GETTING RID OF EINSTEIN RELATIVITY
in [Relativity]
|
Prev: What is the Aether?
Next: Debunking Nimtz
From: Jeckyl on 12 Sep 2007 20:28 "Androcles" <Engineer(a)hogwarts.physics> wrote in message news:cjTFi.199157$p7.180581(a)fe2.news.blueyonder.co.uk... > Fecal Jekyll is just a moronic troll and stupid enough to > respond to one of my posts. I can act childishly too if that's how you want to play. "I know you are, but what am I"
From: Henri Wilson on 13 Sep 2007 18:17 On Wed, 12 Sep 2007 05:54:41 -0700, sean <jaymoseley(a)hotmail.com> wrote: >On 12 Sep, 08:03, "Paul B. Andersen" ><paul.b.ander...(a)hiadeletethis.no> wrote: >> Henri Wilson wrote: >> >> Maybe in your own personal version of emmision theory. But Im saying >> >> that if one can model emmision theory as having light propagate away >> >>from any source always at c relative to any source, then,.... >> >> one can explain MMx and sagnac >> >> > It explains the MMX but not Sagnac.... >> > Sagnac is very complicated. >> >> Sagnac isn't complicated at all. >> It is however bothering to you, since it falsifies emission theory. >I agree that Henri seems to not be able to understand sagnac very >well. I understand Sagnac far better than you ever will. >But I think your wrong when you say sagnac falsifies >emmision theory. If you look at any claim that it does and >study the so called proof by relativists on how it falsifies >emmision theory. >Youll notice that the paths of both beams are incorrectly >calculated which in turn give the false impression that there >is no path difference for emmision theory vis a vis sagnac. >Notice any simulation of emmision theory in the lab frame >for sagnac INCORRECTLY shows the light as travelling in >straight lines in the lab frame. >For emmision theory this is incorrect as it must be straight >in the source frame(like MMx) And if you bothered doing a >simple calculation youd see that a straight line in a >rotating source frame gives a galilean transformation to >a curved line in a lab frame where the source rotates. >Something relativists fail to take into account when >trying to falsely prove taht emmision theory cannot model sagnac > >see also.. >http://www.youtube.com/profile?user=jaymoseleygrb That is a primative and stupid animation, It is wrong and says nothing. >Sean www.users.bigpond.com/hewn/index.htm The difference between a preacher and a used car salesman is that the latter at least has a product to sell.
From: Henri Wilson on 13 Sep 2007 18:22 On Wed, 12 Sep 2007 05:47:09 -0700, sean <jaymoseley(a)hotmail.com> wrote: >On 11 Sep, 01:46, HW@....(Henri Wilson) wrote: >> On Mon, 10 Sep 2007 10:11:42 -0700, sean <jaymose...(a)hotmail.com> wrote: >> >Post 363 >> >> >On 6 Sep, 17:33, bz <bz+...(a)ch100-5.chem.lsu.edu> wrote: >> >> sean <jaymose...(a)hotmail.com> wrote innews:1189089230.764471.53320(a)22g2000hsm.googlegroups.com: >> >> >> > On 30 Aug, 23:55, bz <bz+...(a)ch100-5.chem.lsu.edu> wrote: >> >> >> sean <jaymose...(a)hotmail.com> wrote in news:1188512224.511353.237820 >> >> >> @i13g2000prf.googlegroups.com: >> >> >> >> > If you bothered analysing your sean planets sim, youd >> >> >> > see that its the only way to have light propagating away >> >> >> > from any source at c. >> >> >> >> One of the most fundamental laws of physics is that an object in motion >> >> >> continues in that motion unless acted upon by an outside force. >> >> > You ignore several things here, First of all if emmision theory >> >> > predicts that light always is at c relative to a source >> >> >> That only applies at the moment of emission. If the source changes its >> >> motion after emission, the light does not know or care. >> >Maybe in your own personal version of emmision theory. But Im saying >> >that if one can model emmision theory as having light propagate away >> >from any source always at c relative to any source, then,.... >> >one can explain MMx and sagnac >> >> It explains the MMX but not Sagnac.... >> Sagnac is very complicated. >You say this over and over but wheres your proof? Have you tried an >accuarate mathematical simulation to back up this claim you make? George Dishman and I thoroughly analysed Sagnac last year. ...with lots of diagrams, equations and animations. >No. >The fact is I have, using vector calculations and posted them as >sagnac sims at... >http://www.youtube.com/profile?user=jaymoseleygrb It is a joke compared with mine..... >And these show that as long as you calculate in the source >frame with the speed as c in the source frame one gets >a path difference which is whats observed. There is a reduced path difference. More importantly, rays that start out 90 apart end up displaced sideways at the receiver. >You >can keep repeating your claim that this doesnt work but if you >dont bother double checking by doing your own calculations in the >source frame then all your claims are unsubstantiated And erroneous. >Not to mention the fact that your own explanation relies on wave >particle duality which itself needs magic to explain how the photon >switches from particle to wave. >Sean >www.gammarayburst.com www.users.bigpond.com/hewn/index.htm The difference between a preacher and a used car salesman is that the latter at least has a product to sell.
