Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: Henri Wilson on 16 Oct 2005 18:09 On Sun, 16 Oct 2005 09:23:07 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <H@..> wrote in message >news:bqv2l1lcficiap82dlub1bfn07nna0f0pf(a)4ax.com... >> >> Hmm. Using rotating frames always introduces some confusion. > >For the iFOG where the path is circular, it is >easier. With mirrors where the path is chords, >the lab frame is easier. > >However, isn't it an interesting that Sagnac gives >a hand-waving way to predict the equivalence of >acceleration and gravity? That's a big question. It gets a bit messy. Consider the frame of the first mirror. Light in transit appears to accelerate in that frame. It also changes direction. GR would try to distort space so that it remained at the same speed in a straight line. I don't go along with the equivalence principle at all, anyway. I prefer the define acceleration as d2x/dt2. F/M not a definition.... just a magnitude... >>>>>Also the speed of the light varies with the radius >>>>>because the frame is accelerating. There are lots >>>>>of tricky aspects to consider that come from using >>>>>an accelerating origin. >>>> >>>> The whole thing is very tricky....particularly when nobody has >>>> a real clue as to what a 'photon' actually is. >>> >>>You may not. >> >> You do not either. > >No, but the people who have studied QED do. hoho! funny. >>>constant. That in turn means the time taken is >>>inversely proportional to the speed. Working >>>out the speed becomes complex in the mirror >>>frame. >> >> I thnk you have gotten that wrong. >> I think the speed remains c but the path length changes. > >I think you misunderstood me. The general result >is that in calculating the time taken, the speed >still matters if the distance is constant. I also >think that the speed of emission would remain >equal to c irrespective of the angular velocity >according to Ritz. However, the path length does >not change because "each mirror is moving normally >in the frame of the next". If there is no distance >change and Ritz predicts no speed change, then it >also predicts no change to the time taken, hence a >null output. George, I really think you should stop trying to work out the principle behind sagnac before it sends you crazy. I am content to accept that it is not a test of source dependency. The standard explanations are just those involving a 'local aether frame' and it is quite possible that such a frame DOES exist around Earth. The BaTh is mainly concerned with linear motion. Sagnac involves rotation and aspectsof light that I believe are quite unknown at this stage. >>>I've shown you the maths repeatedly or you can >>>find it on several pages on the web if you don't >>>trust me. It is a simple statement of fact that >>>Ritz produces a null prediction for Sagnac. >> >> I have now told you why Ritz was wrong then. > >Yes, we are nowin agreement :-) there have been 'developments' since Ritz died. > >>>>>If you introduce a new feature like a "photon gyro" >>>>>to produce a non-null result for Sagnac, it may also >>>>>produce a non-null prediction for the MMX. >>>> >>>> It doesn't. >>> >>>You don't know that, you don't have any equations >>>for your "photon gyro" concept so you can't do the >>>calculation to find out. >> >> George, you SR 'explanation' is really just the LET explanation. > >No Henri it isn't even vaguely close. I tried to >explain some of the differences and you seemed to >be getting an inkling of how SR works but you >stopped replying to that thread. See message > news:dfaarq$9ag$1(a)news.freedom2surf.net I know hw SR 'explains' sagnac. It merely states the aether principle that light speed = c. >> 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. SR wrongly assumes that a vertical beam of light in one frame becomes a diagonal beam in another. In LET that is not true either. The diagonal beam in the second frame is not the same beam as the vertical one in the first frame. > >>>> The MMX result is straight forward. >>> >>>Without your "photon gyro" it is, you have no >>>idea how that might affect it. >> >> But you know that the fringes become displaced only when movement is >> actually >> occuring. >> They return to their original positions when the rotation ceases. >> I doubt if Michelson watched while he rotated the apparatus ,...but if he >> had, >> it would have done what any sagnac would do. > >No, there would have been no shift or only a tiny >shift if they used multiple passes along the legs >to increase the effective distance, the reflection >points need to be slightly offset. Too much speculation George. >>>> If the MMX apparatus is rotated, it becomes a sagnac. >>> >>>That depends on the details but