Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: Eric Gisse on 22 Oct 2005 22:42 Robert wrote: > On 22 Oct 2005 03:40:18 -0700, "Eric Gisse" <jowr.pi(a)gmail.com> wrote: > > > > >Henri Wilson wrote: > >> On Fri, 21 Oct 2005 15:26:27 +0100, "George Dishman" <george(a)briar.demon.co.uk> > >> wrote: > >> > >> > > > >> > > >> >The geometry of spacetime. You might want to think > >> >of it as an extension of Pyhtagoras to 4D. > >> > >> George, physically speaking, there is NO spacetime. > > > >Mabey, but it is the best description we have. > > It describes nothing physical. > 'Spacetime' is just a big word designed to impress little boys. You do not understand SR by your own admission so I fail to see how you can decide if it is valid or not. > > > > >> It is just a maths tool. > > > >So? > > > > > >> > >> > > >> >> I think you are just a good old fashioned aetherist. > >> > > >> >I can't help the mistakes you make, you have said > >> >ignorance is your choice. > >> > >> brainwashing is yours then.. > >> > >> >>>Wrong, the aether principle says the speed > >> >>>is c relative to the aether, not the lab. > >> >>>You are still just demonstrating you don't > >> >>>understand either Henri. > >> >> > >> >> So how do YOU explain why light speed travels at anything other than c wrt > >> >> its > >> >> source, in the case of the huge remote interferometer. > >> > > >> >See news:dfaarq$9ag$1(a)news.freedom2surf.net > >> > >> can't get it. > >> > >> > > >> >>>>>> As always, SR reverts to LET when it tries to go physical instead of > >> >>>>>> plain mathematical. > >> >>>>> > >> >>>>>You do that, SR uses geometry alone to derive the > >> >>>>>LTs and is then mathematical. > >> >>>> > >> >>>> SR uses LET geometry. > >> >>> > >> >>>Wrong again. The geometry of LET is 3d Euclidean, the > >> >>>geometry of SR is 4d Riemann with signature (+---). > >> >> > >> >> very funny. > >> > > >> >Fact Henri, but then you have chosen to remain ignorant > >> >of SR, or at least so you told me. > >> > >> Try as I may, I cannot understand bullshit. > > > >It wouldn't compromise your position to understand SR and still have a > >opinion that is counter to it. > > > >The only reason you "cannot understand" is because you are incapable of > >learning anything that doesn't agree with your preconcieved notion of > >how the universe works. > > In this universe, no vertical beam of light in one frame becomes diagonal in > another. So it is vertical in every frame? > If you cannot deduce this yourself then consult my program 'movingframe.exe'. Why should I consult your computer program when you could simply write out the mathematics of the transformation from one frame to another? > > > >> I don't wish to be associated with any theory that is based on the obvious > >> fallacy that a vertical light beam in one frame becomes a diagonal light beam > >> in another. > >> That is plain nonsense. > > > >So its always vertical then? > > > >How about if I emit a photon, then catch up to it? Is it still going > >vertical relative to me? > > You cannot catch up. You don't have enough (mc^2) energy in you body to get the > last atom up to that speed. Odd. You have used superluminal muons in examples before, eg: http://groups.google.com/group/sci.physics.relativity/msg/a7b289a357541458?dmode=source&hl=en Why the change of tune? Are you now saying massive particles CANNOT go faster than the speed of light? Furthermore, E = mc^2 is a prediction of relativity, not the "BaTh". You cannot derive E = mc^2 classically, and I defy you to show me otherwise. > > > >> >Irrelevant, the result is known. That is what 'empirical' > >> >means. > >> > >> If the result agrees with anything you want it to prove, it is for the wrong > >> reasons. > > > >So you are dismissing observational evidence because you know it is > >wrong, even though you can't explain why it is wrong? > > There is no experimental evidence that supports SR. There is no experimental > evidence that refutes the BaTh. What about "willusions" ? The "willusions" match the form of relativity. Why is that? Remember, you have a VERY large amount of posts for me to quote. > > >> >> Well, for tyhe MMX, assumptions might have to be made about the reflection > >> >> of > >> >> normally incident