial.
Your blatant error should be obvious to even you. Yet either you're a
bigger moron than I thought, or as big a liar.
When are you going to say something intelligent?
This is Henry's typical results when he knows that he is beaten.
Of course, everything I say is intelligent. But you are just too big of a
moron to understand it.
If two pulses go into a fixed length circular tube (in opposite directions)
at the same time, and travel for the same duration, then they MUST arrive at
the detector at the same time and so the arrival rate of successive pairs of
pulses MUST be the same. Your claim to the contrary is ridiculous.
At least you've retracted your claim that there are differing numbers of
pulses in the tube going one way to the other. Are you ready to retract
your ridiculous claim that the arrival rates are different when you admit
that every pair of pulses arrives at the same time?
Note that even with what you have admitted now, there can be no phase shift
because the pairs of pulses emitted together arrive together at the same
time. Phase shift occurs when pulses arrive at different times (as, for
example, SR predicts). Emission theory is once again soundly refuted, using
I see I'll have to unplonk you for your own sake. You need help.
1st question: Do you understand how and why the path lengths are different in
the nonR frame?
Of course .. I understand all this far better than you.

Do YOU understand that the number of particles (at any given time) in the
tube/fibre going in one direction is the same number as going in the other?

Do YOU understand that every pulse has its own path?

Do YOU understand that the pulses with the longer paths travel faster than
the ones with the slower paths so that the time to travel the whole path is
the same in either direction?

Do YOU understand that a pair of pulses emitted at the same time in opposite
directions will therefore take the same time to travel along their
respective paths and will arrive at the same time?

Do YOU understand the pairs of pulses arriving at the same time means the
arrival rates are the same for both paths?

Do YOU understand that there is only one pulse on each path, because every
pulse has its own path?
Henry Wilson DSc
2010-02-16 21:57:49 UTC
Post by Inertial
Post by Henry Wilson DSc
I see I'll have to unplonk you for your own sake. You need help.
1st question: Do you understand how and why the path lengths are different in
the nonR frame?
Of course .. I understand all this far better than you.
So you should understand that since the distance between pulses is always 1
metre, there are 124 more pulses in one frame than in the other.
Post by Inertial
Do YOU understand that the number of particles (at any given time) in the
tube/fibre going in one direction is the same number as going in the other?
Definitely not.
Post by Inertial
Do YOU understand that every pulse has its own path?
I certainly do...and I'm glad your brain cells are now showing signs of life.
Post by Inertial
Do YOU understand that the pulses with the longer paths travel faster than
the ones with the slower paths so that the time to travel the whole path is
the same in either direction?
I certainly do.
Post by Inertial
Do YOU understand that a pair of pulses emitted at the same time in opposite
directions will therefore take the same time to travel along their
respective paths and will arrive at the same time?
I certainly do.
Post by Inertial
Do YOU understand the pairs of pulses arriving at the same time means the
arrival rates are the same for both paths?
NO. Where did the 62 pulses go? Did your fairies gobble them up?
Post by Inertial
Post by Henry Wilson DSc
Do YOU understand that there is only one pulse on each path, because every
pulse has its own path?
That's not a good way to look at it.
the paths are the same but the start and end points advance 1 metre per pulse
in the nonR frame.

In that wiki diagram I referred to, the whole pattern is rotating with the
Earth.

What is interesting and where Paul, Tom and Jerry have gone wrong is that the
distance vt and the 62 pulses don't magically disappear in the frame of the
source.

D=============62===============|==============================
=========================================S
__________________________________<-0.465km/s______________________________Ground

The nonR observer sees the above picture and sees it ROTATING with the earth.

An observer at the pole axis and rotating with the earth still sees the above
picture but now it is NOT rotating with the Earth.

Get it now?

Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-16 23:00:33 UTC
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
I see I'll have to unplonk you for your own sake. You need help.
1st question: Do you understand how and why the path lengths are
different
in
the nonR frame?
Of course .. I understand all this far better than you.
So you should understand that since the distance between pulses is always 1
metre, there are 124 more pulses in one frame than in the other.
Post by Inertial
Do YOU understand that the number of particles (at any given time) in the
tube/fibre going in one direction is the same number as going in the other?
Definitely not.
Post by Inertial
Do YOU understand that every pulse has its own path?
I certainly do...and I'm glad your brain cells are now showing signs of life.
Don't be a moron .. I've understood this the whole time .. it appears you
are just learning
Post by Henry Wilson DSc
Post by Inertial
Do YOU understand that the pulses with the longer paths travel faster than
the ones with the slower paths so that the time to travel the whole path is
the same in either direction?
I certainly do.
Post by Inertial
Do YOU understand that a pair of pulses emitted at the same time in opposite
directions will therefore take the same time to travel along their
respective paths and will arrive at the same time?
I certainly do.
Post by Inertial
Do YOU understand the pairs of pulses arriving at the same time means the
arrival rates are the same for both paths?
NO.
See . you don't understand yet. If every pair of pulses arrives together,
how can one of each pair be arriving with a faster arrival rate? Its a

Just to illustrate the obvious that you are missing

Lets say the emitted produces the pair A1 and B1 and send them in opposite
directions. Then a short interval, dt, later produces the pair A2 and B2,
also sent in opposite directions. Then the same interval, dt, later
produces the pair A3 and B3, also sent in opposite directions.

A1 and B1 go around their respective paths in the same time and arrive back
together as a pair at the same time

A2 and B2 go around their respective paths in the same time and arrive back
together as a pair at the same time. They arrive together at the same
interval, dt, after A1 and B1

A3 and B3 go around their respective paths in the same time and arrive back
together as a pair at the same time. They arrive together at the same
interval, dt, after A2 and B2

So the arrival rate of the A-pulses is the same as the arrival rate of the B
pulses, because there is (by your own admission) the same time interval
between each arrival
Post by Henry Wilson DSc
Where did the 62 pulses go? Did your fairies gobble them up?
There are no '62 pulses' .. you only get that figure by double-counting the
number of photons at a given time in one direction and missing some in the
other .. ie by cheating.

