Seeking a correct explanation for Stellar Abberation
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From: Henry Wilson DSc on 12 Apr 2010 17:06 On Mon, 12 Apr 2010 16:11:12 +0200, "Paul B. Andersen" <paul.b.andersen(a)somewhere.no> wrote: >On 08.04.2010 23:14, Henry Wilson DSc wrote: >> On Thu, 08 Apr 2010 15:03:02 +0200, "Paul B. Andersen" >> <paul.b.andersen(a)somewhere.no> wrote: >> >> Your use of the word 'trajectory' is unfortunate. >> >> >> Paul, let me explain. >> >> Consider an object that is moving with random velocities along a laser beam. >> >> Does its trajectory ever deviate from a straight line? >> >> If your answer is 'NO', one must assume you have discovered a very simple way >> to bend a light beam. >> >> All one has to do is move perpendicularly to the beam...and it bends.... > >Good grief, Henry. >You never get tired of making a fool of yourself, do you? :-) > >You throw a ball with a horizontal velocity component v (in ground frame). >Is the trajectory in the ground frame a parabola, Henry? It is indeed. >(If you don't know what 'a trajectory' is, look it up.) > >A laser with a vertical beam is moving along the ground under the ball >with the speed speed v, so that the beam always hits the ball. hahahahhhahaha! .......a kind of tracking device....the laser has to be rotated continuously. Sorry Paul...differrent problem entirely... >Is the ball moving along the laser beam, Henry? > >Bending a laser beam is really simple, isn't it Henry? :-) If you rotate the laser it, kind of, 'is'. .......but in my experiment the laser is not rotated_ is it Paul...and we know that rotation is absolute_ don't we Paul. >Because nothing can move in a straight line in one frame but a >parabola in another. Very true... you occasionally make a correct statement...surprising really! >Or can it, Henry? :-) > >Your stupidity has ceased to amaze, but it is still amusing. :-) .....'rotate the laser'.....hahahahhahhahahhhahaaaaahhaha! Thanks for the laughs, Paul. "If the straightness of the laser beam produces the wrong answer, then just rotate the laser"___Andersen, 2010.. Henry Wilson... ........A person's IQ = his snipping ability.
From: Henry Wilson DSc on 12 Apr 2010 17:46 On Mon, 12 Apr 2010 15:21:33 +0200, "Paul B. Andersen" <paul.b.andersen(a)somewhere.no> wrote: >On 09.04.2010 23:09, Henry Wilson DSc wrote: >> I will simplify my question even further. > >OK. >This (what is happening when we increase the length of the pole) >is actually an interesting problem, so I will respond. > >(Your scenario isn't a simplification, it is much more complicated > than you in your naivety think.) I told you days ago that I know the answer. >> You and your telescope are standing on a small nonrotating asteroid that is >> orbiting the sun with a period of 1 year. Next to you is a short vertical pole >> on top of which is bright object. You point your telescope vertically so the >> object appears in the centre of your eyepiece. > >Let us add that the orbit is the same as the Earth, >and let us call the asteroid 'the Earth'. >The pole is perpendicular to the orbital plane (ecliptic), >so it is pointing towards the ecliptic North pole. >Let us call the bright object on top of the pole 'the star'. >We will call the direction towards the ecliptic pole 'vertical'. >Let v be the orbital speed of the Earth. > >> Q) DOES YOUR TELESCOPE HAVE TO BE TILTED AT DIFFERENT ANGLES AS YOU AND THE >> OBJECT ORBIT THE SUN (to keep the object in view)? >> >> A) No of course not. > >Close enough, if the pole is short. >Not exactly right, you would have to tilt the telescope a bit, No I don't. >but so little that it is many orders of magnitude less than what >can be measured. > >> >> Next, increase the length of the pole to 1 light minute. > >Same answer. >The angle is unmeasurable. >We see the 'star' on the top of the pole at the ecliptic north pole. > >Consider the following carefully. >It may seem obvious, and unnecessary to view it in the Sun-frame, >but it will be very important later: >In the non-rotating Sun-frame, the light that hits the telescope now, >has an angle arctan(v/c) ~= v/c in the Sun-frame. >But this is exactly equal to the aberration between the Sun-frame >and the Earth-frame, so the telescope will have to be vertical. > >Drawn in the Sun-frame: >----------------------- > > *' = position of star when light was emitted > * = position of star now > O = Earth now > \ = path of the light that is received now > > *' * -> v > \ > \ > \ > O -> v > >The light is entering the telescope from 'the left', >but since the telescope is moving to the right, >aberration will be v/c towards the right. >The net effect is that the telescope must be vertical. > >> >> Again: >> Q) DOES YOUR TELESCOPE HAVE TO BE TILTED AT DIFFERENT ANGLES AS YOU ORBIT THE >> SUN (to keep the object in view)? >> >> (I await Paul's answer) >> >> Next, increase the pole's length to precisely 1Ly and answer the same question. > >Now the telescope will have to be tilted an angle v/c toward the right. >The light that is observed now was emitted a year ago, when the 'star' >was in the same position as it is now when it has made exactly one orbit >since the light was emitted. >But NOW IS THE PATH OF THE OBSERVED LIGHT VERTICAL IN THE SUN-FRAME: > >Drawn in the Sun-frame: >----------------------- > > * -> v (star is now back after having made a full orbit) > *' -> v (star where it was a year ago when light was emitted) > | > | > | > | > O -> v (Earth now) > >The light is entering the telescope 'vertically', >but since the telescope is moving to the right, >aberration will be v/c towards the right. >The net effect is that the telescope must be tilted v/c to the right. That does not tie in with your previous diagram in which you said that the telescope has to remain vertical. The emission point *' should still be to the left, as before. You are contradicting yourself. >> How about 100LYs? > >The telescope will always have to be tilted v/c in the direction >of the orbital velocity, which half a year later is in the opposite >direction. The star will appear to move in a circle around the ecliptic >north pole with radius v/c. > >The reason for this is as I now have explained over and over. You have put forward two conflicting theories. >BUT what happens when the length of the pole is gradually increased >is in fact quite complicated. That's why I phrased the question in a simple form. You couldn't even get that right. >Where will we see the star when the distance (length of pole) is 1/2 LY? > >Drawn in the the Sun-frame in a plane perpendicular to orbital velocity: >------------------------------------------------------------------------ > > |< 2AU >| > * *' > / > / > / > / > / > / > O > >The star will appear to be an angle 2AU/0.5LY = 13 arcsecs >off the vertical, towards the Sun. > > >Drawn in the the Sun-frame in a plane parallel to the orbital velocity: >-------------------------------------------------------------------- >As for 1 LY. > >The net effect is that the telescope must be tilted >v/c = 20.5 arcsecs in the direction of the orbital velocity, >and 13 arcsecs toward the sun. > >If we increase the distance (length of pole) to 1.5 LY, the angle >towards the Sun will be 2AU/1.5LY = 4.3 arcsecs. > >So to sum it up: >When we increase the length of the pole, the star will appear to >move away from vertical, one component will be in the direction >of the orbital velocity, and one component will be towards the sun. ...not bad... >The former component will vary a bit with the distance, it will be >exactly v/c = 20.5 arcsecs every odd number of half light years, >but be a bit less under whole numbers of LYs (when the apparent >position is 'behind' the true position) and a bit more above whole >numbers of LYs. The variation will be less with the distance, >for 100+ LY, the variation will be negligible. > >The latter component will have a maximum every odd number of half >light years, and be zero for whole number of light years. It will be >smaller as the distance increases, and for 100.5LY, it will be >a mere 0.062 arcsecs. > >So we can conclude that for distances 100+ LYs, the star at the end >of your pole will move in a circle with 20.5 arcsecs radius around >vertical. Sorry Paul, you haven't analysed the problem correctly or given the right answer. I'll give you a clue. """"If the star emits a laser beam, at what precise direction does the laser have to be pointed for the beam to strike my telescope? At what angle does my telescoope have to be set so the beam will go straight down the centre? > >I have no doubt that you are as confused as ever, >but it was an interesting problem, and there may be >a lurker out there who has learned something. It IS and interesting question....one that also has an interesting answer...but you haven't been able to find it. >But you haven't, Henry. >Right? :-) Henry Wilson... ........A person's IQ = his snipping ability.