light from a sideways moving mirror. In the sagnac, the > >> >> mirrors are at 45, so the light gets a 'kick' at each reflection according > >> >> to > >> >> the BaTh. > >> >> I still think they are equivalent....but I could be wrong. > >> > > >> >There is no 'kick' even in Sagnac. I showed you months > >> >ago that the light arrives at c so when it leaves at c, > >> >the speed is the same. Again, you are just rehashing > >> >old ground. > >> > >> I don't accept that. > >> Too many assumptions are involved. It can never be directly verified. > > > >That is a cop-out. "I have to think too much" is what it boils down to. > > > >You wouldn't understand the explanation anyway. Remember this? "Try as > >I may, I cannot understand bullshit." > > > >> > >> > > >> >>>> No I don't believe sagnac is a test of Ritz. > >> >>> > >> >>>Tough luck, it involves the speed of light emitted > >> >>>from a moving source which is how Ritz differs from > >> >>>other theories. > >> > >> >> > >> >> So what determines the speed of the beam in the large apparatus mention > >> >> above? > >> >> You cannot answer that can you. > >> > > >> >I have already told you many times but that is irrelevant. > >> >The apparatus you describe would measure the speed. Sagnac > >> >measures the speed from a moving source in the lab and the > >> >result is c. > >> > >> No it doesn't . > >> The speed of the source is normal to the next mirror's movement. > >> > >> >>>> It involves rotation and all the > >> >>>> complications that go with that.. > >> >>>> Ritz requires straight lines. > >> >>> > >> >>>No, Ritz is supposed to be a scientific theory > >> >>>for light propagation which means I can apply > >> >>>it to any situation I like. Those are the rules. > >> >> > >> >> Ritz and the BaTh say that light moves at c wrt its source. > >> > > >> >Exactly, yet in Sagnac it moves at c relative to the lab > >> >inertial frame even though the source is moving in that > >> >frame, ergo Ritz is wrong. > >> > >> So was Einstein then. Lorentz was correct. > >> There is at least a 'local aether'. > > > >I am sure you have evidence to support that assertion? > > The Sagnac definitely DOES NOT refute the BaTh. I have just shown George why > this is the case. You missed my point, Henri. Support the assertion that there is a local aether frame. > > >> > >> Yes..so SR breaks down in any physical situation.. > > > >That conclusion does not follow. > > > >How do you even know SR "breaks down", you admit you don't understand > >it! > > I understand what it cannot do. No, you do not. Otherwise you wouldn't make up stupid examples time and again. You wouldn't be asking questions that you know have no answer within the context of a physical theory. > SR cannot explain why simultaneously emitted light pulses from adjacent but > differently moving sources should travel together through space. > > ->S1__________________>p > <-S2 Relativity of simultainety, constancy of the speed of light. It follows exactly from the postulates of the theory. If you understood the theory you would know this. Within the framework of the theory, its perfectly obvious. Unfortunately you cannot work within the framework to find the solution, because you are a goddamn idiot. > > Until you can provide an alternative reason why they should, I will maintain > that only an absolute spatial reference frame could achieve that outcome. You can maintain whatever you want, neither the universe nor anyone else really cares. > > >> >In reality, the speed measured by each observer > >> >is c and SR explains that by the geometry of spacetime. > >> >It is the metric which is physical. > >> > >> Never verified. > > > >Then test it then. > > > >Oh wait, you can't - you don't understand SR. > > "Try as I may, I cannot > understand bullshit." Yes, that is what you said. You are thus unfit to judge the merits of SR. If you do not understand the theory, how can you possibly know what it does and does not predict? > > > >> > >> > > >> >> This is just a mathematical trick. > >> > > >> >Sorry Henri, as long as you choose to remain ignorant of > >> >how SR works, you'll never understand the physical basis. > >> >We went a long way to starting you off but you decided > >> >not to respond so I'll leave it