On 26 Oct, 10:41, HW@....(Clueless Henri Wilson) wrote:
On Fri, 26 Oct 2007 00:48:51 -0700, George Dishman wrote:
On 25 Oct, 23:20, HW@....(Clueless Henri Wilson) wrote:
On Thu, 25 Oct 2007 17:30:45 +0100, "George Dishman" wrote:
"Clueless Henri Wilson" HW@.... wrote in messagenews:2edvh3d5t4kv8u7l2fi50hbu4ifbnpvpqo@4ax .com...

George, let me explain.

George, you don't understand frames.

Henry, you are the one trying to say that fixed points
can move, it is obvious to everyone you have no idea
what a frame is.

You still can't see that the start point
is static in the inertial frame but moving backward in the rotating frame.

Thank you for proving the point.

wHAT?..THAT YOU ARE ACTING DUMB?

Paul, I and many others have repeatedly told you
that you don't understand what a frame is but
instead of listening to the explanations you just
shout abuse. This is just another case where your
ignorance is catching you out.

Not
only that, every previously emitted 'wavecrest' moves backward in proportion.

That is the physics that matters. If that happened
the speed would not be c relative to the source in
the rotating frame, it woud be c relative to your
hypothetical point, which of course is what SR says,
that's why you get the "right answer".

Gord, your not acting at all.....
The point is moving in the rotating frame, the light moves at c+v wrt that
point in the rotating frame....because the source is moving at v in the
nonrotating frame even though the emission point is not..

It is moving at that speed relative to the LAB,
not a POINT in the frame. Learn the difference
between coordinates and the physical objects
to which they relate.

George my server finally fixed the problem and I was abl to upload this:

http://www.users.bigpond.com/hewn/ringgyro.exe

It may be some time until I can look at it, my wife
and I are at an exhibition for the nxt two days and
I have tickets for the Dolphins vs. the Giants at
Wembley on Sunday :-)

OK.

Actually I did comment in another reply after a quick
look.

ROFL, Henry that's a classic: "the fact that the elements
emitted simutaneously do not arrive simultaneously" is
just another way of saying there is a phase difference at
the detector!

Again you miss the point.
SR says that the elements of the rays that reunite were NOT emitted
simultaneously. The ones that DO meet at the detector were emitted with
different phases. However, since the travel times are DIFFERENT in SR, this
phase difference cancels to some extent.

You are losing it Henry, think again. The elements are
emitted in phase and have different travel times so
arrive out of phase, there is no cancellation, the
effect you descibe is what causes the signals to be
out of phase in the first place.

think again George. I'm not losing it. You are Andersen are..
It's only a small second order effect anyway.

No, it is the first order effct. You should be
able to work this out Henry, you are sort of
double-counting.

It's the one involved when you
replace c^2-v^2 with c^2. Don't worry about it.



BaTh says they were emitted simultaneously but differ in phase when they
arrive....due to an intrinsic effect.

No, apply eqn [2] of the theory, ballistic theory
says they are emitted simultaneously and have equal
travel times so arrive simultaneously.

Indeed they do!!!!
And their 'intrinsic oscillation' is out of phase because it goes through 1
cycle every wavelength traveled.
Simple isn't it.

Very. The distance travelled is the distance
moved by a surface of given phase, so the
'intrinsic oscillation' appears when you measure
at a fixed point and is of constant phase at
a point moving at the speed of the photon.

However, even if you try to do what you want, it
isn't "every wavelength traveled", it should be
the distance travelled by a wave in the duration
of a source cycle, that is the error I have been
pointing out consistently throughout. When you
decide to listen, you will be able to correct
your maths error.

Correct, and since they were emitted in phase that
means they arrive in phase.

no George, that's only according to your classical wave theory. it doesn't
apply

Arrival time = emission time plus travel time
in all theories.

George, the leading edge - and indeed the whole waveform - of your 'moving
wiggles' does not change.
The leading edge of a BaTh photon goes trough a cycle for every wavelength
moved.

