
On Tue, 7 Mar 2006 00:11:47 0000, "George Dishman" <[email protected]>
wrote:
>
>"Henri Wilson" <HW@..> wrote in message
>news:[email protected]
>> On 6 Mar 2006 06:05:47 0800, "George Dishman" <[email protected]>
>> wrote:
>>>
>>>Understood. That means light at cv wrt the pocket
>>>has its speed increased. That's not a problem, just
>>>an observation. The point is that you can then deduce
>>>the relationship is linear for v << c and if we call the
>>>speed V (so inititially V = c+v) then
>>>
>>> dV/ds = o(cV)/L
>>>
>>>where 's' is the distance travelled thrrough the pocket
>>>of density 'o', and L is a universal characteristic length.
>>>For uniform density, the speed approaches c exponentially.
>>
>> If we assume that the pocket behaves entirely like a gas then the normal
>> extinction rules should apply. (Note: your equation is dimensionally wrong
>> unless 'o' is a ratio, relative to a reference density).
>
>Yes, I spotted that just after posting but as
>you say it can be treated as the ratio or rolled
>up into constant L.
>
>> However in this case, densities are so low that V and cV hardly change.
>> The
>> effect would be almost linear across the pocket if its density were
>> homogeneous, which would generally not be true. We would have to make
>> assumptions about density distribution to really analyse this.
>
>The neat thing about the differential form is that
>it is reasonable to assume homogeneity over an
>infinitesimal distance ds. Variations are then
>dealt with by the process of integrating.
Not so easy George. 'o' becomes a function of distance.
The only way to solve this kind of integral is via computer.
>> ...but your equation is basically correct. The rate of change would be
>> dependent on pocket density and difference between incoming photon speed
>> and
>> the 'equilibrium light speed' in the pocket.
>>
>> This latter term is very vague. Whereas the 'aether' defines light speed
>> wrt
>> its absolute self, my Haether has NO definite equilibrium speed unless it
>> is
>> considered uniform and infinite.
>
>No, it should be the equilibrium speed over an
>infinitesimal distance, i.e. the speed at which
>there is no rate of change.
All right, you are looking at this from a different point of view.
You are saying that if lots of light enters the pocket with a wide range of
speeds, there will be some light that has just the right speed to be unaffected
by the Haether.
That's not quite what I have in mind.
I would say that the equilibrium speed wrt the 'pocket frame' is always close
to c. It would have a conventional refractive index that would always be very
close to one.
>> In practice, this speed will never be even
>> approached by light entering such a pocket at a markedly different
>> relative
>> speed.
>>
>> I cannot see anything wrong with this theory.
>
>It is selfconsistent and defined by the above
>equation so it only remains to establish the
>value of L.
L would vary with conditions in te pocket.
That's where a computer analysis would come in handy.
>>>
>>>Well you can't expect me to know that. Anyway
>>>you used the term so I used it as well, if you don't
>>>like the term, don't use it.
>>
>> I DO like the term.
>
>Then stop complaining when I use it. You can
>see above what I think it means and since say
>my equation is "basically correct" I don't
>know why you are objecting.
You don't understand what it means. Haether does not radically change light
speed over tens or even hundreds of LYs. It doesn't have to affect speed very
much to drastically alter brightness curves observed further down the track..
Star orbit speeds are never anywhere near 'c'. Light from different orbit
phases travels at c+v and cv wrt the orbit centre. It wouldn't take much
change to bring both of these closer to 'c' and dramatically reduce the size of
the brightness curve at long distances.
>>>> >> Who knows?
>>>> >
>>>> >We do, the photon densities are well defined.
>>>>
>>>> ....even that's very debatable.
>>>> Nobody has a decent theory of what a 'photon' actually is.
>>>
>>>You may not, QED works just fine.
>>
>> Largely empirical and with many unknown quantities.
>
>There are no unknowns, it gives predictions
>more accurate than any other theory.
No comment.
>>>> >I am not an interferometer  read your
>>>> >question again ;)
>>>>
>>>> Nothing wrong with my question George.
>>>
>>>The question suggests _I_ could feel the difference.
>>
>> YOU COULD feel the difference. It is a huge effect.
>
>For a carousel rotating once every 400000
>years? I don't think so.
An interferomoter is sensitive to 1 in 10^12 or less.
>>>> >> I will repeat the old question (which you cannot answer):
>>>> >
>>>> >Answered repeatedly:
>>>> >
>>>...
>>>> >> Why should the pulses travel together through space?
>>>> >
>>>> >Because of the geometry of spacetime. I can tell
>>>> >you the answer, I can't forcefeed understanding.
>>>>
>>>> Don't run for cover behind your religion's fancy jargon, George.
>>>
>>>Don't hide behind igorance Henry, you asked for
>>>an answer and I gave it. It's up to you to learn
>>>geometry if you don't already know it.
>>
>> You have never answered it. You can only quote a disguised version of the
>> second postulate.
>
>The postulate describes the effect, the geometry
>is the cause. You asked for the cause and that's
>what I answered and have answered many, many times.
Who are you trying to kid, George?
>
>>>> >Nope, it only has one in which the jar is at
>>>The same kind yes, but you are incapable of adding
>>>the second frame because the animations for SR and
>>>aether theory are mutually exclusive.
