GETTING RID OF EINSTEIN RELATIVITY
in [Relativity]
|
Prev: What is the Aether?
Next: Debunking Nimtz
From: Henri Wilson on 7 Sep 2007 19:00 On Fri, 07 Sep 2007 12:00:36 -0700, Randy Poe <poespam-trap(a)yahoo.com> wrote: >On Sep 6, 7:13 pm, HW@....(Henri Wilson) wrote: >> On Thu, 06 Sep 2007 00:52:49 -0400, "T.M. Sommers" <t...(a)nj.net> wrote: >> >T.M. Sommers wrote: >> >> Are you saying that photons are not points but have extent and some >> >> internal structure, or are you just talking about wavelength? >> >> If photons were nothing but 'points' they wouldn't be photons... > >This sounds like a theorem or postulate you've added to >standard physics. Can you point to something in the definition >of photon that says such a thing? Why is it that relativists can never understand even the most basic logic? Look up the definition of 'a point', BP. Does it mention the word 'photon'? > - Randy 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 7 Sep 2007 19:13 On Fri, 07 Sep 2007 01:32:37 -0700, sean <jaymoseley(a)hotmail.com> wrote: >On 24 Aug, 16:34, Randy Poe <poespam-t...(a)yahoo.com> wrote: >> On Aug 24, 12:45 am, sean <jaymose...(a)hotmail.com> wrote: >> >> > All evidence. Not likely . Most if not all evidence points >> > to light being a wave only. >> >> You mean all those photon detectors aren't real? >Did I say they werent real? No. You did >Look at the thread `what evidence photons` from a few years ago. In it >I show that one can mechanically explain the apparent particle like >effects >in a pmt and I can statistically describe to a better accuracy the >observed >results of incident `photons` than even QT can. > >> > Can a photon describe a interference >> > pattern? >> > No. >> >> Yes. >I was reffering to the photon as a particle but.. >If you think about it a particle cannot describe a >interference pattern can it? Try getting one from a stream of billiard >balls. >In fact the only way a photon can describe a interference pattern is >to pretend >it is a wave at times. In other words a photons particle properties >cannot describe interference . Its wave properties do. >Which means that in fact wave theory describes interference patterns. >Not photons or particles The true nature of a photon should by now be obvious. run: www.users.bigpond.com/hewn/e-field.exe >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: sean on 10 Sep 2007 13:11 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 > > then it would be impossible for light to be effected by gravity under > > these terms. > > Once light has left the source, gravity, mirrors, lenses, etc can effect > its motion. The 'leaves its source at c' only applies at the moment of > emission and then only when measured with equipment that are co-moving > with the source and hence in the same inertial frame of reference. If the > source suddenly accelerates at right angles and goes flying off, into the > night (you have just thrown your flashlight away) the light that has been > already emitted from that flashlight does NOT suddenly change its motion > in order to maintain a velocity of c wrt the source that it has left far > behind. How do you know. Wheres the proof ? You have none. I discussed this earlier in this thread with George and I pointed out that it is technically almost impossible to test these claims. At least there is no experimental setup that exists that can test this clai you make. If a lab was 10 meters long and a rotating light source on one end was observed from the other end. The actual displacement due to this `path drag` would be minute. Something like a 0.000003 mm deflection over 10 m. Youd never be able to measure that. Nobody has yet. Therefore no proof exists that tests this possibilty one way or the other > > Secondly you ignore what I say repeatedly in the post > > youve just replied to. > > I ignore many things because I am trying to focus your attention on one > small point. And Im trying to focus your mind on one point. That is that the MMx source is moving yet the light that leaves the source still has a speed of c relative to the moving source. Contrary to what you claim taht it isnty moving at c in the lab. > > That emmision theory cannot have light as a particle with mass. > > I don't care if light is a wave or a particle at this moment. It doesn't > matter. The light, once emitted, goes its merry way. Unlike the teenager, > leaving home for the first time, the light does not 'write home' and ask > for more money. The light doesn't care what happens to the source after it > has left home. > > [snipping the rest of the stuff that is unrelated to the point I am trying > to make and snipping your insults. If you want to continue our discussion, > you will avoid insults. I don't bother with those that lack enough self > respect to treat others with respect.] ..... My apologies. I get non stop sleazy insinuations, character smears and insults from the rest of your peers. I forgot to check if you werent up to the same. A quick check shows me that so far you seem to have been OK. Thanks. I appreciate that. JUst dont forget that Im argueing here for a wave only emmision model to explain the various observed phenomena. So please dont expect me to incorporate particle theory in this model. Like for instance.. electrons. > >> The light emitted consists of 'objects'. Once emitted, they must travel > >> in straight