There are not multiple pulses in a single path .. there is 1 pulses per path
(which you acknowledged above). So there is no 62 pulses difference in the
number of pulses in a path. Each has only one pulses in it. And there are
the same number of paths going in one direction as there are in the other
direction, because there are the same number of pulses in the tube going in
one direction as in the other .. and as you admit, each pulses has its own
path.
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Do YOU understand that there is only one pulse on each path, because every
pulse has its own path?
That's not a good way to look at it.
Its a very good way, because that is how it is. You may not like it,
Post by Henry Wilson DSc
the paths are the same
but the start and end points advance 1 metre per pulse
in the nonR frame.
So it is *not* the same path .. its a *different* path. You cannot have two
paths being the same when their start and end points are different. The
*lengths* of each successive path (in a given direction) is the same, of
course.
Post by Henry Wilson DSc
In that wiki diagram I referred to, the whole pattern is rotating with the
Earth.
What is interesting and where Paul, Tom and Jerry have gone wrong is that the
distance vt and the 62 pulses don't magically disappear in the frame of the
source.
There never was any 62 pulse difference in ANY frame. Only in your mind
Post by Henry Wilson DSc
D=============62===============|==============================
=========================================S
__________________________________<-0.465km/s______________________________Ground
The nonR observer sees the above picture
Nope .. there's no 62 pulses. What is seen looks nothing like what you
show. There are the SAME NUMBER OF PULSES TRAVELLING IN EACH DIRECTION at
any time you take the 'picture'. There is NO difference in the number of
pulses that are there. We know that because the distance between the pulses
is fixed and the length of the tube that they fit in is fixed.
Post by Henry Wilson DSc
and sees it ROTATING with the earth.
An observer at the pole axis and rotating with the earth still sees the above
picture but now it is NOT rotating with the Earth.
Get it now?
I've 'gotten it' from the beginning. You still don't. Sad. Its been
explained over and over.

Perhaps you might show some progress when you acknowledge the point above:
the pairs of pulses arriving at the same time means the arrival rates are
the same for both paths.
Paul B. Andersen
2010-02-17 12:10:56 UTC
Post by Henry Wilson DSc
On Mon, 15 Feb 2010 14:13:02 +0100, "Paul B. Andersen"
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
I recognized my mistake and immediately corrected it.
The pulses that leave together arrive together.
If you want to comment, use my later version in which the pulse arrival rates
are different from both directions. The theory should be obvious to anyone with
a higher IQ than inertial.
"The pulses that leave together arrive together", but the pulses are
arriving at a different rate than those they arriving together with.

It sure takes a high IQ to understand that! :-)

BTW, in my previous posting I wrote:
| Yet again you have made a giant fool of yourself.
| I wonder how long it will take this time before you
| realize that you have made a giant, unbelievable stupid blunder.
|
| I bet it will take a very long time, and if it eventually should
| dawn to you, you will never admit it. What will you do then?
| Start a new thread with a modified 'experiment'? :-)

Now we know what you did! :-)
conclusion based on that blunder was wrong, so you had to
make an even bigger blunder to defend your wrong conclusion.

Good to see that you still are able to invent new stupidities
and not only are recycling old ones.
This was definitely one of your better.

Keep it up, the sky is the limit!
--
Paul, still able to be amused by Ralph Rabbidge's stupidites

http://home.c2i.net/pb_andersen/
Henry Wilson DSc
2010-02-17 22:06:01 UTC
On Wed, 17 Feb 2010 13:10:56 +0100, "Paul B. Andersen"
Post by Paul B. Andersen
Post by Henry Wilson DSc
On Mon, 15 Feb 2010 14:13:02 +0100, "Paul B. Andersen"
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
I recognized my mistake and immediately corrected it.
The pulses that leave together arrive together.
If you want to comment, use my later version in which the pulse arrival rates
are different from both directions. The theory should be obvious to anyone with
a higher IQ than inertial.
"The pulses that leave together arrive together", but the pulses are
arriving at a different rate than those they arriving together with.
It sure takes a high IQ to understand that! :-)
Maybe you are having trouble with frames.
Post by Paul B. Andersen
| Yet again you have made a giant fool of yourself.
| I wonder how long it will take this time before you
| realize that you have made a giant, unbelievable stupid blunder.
|
| I bet it will take a very long time, and if it eventually should
| dawn to you, you will never admit it. What will you do then?
| Start a new thread with a modified 'experiment'? :-)
Now we know what you did! :-)
conclusion based on that blunder was wrong, so you had to
make an even bigger blunder to defend your wrong conclusion.
Good to see that you still are able to invent new stupidities
and not only are recycling old ones.
This was definitely one of your better.
Keep it up, the sky is the limit!
==//=================p---------q|====================//====optical tube
________________________earth____<-v__________________(actually curved)

A pulse of light is emitted every 3.333 nanosecs by a source. The pulses, which
are separated by 1 metre, are sent down an optical fibre that is wrapped around
the earth.. There are 4000000 pulses around the equator.

A pulse emitted at point q will travel leftward around the Earth through the
fibre before reaching a detector at point p. During its travel time the Earth
revolves by 62 metres.

We will consider what happens when pulse N is emitted at point q. It will pass
through point p then travel around the Earth before being finally absorbed
again at point p.
When it is emitted, there are 62 pulses between q and p. When pulse N reaches p
the first time, there are say 61 pulses between p and q and 62 pulses have
already passed through point p. There are 4000000-61 pulses around the earth
at that instant between p and q. So another 61 + 4000000 - 41 pulses will reach
p by the time N gets there. That makes 4000062 altogether.
In the other direction, only 3999938 arrive in the same time.

Henry Wilson...

.......provider of free physics lessons
Jerry
2010-02-17 22:55:15 UTC
Post by Henry Wilson DSc
We will consider what happens when pulse N is emitted at point q. It will pass
through point p then travel around the Earth before being finally absorbed
again at point p.
When it is emitted, there are 62 pulses between q and p. When pulse N reaches p
the first time, there are say 61 pulses between p and q and 62 pulses have
already passed through point p. There are 4000000-61  pulses around the earth
at that instant between p and q. So another 61 + 4000000 - 41 pulses will reach
p by the time N gets there. That makes 4000062 altogether.
In the other direction, only 3999938 arrive in the same time.
(sigh)
In the STEADY STATE, how many pulses cross p each second?
In the STEADY STATE, how many pulses cross q each second?

1/3.333e-9 = 300030003 pulses.

Jerry
Henry Wilson DSc
2010-02-18 04:22:40 UTC
On Wed, 17 Feb 2010 14:55:15 -0800 (PST), Jerry
Post by Jerry
Post by Henry Wilson DSc
We will consider what happens when pulse N is emitted at point q. It will pass
through point p then travel around the Earth before being finally absorbed
again at point p.
When it is emitted, there are 62 pulses between q and p. When pulse N reaches p
the first time, there are say 61 pulses between p and q and 62 pulses have
already passed through point p. There are 4000000-61  pulses around the earth
at that instant between p and q. So another 61 + 4000000 - 41 pulses will reach
p by the time N gets there. That makes 4000062 altogether.
In the other direction, only 3999938 arrive in the same time.
(sigh)
In the STEADY STATE, how many pulses cross p each second?
In the STEADY STATE, how many pulses cross q each second?
1/3.333e-9 = 300030003 pulses.
why should it not be?
Post by Jerry
Jerry
Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-17 23:28:56 UTC
Post by Henry Wilson DSc
On Wed, 17 Feb 2010 13:10:56 +0100, "Paul B. Andersen"
Post by Paul B. Andersen
Post by Henry Wilson DSc
On Mon, 15 Feb 2010 14:13:02 +0100, "Paul B. Andersen"
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
I recognized my mistake and immediately corrected it.
The pulses that leave together arrive together.
If you want to comment, use my later version in which the pulse arrival rates
are different from both directions. The theory should be obvious to anyone with
a higher IQ than inertial.
"The pulses that leave together arrive together", but the pulses are
arriving at a different rate than those they arriving together with.
It sure takes a high IQ to understand that! :-)
Maybe you are having trouble with frames.
Post by Paul B. Andersen
| Yet again you have made a giant fool of yourself.
| I wonder how long it will take this time before you
| realize that you have made a giant, unbelievable stupid blunder.
|
| I bet it will take a very long time, and if it eventually should
| dawn to you, you will never admit it. What will you do then?
| Start a new thread with a modified 'experiment'? :-)
Now we know what you did! :-)
conclusion based on that blunder was wrong, so you had to
make an even bigger blunder to defend your wrong conclusion.
Good to see that you still are able to invent new stupidities
and not only are recycling old ones.
This was definitely one of your better.
Keep it up, the sky is the limit!
==//=================p---------q|====================//====optical tube
________________________earth____<-v__________________(actually curved)
So is tube attached to the earth, or is the tube stationary (in the NonR
inertial frame) and the earth moving at <-v wrt as your diagram seems to
indicate?

