Re: Simple Sagnac


<bsr3997@xxxxxxxxxxx> wrote in message
news:1123309156.531720.99610@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
|
| sal wrote:
| > >> Ah, so you intentionally misunderstand.
| > >
| > > Not sure what you are talking about here but I don't intentionally
| > > misunderstand things. It is not my intention to be a troll. I
| > > commend you for the effort you have put into building your site.
I
| > > was just trying to correct a few things that I knew were not quite
| > > right. It wasn't meant as an attack. Sorry if it came off as
one.
| > >
| > > As for my being a crank, I do not claim that there is anything
wrong
| > > with relativity.

You should, because there is.

[snip]

sal:
| > The thing that is wrong was my assertion that "classically" the
effect
| > is inexplicable. (More on that, below.) The use I made of
relativity
| > is fine, however.
| >
| > You should, by the way, notice something: Noplace on that page do I
| > use the term "special relativity". It just happens that the only
math
| > I used was math which one encounters in special relativity.
|
| If it quacks like a duck ....

You've got that right.
I wasn't speeding, officer, it just happens that I was driving at 100
mph. You can't win against bigotry.
Actually Sagnac is extremely simple to explain, the speed of the light
in the rotating frame is c and in the laboratory frame it is c+v, c-v.
Naturally there is a beat frequency in the laboratory frame from
Doppler's equation f' = f (c+v)/c and we have f1-f2 as the beat.
If Einstein's whim
[quote]
we establish by definition that the "time" required by light to travel
from A to B equals the "time" it requires to travel from B to A.
[end quote]
Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/
is to be satisfied we can apply his equations and dilate time
in the laboratory frame, but this is contradictory to his claim
[quote]
If at the points A and B of K there are stationary clocks which,
viewed in the stationary system, are synchronous; and if the clock
at A is moved with the velocity v along the line AB to B, then on its
arrival at B the two clocks no longer synchronize, but the clock
moved from A to B lags behind the other which has remained at B
by (1/2)tv^2/c^2 (up to magnitudes of fourth and higher order),
t being the time occupied in the journey from A to B.
It is at once apparent that this result still holds good if the clock
moves from A to B in any polygonal line, and also when the points
A and B coincide.
If we assume that the result proved for a polygonal line is also valid
for a continuously curved line, we arrive at this result: If one of two
synchronous clocks at A is moved in a closed curve with constant
velocity until it returns to A, the journey lasting t seconds, then by
the clock which has remained at rest the travelled clock on its arrival
at A will be (1/2)tv^2/c^2 slow.
[end quote]
but in full agreement with
[quote]
Take, for example, the reciprocal electrodynamic action of a magnet and
a conductor. The observable phenomenon here depends only on the relative
motion of the conductor and the magnet, whereas the customary view draws
a sharp distinction between the two cases in which either the one or the
other of these bodies is in motion."
[end quote]

As every mariner knows, the aircraft carrier rotates around the compass
needle. The "customary view" is that the needle moves, and is incorrect.

This is exactly the situation we have with Sagnac, with the "customary
view" being that it is the ring that it is in motion. It isn't. The
laboratory
rotates around the ring.
Stand beside Sagnac in operation and your watch will slow down.
This is clearly absurd and there can be no connection between time and
the speed of light.


[snip acceleration]

| With Einstein's SR clock syncing provides one and only one time
| coordinate for a given location in a frame. You yourself have said it
| is absurd to have a clock out of sync with itself, but that is what
you
| end up with when you go full circle. A curve is not a straight line.
| You can only perpetuate that lie so long before it comes around and
| bites your backside.

It already has. sal is miffed. He plonked me, and I did the same to him.
Good riddance to bad rubbish.

[snip CoV]

|
| Whether you can see it or not the curve is still there. When you are
| pushing a new system of measurement because it is more accurate than
| the old, it seems sort of foolish to say that it is better if you
don't
| look too close, or in this case look so close that you don't see the
| big picture.


