Bigbang, a fantasy?
in [Relativity]
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From: eric gisse on 11 Jan 2010 18:46 ...@..(Henry Wilson DSc) wrote: [...] > There is no known definition of 'photon frequency'. It is not an intrinsic > property but is inferred from the equation E = hc/lambda. Here, c/lambda > is nothing other than 'wave arrival rate', a velocity dependent quantity. Let us once again marvel at Ralph's repeated insistence that something not known to him is not known to anybody. [...]
From: Androcles on 11 Jan 2010 19:12 "Henry Wilson DSc" <..@..> wrote in message news:u6bnk591qjbndooefkbukb53hke0k01jpp(a)4ax.com... > On Mon, 11 Jan 2010 21:55:46 +0000, John Kennaugh > <JKNG(a)notworking.freeserve.co.uk> wrote: > >>Androcles wrote: >>> >>>"John Kennaugh" <JKNG(a)notworking.freeserve.co.uk> wrote in message >>>news:RMpWrUI66JSLFwV$@kennaugh2435hex.freeserve.co.uk... >>>> Androcles wrote: > >>>> Simple mathematics. >>>> >>>> X---->v >>>> >>>> >>>> >>>> >>>> >>>> >>>> Y T >>>> >>>> >>>> In order for light leaving X to hit T it has to set out in the >>>> direction >>>> XY where YT = vt. The photons have a component of velocity c in the >>>> direction XY and a component v in the X direction such that the >>>> resultant is in the direction XT. What you have is a velocity triangle >>>> XY = c YT = v so >>>> >>>> the velocity X-T = Sqr( c^2 - v^2) by pythag >>>> So Sqr( c^2 - v^2) = F' x L >>>> But c = Fo x L (L = wavelength) >>>> So F'/Fo = Sqr( c^2 - v^2)/c = Sqr(1 - v^2/c^2) >>>> > >>>> It is possibly easier to see if observed from The IFoR in which the >>>> light is emitted in which, according to emission theory, light travels >>>> every which way at c >>>> >>>> O >>>> >>>> >>>> >>>> >>>> >>>> >>>> Y T >>>> v<--- >>>> >>>> Light leaves O when O is orthogonal w.r.t T but when it reaches T, T is >>>> at Y and has a component of motion away from X which will result in >>>> lower frequency. Of course if the source is in orbit around Y it is >>>> constantly changing its IFoR and will no longer be in the IFoR of O >>>> (the >>>> IFor in which the light was emitted) when light reaches Y. >>>> >>>> Waldron's ballistic theory gives a slightly different result as his >>>> theory is basically a photon theory rather than a wave theory. He >>>> calculates the same change of velocity as above but uses the change in >>>> velocity to calculate the energy of the photon reaching T. He >>>> calculates the Transverse Doppler shift as >>>> >>>> f'/fo = (1 - v^2/(2.c^2)) >>>> >>>> which differs only in forth order v/c term. >>>> >>>This Waldon guy is a fruit cake, bullets do not vanish into or >>>appear from thin aether. >>> >>>Note that emission theory predicts NO transverse Doppler, the >>>opposite of SR. >> >>Wrong! You may have dedicated yourself to showing that light speed is >>source dependent but even if you succeed that is the *basis* of a theory >>not a theory in its own right. Ritz's theory was produced a century ago >>and is flawed by subsequent experiment - not that it matters it was >>totally ignored. > > Those 'subsequent experiments' were flawed. Ballistic theory is alive and > well. > >>Waldron is the only person I know of who has made a >>serious attempt since and you dismiss him without studying his theory. >>That makes you as much of a narrow minded bigot at the relativists you >>attack. >> >>Waldron's theory is a photon theory. Frequency is simply related to the >>energy of a photon by Planck's constant. >> >> >> O' O >> | >> | >> | >> | >> | >> X---->v >> >>You are at X orthogonal w.r.t O. You fire at a target at O. If you are >>moving at v and you aim *at* O you will miss as your bullets have a >>component v. You have to aim at a point O'. Now if v is equal to the >>muzzle velocity Vm you cannot hit O at all no matter which way you point >>the gun. >>It is more obvious in the FoR of X >> >> >> v<---O >> | >> | >> | >> | >> | >> X >> >>OK let us assume v is less than the muzzle velocity and that X is >>sending out an omni-directional array of bullets when he is exactly >>orthogonal w.r.t O which bullet hits O? >> >> v<---O' O >> >> >> >> >> >> X >> >>the bullet hits O at O' if O-O' = vt and X-O' = Vm.t. Where Vm is the >>muzzle velocity. What is the kinetic energy involved w.r.t the collision >>between the bullet and O? It is dependent on v and drops to zero as v >>approaches Vm > > That is a very roundabout way to reach a simple conclusion. > > Photon energy is h x ('wavecrest arrival rate') > > The latter term is obviously dependent on relative photon/observer speed > (c-v). > >>It is far more difficult to imagine - but doesn't change anything if you >>change it to the FoR of O and have X orbiting O i.e. you are on a >>roundabout taking pot shots at the target in the middle. It doesn't >>change anything because what happens to X *after* the bullet has left >>does not affect the bullet so if X changes its direction i.e. follows a >>circular path around O it does not alter the fact that the kinetic >>energy involved in the collision IS velocity dependent in the orthogonal >>case. Energy is equivalent to frequency. > > There is no known definition of 'photon frequency'. It is not an intrinsic > property but is inferred from the equation E = hc/lambda. Here, c/lambda > is > nothing other than 'wave arrival rate', a velocity dependent quantity. > He's trying to change to the subject, H. This Waldron idiot has him brainwashed. The simple fact is he's fallen foul of the tick fairy, he thinks more photons/bullets arrive per hour than left per hour (or less do, it doesn't matter which). Shame really, he started out with some sense. Dementia is a wicked ailment, it robs one of intelligence. I can only feel pity for him (and you with your decapitated crocodiles, lucky white heather, wedge-on worbits and unifuckation).
From: Androcles on 12 Jan 2010 00:20 "Henry Wilson DSc" <..@..> wrote in message news:psdnk55bkksdvtv0u0c4q8rnib8sntt9bh(a)4ax.com... > On Mon, 11 Jan 2010 23:17:14 -0000, "Androcles" > <Headmaster(a)Hogwarts.physics_r> > wrote: > >> >>"John Kennaugh" <JKNG(a)notworking.freeserve.co.uk> wrote in message >>news:s$$pHtHi55SLFwxA(a)kennaugh2435hex.freeserve.co.uk... >>> Androcles wrote: >>>> >>>>"John Kennaugh" <JKNG(a)notworking.freeserve.co.uk> wrote in message >>>>news:RMpWrUI66JSLFwV$@kennaugh2435hex.freeserve.co.uk... >>>>> Androcles wrote: >>>>>> > >> >>That's your problem, not mine. >>Note that emission theory predicts NO transverse Doppler, the >>opposite of SR. -- RIGHT! >> >> >> >>- but doesn't change anything if you >>> change it to the FoR of O and have X orbiting O i.e. you are on a >>> roundabout taking pot shots at the target in the middle. It doesn't >>> change anything because what happens to X *after* the bullet has left >>> does not affect the bullet so if X changes its direction i.e. follows a >>> circular path around O it does not alter the fact that the kinetic >>> energy involved in the collision IS velocity dependent in the orthogonal >>> case. Energy is equivalent to frequency. >> >> >>Simple arithmetic a 12-year-old can manage. >>X emits (fires) 2 bullets each second. >>That's a frequency of 2 * 60 * 60 = 7200 bullets per hour. >>T sees (is hit by) 1 bullet each second (red shift) or 4 bullets each >>second (blue shift). >>Those are frequencies of 2* 7200 = 14400 bullets per hour and >>7200/2 = 3600 bullets per hour. >>After one hour X has fired 7,200 bullets and 3600 have hit T (red shift) >>or 14,400 bullets have hit T (blue shift). What happened to the missing >>bullets (red shift) or where do the extra bullets come from (blue shift)? >> >>Now answer the question, genius. >>Note that emission theory predicts NO transverse Doppler, the >>opposite of SR. -- RIGHT! > > Wait a minute, A, wait a minute. It isn't that easy. Make it quick, I haven't got all day. > > Sure, in the orbiting case, all the bullets that are fired hit the targret > so > the number arriving equals the number fired...no change in frequency at > the > centre. BUT, is the number arriving per second what really matters? Number per hour, H. Or number per day, or number