at that. Nobody can force > >> >you to learn and I don't intend to waste my time trying. > >> > >> Geoge, expressing 3D space and time as a 4D graphic achieves nothing new. > >> It might impress a lot of little kids but that's about all. > > > >METRIC, Henri, METRIC. Not graphic. > > > >Do you even know what a metric is? Can you say ANYTHING qualitative or > >quanitative about the idea of a metric that shows you have had the > >education to understand it? > > a length in a 4D graph of space and time? METRIC, Henri, METRIC. Not graph. I should only have to say it once considering you quoted me saying it! The metric preserves the distance between events, this is basic differential geometry. > > What does that achieve. What would be the point of telling you? You will either argue that my explaination is useless or that you don't, in effect, understand it. If you really want to know, you can read any of the modern treatments of special and general relativity where the metric plays a fundamental role.
From: George Dishman on 23 Oct 2005 03:51 "Robert" <RB@..> wrote in message news:1fjll15gh6g48sdvqhf3jvj582j3hql7ae(a)4ax.com... > On Sat, 22 Oct 2005 11:36:54 +0100, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: > >> >>"Henri Wilson" <H@..> wrote in message >>news:d51kl1tj9vvemc5f4go35ej7e7coi91621(a)4ax.com... >>> On Fri, 21 Oct 2005 15:26:27 +0100, "George Dishman" >>> <george(a)briar.demon.co.uk> >>> wrote: >>> >>>> >>>>"Henri Wilson" <H@..> wrote in message >>>>news:h148l1972bdn5s7mebdb2095netb78ddn8(a)4ax.com... >>> >>>>> Can you not see that you are quoting the aether explanation. You seem >>>>> to >>>>> think >>>>> that space has absolute properties that determine light speed. >>>> >>>>I know you have chosen not to understand SR, you said >>>>so a few weeks ago. That doesn't alter the fact that >>>>the explanation it provides for Sagnac is trivially >>>>simple. >>> >>> Of course it is simple. It has no physical basis. It is just circular >>> maths. >>> >>>>> Imagine a completely remote sagnac interferometer with mirrors 1 >>>>> million >>>>> LYs >>>>> apart. Are you suggesting that the speed of the beam is determined by >>>>> something >>>>> other than its relationship with its source? If so, what might that >>>>> 'something' >>>>> be? >>>> >>>>The geometry of spacetime. You might want to think >>>>of it as an extension of Pyhtagoras to 4D. >>> >>> George, physically speaking, there is NO spacetime. >> >>There is a nice artists impression of a binary >>system on the LISA site home page: >> >>http://lisa.jpl.nasa.gov/ >> >>The middle link illustrates gravitational waves, >>ripples in spacetime: >> >>http://lisa.jpl.nasa.gov/popups/ripples.html > > There isn't any 'spacetime' there. Just 'movement in space'. Observations like that of Hulse and Taylor tell me otherwise. >>Those ripples should carry off energy, a prediction >>confirmed by Hulse and Taylor. >> >>http://astrosun2.astro.cornell.edu/academics/courses//astro201/psr1913.htm >> >>> It is just a maths tool. >> >>Maths tools don't change the orbits of stars, that >>takes something physical. The energy is transported >>as ripples in spacetime so spacetime must be physical. > > Bullshit. > Spacetime doesn't describe any physical quantity. > It is just a bit of maths. More verbiage, but that binary system is losing energy and at the rate the GR predicts. That energy is being transported by ripples in spacetime so it is physical, the maths is just a description of it. > SR is within the capabilities of anyone who cannot see that a vertical > light > beam doesn't suddenly become a diagonal one in a moving frame. Your "movingframe" diagram shows it does. >>> I don't wish to be associated with any theory that is based on the >>> obvious >>> fallacy that a vertical light beam in one frame becomes a diagonal light >>> beam in another. That is plain nonsense. >> >>Let me quote from your own program introduction: >> >>"One frame (rest) is equipped with a laser which >>sends pulses vertically. ... the path of each >>infinitesimal element of each photon is diagonal." >> >>The phrase "each infinitesimal element of each photon" >>is gibberish, a photon is a point particle, but what >>we are concerned with is the paths. > > There can be no