Re: The Emission Theory of Androcles

On Tue, 15 Sep 2009 10:40:56 -0400, Jonah Thomas <jethomas5@xxxxxxxxx> wrote:

hw@..(Henry Wilson, DSc) wrote:
Jonah Thomas <jethomas5@xxxxxxxxx> wrote:
hw@..(Henry Wilson, DSc) wrote:



A moving source would give you a compression, you'd get eccentric
circles instead.

But you could use that eccentricity to tell who was moving. If there
aren't preferred frames then everybody ought to calculate those
things as concentric circles. And that's one of the things SR gives
you.

It doesn't. It simply says it does by postulate.

Sure, but it gives a model that does result in the waves moving in
concentric circles independent of frame, and so far it's largely
compatible with experimental evidence. What more would you want? Well,
it would be nice if it made sense. But apart from that....

Wake up. There is NO experimental evidence.



It was thrown away when the PE effect was discovered.......
ironically> by Einstein himself.

I don't see that. The PE effect is perfectly compatible with light as
waves, isn't it?

NONONONONONONONONO!

That's the problem.

Relativity is incompatible with quantium theory yet idiots like
inetial and demented dougie still try to defend the nonsense.

But apart from relativity, there's no particular reason you can't have
light waves and still have atoms that demonstrate the PE effect, right?

There is. That's been known for a century




It seems to me that you don't have an alternative model, you have a
proposal for an alternative model.

Well I have the basis of a model. I don't spend all day thinking about
it. I don't care how many more UNI students are brainwashed by the
physics establishment. The truth will eventually come out.

Sure, but your model isn't better than their model until you have
something that works.

It does work. It produces the experimentally verified equation for sagnac. Why
do you think 'inertial' and Jerry are so desperate to find a flaw in my theory?
I am a threat to their whole belief system. Poor old Jerry has backed a loser
all his life. Do you think he wants to know about it?


You're getting close.
My definition of wavelength is something like "In the source frame, a
photon moves a certain distance in one 'cycle' of its intrinsic
oscillation (whatever that may be)". That distance is an absolute and
invariant spatial interval....just like the distance between the ends
of a rigid rod..

So turns go with distance independent of time. Light can have different
"wavelengths", it can have a different number of turns per unit
distance, but it's distance that matters independent of the time or the
speed. We can stop talking about frequency. Turns, distance, and time
give everything so far. There's polarization, which people interpret as
linear in any direction perpendicular to the direction of travel, or as
circular, or as elliptical. At this point my interpretation of their
interpretation is that the axis of turning can be any direction in 3D.
When it's the direction of travel it's circular polarization. When it's
perpendicular to the direction of travel you get linear polarization in
the third direction. You'll probably want something analogous for your
turns because what they do fits the reality on some level.

Any theory about the nature of photons has to explain polarization. Maxwell's
wave theory does that but it requires a medium. I believe a photon carries it
own little bit of 'aether' ...and that's where the intrinsic oscillation
occurs. Fields cannot simply happen in 'nothing'. As I've pointed out many
times, space carrying a field must be different from space devoid of fields.
What might that difference be?


that's the other demo. THe stationary wave is put there purely
so> >you> can see the phase difference.

No, this one too. You drew waves that get extended around a
circle.> >At any one spot the wave never changes after it gets drawn.
Those> >waves are frozen once they are drawn.

OK. You have to find a model that requires the emitted light to
experience the same number of cycles per path as there are absolute
wavelengths.

So, you measure the pathlength and that gives you the number of
turns. OK.

Yep.

OK!


During any CHANGE in rotational speed, a change also occurs in the
number of wavelengths in each path. They flow out of one and into
the> other.

Mmmm. You change the rotational speed. The number of turns from the
emitter to the detector is unchanged.

No it isn't. The distance 'vt' changes. That's the distance between
the start and detection points in the inertial frame....according to
both SR and BaTh.

The distance from the emitter to the detector never changes. What
changes is the distance from the emitter at time t0 to the detector at
time t1. OK.

