Groups > sci.physics.relativity > #402969

Re: Why Do SR and BaTh Both Give the Right Answer for a Sagnac Ring Gyro?

From	HGW <hw@....>
Newsgroups	sci.physics.relativity
Subject	Re: Why Do SR and BaTh Both Give the Right Answer for a Sagnac Ring Gyro?
Date	2016-12-26 05:23 +1100
Organization	Aioe.org NNTP Server
Message-ID	<o3p2q5$1mun$1@gioia.aioe.org> (permalink)
References	(1 earlier) <4ead48f7-f281-4cdb-bc02-12316f993bab@googlegroups.com> <o3mpva$tu5$1@gioia.aioe.org> <f0cfbb33-b1f8-423d-b4d4-2704e6f6927f@googlegroups.com> <o3nshf$582$1@gioia.aioe.org> <4189b378-fa45-40b0-b238-2e51bae7fc9e@googlegroups.com>

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On 25/12/16 20:30, Prokaryotic Caspase Homolog wrote:
> On Sunday, December 25, 2016 at 1:29:53 AM UTC-6, HGW wrote:

>>>
>>> You have little or no idea of what is going on, and BaTh is out of your control.
>>
>> Gawd! Dono is right about you. You are proving to be the biggest idiot
>> here...and that's pretty hard to achieve.
>>
>> You obviously do not have the intelligence to even understand what I
>> have said and are wingeing and whining purely for ideological reasons.
>> BaTh's Sagnac analysis assumes only one thing, that light moves at c
>> relative to its source. It uses the non-rotating frame, which is
>> perfectly normal and valid.
>> By the time a particular photon reaches the detector, the latter has
>> moved. Both rays take the same time to reach it but there are more
>> (absolute) wavelengths in one than the other when the thing is rotating.
>> How much simpler can one get?
>> If the rotating frame is used, as Roberts and Andersen attempt to do,
>> they get completely stuffed up by the fact that there is an imaginary
>> change in wavelength.
>
> Henry, here is an exercise for you:
>
> PART 1:
> In rayphases, you have outlined the bidirectional propagation of photons from
> an emission point on a rotating ring. At any one moment, because of ballistic
> c ± v emission, the forwards photon has moved ahead of the detector the same
> distance that the reverse photon has moved behind. The wavelengths of the
> forwards and reverse photons are the same. The phase of the forwards photon
> advances more rapidly than the phase of the reverse photon. Because of the
> differences in rate of phase advancement (I know that you hate the word
> "frequency"), when the photons meet, they are of different phase.

Look, you poor little fellow, I know it requires superior intelligence 
to understand this stuff....but you can at least make an effort.

During an angular acceleration, waves pass from one path to the 
other...so when the ring is rotating, there are already more waves in 
one path than the other. Therefore, at the instant a particular 
'wavecrest' leaves the source, it has the same phase in the two rays... 
but the ones arriving there at that same instant are NOT in phase... 
because the number that pass through the detector from each direction 
during the travel time of a particular element is different in each ray.
At constant rotation speed, the waves leave at the same rate and arrive 
at the same rate but they are already out of phase on arrival because of 
a prior acceleration.

That should be simple enough to understand now that I have explained it 
in such simple terms.

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