The Emission Theory of Androcles
in [Relativity]
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From: Jonah Thomas on 13 Sep 2009 16:19 Helmut Wabnig <hwabnig@ .- --- -. dotat> wrote: > Jonah Thomas <jethomas5(a)gmail.com>wrote: > >Helmut Wabnig <hwabnig@ .- --- -. dotat> wrote: > >> Jonah Thomas <jethomas5(a)gmail.com> wrote: > >> >"Sue..." <suzysewnshow(a)yahoo.com.au> wrote: > >> >> http://personal.ee.surrey.ac.uk/Personal/D.Jefferies/mwmeas.html > >> >> http://en.wikipedia.org/wiki/Speed_of_light#Modern_methods > >> > > >> >Thank you! Those look like useful links, with a wealth of > >information> >buried in the first one like secret treasure. > >> > > >> >I was interested in the claim that with modern nanosecond-scale > >> >timers you can actually measure the delay for light to travel > >known> >distances. That might allow some potentially-new experiments. > >> > >> today we are talking femtoseconds, that is 1000x1000 less than > >nanos.> this is year 2009, not 1909. > > > >So I could imagine a test where you open a shutter (unblack a liquid > >crystal? something faster?) and measure the time it takes for light > >from two different stars to arrive past the shutter to a sensor. One > >red-shifted, one blue-shifted. Relatively small differences in > >lightspeed might be detectable. There's the problem that we'd want > >two stars with different redshifts that are fairly close together, > >since if they are on opposite sides of the sky and reflected together > >that might change their speeds. > > > >It might be enough to have separate "shutters" and sensors and > >timers, and just measure how much slower the signal is that says the > >shutter was opened compared to the light signal. > Why so complicated? > Shutters. > Time measurements in picosec. > > Just put a Fizeau toothed wheel setup > or a Marinov apparatus into the light beam > of the telescope. That sounds good. So, the light comes out of the telescope and you put it through the wheel with a notch in it, and split the light two different ways and measure not the interference but the actual time it takes it to go some distance. Then do the same thing with it going two other directions. The distances and maybe times would be compared to light whose source is at rest wrt the telescope. The stationary laser could shine through a different notch. The major problem may be getting funding, since emission theories are probably considered so thoroughly discredited that it isn't interesting to test them. Maybe it could be done as a control for some other experiment. Like, you have some other experiment that requires you to do all that setup, and then you measure the speeds as a control, and if they are different you correct for the difference so the primary experiment can proceed correctly.
From: Henry Wilson, DSc on 13 Sep 2009 16:10 On Sun, 13 Sep 2009 19:53:11 +1000, "Inertial" <relatively(a)rest.com> wrote: >"Henry Wilson, DSc" <hw@..> wrote in message >news:aqepa5pmhtirsgm8i09hvdtlvl2e4job6h(a)4ax.com... >> On Sun, 13 Sep 2009 15:37:58 +1000, "Inertial" <relatively(a)rest.com> >> wrote: >> >>>"Henry Wilson, DSc" <hw@..> wrote in message >>>news:igpoa55gccc3161c7t51mn7nrk7ongvi79(a)4ax.com... >>>> On Sun, 13 Sep 2009 09:04:34 +1000, "Inertial" <relatively(a)rest.com> >>>> wrote: >> >>>>>That's right .. that's what i just showed. >>>>> >>>>>And the heads of the ropes arrive at the same place at the same time and >>>>>so >>>>>the marks on the rope (the wavelengths) is in phase. >>>> >>>> Are you saying the ropes are flexible? >>> >>>No. I would have used rubber bands if I wanted that. >>>> If one rope is longer than the other, there must be a different number >>>> of >>>> turns >>>> in the two. >>> >>>Turns? You mean the marks on the rope I described .. yes there are. Let >>>say we put the marks on the rop at regular intervals from the head of the >>>rope .. then when the ropes meet the detector point at the same time, the >>>marks on the heads line up. >>> >>>> That means there is a phase shift where they meet. >>> >>>No .. it doesn't. The marks on the rope move WITH THE ROPE. >>> >>>> I really think this is too hard for you... >>> >>>Not at all .. you're the only one who is inconsistent and unable to >>>understand basic physics. OK .. you and Androcles. >> >> You really haven't a clue about this. > >I understand it perfectly. > >> I'll have to explain in very basic >> language. > >Oh please .. show us your delusions again > >> The emission and detection points of a particular wave element are >> separated by >> the distance vt in the inertial frame. (or the rotating frame for that >> matter) > >Yes > >> This means that the path lengths of the oppositely moving halves of that >> element are different > >Yes > >> and so they reunite out of phase. > >No .. as they are in phase at the WAVE FRONT. The part that arrives at the >detector. That's how waves work. there are no wave fronts, idiot. >> THE START AND EMISSION POINTS FOR THE NEXT ELEMENT ARE NOT THE SAME >> POINTS. > >Yeup. > >> That pair are also moving around the ring at v....BUT THEY REMAIN >> SEPARATED by >> the same amount vt. > >Yeup, you're still deluded. > Henry Wilson...www.users.bigpond.com/hewn/index.htm Einstein...World's greatest SciFi writer..