such animal as a 'point particle'. > Such would be indistinguishable from 'nothing'. If that's your opinion, then ... >>You have a small error in your diagram, obviously the >>short section at the tip of each diagonal path needs to >>align with the path because, if you treat the photon as >>extended, then the part farthest from the laser was >>emitted earlier so will be laterally displaced from the >>end nearest the laser in the moving frame by the amount >>the frame moved during the time the element was emitted. > > My 'movingframe.exe' program is better. I've now looked at that too. Consider just the first of your elements. Its path is outlined by the two parallel lines that suddenly appear on the upper diagram. That path is diagonal while its equivalent in the lower diagram is vertical. > Yes, the vertical green dashes are really just there to show how all the > ends > of the diagonal paths remain in vertical alignment. Of course, nobody is disputing that. Your diagram confirms the conventional view with the minor error that your small elements should lie on the diagonal. The two small elements you illustrate as sine wave segments show that quite well. you need to use the purple one at the head of each line in the moving frame diagram while it is the green one in the static frame. > The point I am getting across is that no continuous beam moves up any one > particular diagonal. The point that Einstein made in his illustration of the train is that when considering the path of a single one of your elements, the length is increased because that path is diagonal. His argument applies to each element individually. > The diagonals are lines of infinitesimal thickness. What > goes up them is certainly not light. The diagonals are "rays", hypothetical perpendiculars to the wavefront as in Huygens. >>>>There is no 'kick' even in Sagnac. I showed you months >>>>ago that the light arrives at c so when it leaves at c, >>>>the speed is the same. Again, you are just rehashing >>>>old ground. >>> >>> I don't accept that. >> >>Check for yourself then: >> >>http://www.briar.demon.co.uk/Henri/speed.gif > > It's wrong. You didn't use the mirror frame. You used the screen frame. So use Galilean relativity to switch to the mirror frame and tell me what you get. >>> Too many assumptions are involved. It can never be directly verified. >> >>No assumptions, just what Ritz tells us, the speed of >>emission is c relative to the source. > > that's what Einstein said, too. Nonsense. Why are you wasting our time saying things you know aren't true. >>>>>>> No I don't believe sagnac is a test of Ritz. >>>>>> >>>>>>Tough luck, it involves the speed of light emitted >>>>>>from a moving source which is how Ritz differs from >>>>>>other theories. >>>>> >>>>> So what determines the speed of the beam in the large apparatus >>>>> mention >>>>> above? >>>>> You cannot answer that can you. >>>> >>>>I have already told you many times but that is irrelevant. >>>>The apparatus you describe would measure the speed. Sagnac >>>>measures the speed from a moving source in the lab and the >>>>result is c. >>> >>> No it doesn't . >>> The speed of the source is normal to the next mirror's movement. >> >>Speed is distance divided by time. The distance is >>known from the dimensions of the experiment and the >>rate of rotation. The time difference is measured >>and when you do the sum you find the speed is c. > > The part of the beam that goes from the source to the centre of the mirror > travels at c towards that centre no matter how the apparatus rotates. No, in Ritz it is supposed to be c+mv in the lab frame (where m is a factor that depends on the number of mirrors). > You already agreed that light from the moon - or any other object in > circular > orbit - always travels to Earth at c. No I didn't, what we agreed when discussing the Moon was that there was no transverse Doppler in Ritz. We didn't discuss speed at all. > Sagnac involves the same principle. No it doesn't, the light goes round the cicumference or along chords to the circumference, it never goes to the centre of the table. > So the BaTh explanation is exactly the same as any other. > > Thankyou for clearing this up. My pleasure, shame you had so manyt errors in your description. >>> Geoge, expressing 3D space and time as a 4D graphic achieves nothing >>> new. >> >>It brings our maths into line with what happens in >>reality. Clock ticks are measured to be equally >>spaced in 4D, not in 1D as Newton thought. > > It merely complicates the whole issue unnecesarily and to no advantage. The advantage was that the errors that were a problem in late nineteenth century were resolved. Ritz can't do that as witnessed by Sagnac. No matter how much hand- waving you try, Ritz gives a null prediction which contradicts the observed result. George