per week. > To hit the centre, the orbiting gun has to be pointed BEHIND the target > because > the bullet gets a forward velocity component equal to the peripheral > velocity. Doesn't matter about the aiming point, its the frequency at the arrival point that matters. Anyway, it's a rotating frame and so off limits to you. "DON'T TRY TO USE ROTATING FRAMES." -- Wilson. news:aqqqm35ka2ef6qheidr8b1qh2bi0tv1bgo(a)4ax.com > v<G > \V gun is aimed diagonally at arcsin(v/V) > | > | > | bullet speed towards target = sqrt(V^2-v^2) > T > > (we will ignore the initial extra weight of the bullet because of its > centrifugal acceleration...but it could be significant) > > Now, which photon model are you going to use? This one: http://z.about.com/d/inventors/1/5/4/T/gramophoneLarge.jpg You don't get any sound at all if the needle hits the exact centre. "A rotating frame is not a 'rotating frame'... hahahahhahahahaha!" --Wilson news:mu2nm3d6urgddt8jgdnss0ddualtsrnm7n(a)4ax.com > > If you use my 'serated bullet' (decapitated crocodile) > model, the number of serations (wavecrests) > arriving at the target per second IS dependent on bullet speed....so that > model > DOES predict a transverse redshift. "ROTATING FRAMES FEATURE IMAGINARY EFFECTS. DON'T TRY TO USE THEM."-- Wilson news:drh9e553jdb7u87m75v26nerpghee6bk1p(a)4ax.com Leave rotating frames to engineers, they are way too difficult for mere physicists to understand. It's more modern version: http://www.fahad.com/pics/pioneer-blu-ray.jpg > > However the simple wave model does not, for the reasons given above.... > the > number of waves leaving the 'gun' per sec = number arriving at target per > second...so NO frequency shift. The tick fairy is getting to you. You'll never be an engineer. > So, if a transverse doppler shift is observed as claimed, it certainly > supports > the BaTh....and particularly my 'intrinsic wavelength' principle, where > wavelength is an absolute spatial interval determined by the physical > geometry > of the 'photon particle'. Nobody except you ever called WET BaTh WaSh a viable theory. I've told you, it's SoAp.
From: Androcles on 13 Jan 2010 08:32 "John Kennaugh the snipping bigot" <JKNG(a)notworking.freeserve.co.uk> snipped in message news:n5GLWnIAAcTLFwJJ(a)kennaugh2435hex.freeserve.co.uk... >>X emits (fires) 2 bullets each second. >>That's a frequency of 2 * 60 * 60 = 7200 bullets per hour. >>T sees (is hit by) 1 bullet each second (red shift) or 4 bullets each >>second (blue shift). >>Those are frequencies of 2* 7200 = 14400 bullets per hour and >>7200/2 = 3600 bullets per hour. >>After one hour X has fired 7,200 bullets and 3600 have hit T (red shift) >>or 14,400 bullets have hit T (blue shift). What happened to the missing >>bullets (red shift) or where do the extra bullets come from (blue shift)? >> >>Now answer the question, genius. <no answer> You FAILED! Toodle-loo, bigot, I have intelligent people to talk to.
From: Sue... on 13 Jan 2010 09:13
On Jan 5, 7:44 pm, mluttgens <mluttg...(a)orange.fr> wrote: > On 6 jan, 01:40, mluttgens <mluttg...(a)orange.fr> wrote: > > > On 6 jan, 01:32, mluttgens <mluttg...(a)orange.fr> wrote: > > > > Seehttp://www.spacetelescope.org/new/htmeheic1007.html > > > > Marcel Luttgens > > > Sorry, see http://www.spacetelescope.org/news/html/heic1001.html ============ It is only necessary to consult the archaeological record to see that neither the standard apple nor the typical bite taken from it has changed in the past 6000 years. The mass of one apple bite comes up just a bit short in explaining the disparity between Newton's falling apple and Pound/Rebka's falling photon. But that ignores the possibility that apple trees might have been as tall as Harvard tower back when Adam and Eve shared fruit. It scientists really wanted to know the truth about creation they would be out digging around for a tall petrified apple tree instead of wasting other people's money on elaborate star gazing devices. And that's the truth. http://farm4.static.flickr.com/3382/3326794623_6c8136f699.jpg Sue... > > OK |