Ballistic theory of light demolished
in [Relativity]
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From: Henry Wilson DSc on 9 Aug 2010 19:39 On Sun, 8 Aug 2010 18:34:37 -0700 (PDT), Jerry <Cephalobus_alienus(a)comcast.net> wrote: >On Aug 8, 5:35�pm, ..@..(Henry Wilson DSc) wrote: >> On Sat, 7 Aug 2010 15:28:28 -0700 (PDT), Jerry <Cephalobus_alie...(a)comcast.net> >> wrote: >> >> >The emanations cannot be from "slightly" different levels. >> >The phase differences between light curves measured in >> >differing wave lengths may be on the order of days, for a >> >long-period Cepheid. What stars do you know that are that >> >large? >> >> What the hell are you talking about? If you want to criticize my theory, please >> learn what it says first or you will continue to make a big fool of yourself. >> >> The majority of cepheids appear to be pulsating stars. Their surface radial >> velocities oscillate in very similar fashion to those of a star in an >> elliptical orbit of eccentricity around 0.25 and a yaw angle around 70deg. > >How large does that imply the Cepheid is if it has a period of, >say, two months and an amplitude of 0.7 magnitudes? In that case it is probably an orbiting star...not a pulsating one. >What is the difference in diameters between the Cepheid at max >versus the Cepheid at min? All short period cepheids are likely to be pulsating stars. >How do measurements of Doppler compare with BaTh predictions? Perfectly well. One of the big mysteries surrounding cepheids is the question why their radial velocity curves are almost mirror images of their brightness curves. BaTh explains this with ADoppler. (A photon's wavelength is acceleration dependent and, to a limited extent, to distance traveled) Astronomers are measuring ADoppler spectral shifts and assuming they are conventional VDoppler...resulting in huge errors and estimatiuons. >If this two month Cepheid has a one-week phase difference between >the peaks of the B band versus the K band, how much separation >exists between layers? That dependes on its distance from Earth. >What do the laws governing black body radiation say about the >ability to see that deeply beneath the surface of a glowing gas? Crank, have you ever seen photos of hte sun through diferent filters? You see different levels. >You have never dared to answer these questions in quantitative >fashion. The only times that you attempted to answer in >qualitative fashion, you faked your diagrams, and got caught >each time. That was a harmless joke that I admitted and which just happened to lead to the right explanation. I don't have to fake them any more. The velocity curves BaTh produces are a combination of ADoppler and VDoppler. Most are mirror images of the corressponding brigthness curves with small phase differences due to the VDoppler component. >Jerry Henry Wilson... ........Einstein's Relativity...The religion that worships negative space.
From: Jerry on 9 Aug 2010 21:09 On Aug 9, 6:39 pm, ..@..(Henry Wilson DSc) wrote: > On Sun, 8 Aug 2010 18:34:37 -0700 (PDT), Jerry <Cephalobus_alie...(a)comcast.net> > wrote: > > > > > > >On Aug 8, 5:35 pm, ..@..(Henry Wilson DSc) wrote: > >> On Sat, 7 Aug 2010 15:28:28 -0700 (PDT), Jerry <Cephalobus_alie...(a)comcast.net> > >> wrote: > > >> >The emanations cannot be from "slightly" different levels. > >> >The phase differences between light curves measured in > >> >differing wave lengths may be on the order of days, for a > >> >long-period Cepheid. What stars do you know that are that > >> >large? > > >> What the hell are you talking about? If you want to criticize my theory, please > >> learn what it says first or you will continue to make a big fool of yourself. > > >> The majority of cepheids appear to be pulsating stars. Their surface radial > >> velocities oscillate in very similar fashion to those of a star in an > >> elliptical orbit of eccentricity around 0.25 and a yaw angle around 70deg. > > >How large does that imply the Cepheid is if it has a period of, > >say, two months and an amplitude of 0.7 magnitudes? > > In that case it is probably an orbiting star...not a pulsating one. > > >What is the difference in diameters between the Cepheid at max > >versus the Cepheid at min? > > All short period cepheids are likely to be pulsating stars. Also long period Cepheids. Also intermediate period Cepheids... > >How do measurements of Doppler compare with BaTh predictions? > > Perfectly well. Dismally. > One of the big mysteries surrounding cepheids is the question why their radial > velocity curves are almost mirror images of their brightness curves. False. > BaTh explains this with ADoppler. A false explanation of an incorrect understanding. > (A photon's wavelength is acceleration > dependent and, to a limited extent, to distance traveled) > > Astronomers are measuring ADoppler spectral shifts and assuming they are > conventional VDoppler...resulting in huge errors and estimatiuons. > > >If this two month Cepheid has a one-week phase difference between > >the peaks of the B band versus the K