lines at a constant velocity unless acted upon by an > >> outside force. > > >> What force would 'tie' the light, once emitted, to the source so that > >> it must follow the source? > > >> There is no evidence of any such 'tie' between light, once emitted, and > >> its source. > > You jump to conclusions here. What "evidence" do you have that light > > doesnt > > travel at c in any direction from a source? > > I made no such claim. It DOES travel at c in all directions from where the > source was at the moment of emission. But once emitted, it doesn't care > where the source goes afterwards. This is an odd statement . You say you never claimed that light `doesnt travel at c relative to a source`. And then you contradict yourself in the same sentence above and say that in fact it `doesnt travel at c relative to the source`.?? So what do you want to claim? That it does or it doesnt travel at c relative to the source? > > My evidence that they do is MMx. Notice here that the MMx source is > > rotating > > around the earths axis yet we also know that the light is travelling > > at c on > > both paths. This would not be possible if light travelled away from a > > rotating > > source at different speeds in different directions as you erroneously > > suggest.above > > You misread my suggestion. How did I misread your sugggestion? YOu said... "The light emitted consists of 'objects'. Once emitted, they must travel in straight lines at a constant velocity unless acted upon by an outside force." In other words you say above that once light leaves the MMx source it must travel in straight lines at c + the speed of earths rotation etc, relative to the source. THat means that you think that light travels at different speeds away from a source relative to that source. > >> In a cyclotron, particles traveling around a circular path emit beams > >> of xrays. > >> Those xrays travel in straight lines from the point of emission. They > >> do NOT follow, in any way, the subsequent motion of the emitting > >> particle. > > Two things here.. Firstly,.. whats your proof these xrays > > travel in straight lines?? You have none. > > You put three lead plates with small holes in them along a straight line > leading away from the source. The xrays only reach the detector when all > three plates line up with the holes in a straight line. If the xrays were > traveling along a curved path, the holes would need to line up along a > curved path. > > > Show me the part > > of an experimental setup that checks the direction relative to the > > known > > source and the speed of the xray. You have neither nor do both tests > > exist. > > You have never worked with an xray beam line at a synchrotron. > visit a facility such ashttp://www.camd.lsu.edu/and you will see that > they use the synchrotron to generate xray beams. BEAMS of xrays traveling > in straight lines. If your idea about the emission following the movements > of the source were correct, CAMD would not work as it does. > > > This is a fabrication on your part. > > I don't need to fabricate anything because I know what I am talking about. > > > Secondly,.. you make the assumption that the xrays are produced by > > moving particles/ie ...rotating sources like the MMx or sagnac > > source. But what proof have you that the sources in a cyclotron > > is rotating?? NONE > > The xrays are produced when a beam of particles is forced to CHANGE > directions. When this is done, they emit photons. This is known as > cyclotron radiation. > > > You can only assume the xray `sources` are rotating theoretically. > > IM not familar enough with the technical setup of a cyclotron > > I am. > > > but my reference refers to the `particles` not as atoms but as > > electrons > > Electrons or positrons or protons or heavier particles can be accelerated. > When the beams are bent(forced to change direction of motion) photons are > emitted. > > > As far as Im aware electrons are emr. > > You are quite mistaken. The motion of electrons generates electro magnetic > radiation but electrons are NOT electro magnetic radiation. But, all of > this is beside the point. Sorry Im not sure why I said this .Yes the electrons in wave theory too , arent the emr but they generate the emr . > The point being that once emitted, light doesn't care what happens to the > source. > > [snipping the rest of the side issues] > > .... > > > Lets see if for starters you can supply the lab setup in a cyclotron > > that measures the speed and dierection of the observed xrays.. > > You cant, because no such test was ever made. > > Wrong. The direction is easily determined by three lead plates with holes > in them. OK fair enough. This is the first time anyonse asked me to define a cyclotron in wave only theory so I need to get uo to speed on how they are constructed. Its not quite clear in my reference how this is done. For instance it shows the defelected `electrons` but not where the xrays are generated. I assume the whole cavity generates them or is it just a point at where the electron is deflected at the small outlet hole? Its very vague my reference. > The speed is easily determined by measuring the time between the moment > that the batch of particles gets forced to change direction of motion and > the time the xray pulse hits a target along the beam line. Also, the beam > line can contain two detectors, a known