Right on!

| > Now, as to the analysis from the point of view of the cable -- we
can
| > again note that the acceleration doesn't affect time, and from that
we
| > at once see that it's just the same as the "straight" case, save
that
| > it's bent. The bend itself is irrelevant. (Magnify the picture
| > enough, and you can't even see the bend; and it's in the "magnified"
| > view that we actually take all the derivatives, which are what we're
| > concerned with here.)
| >
| > From there, just picturing it makes it clear that the man walking
| > around the rim carrying a watch must see things just exactly the
same
| > way the man walking down a straight moving cable would see them.
The
| > fact that he's accelerating inward, again, doesn't affect his watch,
| > and doesn't affect his measurements of tangential lengths. So, the
| > result _must_ come out the same as if the cable where laid out
| > straight.
| >
| > If we want to be totally complete in our picture, we can also
imagine
| > that we're using light pulses to synchronize closely spaced clocks
all
| > around the rim of the disk. They're close enough together that we
| > don't have to worry about the acceleration. If we ask ourselves
what
| > will happen, it's pretty clear that we'll get exactly the same
result
| > that way as we would if we did it on the straight moving cable --
when
| > we get to the "other end" we've got what looks like a time skew
| > relative to the starting point from the point of view of someone in
| > the stationary frame.
|
| It is obvious that signals transmitted in opposite directions around
| the disk do not get back to the starting point at the same instant.

You are confusing yourself when you say "starting point"; there are two
starting points, one in the lab frame and one in the ring frame.
The starting point in the ring also moves with the ring, it is also the
end point.
The laboratory rotates around the ring, so the clock at B (lab) has
moved relative to the clock at A (ring, and the light it contains).
In the ring, start and end points are the same.
In the lab, they are displaced.
It is the "customary view" that signals transmitted in opposite
directions
around the disk do not get back to the starting point at the same
instant.
It is obvious that signals transmitted in opposite directions
around the disk DO get back to the starting point at the same instant.



| Saying that it is the same instant when timed by two different clocks
| at the same location doesn't sound very convincing. Classical theory
| can get the two different times measured with one clock. Which sounds
| more reasonable? Now what did we do with that razor? ;)

The heart of special relativity is Einstein's wishful thinking that
light takes
the same "time" both ways. We have
½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v)),
for the light in one direction and
½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c+v))
for the other. The ½ cannot be correct, for that would mean
tau(x',0,0,t+x'/(c-v)) = tau(x',0,0,t+x'/(c+v)).
Nor is the coordinate x', although the distance the light travels, x',
is correct.
The clock in the lab frame moves relatively to the ring and that is
what Einstein's equation refers to.
The equation then becomes (for Sagnac)
½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(0,0,0,t+x'/(c-v)),
and we can disregard the x',y, and z coordinates to give
½[tau(t)+tau(t+x'/(c-v)+x'/(c+v))] = tau(t+x'/(c-v)).
The term t refers to the initial synchronization time of the clocks A
and B,
3:15 pm and can also be set to zero,

½[tau(0)+tau(x'/(c-v)+x'/(c+v))] = tau(x'/(c-v)), and since tau(0)= 0
by initial synchronization,
½tau(x'/(c-v)+x'/(c+v)) = tau(x'/(c-v)).
Let a = x'/(c-v) and b = x'/(c+v),
and we have
½tau(a+b) = tau(a).

But in Sagnac, b follows a, the times are concurrent and Einstein's
equation uses consecutive times, the light having been reflected.
Thus we must double his equation to give
tau(a+b) = 2tau(a)
Since a =/=b, the function tau is not linear as claimed and
the cuckoo transformations cannot be derived.

Sagnac proves SR wrong. Bring in Ockham's axe and execute
special relativity. GR falls with it. Start physics over from Newtonian
Principles.


[snip]

| > > Oh, I see, anything that doesn't agree with your point of view is
| > > silly.
| >
| > No, that's not what I meant at all.
| >
| > I meant it's silly to claim that the Sagnac effect _refutes_
| > relativity in any way.

It isn't silly at all. It is the "customary view" that is silly.
The Sagnac effect refutes special relativity and upholds
Galilean relativity. (Not that sal wil read that, he's got his head
buried in the sand.)
Anything that doesn't agree with his point of view is silly, but his
point of view is the "customary view".
Androcles.

.

Relevant Pages

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