band, how much separation > >exists between layers? > > That dependes on its distance from Earth. > > >What do the laws governing black body radiation say about the > >ability to see that deeply beneath the surface of a glowing gas? > > Crank, have you ever seen photos of hte sun through diferent filters? > You see different levels. The diameters measured at different wavelengths below EUV are identical to tiny fractions of a percent. Only at EUV and beyond do you get disparate measurements of diameter, because at those wavelengths, the emissions do not represent black body radiation. Learn some basic astronomy before spouting off. > >You have never dared to answer these questions in quantitative > >fashion. The only times that you attempted to answer in > >qualitative fashion, you faked your diagrams, and got caught > >each time. > > That was a harmless joke that I admitted and which just happened to lead to the > right explanation. I don't have to fake them any more. The velocity curves BaTh > produces are a combination of ADoppler and VDoppler. Most are mirror images of > the corressponding brigthness curves with small phase differences due to the > VDoppler component. I started feeling that something was -very- wrong, but I wasn't going to accuse you of lying until I was sure. You sensed that I was going to pick up on your bald-faced lies, and decided to head me off. You lied to George, as well, but he was quicker on the uptake than me...possibly because by then your history of posting fraudulent images had become well-known. Jerry
From: Henry Wilson DSc on 10 Aug 2010 17:55 On Mon, 9 Aug 2010 16:03:02 -0700 (PDT), Jerry <Cephalobus_alienus(a)comcast.net> wrote: >On Aug 9, 5:05�pm, ..@..(Henry Wilson DSc) wrote: > >> No....of course, at high speeds, MOST of hte energy is used in maintaining the >> bubble. > >This means, of course, that you believe the resistive forces are >dissipative in nature. In other words, "frictional" rather than >"inertial" resistance. This is completely contrary to fact. What the hell are you raving about now? >You won't admit this, of course... > >Jerry Henry Wilson... ........Einstein's Relativity...The religion that worships negative space.
From: Paul B. Andersen on 10 Aug 2010 17:59 On 10.08.2010 01:27, Henry Wilson DSc wrote: > On Mon, 09 Aug 2010 01:49:14 +0200, "Paul B. Andersen"<someone(a)somewhere.no> > wrote: > >> On 07.08.2010 01:13, Henry Wilson DSc wrote: >>> On Fri, 06 Aug 2010 15:31:33 +0200, "Paul B. Andersen"<something(a)somewhere.no> >>> wrote: >>> > >>>> Either YOU weren't thinking, or you don't know >>>> that inertial particles don't radiate. >>> >>> This is obviosly too hard for you. >> >> Indeed. >> >> Paul: >> "How come a fully qualified physicist and Doctor of science can >> be ignorant of the fact that synchrotron radiation is emitted >> in a narrow cone along the direction of motion of the charged >> particle?" >> >> Ralph Rabbidge: >> "Hahahahha! >> Is that charged particle moving inertially? >> If it is, why should anything it emits have >> a 'preferred direction'?" > > Since the particle is changing direction continuously in a cyclotron, how does > the radiation know which direction to travel in? Are you serious? :-) An aeroplane has a cannon pointing forwards. It shoots while it is turning. How does the bullets know which direction to travel in? > >>> My point was that the direction of radiation must be acceleration and not >>> velocity dependent BECAUSE the particles are NOT inertial. >> >> Yea, right! >> Hilarious, no? :-) >> >> But let's remember your claim: >> "the direction of radiation must be acceleration and not >> velocity dependent" > > Well, let's put that another way. If there is NO acceleration there is NO > radiation whether the particle has a relative velocity or not. So, in that > respect, it is solely acceleration dependent.. So the direction of radiation isn't velocity dependent, because there wouldn't be a radiation if there was no acceleration. You are doing great now, Ralph! :-) > >>>> It probably was the former, so why don't you admit that >>>> in stead of pretending that your question was a sensible one. >>> >>>>>>> That process is dependent solely on the particle's acceleration so: >>>>>> >>>>>> No, it's not solely dependent on the particle's acceleration. It >>>>>> happens when there IS an acceleration, but the radiation depends on >>>>>> other quantities besides the acceleration. Like, the direction the >>>>>> particle is going and its charge. >>>>> >>>>> The particle has NO direction in its own frame. If you were right, radiation >>>>> should be emitted equally in all directions. >>>> >>>> In the particle's frame of reference, radiation _is_ emitted >>>> in all directions, but not equally intense in all directions. >>>> (If we are talking about a single photon, replace 'intensity >>>> in a direction' with 'probability of emission in a direction'). >>>> >>>> In the frame of reference of the particle, let's call it the S' frame, >>>> there is a moving magnetic field. So the velocity vector is