From: Henry Wilson, DSc on 13 Sep 2009 16:21 On Sun, 13 Sep 2009 04:06:12 -0700 (PDT), Jerry <Cephalobus_alienus(a)comcast.net> wrote: >On Sep 13, 5:35�am, Jerry <Cephalobus_alie...(a)comcast.net> wrote: >> On Sep 13, 4:53�am, "Inertial" <relativ...(a)rest.com> wrote: >> >> Henri's presumption is that the waves traced out by the c+v >> particles and the c-v particles have the same wavelength as >> measured in the inertial frame. >> >> That necessarily implies that the c+v particles and the c-v >> particles oscillate at different frequencies. >> >> This effect is clearly demonstrated in Henri's animation as >> well as in my Java re-creation of his VB animation.http://mysite.verizon.net/cephalobus_alienus/henri/HWFantasy.htm >> >> Somehow or other, Henri doesn't have any problem with this. > >I was guilty of a pun here when using the word "wave". In >conventional ballistic theory, the TRAVELING waves emitted by >a moving source are of constant wavelength in all directions. > >In conventional ballistic theory, a stationary observer measures >a frequency shift, but no wavelength shift. > >In Henri's variant of ballistic theory, the "WACHAMACALIT waves" >are of constant wavelength (measured in the inertial frame) in >all directions. > >In Henri's variant of ballistic theory, a stationary observer >measures both a frequency shift and a wavelength shift. The >frequency shift is a "double-whammy" frequency shift that is also >measurable by a moving source observer. This provides the moving >source observer a means of detecting his motion in absolute space. > >Jerry Poor old Crank,....stilll doesn't get it.... There is NO actual frequency to worry about in sagnac. Wavelength is intrinsic to each photon. 'Frequency' is a inferred quantity defined as the number of waves passing a point in the inertial frame per second. Photons possess spatial 'waves', as in inertial's 'rope analogy'. The path lengths of the rays are different, therefore the number of turns of the rope are different in each ray and they meet out of phase. This is terribly simple really. I don't know why you are having so much difficulty understanding it. Henry Wilson...www.users.bigpond.com/hewn/index.htm Einstein...World's greatest SciFi writer..
From: Henry Wilson, DSc on 13 Sep 2009 16:22 On Sun, 13 Sep 2009 15:38:44 +1000, "Inertial" <relatively(a)rest.com> wrote: >"Henry Wilson, DSc" <hw@..> wrote in message >news:klpoa5l61ni29vjgv428qdvn8u4aoa5sip(a)4ax.com... >> On Sun, 13 Sep 2009 09:01:16 +1000, "Inertial" <relatively(a)rest.com> >> wrote: >> >>>"Henry Wilson, DSc" <hw@..> wrote in message >>>news:qp2oa5he52gtd4d0tp2c2gof0ntqd0jdoe(a)4ax.com... >>>> On Sat, 12 Sep 2009 02:24:47 -0400, Jonah Thomas <jethomas5(a)gmail.com> >> >>>> One path is longer than the other. The number of turns in each 'path' is >>>> different no matter how fast you spin the spring. >>>> >>>>>Similarly in the other direction. The first backward wave is extruded >>>>>slowly, so that at time = 0.1 its forward edge is at 9.1 and its leading >>>>>edge is at the emitter at 0.1. Then the second one goes slowly, and at >>>>>time 0.2 the forward edge of the first wave is at 8.2 and the end of the >>>>>second wave is at 0.2. And by the time we get to 1 second, the leading >>>>>edge of the first wave is at 1.0 and the end of the tenth wave is also >>>>>at 1.0. Exactly ten waves for that one too. >>>>> >>>>>If we were wrapping wire around a torus and we had to cover it until we >>>>>could quit, then at 1.1c we'd have to cover part of it twice while for >>>>>0.9c we wouldn't cover the whole thing. But our wire is sliding along >>>>>the torus and it's the first wave of the fast side that covers the same >>>>>area twice, and the area that the first wave of the slwo side doesn't >>>>>get to, the last wave has backed onto. >>>>> >>>>>Ten waves both times. No phase shift. >>>> >>>> No, you don't get the picture at all. >>>> >>>>>> >I imagine the emitter creating a wave that moves at 1.1c while the >>>>>> >emitter itself moves at 0.1c. There are 10 waves present covering the >>>>>> >distance around the circle from the emitter to the detector which is >>>>>> >in basicly the same place. A new one is being created