A Lesson In Sagnac and Rotating frames, For Paul, Tom and Jerry.
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From: Jerry on 17 Feb 2010 11:35 On Feb 17, 9:42 am, "Androcles" <Headmas...(a)Hogwarts.physics_u> wrote: > So you want to know the error in YOUR calculation to show why > Sagnac PROBABLY doesn't work, when all we are looking for is > a change in phase of two signals each with a wavelength measured > in Angstroms, because a fraction of an Angstrom plus a kilometre > is immeasurably different from a kilometre plus an Angstrom. > You've never heard of Vernier, either. > http://en.wikipedia.org/wiki/Vernier_scale > You are immeasurably different from an ignorant lunatic. Sagnac works all right. It just doesn't work the way that you think it does. If you bothered to find where I made my mistake, you'd discover that the conclusion remains unchanged. Coriolis has absolutely NOTHING to do with how Sagnac works. Jerry
From: Henry Wilson DSc on 17 Feb 2010 17:06 On Wed, 17 Feb 2010 13:10:56 +0100, "Paul B. Andersen" <paul.b.andersen(a)somewhere.no> wrote: >On 15.02.2010 23:24, Henry Wilson DSc wrote: >> On Mon, 15 Feb 2010 14:13:02 +0100, "Paul B. Andersen" >> <paul.b.andersen(a)somewhere.no> wrote: >> >>> On 12.02.2010 00:33, Henry Wilson DSc wrote: >>>> Paul and Jerry are standing together on the equator of an Earthlike planet, >>>> next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its >>>> circumference is 40 million metres. >>>> >>>> //===================C======================// optical fibre around equator >>>> _____________v<-_____p___q_______________________surface (actually curved) >>>> >> >> I recognized my mistake and immediately corrected it. >> >> The pulses that leave together arrive together. >> >> If you want to comment, use my later version in which the pulse arrival rates >> are different from both directions. The theory should be obvious to anyone with >> a higher IQ than inertial. > >"The pulses that leave together arrive together", but the pulses are >arriving at a different rate than those they arriving together with. > >It sure takes a high IQ to understand that! :-) Maybe you are having trouble with frames. > >BTW, in my previous posting I wrote: >| Yet again you have made a giant fool of yourself. >| I wonder how long it will take this time before you >| realize that you have made a giant, unbelievable stupid blunder. >| >| I bet it will take a very long time, and if it eventually should >| dawn to you, you will never admit it. What will you do then? >| Start a new thread with a modified 'experiment'? :-) > >Now we know what you did! :-) >You admitted your blunder, but couldn't admit that your >conclusion based on that blunder was wrong, so you had to >make an even bigger blunder to defend your wrong conclusion. > >Good to see that you still are able to invent new stupidities >and not only are recycling old ones. >This was definitely one of your better. > >Keep it up, the sky is the limit! ==//=================p---------q|====================//====optical tube ________________________earth____<-v__________________(actually curved) A pulse of light is emitted every 3.333 nanosecs by a source. The pulses, which are separated by 1 metre, are sent down an optical fibre that is wrapped around the earth.. There are 4000000 pulses around the equator. A pulse emitted at point q will travel leftward around the Earth through the fibre before reaching a detector at point p. During its travel time the Earth revolves by 62 metres. We will consider what happens when pulse N is emitted at point q. It will pass through point p then travel around the Earth before being finally absorbed again at point p. When it is emitted, there are 62 pulses between q and p. When pulse N reaches p the first time, there are say 61 pulses between p and q and 62 pulses have already passed through point p. There are 4000000-61 pulses around the earth at that instant between p and q. So another 61 + 4000000 - 41 pulses will reach p by the time N gets there. That makes 4000062 altogether. In the other direction, only 3999938 arrive in the same time. Henry Wilson... ........provider of free physics lessons
From: Jerry on 17 Feb 2010 17:55 On Feb 17, 4:06 pm, ..@..(Henry Wilson DSc) wrote: > We will consider what happens when pulse N is emitted at point q. It will pass > through point p then travel around the Earth before being finally absorbed > again at point p. > When it is emitted, there are 62 pulses between q and p. When pulse N reaches p > the first time, there are say 61 pulses between p and q and 62 pulses have > already passed through point p. There are 4000000-61 pulses around the earth > at that instant between p and q. So another 61 + 4000000 - 41 pulses will reach > p by the time N gets there. That makes 4000062 altogether. > In the other direction, only 3999938 arrive in the same time. (sigh) In the STEADY STATE, how many pulses cross p each second? In the STEADY STATE, how many pulses cross q each second? For both p and q, the answer had better be 1/3.333e-9 = 300030003 pulses. Jerry