generally it >>>won't. The delay in Sagnac is propotional to the >>>area enclosed by the path but in MMX the light >>>usually returns back along the same path on each >>>leg so the ennclosed area is zero. >> >> Not when the MMX is rotating. A sagnac interferometer is just a Michelson >> one >> set up so that the fringe offset can be observed during rotation. > >For Sagnac the first leg is A->B->C->D->A > > A--->---B > | | > | v > ^ | > | | > D---<---C > >That encloses an area and the empirical result >is that the shift is proportional to the area. > >In MMX the first leg goes from A to B and back: > > A-<--->-B > | > | > | > | > D > >Neither leg encloses any area so empirically we >expect no shift when rotating. Your ideas on a >"photon gyro" would suggest to me that the act >of rotating should produce a shift even in MMX >but that is hand-waving. Until you produce the >equations, we can only guess ... as I said: Well, I'm not convinced that rotating the MMX is not equivalent. > >>>>>That's >>>>>why you can't just guess, you have to write out the >>>>>equations and work each of the experiments. If you >>>>>can't provide the equations, you don't have a theory, >>>>>just a speculation. >>>> >>>> One must begin with speculation. >>> >>>Sure, you have a speculation but not a theory. >> >> well forget the 'photon gyro' business. > >I think we have to until you can write down the >equations so we can both apply them. Degrees of >freedom for a photon are also intimately tied to >thermodynamic considerations and you would need >to do a pile of work to show your idea didn't >fail those experiments. It's not something we >can do here. I'm not going to speculate George. > >It's also somewhat academic since part of the >design of iFOGs is to apply a modulation to >the light and it also exhibits the same time >delay. You might find a way to use a "photon >gyro" to explain the carrier shift but not the >modulation. > >> I think that the path length difference is not affected by source speed > >Indeed, the path length isn't affected, but the >time taken to traverse that path depends on the >emission speed. I think that's obvious. Constant >speed means constant time taken, but the >experiment shows variable time hence Ritz is >wrong. No I don't believe sagnac is a test of Ritz. It involves rotation and all the complications that go with that.. Ritz requires straight lines. I'll stick with star light. >> because >> that is always normal to the next mirror (in that mirror's frame), not 45 >> degrees as you previously thought. > >I never thought that Henri, I said the _light_ >path is at 45 degrees to the mirror hence the >speed in the _lab_ frame is c + v / sqrt(2) >according to Ritz. > >In the non-rotating mirror-centred frame, the >source is moving in a circle but the light >moves on a curved path at variable speed, it >all gets very complicated! It sure does... > >George > 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".
From: Henri Wilson on 16 Oct 2005 18:14 On 15 Oct 2005 17:52:11 -0700, "Eric Gisse" <jowr.pi(a)gmail.com> wrote: > >Henri Wilson wrote: > >[snip] > >I find it interesting that you completely ignored every point George >made about you not having any equations so you can make actual >predictions. geese, those of us who possess natural scientific ability can visualise processes like these. Usually, one must do that before one can formulate a maths theory. Quite often, there is no need to apply any maths because the process is quite obvious to the trained eye. You obviously don't have this ability. So give up physics before you waste any more of your life. I heard they are short of cleaners at the local university.... 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".
From: Henri Wilson on 16 Oct 2005 20:39 On Sun, 16 Oct 2005 22:39:11 +0200, "Paul B. Andersen" <paul.b.andersen(a)deletethishia.no> wrote: >Henri Wilson wrote: >> On Sat, 15 Oct 2005 16:53:23 +0200, "Paul B. Andersen" >> <paul.b.andersen(a)deletethishia.no> wrote: >> >>>A Doppler shifted K2 spectrum is still a K2 spectrum, >>>and cannot be mistaken for anything else than the spectrum >>>emitted from a star with temperature ca. 3800K. >>> >>>I have told you before, do you never learn? >>>The temperature of a star is NOT determined by where >>>the black body spectrum peaks. >>>The spectral class is determined by the relative positions >>>and intensities of the absorption lines, and these are >>>unaffected by a Doppler shift. >> >> >> I don't think that is quite what you wanted to say. > >Yes, that is exactly what I wanted to say. >Each spectral class has a very characteristic pattern >of absorption lines, and this pattern isn't affected >by a Doppler shift. Their relative position and