Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: Henri Wilson on 21 Nov 2005 19:47 On Mon, 21 Nov 2005 23:48:23 +0100, "Paul B. Andersen" <paul.b.andersen(a)hiadeletethis.no> wrote: >Henri Wilson wrote: >> On Thu, 17 Nov 2005 16:54:34 +0100, "Paul B. Andersen" >> <paul.b.andersen(a)hiadeletethis.no> wrote: >> >>>>>Look Henri. >>>>>Pick a distant star, say 500 LY away. >>>>>Point your telescope at it, so that the image is at the centre. >>>>>Measure the absolute angle of your telescope. >>>>>Repeat 6 month later. >>>>>The telescope will now point in a direction 22 arcsecs >>>>>different from the first time. >>>>> >>>>>The parallax is negligible. The light path is the same, >>>>>nameley a straight line from the star to the Earth. >>>>>So why are the angle of the light path different? >>>>>It is caused by the different velocity of the Earth >>>>>at the two occations. We are observing the star from >>>>>two different frames of reference. >>>>>Both frames are moving relative to the star. >>>> >>>> >>>>So what? Ligth leaves the star spherically. >>> >>>Still drunk? >>>Didn't you get it? >>>There is but one light path - the path from the Star to the Earth. >> >> >> Are you under the impression that the star is emitting all its light in one >> particular direction, as with a narrow laser beam? > >Still drunk? >Or are you too stupid to get the simple point? > >The light that does not hit the CCD in our telescope is of no >interest whatsoever. It might as well have been a laser beam, >it would make no difference. >(The width of the beam would have to be at as wide >as the parallax angle, though. But as this is much smaller >than the aberration angle, we can forget it.) > >Ignore the parallax. >We can in our thought example imagine that the star >is in the equatorial plane, and that we are observing >it when the Sun, Earth and star are all in line. >(That means that the star will be behind the Sun >at one occasion, but it doesn't matter in principle) > >Then we have the picture drawn in solar frame: > > * star > | > | the one and only > | light path > | > | > O-> v observer > | > S > | > v<-O > >At one occasion, the observer is moving to the right, >an will have to point his telescope like this: > > / > / light path down the tube > / > ------ >Six month later, the observer is moving to the left, >and will have to point his telescope like this: > > > \ > \ light path down the tube > \ > ------ > >At both occasions, the angle is v/c radians from vertical. >v/c = 10^-4 radians = 11 arcsecs > >The difference is 22 arcsecs. I'm not disputing the fact tat you have to tilt the telescope...but you are very confused concerning the path of the beam. It remains vertical. >>>(We can neglect the small parallax angle which is only 0.0006 of >>> the aberration angle) >>>It is obviously utterly irrelevant that the star emits light >>>in all other directions that don't hit the Earth, >>>so why the hell are you stating this stupidity? >>> >>>The only light path of interest is the one that hits the Earth! >>>The _direction_ of that light path is down the middle of >>>our telescope. >> >> >> You really are funny today. >> >> >>>The direction of that single light path changes throughout the year >>>because the velocity of the frame of reference (Earth) changes >>>throughout the year. >> >> >> Very good Paul. You are improving. >> >> >>>>This is in no way related to our discussion. You are diverting attention from >>>>the fact that SR is proved to be nonsense. >>> >>>It is related to your incredible stupid statements: >>>"Wavefronts really exist only in the source frame." >>>and: >>>"Whatever is moving diagonally isn't light. It is >>>infinitesimal points." >> >> >> That is right. Of course I was refering to the plotting, in my frame, of the >> paths of individual 'points' inside a vertical laser beam as I move >> horizontally past it. > >You are evading the point, Henri. >There is no important difference from your vertical beam. >The observer is moving perpendicular to the light beam from >the star that hits the Earth. That light path from the star is >"vertical" in our "stationary" solar reference system, >which is the same as the "source frame". >But the light path that is going down the observer's >telescope is at an angle v/c radians from vertical. That's what a raw amateur would say. The tilted telescope has to move sideways to keep the incoming vertical beam at its centre. The beam, as a whole, is v/c radians from the telescope tilt... which means it is actually vertical...but moving sideways in the observer frame. Do you know how to move a vertical object