>>
>> The only difference is that the aether has a different rate for each
>> frame.
>
>The difference is that the speeds for the second
>jar must be symmetrical in SR but asymmetrical
>in aether theory. Those are mutually exclusive
>therefore they cannot be the same. The logic is
>simple and inescapable.
>
>> The theories rely on the same: "light speed is determined by a property of
>> 'fixed' space rather than the source".
No comment from George. He's too wrapped in geometry...
>>>> >> Every pulse represents a different
>>>> >> frame.
>>>> >
>>>> >You should find out what a frame is.
>>>>
>>>> YOU SHOULD!!!!!
>>>
>>>I hav already told you what is needed. You
>>>can't do it because your claim is untrue.
>>
>> You just don't want to accept that SR is just an aether theory.
>
>I don't accept things I have proved to be wrong.
Well you haven't proved ME wrong.
>>>> Photons don't like being rotated George. They have 'axes'...built in
>>>> gyros...
>>>
>>>I know, we call it spin or polarisation and that is
>>>controlled in iFOGs. It isn't allowed to contaminate
>>>the operation.
>>
>> No it isn't 'spin', a very vague term at best.
>
>Spin is precisely defined, do some research.
What exactly 'spins'?
>> It could be related to polarization because that is prominent in any
>> reflection..
>
>Look up the FAQ.
I already have a fair idea.
>>>> Measure the speed of a pulse through a long fibre A) when it is straight
>>>> and B)
>>>> when it is coiled tightly.
>>>>
>>>> I have asked Paul Andersen to arrange for this to be peformed by
>>>> students.
>>>> If he doesn't do that I can only assume he is running from the truth.
>>>
>>>Don't be an idiot Henry, it is your hypothesis so get
>>>off your backside and do some work for a change.
>>>If you don't I can only assume you are bone idle.
>>
>> I can't do it. I don't have a laboratory.
>
>You don't need one for my version, a length
>of straight pavement would be fine though it
>might be easier after dark.
How do I make a coil with pavement?
>>>> >Well done Henry, I wondered if you would get it.
>>>> >Here's what I was thinking of. Take one source
>>>> >S and split it using a directional coupler into
>>>> >two long fibres of equal length (the equality
>>>> >isn't critical). Put a loop near the end of each
>>>> >and let the light from the ends overlap onto a
>>>> >screen:
>>>> >
>>>> > _____________________O__ 
>>>> > S ___/  screen
>>>> > \_____________________O__ 
>>>> > 
>>>> >
>>>> >You get a fringe pattern. Now move
>>>> >one loop along the fibre:
>>>> >
>>>> > _____________________O__ 
>>>> > S ___/  screen
>>>> > \__O_____________________ 
>>>> > 
>>>> >
>>>> >
>>>> >The light before the loop moves at c while
>>>> >after the loop is it reduced. The longer
>>>> >period at the reduced speed in the second
>>>> >case will delay the light relative to the
>>>> >first reference fibre so cause the fringes
>>>> >to be displaced in proportion to the
>>>> >movement of the loop.
>>>>
>>>> That's a little too messy....you can't just 'move the loop'.
>>>
>>>Stretch out a bit of string on your desk, make a
>>>loop, put your finger in it and slide it along. the
>>>loop will move just fine. Wrap it round a tin can
>>>and roll it along the desk and you can even
>>>maintain the diameter. Repeat numerous times
>>>in each direction and small stretches will cancel
>>>out, you can plot a scatter diagram and easily
>>>fit a straight line and get an error measurement
>>>too.
>>
>> It would have to be done precisely to maintain a perfectly constant
>> tension.
>
>Use minimal tension to straighten the fibre,
>reduce it in small steps measuring the
>stretching of an elastic band, plot the
>fringes and project the intercept at zero
>tension.
I would wind hundreds of turns around a 5 cm diameter former then roll that
carefully from one end to the other, making sure no turns crossed over any
others.
It might be possible. With my help, you could still become famous George.
>
>>>> Mine is simpler and straight forward for a start although the effect may
>>>> be too
>>>> small to detect easily.
>>>> Even with a 3000 metre fibre, the travel time is only 10 us and the
>>>> difference
>>>> might be only a small fraction of that.
>>>
>>>That's the problem. Try calculating it from the
>>>numbers we had for the KVH iFOG.
>>
>> We don't know the size of the effect though. Only an experiemnt can reveal
>> that.
>
>Yes we do, it is what is need to make Sagnac
>work. That was your original reason for
>proposing it. Just work out the coefficient
>from that.
That will inevitably lead us to the conclusion that SR explains the the iFoG
effect but for the wrong reasons.
>> I should imagine you would need hundreds of turns in the loop to produce a
>> noticeable effect. I think the lengths would have to be longer than 100m
>> too.
>
>Don't imagine, use science. You gave the equation
>some time back so just apply it and calculate how
>many turns for a given radius (say a few cm) would
>be needed to give a shift of one fringe over that
>length. Since c >> v in iFOGs, I think it may be
>surprisingly easy to detect.
But as I pointed out there are two separate effects.
One relates to differential slowing due to drag, the other to the fact that the
average energy in the beams changes radius by different amounts due to slightly
different centrifugal forces.
>
>George
>
HW.
www.users.bigpond.com/hewn/index.htm

