Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: George Dishman on 24 Oct 2005 06:42 <HW@..> wrote in message news:li1ol19gggtqml7m0rdic4srciltnvfjkn(a)4ax.com... > On Sun, 23 Oct 2005 08:51:29 +0100, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: > >> >>"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> > >>>>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. > > George, the ripples are moving in space at a certain time rate. GR predicts the ripples move at the speed of light. >>> 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. > > George there is no BEAM moving up one particular diagonal. There is ONE > INFINITESIMAL ELEMENT of the beam moving up each diagonal. > The Diagonal is merely a line showing the path of one such element. Right, so when you want to know the length of the path, it is the length of that diagonal. That's the basis of Einstein's illlustration. > A line has > no thickness. It cannot constitute light or anything else. See below. >>> 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. > > That's because the upper plot of the beam is in the moving frame. The > lower one > shows the beam in the rest frame. Right, that's why I say your program illustrates that Einstein was correct. >>> 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. > > That is wrong George. No, it's right, the "conventional view" relates to the path taken, not the orientation of elements. In fact when discussing it I usually describe the source as a photographic flash bulb to make it clearer. > The conventional view ignores the fact that the ends of the 'elements' are > emitted at different times, during which the source moves along a little. > All the elements of the beam remain aligned vertically in both frames. > That is what the program is intended to show and it DOES just that. Yes, if you had a series of flashes, that would be true, but the illustration relates to the path length for a single flash so that is not contrary to the conventional view. > The vertical green dashes could be taken as elements of finite lengths if > you > like. Indeed, consider the light from a car indicator. You would have elements with equal length gaps between them. >>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. > > No George, the program is correct. You will have to study it a bit more > closely. <from above> > A line has > no thickness. It cannot constitute light or anything else. Indeed, I made a mistake. You are right about the orientation of the element, the mistake is that the entire element sweeps out an area, the diagonal line trailing behind the element should be wider. >>> 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. > > He said each element moves at c. Correct. Each element in your diagram or the flash from the bulb as I put it moves at the speed of light, because it is light. > Why should it? It isn't a light beam. It isn't anything. It certainly is > not > governed by maxwell's equations. Of course it is Henri. How can you say the light from a laser isn't light? That's just bizarre. > Einstein would be correct if an ether existed. > >>> 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. > > No. they aren't even that. > They are merely loci of points...lines on a graph...with no physical > properties > at all. The locus has a key property - length. >>>>>>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. > > I get exactly the conventional explanation. Light always moves at c from > the > source to where the mirror will be when the beam gets there. > the path length is longer in one direction than the other. In that case you haven't actually tried to do it, you are just saying you did. >>>>> 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. > > Even Paul Andersen agrees that light leaves its source at c (relative to > the > source). Sorry Henri, I don't respond to accusations of what other people might have said. I know Paul understands SR and you admit you don't so I'm not surprised you don't follow him. > Are you seriously suggesting that it leaves at some other speed? Light will be measured to move at c in any inertial frame, that's been known for a century now. >>>>> 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). > > Well that is wrong. That's what Ritz predicts, I agree Ritz is wrong. > It always travels at c relative to the (moving) point representing the > centre > of the next mirror. Yes, that's correct and follows from the above since the next mirror is moving at mv relative to the lab frame. > The path lengths are differnet in either direction ..just as in the > conventional explanation. Right, in the lab frame, but the path length difference matches the speed difference hence to propagation times are the same in Ritz, hence a null prediction >>> 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. > > The source revolves around the first mirror. Exactly, not the centre of the table. > George the mistake you have been making all along is that you have > calculated > the light speed towards the mirror as c + v/root 2. You have ignored the > movement of the mirror. c + v/root 2 is the speed wrt the point where the > mirror was at the time of emission. > The speed of the beam relative to the moving mirror AS IT REACHES the > CENTER OF > the DISPLACED mirror, is always exactly c. Yes, that is exactly what I showed in this diagram: http://www.briar.demon.co.uk/Henri/speed.gif hence there is no "kick" in the speed. You finally agreed with me :-) > I have now proved conclusively that Sagnac does NOT refute the BaTh. No, you have now proved that it predicts a null result. The speed is c and the path in that frame is unchaged by the rotational speed too. >>> 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. > > What I have stated hasn't sunken in to you yet. What you have stated is exactly what I drew on that diagram and it is dated 1st Feb, 2004. You finally caught up! >>... No matter how much hand- >>waving you try, Ritz gives a null prediction which >>contradicts the observed result. > > Ritz gives exactly the same prediction as the conventional one. Nope, you just proved it gives a null prediction. George