sideways Paul? >So what is it that is moving at an angle down the tube >and hitting the CCD? Is it light, or is it just infinitesimal >points? How can the CCD detect infinitesimal points? It is vertical light moving down the sideways moving tilted tube. That should be obvious to all but raw beginners. >>>"WHATEVER IS COMING FROM THAT STAR, MOVING WITH CHANGING >>> DIRECTION, IS NOT LIGHT, IT IS ONLY INFINITESIMAL POINTS." >> >> >> You are quoting me completely out of context and you know it. >> tat has nothing whatsoever to do with the topic or anything I have said. > >No, I am not. >You claimed that the light is moving vertically in the "source frame", >and "whatever is moving diagonally in the observer frame is not light." that's correct. And your telescope example shows that I am right. >The light from the star IS moving vertically in the source frame. >But our observer have to point is telescope "diagonally" to make >the "whatever that is moving" go down his telescope tube. > >So what is the "whatever" that is moving diagonally down the tube? You are forgetting that the tube is tilted...and you are now making a complete fool of yourself. Try this: ^ v L t======.............................................................u<-b.......---O You are at O with a gun. You want to fire a bullet radially at velocity u, at a target (t) that lies at the centre of a pipe (L), which is spinning around you at v. Q1) Is it possible to hit the target if the pipe remains aligned radially? What if you tilt your gun wrt the pipe axis? Q2) If not, at what angle from the radius vector should the pipe be tilted? Q3) How is the concept of 'vertical' defined for you or the pipe? > >You claim it is not light. So what is it then? An infinitesimal element of the vertical beam. >Whatever it is, CCDs detects it as if it were light. Because the CCD is moving sideways so that it will continuously intercept an infinitesimal element of a DIFFERENT section of the vertical beam in successive infinitesimal time intervals. > >>>Now Henri, what is it that hits the CCD in our telescope? >>>Is it light? >>>It cannot be, can it? >>>Because whatever moves along paths with different directions >>>in different frames cannot be light, can it? >> >> >> The star emits a sphere of light Paul. The wavefronts are spherical. Didn't you >> know that. >> When my telescope moves sideways, a different radius vector of the sphere goes >> down the middle of my telescope. What could be more simple? I cannot see why >> you should have any trouble understanding that. > >Please, Henri. >You are not really THIS stupid, are you? :-) A raw beginner might think I am stupid...just as a raw beginner like Einstein would think that raindrops fall diagonally when he looks at them through a moving train window. > >(I am beginning to believe you are. > I didn't think it possible!) > >Don't you understand that if you had two telescopes going in >opposite direction, looking at the same star, they would have to >point in opposite directions as they pass each other? > > * > | > 100 LY > | > / \ > / \ > O->v<-O Yes. That is because the beam is vertical wrt earth's surface and THEY are moving sideways. > >It is the very same one and only light path that has different >angles in the two observers' frames. You fail to see that the successive elements that strike the centre of the telescope do not come from the same infinitesimally narrow section of the vertical beam. I accept that this might be a bit hard for you. >A little movie for you. The "*" is a photon, or a bit of LIGHT. >Two telescope tubes going through each other in opposite directions. > > > enter X * X > tubes / \ / \ > / X \ > / / \ \ > / / \ \ > -ccd- -ccd- > > \ / \ / > X * X > going / \ / \ > down / X \ > / / \ \ > -ccd- -ccd- > > \ X / > and \ / \ / > down X * X > / \ / \ > / X \ > -ccd-ccd- > > \ \ / / > \ X / > \ / \ / > X * X > / \ / \ > -ccccd- > > \ \ / / > \ \ / / > \ X / > \ / \ / > X * X > -ccd- > > \ \ / / > \ \ / / > hit \ \ / / > ccds \ X / > \ / \ / > Xc*dX > >Our single photon - or piece of LIGHT - goes >down both tubes. This single photon has obviously >but one path. This single light path is vertical >in the screen frame, AND the very same light path >is "diagonal" in the two telescope frames. ....which proves the photon as moved vertically wrt Earth. Throw away the Earth and put the telescopes on oppositely moving vehicles. > >The photon - or piece of light- is moving vertically >in the screen frame, AND the very same photon - or piece >of light- is moving diagonally in the telescope frames. ....and the telescope is tilted by the same