intensity >remain the same. According to the standard interpretation of the willusion. > >>>The temperature is then determined by the spectral class. >> >> >> I think what you meant was that if absorption lines were distinguishable and >> recognizeable, their doppler shift would be an indication of the star's radial >> velocity. >> With that knowlegde, the true position of the peak of the radiation curve could >> be established...and so could its temperature. > >No, I wanted to say exactly what I said, namely that when >the spectral class is determined, so is the temperature. >The temperature is NOT determined by the position of >the peak of the black body spectrum. I wonder why not? Would it make the einsteinian based interpretation of the willusion look silly? > >Of course the radial velocity of the star can be deduced >from the Doppler shift, but that was NOT the issue in this case. >As always you are trying to divert the attention from the issue, >namely that a Doppler shift won't make the spectral class or >the temperature of a star appear different. I think the doppler corrected peak of the curve would be a better indicator of temperature. > >Thus you were wrong when insinuating that a spectral class >can appear different because of Doppler I think you are refering to chemical classifications rather than plain temperature. >>>>>And you claim that it is coming from a planet! >>>>>You must have lost your mind completely. >>>> >>>> >>>>Don't lie. >>>>I said it was REFLECTED from the (large) planet's surface. >>> >>>Which is even more stupid - if possible. >>>An A8 spectrum doesn't become a K2 spectrum if you reflect >>>it off a planet. >> >> >> It is reflecting the K2 spectrum then. > >Really? >You claim the that only the B8 star exiats, and that >the K2 star really is a planet. I said that I accepted the presence of another object. I didn't say what it might be. > >So would you please explain what you meant by this comment? > >Of course you cannot. All you have is willusory information. >>> >>>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! >>> >>> >>>>Why is there a brightness INCREASE between supposed major and minor >>>>eclipses? >>> >>>But is it? >>>If we look at the curve in: >>>http://www.macalester.edu/astronomy/research/phys40/mark/ltcurve.htm >>>the drawn curve is pretty much fantasy. It is the points that >>>are the measured values, and the spread reflects the precision >>>of the measurements. And all we can conclude from those points >>>is that the light curve is roughly flat between the primary >>>minima, and any secondary minimum is hard to see. >>>We now know why. >> >> >> I know why. Androcles knows why. >> >> It is distinctly downwardly concave between the two major dips. >> That concavity has been mistakenly identified as a secondary eclipse. > >Can you show me the light curve which >is "distinctly downwardly concave between the two major dips."? > >Of course you cannot, because it only exists in your imagination. >The real observed visible light curve is like this: >http://www.macalester.edu/astronomy/research/phys40/mark/ltcurve.htm >If you look at the measuring points, and not the drawn curve, >it is clear that there is no "distinct concavity" in the light curve, >and the second minimum is not really visible at all. This is what it really looks like: www.users.bigpond.com/hewn/alg2.jpg > >>>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 >>> >>>If this long query doesn't make it through, try this one: >>>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 we can conclude that the observed eclipses are as expected. >> >> >> No. >> We can only conclude that your whole argument above involves circular >> reasoning. >> >> The parameters of the 'two stars' are largely based on your so called 'eclipse >> depth'. >> >> You - and they - have used the parameters of the illusion to justify the >> illusion. > >You pretend not to to get the point, do you? >The conventional explanation predicts that the light curve >should be different at different wavelengths. >The second minimum is observed in the infra-red at 10 um. >It is exactly as it should be according to the conventional explanation. The conventional explanation is based on what is observed in the willusion. >The ballistic theory does NOT predict such a difference. > >The ballistic theory is thus falsified. >Again. The ballistic theory WILL always predict what is observed. However it involves a great deal of trial and error as well as some initial speculation about what might be really happening. It also provides opportunity for discovery. I don't think we can model the rest of the universe on our own solar system. > >Paul 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".