From: Paul B. Andersen on 23 Oct 2005 09:46 Henri Wilson wrote: > On Fri, 21 Oct 2005 13:04:44 +0200, "Paul B. Andersen" > <paul.b.andersen(a)deletethishia.no> wrote: > > >>Henri Wilson wrote: >> >>>On Thu, 20 Oct 2005 15:03:46 +0200, "Paul B. Andersen" >>><paul.b.andersen(a)deletethishia.no> wrote: >>> >>> > > >>>Have you considered that the K2 spectrum might be coming from a different layer >>>of the B8 star. >> >>Definitely not. :-) >>Henri, do you really not understand >>how incredible idiotic this question is? > > > It may sound idiotic to some idiotic people. It _is_ idotic. Abselutely crazy. >>>The BaTh predicts willusions. >>> >>>Note: a "willusion" defines a subgroup of phenomena that are generally >>>classified as "illusions". It applies specifically to observed images of very >>>distant objects such as stars. These images and all data associated with them >>>are distortions of reality due to the fact that information travels from the >>>object to the observer at different speeds. Thus, the information reaching an >>>observer at a particular instant is not generally that which left the star at >>>ONE particular instant. >> >>You are babbling. >>Please answer: >>Does, or does not the BaT predict what >>the observed light curve should be? > > > It does. It must. Ok. So you claim that the BaT predicts what the light curve should look like. >>If it does, does it then predict this light curve: >>http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1978MNRAS.184..523N&data_type=PDF_HIGH&type=PRINTER&filetype=.pdf >> or >>http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=1978MNRAS.184..523N& >> And retrieve the full article. >> >>If it does not, what does it then predict? >>Nothing? The last time I showed you this, Henri Wilson wrote: | It is willusory by definition. | Because light is used for gaining information about the star, it is a | willusion. | The task is to find te truth that causes the willusion. > If it does not, it is simply because not all factirs have yet been considered > (in my program anyway) It does not. The factor you haven't considered is that the BaT is wrong. >>Don't let us forget what your now snipped claim was: >>| That [the brigtness variation is less in IR than in visible] >>| is expained by the fact that the visible is produced in the surface of the >>| stars whereas the IR comes from lower levels where the radial velocities are >>| smaller. >>| Most of these brightness curves are the result of a star being orbited by a >>| WCH, which might be a large planet like "Androcles". The stars wobble around >>| the barycentre of the pair. The IR wobbles less than the visible. >> >> >>>>>The star is wobbling around the barycentre with its orbiting WCH. That centre >>>>>probably lies within the star. If you draw this, you will see that various >>>>>layers within the star have different radial velocities wrt a distant observer. >>>>> >>>>>The situation is further complicated by the star's rotation around its own >>>>>axis. Different regions of each spherical shell will have different radial >>>>>speeds. >>>>> >>>>>In the case of Algol, for instance, the radial velocitiy required to produce >>>>>the willusion is indicative of the main star's rotation around the barycentre >>>>>with its satellite planet, "Androcles". >>>> >>>>This is so obviously idiotic from a number of different reasons, >>>>that I am not sure I will bother to point it out. >>>>But OK, here are some of the reasons: >>>>1. All the black body radiation comes from the photosphere, >>>> and not from "different layers" of the star. >>>>2. If a star is orbiting, but not rotating, every part >>>> of the star will have exactly the same