diagonal in the Earth frame. > >It is obviously mindless babble to claim that the very >same physical object both is and isn't light. The CCD continuously detects an infinitesimal element of differnet light beams from the star in each infinitesimal time interval. There is an infinite number of them. > >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: Jeff Root on 22 Nov 2005 00:50 Paul B. Andersen wrote: > Please, Henri. > You are not really THIS stupid, are you? :-) And I don't believe you are as stupid as you seem. Henri's intelligence has dropped significantly compared to what it was several weeks ago. At the same time, his own estimation of his intelligence and confidence in his abilities have increased. I take these as indications that he is in the manic phase of bipolar disorder. Arguing with him is like arguing with a dog over who gets a bone. Logic and reason are irrelevant. Henri believes whatever he wants to believe. He has no control over what he wants to believe. His beliefs are driven by his mental illness, not by intelligence. Once the manic phase has ended, his intelligence will undoubtedly return to its former level. He will probably also feel like a complete failure again, as depression takes over. > A little movie for you. The "*" is a photon, or a bit of > LIGHT. Two telescope tubes going through each other in > opposite directions. Very nice illustration! Henri replied to it: > You fail to see that the successive elements that strike > the centre of the telescope do not come from the same > infinitesimally narrow section of the vertical beam. If you really want to continue arguing with a mental illness, you need to compare the varying angle of the telescope relative to the angular diameter of the star. Henri thinks the telescope is pointing at different parts of the star over the course of a year, although he doesn't realize that that is what he thinks. He's right that the light paths are different at different times of year, but doesn't see that those different paths can't account for the 22 arcsecond difference in angle of the telescope. Unfortunately, he can't follow quantitative arguments, so your best bet would be to put the numbers in a geometric construction. Of course, his mental illness will still reject it anyway. -- Jeff, in Minneapolis
From: Androcles on 22 Nov 2005 02:49 "Jeff Root" <jeff5(a)freemars.org> wrote in message news:1132638605.644679.35580(a)g43g2000cwa.googlegroups.com... > > Paul B. Andersen wrote: > >> Please, Henri. >> You are not really THIS stupid, are you? :-) > > And I don't believe you are as stupid as you seem. > > Henri's intelligence has dropped significantly compared > to what it was several weeks ago. At the same time, his > own estimation of his intelligence and confidence in his > abilities have increased. I take these as indications > that he is in the manic phase of bipolar disorder. > > Arguing with him is like arguing with a dog over who gets > a bone. Logic and reason are irrelevant. Henri believes > whatever he wants to believe. He has no control over what > he wants to believe. His beliefs are driven by his mental > illness, not by intelligence. > > Once the manic phase has ended, his intelligence will > undoubtedly return to its former level. He will probably > also feel like a complete failure again, as depression > takes over. > >> A little movie for you. The "*" is a photon, or a bit of >> LIGHT. Two telescope tubes going through each other in >> opposite directions. > > Very nice illustration! > > Henri replied to it: > >> You fail to see that the successive elements that strike >> the centre of the telescope do not come from the same >> infinitesimally narrow section of the vertical beam. > > If you really want to continue arguing with a mental > illness, you need to compare the varying angle of the > telescope relative to the angular diameter of the star. > Henri thinks the telescope is pointing at different parts > of the star over the course of a year, although he doesn't > realize that that is what he thinks. He's right that the > light paths are different at different times of year, but > doesn't see that those different paths can't account for > the 22 arcsecond difference in angle of the telescope. > > Unfortunately, he can't follow quantitative arguments, so > your best bet would be to put the numbers in a geometric > construction. Of course, his mental illness will still > reject it anyway. > > -- Jeff, in Minneapolis Kin I see the movie, kin I, kin I? Go on, show me the movie. I wanna see the movie.... wahhhhh..... I wanna see the movie! Androcles.