From: George Dishman on 17 Oct 2005 08:37 "Henri Wilson" <H@..> wrote in message news:vth5l19e1f1v7s4pfq4c2bun0et9p4n6lc(a)4ax.com... > On Sun, 16 Oct 2005 09:23:07 +0100, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: >>"Henri Wilson" <H@..> wrote in message >>news:bqv2l1lcficiap82dlub1bfn07nna0f0pf(a)4ax.com... <snip side issues> >>>>constant. That in turn means the time taken is >>>>inversely proportional to the speed. Working >>>>out the speed becomes complex in the mirror >>>>frame. >>> >>> I thnk you have gotten that wrong. >>> I think the speed remains c but the path length changes. >> >>I think you misunderstood me. The general result >>is that in calculating the time taken, the speed >>still matters if the distance is constant. I also >>think that the speed of emission would remain >>equal to c irrespective of the angular velocity >>according to Ritz. However, the path length does >>not change because "each mirror is moving normally >>in the frame of the next". If there is no distance >>change and Ritz predicts no speed change, then it >>also predicts no change to the time taken, hence a >>null output. > > George, I really think you should stop trying to work out the principle > behind > sagnac before it sends you crazy. The principle is trivially simple, the detector moves while the light is in flight so the beam going against the diriction of rotation takes less time that that going with the rotation. It's only Ritz that has a problem with that, not me :-) > I am content to accept that it is not a test of source dependency. You are content to bury your head in the sand, yes I know. > The standard explanations are just those involving a 'local aether frame' > and > it is quite possible that such a frame DOES exist around Earth. Except that the speed of the Earth through the aether would then enter into the analysis. Let's leave that to others, I'm not interested in aether theory probelms. > The BaTh is mainly concerned with linear motion. Sagnac involves rotation > and > aspectsof light that I believe are quite unknown at this stage. BaT models like any physical theory must give a valid prediction for Sagnac or they are falsified. >>>>>>If you introduce a new feature like a "photon gyro" >>>>>>to produce a non-null result for Sagnac, it may also >>>>>>produce a non-null prediction for the MMX. >>>>> >>>>> It doesn't. >>>> >>>>You don't know that, you don't have any equations >>>>for your "photon gyro" concept so you can't do the >>>>calculation to find out. >>> >>> George, you SR 'explanation' is really just the LET explanation. >> >>No Henri it isn't even vaguely close. I tried to >>explain some of the differences and you seemed to >>be getting an inkling of how SR works but you >>stopped replying to that thread. See message >> news:dfaarq$9ag$1(a)news.freedom2surf.net > > I know hw SR 'explains' sagnac. Then why don't you answer the message. > It merely states the aether principle that light speed = c. 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. >>> 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 (+---). > SR wrongly assumes that a vertical beam of light in one frame becomes a > diagonal beam in another. I thought we discussed that some time ago? Your own graphic showed the change in angle. Henri, I don't want to waste time repeating all that, if you want to discuss it again, can you find where we left off and see if there was any remaining disagreement. >>> I doubt if Michelson watched while he rotated the apparatus ,...but if >>> he >>> had, it would have done what any sagnac would do. >> >>No, there would have been no shift or only a tiny >>shift if they used multiple passes along the legs >>to increase the effective distance, the reflection >>points need to be slightly offset. > > Too much speculation George. No speculation at all Henri, just the empirical result. >>>>> If the MMX apparatus is rotated, it becomes a sagnac. >>>> >>>>That depends on the details but generally it >>>>won't. The delay in Sagnac is propotional to the >>>>area enclosed by the path but in MMX the light >>>>usually returns back along the same path on each >>>>leg so the ennclosed area is zero. >>> >>> Not when the MMX is rotating. A sagnac interferometer is just a >>> Michelson >>> one >>> set up so that the fringe offset can be observed during rotation. >> >>For Sagnac the first leg is A->B->C->D->A >> >> A--->---B >> | | >> | v >> ^ | >> | | >> D---<---C >> >>That encloses an area and the empirical result >>is that the shift is proportional to the area. >> >>In MMX the first leg goes from A to B and back: >> >> A-<--->-B >> | >> | >> | >> | >> D >> >>Neither leg encloses any area so empirically we >>expect no shift when rotating. Your ideas on a >>"photon gyro" would suggest to me that the act >>of rotating should produce a shift even in MMX >>but that is hand-waving. Until you produce the >>equations, we can only guess ... as I said: > > Well, I'm not convinced that rotating the MMX is not equivalent. You should realise there is a fundamental difference in the physical layout of the experiment which means you cannot just assume the result of rotation, you need the equations from your theory to get a prediction. >>>>> One must begin with speculation. >>>> >>>>Sure, you have a speculation but not a theory. >>> >>> well forget the 'photon gyro' business. >> >>I think we have to until you can write down the >>equations so we can both apply them. Degrees of >>freedom for a photon are also