radial velocity >>>> relative to a distant observer. The position of the barycentre >>>> is of no consequence. >>> >>> >>>That is decidedly wrong. Please admit to that in your next post. >>> >>>Every part has the same angular velocity. >> >>Which is zero if the star is not rotating. >> >> >>>Different layers within the star have different radial velocities...and that >>>includes gaseous layers far beyond the extremities of the main body. >> >>This is caused by the rotation _only_. >>The orbital motion has nothing to do with it. > > > Run my little program: > http://www.users.bigpond.com/hewn/radialvs.exe > > (select 'star rotation =1' for tidal lock) So the star is rotating. >>>>3. The rotation of the star will obviously mean that >>>> different parts of the photosphere have different >>>> radial velocity, which will not affect the BB radiation, >>>> but which will broaden the absorption lines slightly. >>>> That's how stellar rotation is measured. >>> >>> >>>Correct. >> >>Of course. >> >>So according to you, a rotating star should >>not emit a black body spectrum. >>It does. > > > Depends how hot it is and how fast it is rotating. In Wondersland, yes. But not in the real world. >>>>>>The BaT predicts no difference in the visible light >>>>>>curve and the 10um light curve and thus is proven wrong. >>>>> >>>>> >>>>>As usual, you are talking nonsense. >>>>> >>>>>You have completely overlooked the common situation in which the main star is >>>>>wobbling around an internal barycentre. In that case, IR should have smaller >>>>>radial speeds than visible. According to my model, that would usually cause >>>>>smaller brightness variation in IR than visible. >>>> >>>>Utter nonsense. >>>>The 10um radiation and the visible light radiation are >>>>coming from the same source. >>> >>> >>>Oh, have you been there? >>> >>> >>> >>>>To claim that these two parts of the spectrum are coming >>> >>>>from two different sources with different radial velocity >>> >>> >>>>is so crazy that you must have lost your mind completely. >>> >>> >>>No Paul, its not lost...just way ahead of you. >>> >>> >>> >>>>>>>>>I don't understand what you mean by 'frequency' here. >>>>>>>>>If you mean light frequency, then that is easy to explain. >>>>>>>> >>>>>>>>So explain it. >>>>>>>> >>>>>>>>Why is the secondary minimum practically unobservable >>>>>>>>in visible light, while it is 0.35 magnitudes deep at 10um, >>>>>>>>exactly as the conventional theory predicts they should be? >>>>>>> >>>>>>> >>>>>>>Who said that? >>>>>> >>>>>>Are you not paying attention, Henri? >>>>>>I have shown you the calculation. >>>>>> >>>>>>Here it is again, all according to conventional theory: >>>>>>We have two stars. >>>>>>Algol A: temperature Ta = 12000K, radius Ra = 2.88 solar radii >>>>>>Algol B: temperature Tb = 4880K, radius Rb = 3.54 solar radii >>>>>> >>>>>>Their relative brightness at the wavelength lambda will be: >>>>>>Ba/Bb = (Ra/Rb)2* W(lambda,Ta)/W(lambda,Tb) >>>>>> >>>>>>where W(lambda,T) is Planck's radiation law. >>>>>>Now we have: >>>>>>(Ra/Rb)2 = 0.66 >>>>>>W(lambda,Ta)/W(lambda,Tb) = >>>>>>(exp(C/(lambda*Tb))-1)/(exp(C/(lambda*Ta))-1) >>>>>>where C = 0.00144 m degree >>>>>> >>>>>>In the visible spectrum lambda = 0.5 um. >>>>>>W(0,5um,Ta)/W(0,5um,Tb) = 40 >>>>>> >>>>>>So their relative visual brightness will be: >>>>>>Ba/Bb = 26. >>>>>>That is A is 26 times brigter than B. >>>>>>The binary is 27 times brighter than B. >>>>>> >>>>>>If we assume that the eclipses are 100%, >>>>>>we get the following brightnesses (B as unit): >>>>>>No eclipse = 27 >>>>>>B eclipses A: 1 (primary) >>>>>>A eclipses B: 26 (secondary) >>>>>> >>>>>>The deepness of the minima in magnitudes will be: >>>>>>Primary: 2.5*log(27) = 3.58 magnitudes >>>>>>Secondary: 2.5*log(27/26) = 0.04 magnitudes. >>>>>> >>>>>>We see that the deepness of the primary minimum fits >>>>>>quite well with what is observed. >>>>>>But the secondary minimum is hardly observable at all >>>>>>in the visible spectrum! >>>>>> >>>>>>So don't we see the secondary minimum, then? >>>>>> >>>>>>Let us calculate what the deepness of the minima would >>>>>>be in the infra-red, lambda = 10um. >>>>>>We use the same method as above: >>>>>> >>>>>>Ba/Bb = (Ra/Rb)2* W(10um,Ta)/W(10m,Tb) = 1.8 >>>>>> >>>>>>No