I'll assume that you just drew the diagram poorly, and that the tube is
moving with the earth. And that points p and q are moving relative to the
tube (and the earth) and so are fixed in the NonR inertial frame.

Yes?
Post by Henry Wilson DSc
A pulse of light is emitted every 3.333 nanosecs by a source.
I'll assume that the source is moving with the earth (and tube) as well.

Yes?
Post by Henry Wilson DSc
The pulses, which are separated by 1 metre, are sent down an
optical fibre that is wrapped around the earth.. There are
4000000 pulses around the equator.
OK

I'll assume they are only going in one direction in this example.

Yes?
Post by Henry Wilson DSc
A pulse emitted at point q will travel leftward around the Earth through the
fibre before reaching a detector at point p. During its travel time the Earth
revolves by 62 metres.
OK
Post by Henry Wilson DSc
We will consider what happens when pulse N is emitted at point q.
I'll assume the source was moving along with the earth, and when it arrives
at point q it emits a pulse.

Yes?
Post by Henry Wilson DSc
It will pass
through point p then travel around the Earth before being finally absorbed
again at point p.
I'm assuming that point p is where the source is at the time the pulses
arrives back at a detector.

Yes?
Post by Henry Wilson DSc
When it is emitted, there are 62 pulses between q and p.
OK .. ones that were emitted before the source was at q, yes.
Post by Henry Wilson DSc
When pulse N reaches p
the first time, there are say 61 pulses
between p and q and 62 pulses have
OK.
Post by Henry Wilson DSc
There are 4000000-61 pulses around the earth
at that instant between p and q.
I'll assume you mean the there are 61 in small gap from q to p, and
4000000-61 in the larger gap from p back to q.
Post by Henry Wilson DSc
So another 61 + 4000000 - 41 pulses will reach
41? Don't you mean 61?
Post by Henry Wilson DSc
p by the time N gets there. That makes 4000062 altogether.
Yes.
Post by Henry Wilson DSc
In the other direction, only 3999938 arrive in the same time.
Yes.

For a detector that is at a fixed point q in the NonR inertial frame and
does *not* move with the source and tube.

This is a very different scenario from what you described previously (and
different to Sagnac) where the detector moves with the source and tube.

Clearly if the source is moving with the earth, and the detector is not, but
is fixed in a non-rotating inertial frame, you can work out the rotation
rate just be seeing how the source moves away from the detector .. you don't
need to go counting pulses.

setup, where you had a tube attached to the earth, and a source+detector at
same location on the tube ??
Henry Wilson DSc
2010-02-18 23:13:08 UTC
Post by Inertial
Post by Henry Wilson DSc
On Wed, 17 Feb 2010 13:10:56 +0100, "Paul B. Andersen"
Post by Paul B. Andersen
Good to see that you still are able to invent new stupidities
and not only are recycling old ones.
This was definitely one of your better.
Keep it up, the sky is the limit!
==//=================p---------q|====================//====optical tube
________________________earth____<-v__________________(actually curved)
So is tube attached to the earth, or is the tube stationary (in the NonR
inertial frame) and the earth moving at <-v wrt as your diagram seems to
indicate?
The tube is attached to the Earth.
Post by Inertial
I'll assume that you just drew the diagram poorly, and that the tube is
moving with the earth. And that points p and q are moving relative to the
tube (and the earth) and so are fixed in the NonR inertial frame.
The points p and q are 'plotted' on the nonR frame. They represent the
positions of the source/detecter when ONE PARTICULAR pulse is emitted and
detected. The whole configuration rotates with the Earth so that when the next
pulse is emitted, p1,q1 would be slightly displaced from p and q...(again,
drawn in the nonR frame). p62,q62 would be 1 metre from p,q.