intimately tied to >>thermodynamic considerations and you would need >>to do a pile of work to show your idea didn't >>fail those experiments. It's not something we >>can do here. > > I'm not going to speculate George. Agreed. >>It's also somewhat academic since part of the >>design of iFOGs is to apply a modulation to >>the light and it also exhibits the same time >>delay. You might find a way to use a "photon >>gyro" to explain the carrier shift but not the >>modulation. >> >>> I think that the path length difference is not affected by source speed >> >>Indeed, the path length isn't affected, but the >>time taken to traverse that path depends on the >>emission speed. I think that's obvious. Constant >>speed means constant time taken, but the >>experiment shows variable time hence Ritz is >>wrong. > > 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. > 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. George
From: Paul B. Andersen on 17 Oct 2005 10:03
Henri Wilson wrote: > On Sun, 16 Oct 2005 22:39:11 +0200, "Paul B. Andersen" > <paul.b.andersen(a)deletethishia.no> wrote: > > >>Henri Wilson wrote: >> >>>On Sat, 15 Oct 2005 16:53:23 +0200, "Paul B. Andersen" >>><paul.b.andersen(a)deletethishia.no> wrote: >>> > > >>>>A Doppler shifted K2 spectrum is still a K2 spectrum, >>>>and cannot be mistaken for anything else than the spectrum >>>>emitted from a star with temperature ca. 3800K. >>>> >>>>I have told you before, do you never learn? >>>>The temperature of a star is NOT determined by where >>>>the black body spectrum peaks. >>>>The spectral class is determined by the relative positions >>>>and intensities of the absorption lines, and these are >>>>unaffected by a Doppler shift. >>> >>> >>>I don't think that is quite what you wanted to say. >> >>Yes, that is exactly what I wanted to say. >>Each spectral class has a very characteristic pattern >>of absorption lines, and this pattern isn't affected >>by a Doppler shift. Their relative position and intensity >>remain the same. > > > According to the standard interpretation of the willusion. This is a FACT. This IS how the spectral class of a star is determined. >>>>The temperature is then determined by the spectral class. >>> >>> >>>I think what you meant was that if absorption lines were distinguishable and >>>recognizeable, their doppler shift would be an indication of the star's radial >>>velocity. >>>With that knowlegde, the true position of the peak of the radiation curve could >>>be established...and so could its temperature. >> >>No, I wanted to say exactly what I said, namely that when >>the spectral class is determined, so is the temperature. >>The temperature is NOT determined by the position of >>the peak of the black body spectrum. > > > I wonder why not? Of course your method is fully possible and OK. When I said it isn't done this way, I didn't mean that it _could_ not be done. But it is much more difficult to precisely determine the position of the peak of the black body spectrum than to determine the spectral class. See below. > Would it make the einsteinian based interpretation of the willusion look silly? The result would be exactly the same - but in most cases the temperature would be determined with much less precision. >>Of course the radial velocity of the star can be deduced > >>from the Doppler shift, but that was NOT the issue in this case. > >>As always you are trying to divert the attention from the issue, >>namely that a Doppler shift won't make the spectral class or >>the temperature of a star appear different. > > > I think the doppler corrected peak of the curve would be a better indicator of > temperature. Of course it is the best indicator of temperature. (You also have to correct for the reddening of the black body spectrum in the atmosphere, though). And this IS how the the temperature originally was determined. A lot of close stars were observed, the spectrum AND the temperature (determined by the peak) was measured. We have learned from these observations that the correlation between the spectral class and temperature is one. This is quite natural, because it is the temperature that determines the absorption lines. In cool M stars, there are molecules and non ionized gases, with their characteristic spectral lines. In hot O stars, there are no molecules, and all the gas is ionized. This give few spectral lines. So we can conclude that the spectral class is a very good indication of the temperature. It is generally much easier to determine the spectral class of a star, than it is to determine where the spectrum peaks. And the difference is more pronounced the fainter the star is. I will not insist that determining the spectral class by recognizing the pattern of the absorption lines is the only used way to determine a star's temperature, though. The most used way is probably to measure the colour index, also called B-V value. This is found by measuring the apparent magnitude with a blue passband filter (B), and comparing this to the apparent magnitude with a passband filter in the middle of the visual range (green-yellow) (V). The reason why this method is much used is that it is easy to do. Just take two pictures with two filters. I assume you will understand why this is a good indication of temperature. It is nicely explained here: http://spiff.rit.edu/classes/phys445/lectures/colors/colors.html This method doesn't take the Doppler shift into consideration, though. But few stars are so heavily Doppler shifted