From: Henri Wilson on 13 Sep 2007 18:29 On Wed, 12 Sep 2007 15:35:05 +0000 (UTC), bz <bz+spr(a)ch100-5.chem.lsu.edu> wrote: >bz <bz+spr(a)ch100-5.chem.lsu.edu> wrote in >news:Xns99A95FF903997WQAHBGMXSZHVspammote(a)130.39.198.139: > >> If the source changes motion or even ceases to exist after the light has >> been emitted, the light neither knows nor cares. > > >In fact, as the 'source' is most often an ion, atom or molecule in an >excited state, > >and the act of emission makes the source 'cease to exist' [in the excited >state] > >then there is no longer a 'reference' to which to tie the light, once it has >been emitted, > >so there is no way for the light to know what it's 'mother' does, once it has >left home. The claim is that a photon initially leaves its source at c relative to the source frame, at the instant of emission. That frame doesn't change just because the source goes somewhere else.... Can't you undertand that? WHY it leaves its source at c is something that physicists should be and would be investigating if Einsteiniana hadn't raised its ugly head.. www.users.bigpond.com/hewn/index.htm The difference between a preacher and a used car salesman is that the latter at least has a product to sell.
From: Henri Wilson on 13 Sep 2007 18:42
On Wed, 12 Sep 2007 23:07:11 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <HW@....> wrote in message >news:oh4hd3tvl3i1aagscrj93i217k7pc94flu(a)4ax.com... >> On Fri, 31 Aug 2007 10:14:38 +0100, "George Dishman" >> <george(a)briar.demon.co.uk> wrote: >... > >>>> George, my simulations are accurate. They also match most star curves. >> George, you admit that the SR sagnac diagram has the rays closing on the >> source >> at c+/-v, when viewed in the rest frame. > >Changing the subject again Henry? We were >talking about the ballistic version. > >> Do you not then agree that to a third orbserver, the light from a distant >> orbiting star would 'close on' another star at c+v.sin(at)? > >No, it would close at c is SR. You got that wrong. >> TWLS is dead constant for the simple reason that light is ballistic. >> That is, it moves at c wrt its source and everything at rest wrt the >> source. >> So naturally tAB=tBA in any TWLS experiment and in that case, TWLS also = >> OWLS >> = c > >Irrelevant, Sagnac measures the one-way speed. the fact that you don't understand doesn't make it irrelevant. > >>>Liar, you say the light moves at c wrt the source and >>>you use the Galilean Transforms so the CLOSING VELOCITY >>>is c in all frames in ballistic theory, you do NOT use >>>closing velocities of c+/-v. >> >> ...but that's not what your sagnac diagram shows... > >You need to be clear about which diagram you >mean, I have done both SR and ballistic. Both >are shown correctly. I am refering to the standard sagnac ring diagram. The rays are shown to move at c+/-v wrt the source. Don't deny it George. >> I certainly did not. >> I said it is almost impossible to tell fom a light curve if a star is >> genuinely >> eclipsing or not. > >Fine, the point remains, the binary we looked >at in detail is 90 degrees out of phase with >ADoppler and all the others have been studied >to see what can be learnt from the detailed >shape of the curves regrading their separation >and to what degree they are 'overcontact'. A >significant phase error would have been headline >news. The phasing of all cepheid brightness and velocity curves is just as the BaTh predicts. >> Only spectral data wil reveal the truth and even that may be >> suspect. Many stars classified as 'eclipsing' would not have been >> investigated >> further. > >How naive. there are lots of stars out there George...and only a limited number of misled astronomers. >>>>>> >> >For a >> It is very useful to have a way of checking the programming. >> Your method is much harder than mine...and only maginally faster. > >You have no idea how to use it then, it >should reduce the calculation time by a >factor of about 100,000 from the figures >you gave me. It doesn't, It is only about twice as fast. >>>>>> So instead of arguing you should be trying to apply it >>>Ah "forces unknown". Henry, in science all we do >>>is apply the equations. If you do that in the >>>above scenario the signal arrives earlier because >>>of gravity and earlier because of the shadowing >>>when on the far side of the planet and ealier >>>because of the reflected light on this side. Three >>>earlies don't make a late. >> >> George, no matter what the process, the final velocity will be the same as >> the >> initial one. > >Not with your new "radiation pressure" model, >it gets increased on both sides of the planet. > >> However D/(c+v) + D/(c-v) = 2Dc/(c^2-v^2).....so the time taken is always >> greater than D/c. > >Nope, without the planet it is > > 2 * D/(c-v1) > >where v1 is due to direct sunlight. With the planet >it becomes > > D/c + D/(c-v1+v2) > >where v2 is due to the reflected light. It is obvious >even qualitatively that both effects cause it to >arrive earlier and that is on top of the ballistic >effect which also makes it earlier. Forget it Henry, >it is a non-solution. Either you don't understand or you are having difficulty in conveying what you mean.... >>>The point is that we don't detect the Shapiro delay >>>for whatever reason so you have no orbital phase >>>reference with which to distinguish VDoppler from >>>ADoppler so you cannot prove anything either way >>>using that system. >> >> hahaha! >> George, you don't seem to understand what a 'simulation' is....very >> strange for >> any modern engineer.... > >Oh I know vastly better than you ever will, and >I also know that since you only use a subset of >the available data (e.g. no pulse amplitudes) >you need a phase reference. By all means do a >proper analysis if you think you can but so far >the only systems we have studied have shown no >ADoppler whatsoever. Don't keep slapping your >gums, just put up or shut up. Geoge, all my curves are ADoppler. I have no trouble in matching observed ones. >George > www.users.bigpond.com/hewn/index.htm The difference between a preacher and a used car salesman is that the latter at least has a product to sell. |