From: Inertial on 17 Feb 2010 18:28 "Henry Wilson DSc" <..@..> wrote in message news:o1oon595e9kfg9m9hnvdtvflb4b8naa6p1(a)4ax.com... > On Wed, 17 Feb 2010 13:10:56 +0100, "Paul B. Andersen" > <paul.b.andersen(a)somewhere.no> wrote: > >>On 15.02.2010 23:24, Henry Wilson DSc wrote: >>> On Mon, 15 Feb 2010 14:13:02 +0100, "Paul B. Andersen" >>> <paul.b.andersen(a)somewhere.no> wrote: >>> >>>> On 12.02.2010 00:33, Henry Wilson DSc wrote: >>>>> Paul and Jerry are standing together on the equator of an Earthlike >>>>> planet, >>>>> next to an optical fibre that encircles the planet, WHICH IS NOT >>>>> ROTATING. Its >>>>> circumference is 40 million metres. >>>>> >>>>> //===================C======================// optical fibre around >>>>> equator >>>>> _____________v<-_____p___q_______________________surface (actually >>>>> curved) >>>>> >>> >>> I recognized my mistake and immediately corrected it. >>> >>> The pulses that leave together arrive together. >>> >>> If you want to comment, use my later version in which the pulse arrival >>> rates >>> are different from both directions. The theory should be obvious to >>> anyone with >>> a higher IQ than inertial. >> >>"The pulses that leave together arrive together", but the pulses are >>arriving at a different rate than those they arriving together with. >> >>It sure takes a high IQ to understand that! :-) > > Maybe you are having trouble with frames. > > >> >>BTW, in my previous posting I wrote: >>| Yet again you have made a giant fool of yourself. >>| I wonder how long it will take this time before you >>| realize that you have made a giant, unbelievable stupid blunder. >>| >>| I bet it will take a very long time, and if it eventually should >>| dawn to you, you will never admit it. What will you do then? >>| Start a new thread with a modified 'experiment'? :-) >> >>Now we know what you did! :-) >>You admitted your blunder, but couldn't admit that your >>conclusion based on that blunder was wrong, so you had to >>make an even bigger blunder to defend your wrong conclusion. >> >>Good to see that you still are able to invent new stupidities >>and not only are recycling old ones. >>This was definitely one of your better. >> >>Keep it up, the sky is the limit! ok .. so now you have: > ==//=================p---------q|====================//====optical tube > ________________________earth____<-v__________________(actually curved) So is tube attached to the earth, or is the tube stationary (in the NonR inertial frame) and the earth moving at <-v wrt as your diagram seems to indicate? I'll assume that you just drew the diagram poorly, and that the tube is moving with the earth. And that points p and q are moving relative to the tube (and the earth) and so are fixed in the NonR inertial frame. Yes? > A pulse of light is emitted every 3.333 nanosecs by a source. I'll assume that the source is moving with the earth (and tube) as well. Yes? > The pulses, which are separated by 1 metre, are sent down an > optical fibre that is wrapped around the earth.. There are > 4000000 pulses around the equator. OK I'll assume they are only going in one direction in this example. Yes? > A pulse emitted at point q will travel leftward around the Earth through > the > fibre before reaching a detector at point p. During its travel time the > Earth > revolves by 62 metres. OK > We will consider what happens when pulse N is emitted at point q. I'll assume the source was moving along with the earth, and when it arrives at point q it emits a pulse. Yes? > It will pass > through point p then travel around the Earth before being finally absorbed > again at point p. I'm assuming that point p is where the source is at the time the pulses arrives back at a detector. Yes? > When it is emitted, there are 62 pulses between q and p. OK .. ones that were emitted before the source was at q, yes. > When pulse N reaches p > the first time, there are say 61 pulses > between p and q and 62 pulses have > already passed through point p. OK. > There are 4000000-61 pulses around the earth > at that instant between p and q. I'll assume you mean the there are 61 in small gap from q to p, and 4000000-61 in the larger gap from p back to q. > So another 61 + 4000000 - 41 pulses will reach 41? Don't you mean 61? > p by the time N gets there. That makes 4000062 altogether. Yes. > In the other direction, only 3999938 arrive in the same time. Yes. For a detector that is at a fixed point q in the NonR inertial frame and does *not* move with the source and tube. This is a very different scenario from what you described previously (and different to Sagnac) where the detector moves with the source and tube. Clearly if the source is moving with the earth, and the detector is not, but is fixed in a non-rotating inertial frame, you can work out the rotation rate just be seeing how the source moves away from the detector .. you don't need to go counting pulses. So what does this have to do with your ridiculous claims about your ORIGINAL setup, where you had a tube attached to the earth, and a source+detector at same location on the tube ??
From: Henry Wilson DSc on 17 Feb 2010 23:22
On Wed, 17 Feb 2010 14:55:15 -0800 (PST), Jerry <Cephalobus_alienus(a)comcast.net> wrote: >On Feb 17, 4:06�pm, ..@..(Henry Wilson DSc) wrote: > >> We will consider what happens when pulse N is emitted at point q. It will pass >> through point p then travel around the Earth before being finally absorbed >> again at point p. >> When it is emitted, there are 62 pulses between q and p. When pulse N reaches p >> the first time, there are say 61 pulses between p and q and 62 pulses have >> already passed through point p. There are 4000000-61 �pulses around the earth >> at that instant between p and q. So another 61 + 4000000 - 41 pulses will reach >> p by the time N gets there. That makes 4000062 altogether. >> In the other direction, only 3999938 arrive in the same time. > >(sigh) >In the STEADY STATE, how many pulses cross p each second? >In the STEADY STATE, how many pulses cross q each second? > >For both p and q, the answer had better be >1/3.333e-9 = 300030003 pulses. why should it not be? >Jerry Henry Wilson... ........provider of free physics lessons |