still >>>> relevant, but here it is the velocity of the B field. >>>> The moving magnetic field is perpendicular to the velocity. >>>> In the S' frame only an electric field can accelerate the particle. >>>> When we transform the moving magnetic field to the S' frame, >>>> the electric field is E' = -v*gamma*B, and the direction is >>>> perpendicular to the B field and the velocity. >>>> So the acceleration is perpendicular to same. >>>> >>>> The radiation intensity distribution is: >>>> I(phi') = Io*cos^2(phi') >>>> where phi' is the angle from the velocity vector. >>>> >>>> Note that for phi' = +/- pi/4, we get >>>> I = Io/2, the intensity is half at these angles. >>>> This is often called the beam width. >>>> >>>> So it will radiate almost in all directions, but mostly >>>> in both directions along the velocity vector, and nothing >>>> perpendicular to it. >>> >>> ....presumably because the acceleration component is zero in that direction. >>> >>>> The radiation diagram will be something like this: >>>> (should be two circles) >>>> . . >>>> * * * * >>>> * * * * >>>> ----*-----------C-----------*-> v >>>> * * * * >>>> * * * * >>>> * * >>>> >>>> But we are observing the radiation in the lab frame S, >>>> where the particle is moving at the speed v. >>>> >>>> The intensity at the angle phi = 0 (in the forward direction) >>>> will be changed by the square of the Doppler effect: >>>> I(0) = Io*(1+v/c)/(1-v/c) >>>> equivalently will the intensity in the opposite direction be: >>>> I(pi) = Io*(1-v/c)/(1+v/c) >>>> >>>> (If you think in photons will the energy of each photon >>>> be Doppler shifted, and the frequency of arrival >>>> of photons will also be Doppler shifted, thus the square.) >>>> >>>> In addition to this effect will we have aberration: >>>> cos(phi) = (cos(phi')+ v/c)/(1 + (v/c)cos(phi')) >>>> >>>> A beam (or photon) in an off axis direction in S' >>>> will be deflected towards the velocity vector in S. >>>> >>>> It can be shown that the combined effect will lead >>>> to a beam width (half intensity) ~= 1/gamma (radians) >>>> in the forward direction. >>>> >>>> For high gammas will we have a very narrow beam >>>> along the velocity vector and very little in the backwards >>>> direction. >>>> >>>> Note that this is a prediction of SR, and measurements >>>> are in accordance with the prediction. >>>> >>>> Synchrotron radiation is yet another confirmation of SR. >>> >>> hahahhhaha! >>> HAHAHAHHHAHHAHAHA! >> >> And where is the stupidity that uses to follow yours "hahahhhaha!"? >> Could you not think of one? :-) > > I suppose you also claim that the christian bible is confirmation of he > existence of a 'heaven'. There it was! :-) > >> It is a _fact_ that synchrotron radiation is emitted >> in a narrow beam along the velocity vector. >> This is observed right now in a number of storage ring >> around the world, some of which are built with the sole >> purpose of producing synchrotron radiation. >> >> It is a _fact_ that SR predicts such a beaming effect, >> and that its predictions are in accordance with the >> measurements. >> >> So Henry Wilson, you were wrong when claiming: >> "the direction of radiation must be acceleration and not >> velocity dependent" > > Hahahhahhaahahahahahha! The "Hahahhahhaahahahahahha!" again! So what follows? > Stop it Paul, your killing me.......Hahahahahhahhahhaha! > > The bloody particles are continuously changing direction. > > HAHAHHAHHAHAHHHHAHA! Thanks for the demonstration, Ralph! :-) >>>>> So I am right. It is dependent solely on the particle's acceleration. >>>> >>>> No. Observations in the lab frame depend >>>> strongly on the particle's velocity in that frame. >>> >>> I was implying that the angular distribution was dependent on acceleration. >>> >>> It is logical according to the WFT that the intensity of radiation might >>> include a relative velocity term (wrt the field) as well as acceleration. >> >> So the WFT doesn't change only from posting to posting, it changes >> within the same posting. >> >> BTW, strange theory that "might have a term.." :-) > > Like all radically new and world shattering theories, the WFT is still under > development. It opens up a whole new world of opportunity for young physicists > to make real names for themselves. Quite. It is obviously under development since it changes all the time, and states the opposite within the same posting of yours. But the math of the ever changing theory is known and fixed. It is identical to the the math of Maxwell an SR. Strange, isn't it? :-) > > Even little eric is too dumbfounded to contribute his usual inane drivel.... > >>>> But the acceleration of the particles in the bends is much higher >>>> than it is in the RF-cavities, and the radiation is _vastly_ more >>>> intense. And since the acceleration is perpendicular to the velocity, >>>> it doesn't add to the kinetic energy of the particles. The radiation >>>> energy is taken from the kinetic energy of the particles, so they >>>> loose kinetic energy. >>> >>> Not according to WFT. They lose energy by partly