From: Paul B. Andersen on 24 Oct 2005 08:17 HW@.. wrote: > On Sun, 23 Oct 2005 15:46:51 +0200, "Paul B. Andersen" > <paul.b.andersen(a)deletethishia.no> wrote: > > >>Henri Wilson wrote: >> >>>On Fri, 21 Oct 2005 13:04:44 +0200, "Paul B. Andersen" >>><paul.b.andersen(a)deletethishia.no> wrote: >>> > > >>>>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. > > > I used to consider that..but soon realised that it cannot be wrong. The point is that you now have admitted that the BaT predicts what the light curve should look like, despite the fact that you called this light curve a "willusion". But according to you, the BaT predicts a light curve quite different from this: 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. Wrong prediction -> theory falsified. BaT is falsified. > > >>>> >>>>>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. > > > Most - ney ALL - stars rotate. Indeed they do. > Are you able to run my program? It takes only about five seconds to download. > You will enjoy it. > A picture speaks a thousand words. I do not need a picture to understand that different parts of a rotating star have different radial velocities. > >>>>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. > > > Can you not see that the light from the edges would be doppler shifted both > ways. It the star was rotating fast enough that would broaden the radiation > curve away from precise black body. You have no sense of proportions, have you? Of course a rotating star will broaden the black body spectrum, but only to such a small degree that it never will be detectable. As you so correctly said above, all stars are rotating. And all stars radiates a black body spectrum. The broadening of the BB spectrum due to the stars rotation is never observable. >>>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? > > > You still haven't run the program. I have. Different parts of the star have different radial velocity _because the star is 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. > > > Bull. I think you know that the BaT predicts the same light curve for 10um as for visible light. >>Another falsification of the BaT. > > > in your dreams. In the real world. Paul
From: bz on 24 Oct 2005 08:59 HW@.. wrote in news:li1ol19gggtqml7m0rdic4srciltnvfjkn(a)4ax.com: > I suppose one could talk about the point where moving scissor blades > meet... Is IT distinguishable from nothing? > > Finally, you mention a phenomina that CAN move faster than light. Take two intersecting laser beams and move the point of intersection by moving them like a pair of scissor blades. The point of intersection CAN move at any velocity you like. -- 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: bz on 24 Oct 2005 09:09 HW@.. wrote in news:li1ol19gggtqml7m0rdic4srciltnvfjkn(a)4ax.com: > The conventional view ignores the fact that the ends of the 'elements' > are emitted at different times, during which the source moves along a > little. All the elements of the beam remain aligned vertically in both > frames. That is what the program is intended to show and it DOES just > that. > > Henri, you and I have discussed this before. If the ends of the photon are emitted at different times, they will be in different locations relative to the vertical in the other frame. They appear to be 'slanted' just as the beam appears to be slanted. Think of shooting a harpoon (with a trailing line) at a whale that is off the portside of a moving ship. The harpooner has been leading the whale and fires at the moment that the gun is exactly perpendicular to the side of the ship and to the motion of the ship. Neglect air resistance and gravity. To the viewers on the ship, the harpoon appears to move in a straight line, perpendicular to the course of the ship. To the viewers floating in a 'stationary' rowboat, near the point where the whale and harpoon are about to meet, the harpoon and line follow a diagonal. -- 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: bz on 24 Oct 2005 09:20
HW@.. wrote in news:8o4ol1tufvl4t3q63in7e3melrpcja450g(a)4ax.com: > On Sun, 23 Oct 2005 15:33:24 +0000 (UTC), bz > <bz+sp(a)ch100-5.chem.lsu.edu> wrote: > >>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. > > Your argument is not related to the problem geesey raised and I > answered. So? I raised a new point(tangentially related). You fail to refute it. > The question is, "if someone emits a pulse of light, can they ever catch > up with it?" The question is 'can matter ever travel fast enough to catch photons'? In a SR/GR/EEP universe, the answer is 'no'. In a BaTer's universe, the answer must be 'yes'. But you appear to say it is NO. This is a logical contradiction. BaT can't fly. > I have replied by saying NO. OK in SR, bad for BaT. > Even if all the person's MC^2 energy was > available to accelerate the last molecule, it could never reach it. I agree. OK in SR, bad for BaT. > Try rocket propulsion theory....remember the exhaust accelerates too, > mainly in the same direction as the pulse. ????. The exhaust accelerates in all directions, the rocket chamber confines the accelerated paricles and only allows them to escape in the direction opposite to the light pulse you are trying to catch. -- 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 |