that it will affect the measurements much. There are other methods as well. Most are variations of the colour index method. But determining the position of the peak is very seldom used, simply because it is practically difficult to do with any precision. > > >>Thus you were wrong when insinuating that a spectral class >>can appear different because of Doppler > > > I think you are refering to chemical classifications rather than plain > temperature. No. A - say - G2 star can have different chemical compositions and still be a G2 star. Sure the spectra of a population I (metal rich) and a population II (metal poor) star are different, that's how we can discriminate between them. But the differences are small compared to what they have in common, so if they have the same temperature, the spectral class will still be the same. You cannot flee from the fact that you were wrong when insinuating that a spectral class can appear different because of Doppler >>>>>>And you claim that it is coming from a planet! >>>>>>You must have lost your mind completely. >>>>> >>>>> >>>>>Don't lie. >>>>>I said it was REFLECTED from the (large) planet's surface. >>>> >>>>Which is even more stupid - if possible. >>>>An A8 spectrum doesn't become a K2 spectrum if you reflect >>>>it off a planet. >>> >>> >>>It is reflecting the K2 spectrum then. >> >>Really? >>You claim the that only the B8 star exiats, and that >>the K2 star really is a planet. > > > I said that I accepted the presence of another object. > I didn't say what it might be. Fleeing again? Henri Wilson wrote: | A, I have supported you on this. In Algol's case, the WCH happens | to be the | large planet 'Androcles'. Do you have any objections to that? I had an objection to that. That's why we are having this conversation. >>So would you please explain what you meant by this comment? >> >>Of course you cannot. > > > All you have is willusory information. Like I said. You cannot. It is of course ridiculous to claim that the observed K2 spectrum is a B8 spectrum reflected off a planet. >>>>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! >>>> >>>> >>>> >>>>>Why is there a brightness INCREASE between supposed major and minor >>>>>eclipses? >>>> >>>>But is it? >>>>If we look at the curve in: >>>>http://www.macalester.edu/astronomy/research/phys40/mark/ltcurve.htm >>>>the drawn curve is pretty much fantasy. It is the points that >>>>are the measured values, and the spread reflects the precision >>>>of the measurements. And all we can conclude from those points >>>>is that the light curve is roughly flat between the primary >>>>minima, and any secondary minimum is hard to see. >>>>We now know why. >>> >>> >>>I know why. Androcles knows why. >>> >>>It is distinctly downwardly concave between the two major dips. >>>That concavity has been mistakenly identified as a secondary eclipse. >> >>Can you show me the light curve which >>is "distinctly downwardly concave between the two major dips."? >> >>Of course you cannot, because it only exists in your imagination. >>The real observed visible light curve is like this: >>http://www.macalester.edu/astronomy/research/phys40/mark/ltcurve.htm >>If you look at the measuring points, and not the drawn curve, >>it is clear that there is no "distinct concavity" in the light curve, >>and the second minimum is not really visible at all. > > > This is what it really looks like: > www.users.bigpond.com/hewn/alg2.jpg Thanks for confirming my words. You cannot show the light curve which is "distinctly downwardly concave between the two major dips", because it only exists in your imagination. >>>>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 >>>> >>>>If this long query doesn't make it through, try this one: >>>>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 we can conclude that the observed eclipses are as expected. >>> >>> >>>No. >>>We can only conclude that your whole argument above involves circular >>>reasoning. >>> >>>The parameters of the 'two stars' are largely based on your so called 'eclipse >>>depth'. >>> >>>You - and they - have used the parameters of the illusion to justify the >>>illusion. >> >>You pretend not to to get the point, do you? >>The conventional explanation predicts that the light curve >>should be different at different wavelengths. >>The second minimum is observed in the infra-red at 10 um. >>It is exactly as it should be according to the conventional explanation. > > > The conventional explanation is based on what is observed in the willusion. You do understand how stupid this answer is, don't you? Look at the light curve in this again: 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. Can you please explain in what way this light curve is illusory? >>The ballistic theory does NOT predict such a difference. >> >>The ballistic theory is thus falsified. >>Again. > > > The ballistic theory WILL always predict what is observed. Isn't it rather stupid to keep asserting what is proven false? > However it involves a great deal of trial and error as well as some initial > speculation about what might be really happening. It also provides opportunity > for discovery. > > I don't think we can model the rest of the universe on our own solar system. You are babbling. You know very well that the ballistic theory does not predict a frequency dependent light curve. But the light curve of Algol IS frequency dependent exactly as predicted by conventional theory. The ballistic theory is falsified. Again. Paul |