From: jgreen on 22 Nov 2005 04:34 Henri Wilson wrote: > On Sat, 19 Nov 2005 10:18:10 -0000, "George Dishman" <george(a)briar.demon.co.uk> > wrote: > > > > >"Henri Wilson" <HW@..> wrote in message > >news:7tgsn11flp2iggll8is15lrdpqck37voaj(a)4ax.com... The sagnac is "zeroed" to read the fringe at 12 o'clock, with the airframe travelling north, OK? When the plane alters course (to say NE), the fringe is logged at (say) 2 o'clock, as an accelleration (in rotation of the craft) took place. Now the observed fringe shift will REMAIN at 2 o'clock, UNLESS a reverse rotation takes place. It was the change of direction (read rotational accelleration) which caused the alteration to the shift position. I predict that, in ACCORDANCE with c'=c+v, while the plane maintains the NE heading, there will be no FURTHER shift (2 o'clock maintained) The end Jim G c'=c+v (G'day George---------just watching :-) )
From: George Dishman on 22 Nov 2005 04:51
"Henri Wilson" <HW@..> wrote in message news:9vj2o15r5gf1eaenenj9ddjtggsrlmn2fe(a)4ax.com... > On Sat, 19 Nov 2005 10:18:10 -0000, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: > >> >>"Henri Wilson" <HW@..> wrote in message >>news:7tgsn11flp2iggll8is15lrdpqck37voaj(a)4ax.com... > >>>>In that case you are wrong if I understand your diagram. >>>>When the rotation is at constant speed, the path length >>>>is also constant at the length of line AD or roughly: >>>> >>>> L' = L + v * t / sqrt(2) >>>> >>>>When the rotation is at constant acceleration, the path >>>>length is also constant at the length of line AE or >>>>roughly: >>>> >>>> L' = L + (v * t + a * t^2 / 2) / sqrt(2) >>>> >>>> _________ >>>> / >>>> / ^ >>>> __________/ | speed >>>> | >>>> ______________________ >>>> >>>> ------> >>>> time >>>> >>>>Your diagram predicts there would be an output like >>>>this: >>>> >>>> __ >>>> | | ^ >>>> | | | output >>>> | | | >>>> | | >>>> __________|__|________ >>>> >>>> ------> >>>> time >>> >>> No it doesn't. >>> >>> Assume that fringe movement is +ve for +ve acceleration and -ve for -ve >>> acceleration. >>> There is no -ve acceleration in the above diagram showing a +ve speed >>> change.. >> >>> Therefore fringe displaement moves in only one direction during the >>> acceleration period. >> >>The fringes ARE DISPLACED in only one direction during >>theacceleration period as shown by your diagram. When >>the acceleration is zero, the factor a t^2 / 2 is zero >>so there is zero displacement, not zero rate-of-change >>of displacement. >> >>> It doesn't suddenly revert ot zero when acceleration ceases. >> >>Your diagram says it does, if 'a' is zero then the >>length difference is zero. >> >>> SO: >>> _________ >>> | ^ >>> | | output >>> | | >>> | >>> __________|__________ >>> >>> ------> >>> time >>> >>> Is the correct result. >> >>Let's add a negative acceleration period: >> >> _________ >> / \ >> / \ ^ >> __________/ \______ | speed >> | >> _______________________________ >> >> ------> >> time >> >>Your diagram predicts there would be an output like >>this: >> >> __ >>+ve | | ^ >> | | | output >> | | | >> 0 ________|__|__________________ >> | | >> ------> | | >> time | | >>-ve |__| >> >>This is basically a graph of the "a t^2 / 2" factor. > > How are you defining 'output'. > > I'm using 'output' to mean 'fringe displacement' from the central > (non-rotating) position. Maybe you are using it as rotation angle. Well strictly it would be the voltage out of the equipment but that is derived from the photodiode and that is really the same as fringe displacement. > In either case, I think I disagree with you. Possibly, but below you say "Yes, I suppose that its right. It is proportional to the path length difference." and later "yes, something like that." so I don't see how you disagree. This next paragraph appears to be in conflict with those later agreements. You need to clarify this for me because if "the fringe displacement is a t^2 / 2 as you show in your diagram." then the fringes are _displaced_ while a is non-zero, i.e. "during acceleration periods" rather than moving. George > Fringes move only during acceleration periods. > If a +ve acceleration is followed by an identical -ve one, the rotation > speed > is back to where it was....and so is fringe displacement. > >>>>There is no method to provide physical integration >>>>because the number of wavelengths in the path does >>>>not affect the time difference between wavefront >>>>arrivals in the two beams which is what produces >>>>the output. >>> >>> The 'change in fringe displacement' is effectively an integration of the >>> path >>> length increase during the acceleration period. >> >>No, the fringe displacement is a t^2 / 2 as you show >>in your diagram. > > Yes, I suppose that its right. It is proportional to the path length > difference > of the two beams.. > >> >>>>Actually I think your diagram is oversimplified but >>>>we can go with it for the moment, it is close enough. >>> >>> It shows what happens during a constant acceleration. In practice, >>> acceleration >>> would vary with time. >> >>Indeed. I can't show a quadratic start and end to >>each period of acceleration but if I could the >>resulting output would look like this: >> >> +ve __ >> / \ >> / \ >> / \ >> 0 ___/ \_______ _____ >> \ / >> \ / >> \ / >> -ve \__/ >> >>Since the acceleration is changing, you now have to >>understand the 'a' in your diagram to be the mean >>acceleration during the flight time. > > yes, something like that. |