distance apart and the time for > the pulse to travel between those detectors can be measured. You obviously know a good deal about cyclotrons. I admit need to find out more about cyclotrons . For instance my reference shows the electrons being fired out the small hole? Or is that the xrays. Its not clear. Any ways in a wave only theory I have to explain electrons as waves only so its going to be difficult for you to argue that the source is particles. For instance I notice that the setup of the cyclotron bears close resemblance to the radio transmittter. Both have oscillating magnetic fields as the `driver` of the radiations and both in the standard model describe these as streams of electrons. But wave theory as I point out in my "redshift without expansion " describes the oscillating field in the transmitter as oscillating magnitic fields of the atoms in the wire which produce the emr radiation. No electrons needed. So Id say the cavity in the cyclotron is the source and it is the oscillating magnetic field in the cavity that generates the xrays. Just as the the wire `cavity` in the radio transmitter generates the radio waves. I assume the cyclotron cavity oscillates at higher frequencies hence the higher frequency emmision.Thats whats suggested in my ref anyways. As with the transmitter the cavity as a whole is the source. And it doesnt move in the lab. It is in the same frame as the observor. Hence contrary to what you thought no bent lines in the x-radiation are predicted by wave only emmision theory because the source isnt moving in the lab > >> He did NOT say 'break apart', he said 'distort'. > >> In the MMX, all it takes is moving part of the apparatus by a small > >> fraction of the wavelength of light in order to create a detectable > >> indication. > > >> THAT is why he said 'distort'. Spinning the apparatus would cause > >> changes in the instrument that would make it useless for the purpose > >> you propose to use it for. > > I assume by this that you and george imply that a rotating MM > > experiment > > physically stretches one arm moreso than another and doesnt give a > > null > > result? If so please supply a reference for such an observation, > > because > > I dont believe you have any proof to back up this claim.(notice that > > will not be able to supply this as none exists. Youve just fabricated > > this evidence) > > If you ever spin anything (a car tire for example) you will notice that > getting everything 'in dynamic balance' is NOT trivial. Any experimental > apparatus consists of many different parts, each with their own mass, > density and, in this case, distance from the point about which the > apparatus is to be rotated. > > The MMX apparatus must maintain all dimensions within fractions of the > wavelength of light. > > ANY slight imbalance or difference in the distances from the center would > produce different torques and invalidate the experiment. Have you ever > seen an MMX apparatus? > > And one final point. The MMX apparatus IS spinning at a slow rate of spin, > 1 revolution per day. I dispute these points on several counts. First MMx DID spin the setup and it didnt create distortion in the arms. So it is possible.Second, if this experiment hasnt been tried in your opinion, it doesnt mean that one can assume that the outcome would be that light wouldnt travel at c relative to the spinning MMx source,.. which leads to my third point; As you point out above. A spinning MMx setup has been tried where no distortion occurs and it does show that light from a spinning source still travels away at c relative to the source. Its called the MM experiment. And it spins 1/day. > > And secondly assuming the remote chance that it does distort such an > > apparatus and not allow us to check what I propose may be observed > > (ie whether or not a rotating MMx that isnt distorted does still > > have light at c on both paths)...THen how is it that you or george > > can say with certainty that a rotating MMx source does not allow > > light > > to propagate down both arms at c? After all, youve just claimed > > that any experiment to test this is impossible. > > The math is not subject to the physical limitations of the equipment. > > [snipping the rest because it has nothing to do with **the point** that > > once light is emitted, subsequent motion of the source can not and does not > effect the already emitted light This is wrong and unsubstantiated. Ive pointed out many times that the MMx results contradict this claim you and others make. Because we know that when light leaves the MMx source it still travels in all directions at c relative to the source. THis is only possible if the subsequent motion of the source did effect the already transmitted light. Sean www.gammarayburst.com
From: Henri Wilson on 10 Sep 2007 20:46 On Mon, 10 Sep 2007 10:11:42 -0700, sean <jaymoseley(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. >> > then it would be impossible for light to be effected by gravity under 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: Paul B. Andersen on 12 Sep 2007 03:00
sean 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 MMX, yes, Sagnac, no. The MMX confirmes SR and emission theory, and falsifies a 'Michelson type' ether theory. Sagnac confirms SR and a 'Michelson type' ether theory, and falsifies emission theory. Only SR passes both tests. Paul |