eclipse = 2.8 >>>>>>B eclipses A: 1 (primary) >>>>>>A eclipses B: 1.8 (secondary) >>>>>> >>>>>>The deepness of the minima in magnitudes will be: >>>>>>Primary: 2.5*log(2.8) = 1.12 magnitudes >>>>>>Secondary: 2.5*log(2.8/1.8) = 0.48 magnitudes. >>>>>> >>>>>>Observation of the secondary minimum at 10um can be found in; >>>>>> >>>>>>http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1978MNRAS.184..523N&data_type=PDF_HIGH&type=PRINTER&filetype=.pdf >>>>>>or: >>>>>>http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=1978MNRAS.184..523N& >>>>>>And retrieve the full article. >>>>>> >>>>>>The observed deepness of the secondary minimum is ca. 0.35. >>>>>>A little less deep than what I calculated it should be. >>>>>>However, since B is larger than A, the eclipse will not be 100%, >>>>>>and the minimum _should_ be less deep. >>>>>> >>>>>>So I repeat my question: >>>>>>Why is the secondary minimum practically unobservable >>>>>>in visible light, while it is 0.35 magnitudes deep at 10um, >>>>>>exactly as the conventional theory predicts they should be? >>>>>> >>>>>>You said it was easy to explain. >>>>>> >>>>>>So explain it. >>>>> >>>>> >>>>>Well maybe it is not all that easy but I can offer a suggestion. >>>>>Do you agree that IR would have its origin inside the star wheras visible is >>>>>more likely to come from the very outer layers. >>>> >>>>No. >>> >>> >>>Why not? >>> >>> >>> >>>>>If so, consider the comparative radial speeds of the IR and visible 'layers' >>>>>wrt a distant observer FOR DIFFERENT POSITIONS OF THE BARYCENTRE. >>>> >>>>Mindless babble. >>> >>> >>>Not everything you don't understand is mindless babble Paul. >> >>I know that this is mindless babble >>because I understand it. > > > run my little program: > http://www.users.bigpond.com/hewn/radialvs.exe > > If you cannot see that the green dots have a higher radial velocities wrt the > observer than the yellow ones then there is something wrong with you. Of course, the star is rotating. Just about _all_ stars are rotating. So what? The secondary minimum of Algol is practically unobservable in visible light, while it is 0.35 magnitudes deep at 10um, exactly as the conventional theory predicts they should be. The BaT does _not_ predict that the light curve in 10um and visible light should be different. Another falsification of the BaT. Paul
From: bz on 23 Oct 2005 11:33 Robert <RB@..> wrote in news:t9lll1lf8gesdltqsprkbqggf5pbq3t3sa(a)4ax.com: >>How about if I emit a photon, then catch up to it? Is it still going >>vertical relative to me? > > You cannot catch up. You don't have enough (mc^2) energy in you body to > get the last atom up to that speed. > > Robert AKA Henri, That should be no problem for a BaTer. All you need to do is to get up to .2 c (particles much better than that all the time in accelerators), and catch up to a photon emitted by a particle going at -0.9c. The particle that emits the photon (going in your direction) is going away from you in the opposite direction from the direction of your travel. By the BaT, c'=c+0.9c = 0.1c So, if you are going 0.2 c you should have no trouble catching a photon that is only going 0.1 c relative to you, should you? Henri, I am afraid you can't have it both ways. You can't say that massive bodies can't go faster than c while maintaining that photons move at c'=c+v. Your approach leads to logical contradictions. -- bz please pardon my infinite ignorance, the set-of-things-I-do-not-know is an infinite set. bz+sp(a)ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap
From: HW on 23 Oct 2005 17:52
On Sun, 23 Oct 2005 02:00:11 GMT, The Ghost In The Machine <ewill(a)sirius.tg00suus7038.net> wrote: >In sci.physics, Eric Gisse ><jowr.pi(a)gmail.com> > wrote >on 22 Oct 2005 17:26:21 -0700 ><1130027181.292522.93000(a)o13g2000cwo.googlegroups.com>: >> >> Robert wrote: >> >> [snip] >> >> So...are we supposed to call your Robert now, Henri? >> > >Either that, or Henri's starting a club. :-) I accidentally wiped everything off my browser. Sorry about that. Not sure who 'Robert' is. HW. www.users.bigpond.com/hewn/index.htm see: www.users.bigpond.com/hewn/variablestars.exe "Sometimes I feel like a complete failure. The most useful thing I have ever done is prove Einstein wrong". |