Yes, these are the emission and detection points (in the inertial frame) for a
particular (infinitesimal) element of a ray.
You would find this is pretty simple stuff if you would bother to calculate
what the distance 'vt' is for different rotational speeds.

You still haven't understood that mathpages diagram.

What about the time it takes to
get from the emitter to the receiver? The time is the distance
divided by the speed. So when it isn't moving the time is d/c. When
it's moving at v then the time is

t=(d+vt)/(c+v)

In the inertial frame
2piR + vt = (c+v)t .........(one ray)

Or 2piR - vt = (c-v)t......(other ray)

So t = 2piR /c

Yes. t is the same either way.

The distance goes up by the amount the detector turns, and speed goes
up by the amount the detector turns.

t-vt/(c+v) = d/(c+v)
t(c+v) -vt = d
ct = d
t = d/c

The time it takes to get to the detector is independent of v. It
takes the same time no matter how fast it spins.

That's correct. THat is easily derived if you use the rotating frame.
However it isn't as simple as it appears.

Why would the number of turns it takes to get to the detector be
different when the number of turns in that distance is constant and
the time it takes to arrive is constant?

That's the big question...and when you answer it, you'll be awarded a
Nobel prize. Don't forget to mention my name will you.

[sigh] I almost took you seriously saying you didn't know. Your answer
is that the number of turns is different although it takes the same time
to arrive both ways.

So, let's pretend for the moment that the light is a series of rotating
particles, and the particles themselves are in phase. One of them leaves
the emitter, then the next one, and then the one after that. Each of
them is at a different part of its rotation cycle when it leaves, so
they will each be at a different part of the cycle when they arrive
anywhere. It is like waves that roll onto the beach, the wave crests do
not stay frozen. If one particle moves at c+v and the other at c-v, and
the first goes a distance d(c+v) and the second at a distance d(c-v),
they will not be in phase at the end because one of them has rotated 2dv
times farther than the other. Even though they start out at the same
place in their rotation they don't end up that way because they are
rotating at different speeds, proportional to their speed and
proportional to the distance they cover.

OK, I can see it that way.

The detector doesn't care much about the element itself. It is receiving
'waves' while the element is in transit from E to D.
The number of waves counted is:pathlength/wavelength.

It looks to me like when we assume that the speed of the light in the
two directions is c+v and c-v and that speed stays constant at c+v
and at c-v the whole distance, we should get no interference. But
when the speed of light is the vector sum of cD+vV where V is a unit
vector in the direction of the source and v is the speed of the
source, and D is the direction that will give us a vector sum in the
direction we're interested in, then we get precisely the amount of
interference we'd expect by classical or by SR methods, the amount
that is experimentally observed.

Yes, the classical explanation is indeed very attractive...except that
an aether does not exist and there is no obvious explanation as to why
the rays should move at c+v and c-v wrt the source.

If the frequency is constant then it goes one way. If the wavelength is
constant it goes differently.

As I said, it it not the frequency of the particular element that matters. It
is the number of waves counted while the element is in transit.

How do we choose between those? Well, I have assumed that the particles
are in phase when they leave the emitter. The two that leave in two
opposite directions and at different speeds are at the same angle in
their rotation. If one of them rotates slower than the other and travels
slower, then when the time comes to send the second particles the slow
ones will be closer together. So if you look at the distance between the
waves, it will be closer for the slow side! Even though we said
"wavelength" was the same! On the slow side the particles are moving
slower and rotating slower so they each rotate once in the same
distance, but the apparent wavelength will be shorter!

No it wont. It is an absolute distance. You are using the wrong model.

So number of turns is something distinct from traditional wavelength.
You have to give it a new name or people will confuse it with wavelength
and they will misunderstand.

There has never been a proper definition of eitIher 'wavelength' OR 'frequency'
in the case of light. Wave theories appeared to explain diffraction and quit a
few other properties of light but quantum mechanics put an end to all that. In
reality nobody knows a bloody thing about the nature of a photon or about light
in transit. Physics is very much in its infancy.... and Einstein hasn't helped
one iota.



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

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

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