sci.physics.relativity

On 4 Oct, 10:44, HW@....(Clueless Henri Wilson) wrote:
> On Thu, 04 Oct 2007 00:27:38 -0700, George Dishman wrote:
> >On 4 Oct, 00:09, HW@....(Clueless Henri Wilson) wrote:
> >> On Wed, 3 Oct 2007 22:53:12 +0100, "George Dishman" wrote:
> >> >Since the period of the source is the same
> >> >for both paths (it is the same source) and both
> >> >travel times are (2 pi R / c), each path must have
> >> >the same number of waves.
>
> >> George, the source emits N waves per second at c wrt itself.
>
> >Right but let's make it clearer:, the source
> >emits N waves per second in all frames. The
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
> >emission speed of the waves is c in the
> >co-rotating frame.
>
> George, George, the source certainly emits N waves per second IN ALL FRAMES.

Henry, Henry, I just said that !

> I guess 'Rate' is absolute in BaTh.

Yes, in this case. Consider the equation t' = t
in the Galilean Transforms that you assume in
ballistic theory.

> THAT DOESN'T MEAN AN OBSERVER IN ANOTHER FRAME WILL RECEIVE THOSE WAVES AT THAT
> RATE.

Right, BUT THE DETECTOR IS IN THE SAME FRAME AS
THE SOURCE, both are actually the same beam
splitter in the Sagnac configuration.

> >> the number of waves arriving at an observer is N(c+v)/c, where v is the
> >> observer's speed towards the source.
>
> >Only for an observer in the inertial frame.
>
> If it wasn't inertial I wouln't write 'v'.

Yep, just making sure you didn't lose sight of the
frame you are using at this point.

> >> Whilst, period is absolute, observed 'frequency' is not.
>
> >The frequency at the detector is the same
> >as the emitted frequency.
>
> It is if there is no acceleration during the time the light is traveling.

Yes, we are analysing the constant velocity case.

> >> All you are saying is that the travel times are both equal to the same number
> >> of absolute time units, WHICH FOR CONVENIENCE IS DEFINED AS THE PERIOD OF
> >> EMISSION OF THIS PARTICULAR LIGHT.
>
> >Not for convenience henry, the RELATIVE PHASE of
> >the two rays when they impact the detector is what
> >determines the brightness (in phase = bright, 180
> >degrees = dark) and the relative phase is
>
> > phase = 360 * time_difference / period
>
> No George,

Yes Henry, for goodness sake learn some basic concepts.

> phase differences occur due to different numbers of wavelengths in
> each path.

Regardless of what causes it, what I just wrote
IS the phase difference.

> >> In other words, your argument is simply, "the travel times are the same,
> >> therefore the travel times are the same..."
>
> >No, "The travel times are the same therefore
> >the rays are in phase therefore the detector
> >is always at the centre of a bright fringe
> >therfore the fringes do not move."
>
> You are silly George.
> The travel times are indeed the same when there is NO ACCELERATION....

You are clueless Henry, we are discussing the situation
during constant .

> and the
> fringes don't move...

Right, but ballistic theory says they are not
DISPLACED.

> that is what is observed...

They are observed to be dsplaced by a constant
amount.

> When there is acceleration, the travel times are NOT the same and the numbers
> of wavelengths in each path changes. One goes up the other goes down......The
> path lengths also change.

When there is angular acceleration, there is Doppler
shift because the path lengths change therefore you
get beats and fring _movement_, but when the acceleration
stops and you allow enough travel time for light emitted
at the new constant speed to travel to the detector, the
displacement reverts to zero in ballistic theory. The
fring DISPLACEMENT is proportional to the angular
ACCELERATION.

> Get it now?

I always did, you still don't.

> >> >To use the path length difference, divide by the
> >> >distance moved by the wave in a given time, not
> >> >the wavelength, because you are measuring the
> >> >path length from where the light was emitted, not
> >> >where the source is when the light is detected. You
> >> >then get a consistent answer from all three methods.
>
> >> George, wake up.....and give up. Einstein is dead...for the second time....
>
> >No, you just screwed up the algebra for the umpteenth
> >time. You can talk philosophy but you can't do physics.
>
> Give it up George.

Give up simple algebra? I don't think so, though
I might give up trying to educate you past eigtth
grade maths.

George