GETTING RID OF EINSTEIN RELATIVITY
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From: bz on 13 Sep 2007 22:47 HW@....(Henri Wilson) wrote in news:p1eje39gkq0ujf6n3tn0j12vers50aub6n(a)4ax.com: > 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 is NOT Sean's claim. He claims that the light must continue to maintain a velocity of c wrt the source[not the sources iFoR] AFTER emission, no matter how the source moves. > That frame doesn't change just because the source goes somewhere > else.... > > Can't you undertand that? I understand that, Sean doesn't. > > 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.. Einstein said that it DOES leave the point of emission [as measured in an iFoR co-moving with the source at the instant of emission] at c. Einstein also said that the exact same photon moves at c as measured in ANY iFoR. Henri, I am surprised at you. You should know by now that science doesn't investigate 'why' questions. Haven't you learned that yet? Religion and philosophy ask and answer the 'why' questions. Science asks and attempts to answer 'how' things behave questions. -- bz please pardon my infinite ignorance, the set-of-things-I-do-not-know is an infinite set. bz+spr(a)ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap
From: George Dishman on 14 Sep 2007 06:59 Henri Wilson wrote: > 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. I got it right, you may be thinking of ballistic theory instead of SR. > >> 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. You said ".. in any TWLS experiment .." but Sagnac isn't a two-way test. What you say is true for MMX for example which _is_ a two way measurement, it measures th sum of the times, not the difference. > >>>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. No, you said ". that's not what YOUR sagnac diagram shows .." (my emphasis). > The rays are shown to move > at c+/-v wrt the source. > Don't deny it George. Sorry Henry, I have corrected that error so many times, I'm not going to repeat it. You know you are wrong but if you insist on wishing to be ignorant, that's your choice. > >> 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. Not true, when you analyse it properly you will find there is no ADoppler, only VDoppler. > >> 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. Only a fraction are eclising binaries and for those that have had their luminosity curves studied, you will find most have also had their spectra recorded to the point of confirming that they are spectroscpic binaries. > >> 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. Then you haven't made best use of it. It sounds as though you are still iterrating along the light path. > >>>>>> 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.... Probably I didn't understand because you had trouble conveying what you meant. Try defining what D and v are in your equation, I may have guessed incorrectly. > >>>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. Sure, but all your curves are matched to the temperature and radius variation, not any ADoppler part so they are universally wrong. Even once you correct that error, it still means nothing, what you have to do to produce a proof is show that it _is_ ADoppler, not just that it _might_ be. George
From: Paul B. Andersen on 14 Sep 2007 10:23 sean wrote: > On 12 Sep, 08:03, "Paul B. Andersen" > <paul.b.ander...(a)hiadeletethis.no> wrote: >> Henri Wilson wrote: >>> 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. 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. Which laws of nature do you invoke to make the light go in curved paths in an inertial frame? :-) (The effects of gravity are negligible in Sagnac) > 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 You got it backwards. When pictured in the rotating frame, the path of the light beam going with the rotation is slightly concave while the other one is slightly convex. In this frame the light is according to the emission theory going at c, so it is the difference in the path-lengths that must be responsible for any difference in the times. There is a difference, but it is order of magnitudes too small to account for the first-order Sagnac effect. > see also.. > http://www.youtube.com/profile?user=jaymoseleygrb > Sean Sagnac falsifies emission theory. No question about it. Paul