while the >>>>>> >oldest one travels just enough faster than the detector that it is >>>>>> >completely consumed by the time the new wave is completely created. >>>>>> > >>>>>> >Meanwhile, the emitter creates a second wave that moves at 0.9c while >>>>>> >the emitter moves away at 0.1c. There are 10 waves present covering >>>>>> >the distance around the circle from the emitter backward to the >>>>>> >detector. A new wave is being created while the oldest one travels >>>>>> >just fast enough into the incoming detector that it is completely >>>>>> >consumed by the time the new wave is completely created. >>>>>> > >>>>>> >Yes, wavelength is absolute and frame independent. If you can look at >>>>>> >them, you can measure them. They're in place at any moment of time, >>>>>> >ready to be measured. In the absence of length contraction everybody >>>>>> >will measure them the same length. >>>>>> > >>>>>> >And frequency at the source is also absolute and frame independent. >>>>>> >You can watch the source create its periodic motion, and everybody >>>>>> >gets the same result apart from things like doppler shift which can >>>>>> >be easily corrected. >>>>>> >>>>>> What does 'frequency' mean when applied to light? >>>>> >>>>>I'm assuming something periodic going on, and in that case the frequency >>>>>is the number of times the periodic thing happens per unit time. In my >>>>>example that's ten times per second. >>>> >>>> Nope. Frequency is implied as the 'number of wavecrests' arriving per >>>> second. >>>> f = hc/lambda. >>> >>>If light is a little particle, what is a wavecrest? >> >> A photon is like a'serated bullet' due to its intrinsic oscillation... >> whatever >> that is. >> Its inferred 'frequency' is the number of serations arriving per second. > >So .. If light is a little particle, what is a wavecrest? A photon is a LONG particle....like your rope.... The wavecrests are defined by each twist. >>>> I reckon a photon also has an intrinsic oscillation and a natural >>>> frequency but >>>> that it not the one that we normally detect. >>> >>>Then what is being detected? Do you have ANY idea? >> >> Yes. > >Obviously not one you're able to express. > Henry Wilson...www.users.bigpond.com/hewn/index.htm Einstein...World's greatest SciFi writer..
From: Henry Wilson, DSc on 13 Sep 2009 16:29
On Sun, 13 Sep 2009 15:09:50 +1000, "Inertial" <relatively(a)rest.com> wrote: >"Henry Wilson, DSc" <hw@..> wrote in message >news:cupoa5ljmlfmu5o0le74firrmeid8sf497(a)4ax.com... >> On Sat, 12 Sep 2009 20:09:28 -0400, Jonah Thomas <jethomas5(a)gmail.com> >> wrote: >>>OK. The number of wavecrests leaving the source in my example is ten per >>>second, in any frame. > >That's correct > >> The number of wavecrests that pass any stationary point marked on the >> nonrotating ring is NOT ten. > >Yeup > >But the detector in Sagnac is NOT a stationary point Let me try to explain in even mire basic terms. (I will try to imagine I am talking to a Chimp) In the inertial frame, the emission and detection points OF A PARTICULAR PHOTON ELEMENT are stationary and are separated by the distance vt. (even your SR agrees) The emission and detection points OF the NEXT PARTICULAR PHOTON ELEMENT are also stationary.....BUT THEY ARE DISPLACED FROM THE PREVIOUS ONES. Both points are progressively 'moving' because different elememts are being progressivlely emitted but they are always separated by 2piR+vt and 2piR-vt for the two rays. >It moves and so also gets 10 per second > >> If you can understand the SR 'explanation' you should be able to >> understand the >> BaTh one too. There is basically very little difference. > >Except the SR one predicts the observed result and ballistic doesn't. But >other than that minor difference ...;) > >> Androcles is totally confused about Sagnac. He still thinks the detector >> is not >> rotating with the apparatus. > >You're such a hypocrite .. that is exactly what your analysis assumes as >well. You've even said that how the ray arrive at the detector in Sagnac is >not relevant. BAHAHAHAHA You simply don't have the intelligece for this. Go away. Henry Wilson...www.users.bigpond.com/hewn/index.htm Einstein...World's greatest SciFi writer.. |