Paul, Tom and Jerry have always wrongly claimed that the points p and q are
together in the rotating frame...and that BaTh is therefore refuted.
These point are still those that a rotating observer would 'see' drawn on the
non R frame, if that were possible. Thus p and q are always separated even
though they would 'move' in the R frame. This is true even in the rotating
frame of the pulses themselves, except that p and q would be much closer
together.
Post by Inertial
Yes?
Post by Henry Wilson DSc
A pulse of light is emitted every 3.333 nanosecs by a source.
I'll assume that the source is moving with the earth (and tube) as well.
Yes?
Post by Henry Wilson DSc
The pulses, which are separated by 1 metre, are sent down an
optical fibre that is wrapped around the earth.. There are
4000000 pulses around the equator.
OK
I'll assume they are only going in one direction in this example.
Yes?
Post by Henry Wilson DSc
A pulse emitted at point q will travel leftward around the Earth through the
fibre before reaching a detector at point p. During its travel time the Earth
revolves by 62 metres.
OK
Post by Henry Wilson DSc
We will consider what happens when pulse N is emitted at point q.
I'll assume the source was moving along with the earth, and when it arrives
at point q it emits a pulse.
It is always at a point we label 'q' even though points q(n) move around the
nonR frame in a circle.
Post by Inertial
Yes?
Post by Henry Wilson DSc
It will pass
through point p then travel around the Earth before being finally absorbed
again at point p.
I'm assuming that point p is where the source is at the time the pulses
arrives back at a detector.
Yes?
Post by Henry Wilson DSc
When it is emitted, there are 62 pulses between q and p.
OK .. ones that were emitted before the source was at q, yes.
Post by Henry Wilson DSc
When pulse N reaches p
the first time, there are say 61 pulses
between p and q and 62 pulses have
OK.
Post by Henry Wilson DSc
There are 4000000-61 pulses around the earth
at that instant between p and q.
I'll assume you mean the there are 61 in small gap from q to p, and
4000000-61 in the larger gap from p back to q.
Post by Henry Wilson DSc
So another 61 + 4000000 - 61 pulses will reach
41? Don't you mean 61?
I did.
Post by Inertial
Post by Henry Wilson DSc
p by the time N gets there. That makes 4000062 altogether.
Yes.
Post by Henry Wilson DSc
In the other direction, only 3999938 arrive in the same time.
Yes.
For a detector that is at a fixed point q in the NonR inertial frame and
does *not* move with the source and tube.
This is a very different scenario from what you described previously (and
different to Sagnac) where the detector moves with the source and tube.
But it isn't. The rotating observer still sees the two points p and q as being
separated. He marks them on the nonR frame, which appears to move backwards.
Post by Inertial
Clearly if the source is moving with the earth, and the detector is not, but
is fixed in a non-rotating inertial frame, you can work out the rotation
rate just be seeing how the source moves away from the detector .. you don't
need to go counting pulses.
You cannot actually mark the bloody points on the nonR frame. You would need a
bloody big sheet of paper.
Post by Inertial
setup, where you had a tube attached to the earth, and a source+detector at
same location on the tube ??
Read this and you might understand
Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-18 23:46:37 UTC
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
On Wed, 17 Feb 2010 13:10:56 +0100, "Paul B. Andersen"
Post by Paul B. Andersen
Good to see that you still are able to invent new stupidities
and not only are recycling old ones.
This was definitely one of your better.
Keep it up, the sky is the limit!
==//=================p---------q|====================//====optical tube
________________________earth____<-v__________________(actually curved)
So is tube attached to the earth, or is the tube stationary (in the NonR
inertial frame) and the earth moving at <-v wrt as your diagram seems to
indicate?
The tube is attached to the Earth.
Post by Inertial
I'll assume that you just drew the diagram poorly, and that the tube is
moving with the earth. And that points p and q are moving relative to the
tube (and the earth) and so are fixed in the NonR inertial frame.
The points p and q are 'plotted' on the nonR frame. They represent the
positions of the source/detecter when ONE PARTICULAR pulse is emitted and
detected.
p when emitted, q when detected at different times
Post by Henry Wilson DSc
The whole configuration rotates with the Earth so that when the next
pulse is emitted, p1,q1 would be slightly displaced from p and q...(again,
drawn in the nonR frame). p62,q62 would be 1 metre from p,q.
yes
Post by Henry Wilson DSc
Paul, Tom and Jerry have always wrongly claimed that the points p and q are
together in the rotating frame...
The points where the photon are emitted and detected in the rotating frame
are together in the rotating frame

The points where the photon are emitted and detected in the non-rotating
frame are not together in the non-rotating frame
Post by Henry Wilson DSc
and that BaTh is therefore refuted.
It is. Your lies and stupidity don't make it true
Post by Henry Wilson DSc
These point are still those that a rotating observer would 'see' drawn on the
non R frame, if that were possible. Thus p and q are always separated even
though they would 'move' in the R frame. This is true even in the rotating
frame of the pulses themselves, except that p and q would be much closer
together.
Post by Inertial
Yes?
Post by Henry Wilson DSc
A pulse of light is emitted every 3.333 nanosecs by a source.
I'll assume that the source is moving with the earth (and tube) as well.
Yes?
Post by Henry Wilson DSc
The pulses, which are separated by 1 metre, are sent down an
optical fibre that is wrapped around the earth.. There are
4000000 pulses around the equator.
OK
I'll assume they are only going in one direction in this example.
Yes?
Post by Henry Wilson DSc
A pulse emitted at point q will travel leftward around the Earth through the
fibre before reaching a detector at point p. During its travel time the Earth
revolves by 62 metres.
OK
Post by Henry Wilson DSc
We will consider what happens when pulse N is emitted at point q.
I'll assume the source was moving along with the earth, and when it arrives
at point q it emits a pulse.
It is always at a point we label 'q'
No .. it is not at q is a fixed point in the non-rotating frame. A fixed
point on the tube in the *rotating* frame cannot always be at a fixed point
in the non-rotating frame
Post by Henry Wilson DSc
even though points q(n) move around the
nonR frame in a circle.
There are multiple q_n points. the source moves from one to the next. It
is not always at the ONE point you labeled q. To claim it is is just
Post by Henry Wilson DSc
Post by Inertial
Yes?
Post by Henry Wilson DSc
It will pass
through point p then travel around the Earth before being finally absorbed
again at point p.
I'm assuming that point p is where the source is at the time the pulses
arrives back at a detector.
Yes?
Post by Henry Wilson DSc
When it is emitted, there are 62 pulses between q and p.
OK .. ones that were emitted before the source was at q, yes.
Post by Henry Wilson DSc
When pulse N reaches p
the first time, there are say 61 pulses
between p and q and 62 pulses have
OK.
Post by Henry Wilson DSc
There are 4000000-61 pulses around the earth
at that instant between p and q.
I'll assume you mean the there are 61 in small gap from q to p, and
4000000-61 in the larger gap from p back to q.
Post by Henry Wilson DSc
So another 61 + 4000000 - 61 pulses will reach
41? Don't you mean 61?
I did.
Post by Inertial
Post by Henry Wilson DSc
p by the time N gets there. That makes 4000062 altogether.
Yes.
Post by Henry Wilson DSc
In the other direction, only 3999938 arrive in the same time.
Yes.
For a detector that is at a fixed point q in the NonR inertial frame and
does *not* move with the source and tube.
This is a very different scenario from what you described previously (and
different to Sagnac) where the detector moves with the source and tube.
But it isn't.
It is TOTALLY difference as you have the detector point always at 'p' so it
is FIXED in the non-rotating frame
Post by Henry Wilson DSc
The rotating observer still sees the two points p and q as being
separated. He marks them on the nonR frame, which appears to move backwards.
Which has NOTHING TO DO with what happens in Sagnac .. because there is no
detector that is always at fixed point p and no source that is always at
fixed point p. Describing what happens over time at those fixed points in
the non-rotating frame has NOTHING AT ALL TO DO with the Sagnac effect or
anything you have been claiming.
Post by Henry Wilson DSc
Post by Inertial
Clearly if the source is moving with the earth, and the detector is not, but
is fixed in a non-rotating inertial frame, you can work out the rotation
rate just be seeing how the source moves away from the detector .. you don't
need to go counting pulses.
You cannot actually mark the bloody points on the nonR frame.
Of course you could.
Post by Henry Wilson DSc
You would need a
bloody big sheet of paper.
Then use one. This is a gedanken.
Post by Henry Wilson DSc
Post by Inertial
setup, where you had a tube attached to the earth, and a source+detector at
same location on the tube ??
Read this and you might understand
I have read .. I understand .. you're a liar who purports to understand
physics and has no idea about it.
Henry Wilson DSc
2010-02-19 01:11:39 UTC
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
On Wed, 17 Feb 2010 13:10:56 +0100, "Paul B. Andersen"
Post by Paul B. Andersen
Good to see that you still are able to invent new stupidities
and not only are recycling old ones.
This was definitely one of your better.
Keep it up, the sky is the limit!
==//=================p---------q|====================//====optical tube
________________________earth____<-v__________________(actually curved)
So is tube attached to the earth, or is the tube stationary (in the NonR
inertial frame) and the earth moving at <-v wrt as your diagram seems to
indicate?
The tube is attached to the Earth.
Post by Inertial
I'll assume that you just drew the diagram poorly, and that the tube is
moving with the earth. And that points p and q are moving relative to the
tube (and the earth) and so are fixed in the NonR inertial frame.
The points p and q are 'plotted' on the nonR frame. They represent the
positions of the source/detecter when ONE PARTICULAR pulse is emitted and
detected.
p when emitted, q when detected at different times
actually it is the other way round. q is emission point.
Post by Inertial
Post by Henry Wilson DSc
The whole configuration rotates with the Earth so that when the next
pulse is emitted, p1,q1 would be slightly displaced from p and q...(again,
drawn in the nonR frame). p62,q62 would be 1 metre from p,q.
yes
Post by Henry Wilson DSc
Paul, Tom and Jerry have always wrongly claimed that the points p and q are
together in the rotating frame...
The points where the photon are emitted and detected in the rotating frame
are together in the rotating frame
seemingly so but wrong...which is why people make mistakes when they use
rotating frames.