Sorry Henry, you are talking nonsense. As I have
pointed out several times, a standing wave is the
combination of two travelling waves and the speed
of each is that of a point of fixed phase.

I have defined wavelength in this context.

Rubbish, the wavelength is defined as the distance
between repetitions of the periodic waveform at a
given instant.

The absolute wavelength of light is the distance moved in the source frame
during one cycle of its 'intrinsic oscillation'. Since it moves at c relative
to the source (in the source frame), lambda = c/nu. Lambda is an absolute
length and the same in all frames.

Yes but the distance moved in that time in any other
frame is not the same as the wavelength. Your web
page mistake is that you use the wavelength for the
distance moved but in the inertial frame, that is
wrong.

'nu' should not be confused with the 'inferred frequency', which is the number
of wavelengths arriving per second...or nu(c+v)/c.

Wrong, the phase difference is pathlength / distance_per_cycle,
your algebra is plucked out of thin air and is not correct.

George, in BaTh the 'distance per cycle' is absolute and the same in all
frames.

No, you are thinking of the wavelength which as you
say is frame invariant. The distance moved per cycle
is (c+v)/f whereas the wavelength is just c/f. That
small error is why your maths is wrong.

Wrong.
see above.

What you say above confirms what I said.

the equations and answer is given at:http://www.users.bigpond.com/hewn/ringgyro.htm

See above, your maths is wrong.

See above. It is not.

See above, your own definition confirms it is.

Androcles wants to use frequency instead of wavelength and is yet to come
up
with a prediction of fringe shift in spite of all his raving.

The correct approach is to form a set of simultaneous
equations for the motion of a phase front of the light
based on the motion of the source (beam splitter) and
for the motion the detector. Solving that gives the
arival time of the phase front at the detector and the
approach allows for arbitrary variations of source speed.

the equations and answer is given at:http://www.users.bigpond.com/hewn/ringgyro.htm

See above, your maths is wrong.

See above. It is not.







A simplification of that is suitable for constant speed
where the travel time is also constant hence detection
time can be found simply as emission time plus travel
time. What you find is that the arrivial time for both
beams is the same hence the detector sees the source
Wrong as usual, the postulate is derived from Maxwell's
Equations each of which is experimentally confirmed, and
the one-way speed is confirmed as c experimentally by the
Sagnac experiment. You don't have the ability to understand
the maths involved.

Maxwell's equations use the absolute aether as a speed reference.

No they don't, they use the observer as the
reference, the speed of the aether relative
to the observer does not appear in the equations.

It has always been too small to worry about.

It is fundamental, maxwell's Equation do
not use a medium.

As I said, it appears you don't have the ability
to understand the maths involved.

George, you are way behind..

No, you just don't have the maths ability.

They don't
apply to photon particles.

Obviously but aggregating photons must produce
Maxwell's Equations.

In a medium. Speeds must always be specified relative to something George.

Nope, no medium, the speeds are relative to the observer.

ie., MAGIC, to adjust both light speeds to
be 'c'.

There is no adjustment clueless, the light is EMITTED moving
at c in the inertial frame.

It requires that the two rays move at c+v and c-v wrt the source.

Nope, they move at c relative to the source.

It states that..... but uses c+/-v in the equations.

Nope, it uses c. It appears you don't have the ability
to understand the maths involved.

Travel time is distance/speed.

Obviously, and the time, distances and speeds must
all be experessed in the same coordinate scheme.

t=2piR/(c+v)

What does the 'c+v' represent, George?

Oh good grief, how basic does this need to be
for you to understand?

"c" is the speed of the light, it tells you
the distance the light moves in a given time.

"v" is the speed of the splitter, it tells you
the distance the splitter moves in a given time.

"c+v" represents the sum of two speeds, it is
the sum of the distance move by the light and
the distance moved by the splitter and is also
the increase in their separation in a given time.
It is NOT the speed of either the light or the
splitter.

George