Re: c+v and Sagnac

On Thu, 3 Sep 2009 13:32:54 +0100, "Androcles" <Headmaster@xxxxxxxxxxxxxxxxxx>
wrote:


"Henry Wilson, DSc" <hw@..> wrote in message
news:q20v955tkn08pqbpmu9ei8csjf5ou8jpip@xxxxxxxxxx
On Thu, 3 Sep 2009 09:13:07 +1000, "Inertial" <relatively@xxxxxxxx> wrote:


No, it doesn't agree. In the SR analysis, the distances travelled by
the
wavefronts are different, and as both rays travel at c in the inertial
frame, the time taken is not the same.

You finally got something right.

I've gotten everything right. You're finally being honest and
acknowledging
some of it

But you don't accept that the above condition means that the rays travel
at
c+v and c-v wrt the source, thus contradicting SR itself.

So there are two different 'end
points' in the inertial frame in the SR analysis, one for each
wavefront.

IDIOT!

According to the SR analysis in the inertial non-rotating frame, the rays
travel at the SAME speed over DIFFERENT distances, so they arrive at
DIFFERENT times, and then there must be two DIFFERENT points of arrival as
the detector is MOVING.

There can only be one 'point' of arrival. It is the 45 mirror. ...but I
now
realise you have accepted the notion of static detection points and were
refering to those in the SR case. I apologise for calling you an idiot
over
this.... but not over anything else..

Now I KNOW you are drunk. The idiot will always be an idiot.

Yes, I'm afraid she will.

If you didn't already know that, then clearly you've not read and
understood
the SR analysis of Sagnac. That explains a lot.

I've been trying to explain it to YOU.
You seem to have learnt something at last.

Glad you said "seem". However, your perceptions are wrong, he hasn't
learnt anything.

No she hasn't


The inertial frame doesn't rotate and is at rest. The frequencies of both
rays
are therefore doppler shifted in opposite directions in that frame. What
could
be simpler?

Very true, but they still arrived at the beam splitter with zero phase
angle.
That's what confuses most people. The phase angle between the beams
varies with length AFTER the beam splitter.

Of course. There is only one beam before the splitter.



Yes we are .. and in the inertial frame the detector is moving. It has a
non-zero instantaneous velocity at the detection point. Your fixed
observer
has a zero velocity.

So has SR's. What's wrong with that?

The observer represents a point at
rest in the inertial frame....just like the one SR uses.

The observer we care about is the detector and it is moving in the frame.
If you are using a point at rest in the inertial frame, then it is NOT the
detector of the Sagnac experiment.

It is the point where the particular wave element will reach the detector.

Just run: www.users.bigpond.com/hewn/rayphases.exe

Clearly that shows changing the path LENGTH changes the phase angle.
In the real experiment the path length is adjusted to give a zero shift when
not
rotating.

the path lengths are equal when there is no rotation.

Changing the SPEED of rotation would not change the phase at all in your
model, but in Sagnac it does.

You obviously didn't understand it. Rotation speed DOES change the fringe
displacement in my model. That is clearly shown. The simple maths is provided
at www.users.bigpond.com/hewn/ringgyro.htm

again and you will see how the frquencies are the same in the rotating
frame
but different in the inertial frame. There are MORE wavelengths in one
path
than the other.

Ever see a strobe light on a record player turntable?
http://www.youtube.com/watch?v=SSRHfYECPog
Changing the speed from 45 RPM to 33 RPM changes to the wavelength
of the black/white markers on the next band. But no way do you change
the number of marks (or 'ticks'). Tick fairies are not allowed.
Frequency is constant, its 50 Hz (or 60 Hz USA). "Wavelength" changes
with speed.

nope. Photon wavelength is absolute and invariant.

If you see the speed of the turntable change, you are seeing a changing
phase angle. If you see the spot change its heading by first speeding up
and then slowing the turnable back to its original speed, you have a
compass. You can do this by having North East South West markers
around the turntable and spinning it at 3000 RPM (50 revs per second).

This is not related to wavelength or sagnac.

I'm refering to the final meeting
point of the elelment that was emitted at the particular start point.
Both those points are at rest in the inertial frame.

Yes .. the points clearly are, I've not denied that .. but the detector
and
source are not. What happens at those points needs to be adjusted (eg
doppler shift) for the velocity of source and detector. The detector does
not remain at that fixed point.

So what?
It is the number of wavelengths in each ray BETWEEN THE EMISSION AND END
POINTS
that matters.

Nope. Where are the start and end points around a record player turntable?
What matters is phase.

You are talkiing about the phase difference between the strobe and the
turntable. This is not related to what happens in a ring gyro.




Ever see a strobe light on a record player turntable?
http://www.youtube.com/watch?v=SSRHfYECPog
Your argument is flawed. You are making a terrible fool of yourself, just as
Tusseladd is with his hopeless math. SR isn't needed for Sagnac or record
player turntables with strobes. All that is needed is the understanding that
the moving black/white 'ticks' which APPEAR stationary can be replaced by
light 'ticks' to eliminate the mechanical turntable and all the crap about
which
arrives first, changing wavelengths, number of wavelengths etc. is just so
much bull***. Phase angle is all that matters.
Mechanical gyroscope compasses have been replaced by optical gyroscope
compasses but the word "gyroscope" has more magic to it than "compass"
since it is less understood.

Rotation is absolute and therefore can theoretically be detected. A ring gyro
does it by measuring small departures from light's behavior in the nonrotating
situation.


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

Einstein...World's greatest SciFi writer..
.

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