Points p and q are 'drawn' on the nonR frame, which spins backwards in the R
frame. So they remain apart to a viewer in the rotating frame.
Post by Inertial
The points where the photon are emitted and detected in the non-rotating
frame are not together in the non-rotating frame
correct.
Post by Inertial
Post by Henry Wilson DSc
and that BaTh is therefore refuted.
It is. Your lies and stupidity don't make it true
Sorry, you are no smarter than paul, tom or jerry.
Post by Inertial
Post by Henry Wilson DSc
These point are still those that a rotating observer would 'see' drawn on the
non R frame, if that were possible. Thus p and q are always separated even
though they would 'move' in the R frame. This is true even in the rotating
frame of the pulses themselves, except that p and q would be much closer
together.
Post by Inertial
Yes?
Post by Henry Wilson DSc
A pulse of light is emitted every 3.333 nanosecs by a source.
I'll assume that the source is moving with the earth (and tube) as well.
Yes?
Post by Henry Wilson DSc
The pulses, which are separated by 1 metre, are sent down an
optical fibre that is wrapped around the earth.. There are
4000000 pulses around the equator.
OK
I'll assume they are only going in one direction in this example.
Yes?
Post by Henry Wilson DSc
A pulse emitted at point q will travel leftward around the Earth through the
fibre before reaching a detector at point p. During its travel time the Earth
revolves by 62 metres.
OK
Post by Henry Wilson DSc
We will consider what happens when pulse N is emitted at point q.
I'll assume the source was moving along with the earth, and when it arrives
at point q it emits a pulse.
It is always at a point we label 'q'
No .. it is not at q is a fixed point in the non-rotating frame. A fixed
point on the tube in the *rotating* frame cannot always be at a fixed point
in the non-rotating frame
Post by Henry Wilson DSc
even though points q(n) move around the
nonR frame in a circle.
There are multiple q_n points. the source moves from one to the next. It
is not always at the ONE point you labeled q. To claim it is is just
q(n) is where the emitter was when pulse(n) was emitted. It is marked on the
nonR frame, where it is static.
When viewed in the R frame, both q(n) and p(n) remain separated but appear to
spin backwards.
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Yes?
Post by Henry Wilson DSc
It will pass
through point p then travel around the Earth before being finally absorbed
again at point p.
I'm assuming that point p is where the source is at the time the pulses
arrives back at a detector.
Yes?
Post by Henry Wilson DSc
When it is emitted, there are 62 pulses between q and p.
OK .. ones that were emitted before the source was at q, yes.
Post by Henry Wilson DSc
When pulse N reaches p
the first time, there are say 61 pulses
between p and q and 62 pulses have
OK.
Post by Henry Wilson DSc
There are 4000000-61 pulses around the earth
at that instant between p and q.
I'll assume you mean the there are 61 in small gap from q to p, and
4000000-61 in the larger gap from p back to q.
Post by Henry Wilson DSc
So another 61 + 4000000 - 61 pulses will reach
41? Don't you mean 61?
I did.
Post by Inertial
Post by Henry Wilson DSc
p by the time N gets there. That makes 4000062 altogether.
Yes.
Post by Henry Wilson DSc
In the other direction, only 3999938 arrive in the same time.
Yes.
For a detector that is at a fixed point q in the NonR inertial frame and
does *not* move with the source and tube.
This is a very different scenario from what you described previously (and
different to Sagnac) where the detector moves with the source and tube.
But it isn't.
It is TOTALLY difference as you have the detector point always at 'p' so it
is FIXED in the non-rotating frame
points p(n) and q(n) move around the nonR frame by an amount proportional to n.
Post by Inertial
Post by Henry Wilson DSc
The rotating observer still sees the two points p and q as being
separated. He marks them on the nonR frame, which appears to move backwards.
Which has NOTHING TO DO with what happens in Sagnac .. because there is no
detector that is always at fixed point p and no source that is always at
fixed point p. Describing what happens over time at those fixed points in
the non-rotating frame has NOTHING AT ALL TO DO with the Sagnac effect or
anything you have been claiming.
Every rotating frame has an associated inertial one.
this is not the sole property of Einstein.
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Clearly if the source is moving with the earth, and the detector is not, but
is fixed in a non-rotating inertial frame, you can work out the rotation
rate just be seeing how the source moves away from the detector .. you don't
need to go counting pulses.
You cannot actually mark the bloody points on the nonR frame.
Of course you could.
Post by Henry Wilson DSc
You would need a
bloody big sheet of paper.
Then use one. This is a gedanken.
Post by Henry Wilson DSc
Post by Inertial
setup, where you had a tube attached to the earth, and a source+detector at
same location on the tube ??
Read this and you might understand
I have read .. I understand .. you're a liar who purports to understand
physics and has no idea about it.

Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-19 01:32:33 UTC
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
On Wed, 17 Feb 2010 13:10:56 +0100, "Paul B. Andersen"
Post by Paul B. Andersen
Good to see that you still are able to invent new stupidities
and not only are recycling old ones.
This was definitely one of your better.
Keep it up, the sky is the limit!
==//=================p---------q|====================//====optical tube
________________________earth____<-v__________________(actually curved)
So is tube attached to the earth, or is the tube stationary (in the NonR
inertial frame) and the earth moving at <-v wrt as your diagram seems to
indicate?
The tube is attached to the Earth.
Post by Inertial
I'll assume that you just drew the diagram poorly, and that the tube is
moving with the earth. And that points p and q are moving relative to the
tube (and the earth) and so are fixed in the NonR inertial frame.
The points p and q are 'plotted' on the nonR frame. They represent the
positions of the source/detecter when ONE PARTICULAR pulse is emitted and
detected.
p when emitted, q when detected at different times
actually it is the other way round. q is emission point.
Whatever .. swa pthe letters and you have the same point
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
The whole configuration rotates with the Earth so that when the next
pulse is emitted, p1,q1 would be slightly displaced from p and q...(again,
drawn in the nonR frame). p62,q62 would be 1 metre from p,q.
yes
Post by Henry Wilson DSc
Paul, Tom and Jerry have always wrongly claimed that the points p and q are
together in the rotating frame...
The points where the photon are emitted and detected in the rotating frame
are together in the rotating frame
seemingly so
NO .. it is 'so' .. not 'seemingly so'. They are physcially at the one
location in that frame at all times. They are physically both at the same
location at all times in the NonR frame .. but that combined location
changes over time. There is never a time in either frame when the source
and detector are separated as you claim.
Post by Henry Wilson DSc
but wrong...
Yes .. you are wrong .. we know that
Post by Henry Wilson DSc
which is why people make mistakes when they use
rotating frames.
You should stop trying to do anlaysis of Sagnac as you clearly cannot handle
what happens when things rotate.
Post by Henry Wilson DSc
Points p and q are 'drawn' on the nonR frame, which spins backwards in the R
frame. So they remain apart to a viewer in the rotating frame.
They make no difference to what happens in a Sagnac device (not in your
tube-around-the-earth device.
Post by Henry Wilson DSc
Post by Inertial
The points where the photon are emitted and detected in the non-rotating
frame are not together in the non-rotating frame
correct.
Post by Inertial
Post by Henry Wilson DSc
and that BaTh is therefore refuted.
It is. Your lies and stupidity don't make it true
Sorry, you are no smarter than paul, tom or jerry.
I don't claim to be .. we are simply ALL smarter than you. And honest.
Something you'll never be.
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
These point are still those that a rotating observer would 'see' drawn
on
the
non R frame, if that were possible. Thus p and q are always separated even
though they would 'move' in the R frame. This is true even in the rotating
frame of the pulses themselves, except that p and q would be much closer
together.
Post by Inertial
Yes?
Post by Henry Wilson DSc
A pulse of light is emitted every 3.333 nanosecs by a source.
I'll assume that the source is moving with the earth (and tube) as well.
Yes?
Post by Henry Wilson DSc
The pulses, which are separated by 1 metre, are sent down an
optical fibre that is wrapped around the earth.. There are
4000000 pulses around the equator.
OK
I'll assume they are only going in one direction in this example.
Yes?
Post by Henry Wilson DSc
A pulse emitted at point q will travel leftward around the Earth
through
the
fibre before reaching a detector at point p. During its travel time
the
Earth
revolves by 62 metres.
OK
Post by Henry Wilson DSc
We will consider what happens when pulse N is emitted at point q.
I'll assume the source was moving along with the earth, and when it arrives
at point q it emits a pulse.
It is always at a point we label 'q'
No .. it is not at q is a fixed point in the non-rotating frame. A fixed
point on the tube in the *rotating* frame cannot always be at a fixed point
in the non-rotating frame
Post by Henry Wilson DSc
even though points q(n) move around the
nonR frame in a circle.
There are multiple q_n points. the source moves from one to the next. It
is not always at the ONE point you labeled q. To claim it is is just
q(n) is where the emitter was when pulse(n) was emitted. It is marked on the
nonR frame, where it is static.
When viewed in the R frame, both q(n) and p(n) remain separated but appear to
spin backwards.
And that make no difference to what happens with the pulses. The emission
rate of the pulses in each direction is the same, their arrival rate a the
detector from each direction is the same. The time they take to travel
around the tube is the same. The number of pulses in the tube in each
direction at any time is the same. Pulses that are emitted together are
detected together.

