Re: c+v and Sagnac

On Thu, 3 Sep 2009 20:22:27 +1000, "Inertial" <relatively@xxxxxxxx> wrote:



"Henry Wilson, DSc" <hw@..> wrote in message
news:jb3v95dl63n0j2qt0oht47ms6qafs6ehse@xxxxxxxxxx
On Thu, 3 Sep 2009 19:13:22 +1000, "Inertial" <relatively@xxxxxxxx> wrote:



That they are Doppler shifted back again to be the same frequency at the
detector .. which is all that matters for Sagnac results

It is not at all important.

So how the rays arrive at the detector are not at all important to Sagnac?
That is a very good indication of your stupidity.

It just ensures that the fringe displacement

Which is zero

remains stable for a particular rotation speed. What IS important is the
number
of wavelengths in each path.

No .. it doesn't have any bearing on it. unless you are measuring from
where the source is when the rays strike the detector. There is the same
number of wavelength from the detector back to where the source is now. I'e
shown you before that that is how your work with waves

Herer's Sagnac flattened out for you (cause its easier to draw that way in
ascii, could do the same in a circle). The top line is the rotating frame,
the bottom the inertial frame. S is the source, D is the detector for the
two rays. If you roll this up, you'd have the S and the tow D's at the same
location. Here's a view of the wave fronts as the rays go around

-D-----------S-----------D---------
-------------s---------------------

--D--------<--S-->--------D--------
-------------s---------------------

---D-----<--<--S-->-->-----D-------
-------------s---------------------

----D--<--<--<--S-->-->-->--D------
-------------s---------------------

-----D--<--<--<--S-->-->-->--D-----
-----d-------s---------------d-----

If you look at the s and d points, then the distance from the LHS d to the s
is shorter than the s to the RHS d, as you say. There are therefore fewer
wavelength. And relative to the bottom inertial frame, the LHS ray travels
faster then the RHS ray.

But, as one measures wavelength backward from the leading wave of the rays,
it doesn't make any difference to the phase. The waves arrived at the D
points at the same time and in phase and at the same speed and with the same
number of wavelength distance from where the S is now.

This is exactly the same in ballistic theory when you wrap this in a circle.

You are making gteh same mistake as Andersen, Roberts and Jerry.

You think a rortaing gyro viewed in the rotating frame is the same as a
nonrotating gyro in the inertial frame.

Rotating frames contain traps for the ignorant. Avoid them or pay the price.


non-zero instantaneous velocity at the detection point. Your fixed
observer
has a zero velocity.

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

There is no fixed observer in the SR analysis.

You obviously haven't even looked at the diagram to which I referred.

Of course I have

Well all the criticism you are leveling at MY analysis would equally apply to
that of SR. You are a hypocrite. You have't a clue what you are talking about.


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

And they reach it in phase

It does not.

Yes it does .. see the diagram above. It is IMPOSSIBLE for a pair of rays
that leave the source at the same speed and frequency and phase, and then
arrive at the destination at the same time with the same speed and frequency
and NOT be in phase. That you suggest otherwise shows monumental stupidity
or dishonesty.

You have tries to use a rotating frame and made an absolute mess of it.
All you have done is reproduce a nonrotating gyro.

The path lengths are different

Yes they are

and wavelength is invariant.

Fine

The travel times are the same

Yes

and the frequencies are different.

Not at the detector

Nobody cares about what happens at the instant it reaches the detector. It is
what occurs along the whole path that matters.

What more evidence do you need?

A valid analysis from you .. don't see that comgin. you're to stubbornly
convinced you are right you can't see the truth. Sad.

I am right. I get the right answer.

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

I've seen it before

You haven't understood it.

again and you will see how the frquencies are the same in the rotating
frame
but different in the inertial frame.

I know that .. I've been telling YOU that

Bull.

Where have I said otherwise?

There are MORE wavelengths in one path
than the other.

I've also told you that. But the rays are in sync when they arrive at the
detector (in ballistic analysis)

Have a good look...

I don't need to look again

They are not in synch


So .. it changes your result because the rays have travelled for the same
time and arrive at the same speed as they left the source, with the same
frequency they had when they left the source, and in the same phase as
when
they left the source.

See, this is called frame jumping.

It has nothing to do with frame jumping. it is FACT. You cannot possibly
deny it without looking a bigger fool than you already appear

Look, I'm way ahead of you on this. I have explained it many times before to
other people as brainwashed as you. I know your silly arguments backwards. I'm
not wasting any more time on this. I have presented the facts, I have animated
a ring gyro. but ther is nothing I can do to make your brain function. I'm
sorry.


I did

and see how stupid it was.

It is perfectly correct

And if you trace wavelengths back to where the source is NOW you
get the same number of wavelengths and in phase.

Where the source is NOW is not where the wave element was emitted.

It is where you count back to. Do you need another diagram?

If you trace it back to
where the source ws when the waves were emitted in teh inertial frame, you
do NOT get them in phase.

That's what I'm trying to tell you.

Not in phase at that source point.

But they ARE in phase at the inertial destination point when the detector
and the rays arrive there

How? The path lengths are different and the wavelengths are the same. can't you
count?




Not at all. Nor am I making a fool of you. You're doing a fine job of
that
on your own.

My analysis is perfectly straightforward

and totally wrong

and sound physics.

sound nonsense

It produces the correct answer.

because you cheat

Only a real idiot would even attempt to criticize it.

Only a real idiot would pretend its valid when you've had it pointed out to
you for YEARS that is it wrong. You are simply stubborn, stupid and a liar.

You simply don't have the brains to discuss physics at even a basic level.

I've proven my point. All can see what a fool you are.

I'm the one who has produced the right explanation of the sagnac effect, not
you or Einstein.

Bye



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

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

Relevant Pages

  • Re: c+v and Sagnac
    ... So the wavelength is NOT frame independent. ... So sometimes you say wavelength is frame indepedent and sometimes not. ... Wavelength is NOT frame dependent (ie., different OBSEVER SPEEDS). ... It arrives at a moving detector, and at a given time the detector will be at ...
    (sci.physics.relativity)
  • Re: c+v and Sagnac
    ... are therefore doppler shifted in opposite directions in that frame. ... So how the rays arrive at the detector are not at all important to Sagnac? ... And relative to the bottom inertial frame, ...
    (sci.physics.relativity)
  • Re: c+v and Sagnac
    ... Intrinsic wavelength changes whenever a photon changes speed. ... It arrives at a moving detector, and at a given time the detector will be at a given point. ... In the inertial frame, they don't have the same frequency. ... So tracing the path of rays in the inertial frame, where we know the rays are in phase, you get a different number of wavelengths. ...
    (sci.physics.relativity)
  • Re: c+v and Sagnac
    ... Intrinsic wavelength changes whenever a photon changes speed. ... So the wavelength is NOT frame independent. ... There is no static detector in the SR analysis ... Because SR is talking about the difference in arrival time at those point. ...
    (sci.physics.relativity)
  • Re: c+v and Sagnac
    ... That is a lie, of course. ... Then you will see that the rays are shown to move at c+v and c-v wrt the ... The detector is not at rest in the inertial frame, ...
    (sci.physics.relativity)
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