neutralising the surrounding >>> applied field, which subsequently emits radiation.. >> >> So both the particle and the field looses energy, the field even twice, >> first by being partly neutralised, secondly by emitting radiation? :-) > > Do you deny that a fast moving charge would create...nay constitute... a > current that would oppose the applied field. > >>>> In its rest frame, the particle "sees" an electric field E = v*gamma*B. >>>> When v ~= c, gamma is thousands and B is very high, that's >>>> a gigantic electric accelerating field! >>> >>> ...requrired because of the Wilson Reverse Field Bubble that buids up around >>> the particle and reduces the applied field. >>> >>>> That's why the next generation of high energy accelerators probably >>>> will be several km long linear accelerators. >>> >>> If they design it according to relativistic physics it will automatically >>> include the WFT because the maths are the same. The Bubble equation includes a >>> gamma term. >> >> Ah. The math, which includes the Lorentz transform, >> is the same, even if the WFT changes all the time, >> and not even Henry Wilson knows what it says! >> >> According to him: >> "The direction of radiation must be acceleration >> and not velocity dependent, but it is logical according to the WFT >> that the intensity of radiation might include a relative velocity >> term (wrt the field) as well as acceleration. " > > Tell me, what is the direction of the radiation in the Ds of a cyclotron. A bit slow, Ralph? Since the particles are changing direction all the time, the radiation gets dizzy, and don't know where to go, so it isn't necessarily tangential to the particle beam. Or is it? > >> "According to the WFT, doesn't this electrostatic >> ACCELERATION also cause synchrotron radiation? It probably does... >> but the gaps are too small for it to be noticed. >> I would be inclined to say that it does not because all the applied >> energy goes into the particle's increased KE." >> >> But the math is the same as SR's. :-) >> So even Henry Wilson knows that the predictions of SR are correct, >> and that SR therefore is confirmed by synchrotron radiation. > > Paul, you're rambling on without saying anything of importance. You are trying > to find flaws in my theory but obviously cannot. Yea. Right. Obviously! :-) > The WFT is in its infancy. I have made the BIG breakthrough but it is not fully > developed by any means. If you tried to be a little bit more positive you might > be able to share in this great adventure of discovery. It's comforting that your megalomania prohibits you from understanding how pathetic you are. Your embarrassment would otherwise be unbearable. >> I will remember that the math of WFT includes >> Lorentz transform! :-) Because it does, doesn't it? "If they design it according to relativistic physics it will automatically include the WFT because the maths are the same." Good grief! :-) -- Paul http://home.c2i.net/pb_andersen/
From: Darwin123 on 10 Aug 2010 19:18
On Aug 5, 6:18 pm, ..@..(Henry Wilson DSc) wrote: > On Wed, 4 Aug 2010 17:00:25 -0700 (PDT), Jerry <Cephalobus_alie...(a)comcast.net> > wrote: > > > > >On Aug 4, 5:08 pm, ..@..(Henry Wilson DSc) wrote: > >> On Wed, 4 Aug 2010 12:00:33 -0700 (PDT), Jerry <Cephalobus_alie...(a)comcast.net> > >> wrote: > > >> >On Aug 3, 3:59 pm, ..@..(Henry Wilson DSc) wrote: > > >> >> The trouble with you blokes is that you think that because fields cannot be > >> >> seen or felt, they have no physical structures. > > >> >> I say the stuff fields are made of emits no EM and passes straight through > >> >> ordinary matter, which after all is 99.99999999999% empty space. > > >> >A most interesting statement. > > >> >Tell me... > > >> >Assume a perfectly machined, circular disk magnet with a uniform > >> >magnetic field lined up along the disk axis. > > >> >I claim that I set the disk spinning along its axis on perfectly > >> >machined, noiseless, vibrationless bearings inside an opaque box > >> >in vacuum. > > >> >Without being allowed to move, touch, or x-ray the box (all > >> >similar such technologies are also prohibited) please explain to > >> >me how, from measurements of the external magnetic flux or other > >> >such electrical or magnetic measurements, you may determine > >> >whether the disk is in fact spinning. > > >> Good question...tell me your answer...and why it is relevant here.... > > >Nope. YOU need to answer. As to why it is relevant, you stated > > > "The trouble with you blokes is that you think that because fields > > cannot be seen or felt, they have no physical structures. I say the > > stuff fields are made of emits no EM and passes straight through > > ordinary matter, which after all is 99.99999999999% empty space." > > >The above reveals a lot about what you understand about fields. > > Nobody knows anything about the physical structure of fields. All physics knows > is the maths of how they operate. Ahh! And there is more, yes? And you know what that is, yes? And you know the maths by which fields operate, yes? > |