None of your claims to the contrary on any of those points have any basis in
fact or in logic
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Yes?
Post by Henry Wilson DSc
It will pass
through point p then travel around the Earth before being finally absorbed
again at point p.
I'm assuming that point p is where the source is at the time the pulses
arrives back at a detector.
Yes?
Post by Henry Wilson DSc
When it is emitted, there are 62 pulses between q and p.
OK .. ones that were emitted before the source was at q, yes.
Post by Henry Wilson DSc
When pulse N reaches p
the first time, there are say 61 pulses
between p and q and 62 pulses have
OK.
Post by Henry Wilson DSc
There are 4000000-61 pulses around the earth
at that instant between p and q.
I'll assume you mean the there are 61 in small gap from q to p, and
4000000-61 in the larger gap from p back to q.
Post by Henry Wilson DSc
So another 61 + 4000000 - 61 pulses will reach
41? Don't you mean 61?
I did.
Post by Inertial
Post by Henry Wilson DSc
p by the time N gets there. That makes 4000062 altogether.
Yes.
Post by Henry Wilson DSc
In the other direction, only 3999938 arrive in the same time.
Yes.
For a detector that is at a fixed point q in the NonR inertial frame and
does *not* move with the source and tube.
This is a very different scenario from what you described previously (and
different to Sagnac) where the detector moves with the source and tube.
But it isn't.
It is TOTALLY difference as you have the detector point always at 'p' so it
is FIXED in the non-rotating frame
points p(n) and q(n) move around the nonR frame by an amount proportional to n.
And your point p (and q) is one point fixed in the NonR frame. Your
explanation is exposed as invalid (yet again)
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
The rotating observer still sees the two points p and q as being
separated. He marks them on the nonR frame, which appears to move backwards.
Which has NOTHING TO DO with what happens in Sagnac .. because there is no
detector that is always at fixed point p and no source that is always at
fixed point p. Describing what happens over time at those fixed points in
the non-rotating frame has NOTHING AT ALL TO DO with the Sagnac effect or
anything you have been claiming.
Every rotating frame has an associated inertial one.
this is not the sole property of Einstein.
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Clearly if the source is moving with the earth, and the detector is not, but
is fixed in a non-rotating inertial frame, you can work out the rotation
rate just be seeing how the source moves away from the detector .. you don't
need to go counting pulses.
You cannot actually mark the bloody points on the nonR frame.
Of course you could.
Post by Henry Wilson DSc
You would need a
bloody big sheet of paper.
Then use one. This is a gedanken.
Post by Henry Wilson DSc
Post by Inertial
setup, where you had a tube attached to the earth, and a source+detector at
same location on the tube ??
Read this and you might understand
I have read .. I understand .. you're a liar who purports to understand
physics and has no idea about it.
No point .. you're a lying scum. Reading your lies again doesn't make them
true.
Henry Wilson DSc
2010-02-19 09:38:15 UTC
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
The whole configuration rotates with the Earth so that when the next
pulse is emitted, p1,q1 would be slightly displaced from p and q...(again,
drawn in the nonR frame). p62,q62 would be 1 metre from p,q.
yes
Post by Henry Wilson DSc
Paul, Tom and Jerry have always wrongly claimed that the points p and q are
together in the rotating frame...
The points where the photon are emitted and detected in the rotating frame
are together in the rotating frame
seemingly so
NO .. it is 'so' .. not 'seemingly so'. They are physcially at the one
location in that frame at all times. They are physically both at the same
location at all times in the NonR frame .. but that combined location
changes over time. There is never a time in either frame when the source
and detector are separated as you claim.
Now you have joined Paul, Tom and Jerry in their ignorance.
You cannot seem to understand that the points p and q are drawn on the nonR
frame and represent the position in that frame of the source/detector at TWO
DIFFERENT TIMES.
So an observer rotating with the source sees each 'p and q' separated but
moving.
Post by Inertial
Post by Henry Wilson DSc
but wrong...
Yes .. you are wrong .. we know that
You are as ignorant as paul, tom and jerry.
Post by Inertial
Post by Henry Wilson DSc
which is why people make mistakes when they use
rotating frames.
You should stop trying to do anlaysis of Sagnac as you clearly cannot handle
what happens when things rotate.
You should give up physics altogether because you don't have the aptitude for
it.
Post by Inertial
Post by Henry Wilson DSc
Points p and q are 'drawn' on the nonR frame, which spins backwards in the R
frame. So they remain apart to a viewer in the rotating frame.
They make no difference to what happens in a Sagnac device (not in your
tube-around-the-earth device.
I've explained that. They do.
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
The points where the photon are emitted and detected in the non-rotating
frame are not together in the non-rotating frame
correct.
Post by Inertial
Post by Henry Wilson DSc
and that BaTh is therefore refuted.
It is. Your lies and stupidity don't make it true
Sorry, you are no smarter than paul, tom or jerry.
I don't claim to be .. we are simply ALL smarter than you. And honest.
Something you'll never be.
Sorry, you are probably more ignorant than paul, tom or jerry.
Post by Inertial
Post by Henry Wilson DSc
q(n) is where the emitter was when pulse(n) was emitted. It is marked on the
nonR frame, where it is static.
When viewed in the R frame, both q(n) and p(n) remain separated but appear to
spin backwards.
And that make no difference to what happens with the pulses. The emission
rate of the pulses in each direction is the same, their arrival rate a the
detector from each direction is the same. The time they take to travel
around the tube is the same. The number of pulses in the tube in each
direction at any time is the same. Pulses that are emitted together are
detected together.
In the nonR frame, the travel times are the same but the arrival rate is
different from the two directions.
In the rotating frame, the same applies because the emission point(s) appeear
to be moving away backwardds and there is a similar doppler shift.
Post by Inertial
None of your claims to the contrary on any of those points have any basis in
fact or in logic
None of my claims can be understood by anyone who doesn't have exceptional
physics ability.
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
This is a very different scenario from what you described previously (and
different to Sagnac) where the detector moves with the source and tube.
But it isn't.
It is TOTALLY difference as you have the detector point always at 'p' so it
is FIXED in the non-rotating frame
points p(n) and q(n) move around the nonR frame by an amount proportional to n.
And your point p (and q) is one point fixed in the NonR frame. Your
explanation is exposed as invalid (yet again)
Each pulse has its own 'p and q'. In the R frame, they move away to the right
but remain separated by 62 pulses.
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
setup, where you had a tube attached to the earth, and a source+detector at
same location on the tube ??
Read this and you might understand
I have read .. I understand .. you're a liar who purports to understand
physics and has no idea about it.
No point .. you're a lying scum. Reading your lies again doesn't make them
true.
You are ignorant...like paul, tom and jerry

Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-19 12:00:40 UTC
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
The whole configuration rotates with the Earth so that when the next
pulse is emitted, p1,q1 would be slightly displaced from p and q...(again,
drawn in the nonR frame). p62,q62 would be 1 metre from p,q.
yes
Post by Henry Wilson DSc
Paul, Tom and Jerry have always wrongly claimed that the points p and
q
are
together in the rotating frame...
The points where the photon are emitted and detected in the rotating frame
are together in the rotating frame
seemingly so
NO .. it is 'so' .. not 'seemingly so'. They are physcially at the one
location in that frame at all times. They are physically both at the same
location at all times in the NonR frame .. but that combined location
changes over time. There is never a time in either frame when the source
and detector are separated as you claim.
Now you have joined Paul, Tom and Jerry in their ignorance.
No ignorance on my part, you lying piece of shit
Post by Henry Wilson DSc
You cannot seem to understand that the points p and q are drawn on the nonR
frame and represent the position in that frame of the source/detector at TWO
DIFFERENT TIMES.
Of course I understand. you lying piece of shit
Post by Henry Wilson DSc
So an observer rotating with the source sees each 'p and q' separated but
moving.
Irrelevant, you lying piece of shit
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
but wrong...
Yes .. you are wrong .. we know that
You are as ignorant as paul, tom and jerry.
Fuck off you lying piece of shit
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
which is why people make mistakes when they use
rotating frames.
You should stop trying to do anlaysis of Sagnac as you clearly cannot handle
what happens when things rotate.
You should give up physics altogether
Nope .. I understand it , not like you, you lying piece of shit
Post by Henry Wilson DSc
because you don't have the aptitude for
it.
Of course I do .. I can shit all over your pathetic excuse for a brain, you
lying piece of shit
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Points p and q are 'drawn' on the nonR frame, which spins backwards in
the
R
frame. So they remain apart to a viewer in the rotating frame.
They make no difference to what happens in a Sagnac device (not in your
tube-around-the-earth device.
I've explained that. They do.
No .. its just another one of your pathetic little lies, you lying piece of
shit
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
The points where the photon are emitted and detected in the non-rotating
frame are not together in the non-rotating frame
correct.
Post by Inertial
Post by Henry Wilson DSc
and that BaTh is therefore refuted.
It is. Your lies and stupidity don't make it true
Sorry, you are no smarter than paul, tom or jerry.
I don't claim to be .. we are simply ALL smarter than you. And honest.
Something you'll never be.
Sorry, you are probably more ignorant than paul, tom or jerry.
Even the most ignorant amogst us is smarter that you, you lying piece of
shit
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
q(n) is where the emitter was when pulse(n) was emitted. It is marked on the
nonR frame, where it is static.
When viewed in the R frame, both q(n) and p(n) remain separated but
appear
to
spin backwards.
And that make no difference to what happens with the pulses. The emission
rate of the pulses in each direction is the same, their arrival rate a the
detector from each direction is the same. The time they take to travel
around the tube is the same. The number of pulses in the tube in each
direction at any time is the same. Pulses that are emitted together are
detected together.
In the nonR frame, the travel times are the same but the arrival rate is
different from the two directions.
Wrong, you lying piece of shit. I've already proven that you re .. you just
keep on lying because you are lying piece of shit
Post by Henry Wilson DSc
In the rotating frame, the same applies because the emission point(s) appeear
to be moving away backwardds and there is a similar doppler shift.
No .. the emission point is fixed in the rotating frame, you lying piece of
shit
Post by Henry Wilson DSc
Post by Inertial
None of your claims to the contrary on any of those points have any basis in
fact or in logic
None of my claims can be understood by anyone who doesn't have exceptional
physics ability.
BAHAHA .. I understand your claims just fine. And they are exactly what one
would expect from a lying piece of shit like you
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
This is a very different scenario from what you described previously (and
different to Sagnac) where the detector moves with the source and tube.
But it isn't.
It is TOTALLY difference as you have the detector point always at 'p' so it
is FIXED in the non-rotating frame
points p(n) and q(n) move around the nonR frame by an amount
proportional
to n.
And your point p (and q) is one point fixed in the NonR frame. Your
explanation is exposed as invalid (yet again)
Each pulse has its own 'p and q'.
And their own path .. that's what I told you, you lying piece of shit
Post by Henry Wilson DSc
In the R frame, they move away to the right
but remain separated by 62 pulses.
They are as meaningless as you are, you lying piece of shit
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
setup, where you had a tube attached to the earth, and a
source+detector
at
same location on the tube ??
Read this and you might understand
I have read .. I understand .. you're a liar who purports to understand
physics and has no idea about it.
No point .. you're a lying scum. Reading your lies again doesn't make them
true.
You are ignorant...like paul, tom and jerry
Fuck off you lying piece of shit
Paul B. Andersen
2010-02-23 11:28:28 UTC
Post by Henry Wilson DSc
On Wed, 17 Feb 2010 13:10:56 +0100, "Paul B. Andersen"
Post by Paul B. Andersen
Post by Henry Wilson DSc
On Mon, 15 Feb 2010 14:13:02 +0100, "Paul B. Andersen"
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
I recognized my mistake and immediately corrected it.
The pulses that leave together arrive together.
If you want to comment, use my later version in which the pulse arrival rates
are different from both directions. The theory should be obvious to anyone with
a higher IQ than inertial.
"The pulses that leave together arrive together", but the pulses are
arriving at a different rate than those they arriving together with.
It sure takes a high IQ to understand that! :-)
Maybe you are having trouble with frames.
Post by Paul B. Andersen
| Yet again you have made a giant fool of yourself.
| I wonder how long it will take this time before you
| realize that you have made a giant, unbelievable stupid blunder.
|
| I bet it will take a very long time, and if it eventually should
| dawn to you, you will never admit it. What will you do then?
| Start a new thread with a modified 'experiment'? :-)
Now we know what you did! :-)
conclusion based on that blunder was wrong, so you had to
make an even bigger blunder to defend your wrong conclusion.
Good to see that you still are able to invent new stupidities
and not only are recycling old ones.
This was definitely one of your better.
Keep it up, the sky is the limit!
==//=================p---------q|====================//====optical tube
________________________earth____<-v__________________(actually curved)
A pulse of light is emitted every 3.333 nanosecs by a source. The pulses, which
are separated by 1 metre, are sent down an optical fibre that is wrapped around
the earth.. There are 4000000 pulses around the equator.
A pulse emitted at point q will travel leftward around the Earth through the
fibre before reaching a detector at point p. During its travel time the Earth
revolves by 62 metres.
We will consider what happens when pulse N is emitted at point q. It will pass
through point p then travel around the Earth before being finally absorbed
again at point p.
When it is emitted, there are 62 pulses between q and p. When pulse N reaches p
the first time, there are say 61 pulses between p and q and 62 pulses have
already passed through point p. There are 4000000-61 pulses around the earth
at that instant between p and q. So another 61 + 4000000 - 41 pulses will reach
p by the time N gets there. That makes 4000062 altogether.
In the other direction, only 3999938 arrive in the same time.
Thus Spoke Doctor Ralph Rabbidge. :-)

As I said, the sky is the limit!
Close, but not quite there.
Go for it!
--
Paul

http://home.c2i.net/pb_andersen/
Henry Wilson DSc
2010-02-23 21:30:22 UTC
On Tue, 23 Feb 2010 12:28:28 +0100, "Paul B. Andersen"
Post by Henry Wilson DSc
On Wed, 17 Feb 2010 13:10:56 +0100, "Paul B. Andersen"
Post by Paul B. Andersen
Post by Henry Wilson DSc
On Mon, 15 Feb 2010 14:13:02 +0100, "Paul B. Andersen"
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
I recognized my mistake and immediately corrected it.
The pulses that leave together arrive together.
If you want to comment, use my later version in which the pulse arrival rates
are different from both directions. The theory should be obvious to anyone with
a higher IQ than inertial.
"The pulses that leave together arrive together", but the pulses are
arriving at a different rate than those they arriving together with.
It sure takes a high IQ to understand that! :-)
Maybe you are having trouble with frames.
Post by Paul B. Andersen
| Yet again you have made a giant fool of yourself.
| I wonder how long it will take this time before you
| realize that you have made a giant, unbelievable stupid blunder.
|
| I bet it will take a very long time, and if it eventually should
| dawn to you, you will never admit it. What will you do then?
| Start a new thread with a modified 'experiment'? :-)
Now we know what you did! :-)
conclusion based on that blunder was wrong, so you had to
make an even bigger blunder to defend your wrong conclusion.
Good to see that you still are able to invent new stupidities
and not only are recycling old ones.
This was definitely one of your better.
Keep it up, the sky is the limit!
==//=================p---------q|====================//====optical tube
________________________earth____<-v__________________(actually curved)
A pulse of light is emitted every 3.333 nanosecs by a source. The pulses, which
are separated by 1 metre, are sent down an optical fibre that is wrapped around
the earth.. There are 4000000 pulses around the equator.
A pulse emitted at point q will travel leftward around the Earth through the
fibre before reaching a detector at point p. During its travel time the Earth
revolves by 62 metres.
We will consider what happens when pulse N is emitted at point q. It will pass
through point p then travel around the Earth before being finally absorbed
again at point p.
When it is emitted, there are 62 pulses between q and p. When pulse N reaches p
the first time, there are say 61 pulses between p and q and 62 pulses have
already passed through point p. There are 4000000-61 pulses around the earth
at that instant between p and q. So another 61 + 4000000 - 61 pulses will reach
p by the time N gets there. That makes 4000062 altogether.
In the other direction, only 3999938 arrive in the same time.
(apologies, the '41' was an obvious typo)
Thus Spoke Doctor Henry Wilson. :-)
As I said, the sky is the limit!
Close, but not quite there.
Go for it!
I'm glad you liked it. Did you learn anything?

Henry Wilson...

.......provider of free physics lessons