From: Henri Wilson on 15 Sep 2007 19:01 On Fri, 14 Sep 2007 03:59:50 -0700, George Dishman <george(a)briar.demon.co.uk> wrote: > >Henri Wilson wrote: >> 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. > >I got it right, you may be thinking of ballistic theory >instead of SR. Most of your colleagues use the term 'closing speed' in this contaxt. It does not have to equal c. >> >> 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. > >You said ".. in any TWLS experiment .." but Sagnac >isn't a two-way test. What you say is true for MMX for >example which _is_ a two way measurement, it measures >th sum of the times, not the difference. Sagnac measures rotation. It does that by exploiting the phenomenon that light does not reflect from a moving mirror at the incident angle and speed. >> >> ...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. > >No, you said ". that's not what YOUR sagnac diagram >shows .." (my emphasis). 'YOUR' here means 'relativist' >> The rays are shown to move >> at c+/-v wrt the source. >> Don't deny it George. > >Sorry Henry, I have corrected that error so many >times, I'm not going to repeat it. You know you >are wrong but if you insist on wishing to be >ignorant, that's your choice. George, your 'correction' is not a correction at all. The plain fact is, the times are calculated to be d/c+v and d/c-v....distance traveled over light speed. Andersen has written this mistake many times for the world to see. It shows that Sagnac refutes SR. >> >> there are lots of stars out there George...and only a limited number of misled >> astronomers. > >Only a fraction are eclising binaries and for those >that have had their luminosity curves studied, you >will find most have also had their spectra recorded >to the point of confirming that they are spectroscpic >binaries. That's right...and most eclipsing spectroscopic binaries are close and have short periods. That means their EM spheres overlap and little or no brightness variation is expected..only that due to VDoppler. Paul asked me many times to explain why HD80715 isn't a variable. I have done that. >> >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. > >Then you haven't made best use of it. It sounds as >though you are still iterrating along the light path. No, your 'doppler' method requires fewer sample points but many more subsequent calculations at each one. On emy computer, each set of curves is produced in less that a second with either method....so it doesn't matter which one I use. It's nice to have agreement between the two....just for checking the methods.... >> > >> >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.... > >Probably I didn't understand because you had trouble >conveying what you meant. Try defining what D and v >are in your equation, I may have guessed incorrectly. A---------------------S---------------------B If an object falls under gravity from A towards the star and then on onto B, Its final speed will equal that at A. The time taken is calculated by integrating dD/v(t). Any force that efectively reduces the garvity pull will result in an increased travel time. >> >>>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. > >Sure, but all your curves are matched to the temperature >and radius variation, not any ADoppler part so they are >universally wrong. Even once you correct that error, it >still means nothing, what you have to do to produce >a proof is show that it _is_ ADoppler, not just that it >_might_ be. The temperature and radius variations are largely mythical. >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.
From: Henri Wilson on 15 Sep 2007 19:09
On Fri, 14 Sep 2007 16:23:32 +0200, "Paul B. Andersen" <paul.b.andersen(a)hiadeletethis.no> wrote: >sean wrote: >> On 12 Sep, 08:03, "Paul B. Andersen" >> <paul.b.ander...(a)hiadeletethis.no> wrote: >>> Henri Wilson wrote: >>>> 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. 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. > >Which laws of nature do you invoke to make >the light go in curved paths in an inertial frame? :-) >(The effects of gravity are negligible in Sagnac) > >> 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 > >You got it backwards. >When pictured in the rotating frame, the path of >the light beam going with the rotation is slightly >concave while the other one is slightly convex. >In this frame the light is according to the emission >theory going at c, so it is the difference in >the path-lengths that must be responsible for any >difference in the times. There is a difference, >but it is order of magnitudes too small to account >for the first-order Sagnac effect. Sean is a little confused about sagnac. For one thing his animation shows the two rays ending up at different angles when in fact they remain parallel...or they do IF IT IS ASSUMED THTAT THEY REFLECT FROM THE MOVING MIRRORS AT THE INCIDENT ANGLES AND SPEEDS.....which they don't..... There IS a path length difference that I think is 1/2*root2 times the claimed one. Additionally, two rays that start out 90 apart end up displaced sideways...that might not cause a fringe shift...but that is not certain. > >> see also.. >> http://www.youtube.com/profile?user=jaymoseleygrb >> Sean > >Sagnac falsifies emission theory. >No question about it. Paul, your own maths show clearly that Sagnac disproves SR. The two travel times YOU calculated as (distance/light speed) are (D+d)/(c+v) and (D-d)/(c-v). What does that say about light always traveling at c? >Paul 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. |