Re: The Emission Theory of Androcles

"Jonah Thomas" <jethomas5@xxxxxxxxx> wrote in message news:20090911041619.22a7a49d.jethomas5@xxxxxxxxxxxx
hw@..(Henry Wilson, DSc) wrote:
Jonah Thomas <jethomas5@xxxxxxxxx> wrote:
>hw@..(Henry Wilson, DSc) wrote:

>> Here is the simple explanation of a four mirror Sagnac.
>> http://www.users.bigpond.com/hewn/sagnac.jpg
>
>I see! And that's what Androcles was hinting at too! Very good! I'm
>laughing, that's delightful.
>
>But now I am having trouble with it. The wavelength should stay the
>same, yes. But the frequency stays the same too.

This is where the relativist rabble goes wrong. In the NON-ROTATING
FRAME the frequencies of the rays are doppler shifted in opposite
directions.

As SR would agree, also the wavelengths.

We are using that frame for our analysis. This is very
basic physics ...but clearly too hard for the relativist mentality.

Not at all. SR analysis works very simply and well. No time dilations or length contractinos invovled .. jsut light travelling at c in the inertial frame

OK, I'll try to stay with the nonrotating frame. I'm trying to
understand this, it's just easy for me to mess us.

True .. messing up isn't hard.

>The wavelength is the
>same because you don't measure wavelength back toward the source when
>it emitted the wave, you measure it in the direction of the
>wavefront. So in a time interval t units long, one side emits n
>cycles at speed c+v and the other side emits n cycles at speed c-v.
>Both arrive at the sensors at the same time. During the time for one
>wave to pass from the c+v side, one wave will pass from the c-v side
>too, slower. I don't see that this gives us a phase shift or a
>frequency difference or anything for an interferometer to pick up.

...because you are jumping from one frame to another. If you try to
use the rotating frame, there is an imaginary time factor, that I
tried to explain before.

There's no imaginary time involved.

In the rotating frame, the emission point of a particular element
MOVES BACKWARDS.

No .. the emission point is fixed in the rotating frame. in the non-rotating frame the emission point (the source) has an instantaneous velocity and moves after the light is emitted.

OK, let me try this again. I'll put ridiculous numbers on it that I hope
are easy to work with.

Let's say that our light is 10 Hertz and the path is 1 light-second
long.

And then we get it rotating at c/10. Now the light in the forward
direction travels at 1.1c while the light in the back direction travels
at 0.9c.

Yeup .. in the non-rotating frame.

But the apparatus itself is moving at 0.1c, so in the same time
that it previously took for the light to go from source to target it
still goes from source to target in both directions.

Yeup

But the light in
one direction travels 10% farther, while the light in the other
direction travels only 90% as far.

Ok

How many cycles have they gone? The same number, 10 cycles.

According to the moving detector, yes. As at the detector (and source) the frequency of both waves is the same

The light is
10 hertz, so in 1 second they each send out 10 waves.

Yeup .. the source has sent out 10 waves in that time.

But each wave on
one side is stretched an extra 10% while each wave on the other side is
just 90% as long.

Not if we're talking about waves, no. The wavelength is the same.. count back 10 wavelength from the (in phase) waves that arrive at the detector .. see where you get. Then it will make sense

Then we let the waves interfere. They start out at the
same time. One of them is 10% longer than a lightwave from a stationary
source, but it also travels 10% faster. The other is 90% the length and
it travels at 90% of the speed. Won't these interfere just exactly like
they would if they both started at the same time and both were the same
length and both traveled at c?

Yeup .. on phase shift, constructive interference.

I just don't see where the phase shift comes from. I'm missing it.

It doesn't .. that's the point

But if the light doesn't just travel around the circle at c+v the whole
way in one direction and c-v the whole way in the other direction, then
it can work fine. And you haven't said anything so far that indicates
your theory needs light to stay the same speed after it reflects off a
mirror.

If you manage somehow to change the velocities to be the same as they'd be for light in an aether, you'd be fine

I and George Dishman looked at the reflection problem very intensely
some years ago. It is not the issue.
The point missed by most people is that the emission and detection
point of a particular wave element are not the same.

The detection point when both waves arrive at the same time IS the same. . hence no phase shift

SR uses this

Because in SR the wave fronts do NOT arrive at the detector at the same time

...so I can't understand why its followers want to complaiin when I
do.

Because in ballistic theory, the waves arrive at the same time. Hence no phase shift. Its so godamn simple

I think I accounted for that. That's why in one direction you travel 10%
farther to reach the end and in the other direction you travel only 90%
as far. Because the detection point has moved 10% since the wave left
the emission point.

There is only a small difference between the SR and BaTh explanation.

SR says the rays both move at c and there is a difference in distance
and time traveled. BaTh says the travel times are the same, the
distances are different but wavelength is the same in both...and
therefore there are more waves in one ray than the other. They flow in
or out during a speed change.

That's the step I'm missing.

There's no step missing .. he's bullshitting you. Therer is no speed change ... the apparatus rotates at a constant angular velocity.

It looks to me like the same number of
waves. Say you have one wave in each direction starting at time zero,
they should both reach the end at time 0.1 second. Exactly nine more
should reach the end by time 1 second, in both directions.

Yeup. So no phase shift

Alternatively, BaTh says the frequencies are doppler shifted
oppositely in the inertial frame and since then travel times are the
same, there is aohase difference when they reunite.

Except that at the detector there is NO frequency difference, the waves arive at the smae speed after the same time with the same frequency. No phase difference

If they were traveling at the same speed when they reunited then I'd see
the doppler effect. But they're still traveling at different speeds and
so I imagine them unrolling in exact overlap. No, wait. They overlap
exactly as they cross the finish line, neck and neck at the start and
the shorter one goes through slower. But the interference pattern they
make past that point? The short, slow wave matched against the long fast
one? Is that what I was missing?

I don't think you're missing all that much. Henry's analysis is wrong.. Everyone (even some of hte other crackpots) recognizes that. Everyone except poor old Henry, who i'm sure MUST know what his mistake is by now, but still claims he's right. That's called lying.

>Just as they misinterpreted yours, I think you're misinterpreting
>them. They can have the rays move at c but one of them has to travel
>farther because of the movement of the mirrors etc.

No, they wriggle out of the problem by claiming that the separation
soeed of the light from the source is indeed c+v.

I don't want to talk about SR. It looks like it's real easy to make
mistakes with SR, and my intuition is no good there either.

Its trivially easy to see the phase difference in SR. In the non-rotating frame the light travels at c in opposite directions .. one ray travels a longer distance than the other, so they arrive at different times, so there is a phase shift.

Emission
theory is so much simpler and easier,

Its trivially easy to see no phase difference in emission theory in the rotating frame. The light travels at c in opposite directions .. both rays travel the same distance, so they arrive at the same times, so there is no phase shift.

Its almost as easy to see no phase difference in emission theory in the non-rotating frame. The light travels a shorter distance, but slower, in one direction and a longer distance, but faster, in the other. Both both rays end up travelling for the same time, so they arrive at the same time, so there is no phase shift.

and I'm still having trouble with
it. I'll try to figure out the easy version first, and only take up SR
if emission theory fails.

SR is simpler. Constant speed c. Easy.


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