From: Henry Wilson DSc on 12 Apr 2010 17:50 On Mon, 12 Apr 2010 13:25:00 -0700 (PDT), GogoJF <jfgogo22(a)yahoo.com> wrote: >On Apr 6, 1:53�pm, GogoJF <jfgog...(a)yahoo.com> wrote: >> On Mar 6, 7:19�pm, train <gehan.ameresek...(a)gmail.com> wrote: >> >> > According to the special theory of relativity, the aberration only >> > depends on the relative velocity v between the observer and the light >> > from the star. >> >> >http://www.mathpages.com/rr/s2-05/2-05.htm >> >> > "relative velocity v between the observer and the light from the >> > star." >> >> > Whic is always c , right? >> >> Looked on Wiki about "Aberration of light"- and printed it out in 12 >> pages. �On page 3, it talks of "apparent and true positions". �It uses >> the well known diagram illustrating stellar aberration. �Along with >> the diagram, it supposes that the star is sufficiently distant, so >> that all light from the star travels in parallel paths to the Earth >> observer, regardless of where the Earth is in its orbit. >> >> And this is where the problem starts: >> >> Wiki: �On the left side of figure 1, the case of infinite light speed >> is shown. �S represents the spot where the star light enters the >> telescope, and E the position of the eye piece. >> >> Gogo: �Then wiki very subtly poses this question: >> >> Wiki: �If light moves instantaneously, the telescope does not move, >> and the true direction of the star relative to the observer can be >> found by following the line ES. �However, if light travels at finite >> speed, the Earth, and therefore the eye piece of the telescope, moves >> from E to E' during the time it takes light to travel from S to E. >> Consequently, the star will no longer appear in the center of the eye >> piece. �The telescope must therefore be adjusted. >> >> Gogo: �But the truth of the matter, there is no adjusting- there is >> nothing that happens or exists which would suggest that any observable >> or instrumental change has been performed, other than by idea, belief, >> or faith that it was and is done this way. �Aberration of light is a >> purely man-made concept! >> >> Wiki: �Consequently, the star will no longer appear in the center of >> the eye piece. >> >> Gogo: �This is complete baloney. �This is pure theory- science >> fiction. �In practice, this does not accur. >> >> Wiki: �The telescope must therefore be adjusted. >> >> Gogo: �Again, there is no adjusting going on here. �This happens only >> "what if" light was finite. >> >> Wiki: �When the telescope is at position E it must be oriented toward >> S' so that the star light enters the telescope at spot S'. >> >> Gogo: �Spot S', do you see how ridiculous this sounds, logically? >> Again, this is 100% one way or 100% the other. �Either light is >> instant, or it is finite. �This illustration or principle is all or >> nothing. �There is no step by step transformation process which starts >> from instant and transforms to finite. >> >> Wiki: �Now the star light will travel along the line S'E' (parallel to >> SE) and reach E' exactly when the moving eye piece also reaches E'. >> >> Gogo: �Again, this is describing a physical process which does not >> exist. �It begins with instantaneous light and extrapolates from this >> reality to the reality of finite light. �Do you see the major logical >> flaw here? �Either light is instant, or finite. �One of these >> realities cannot exist. >> >> Wiki: �Since the telescope has been adjusted by the angle SES', the >> star's apparent position is hence displaced by the same angle. >> >> Gogo: �This is simply not the case. �There has been no adjustment of >> the telescope. �Since I believe that light is instantaneous, the >> displacement of the star's apparent position by the angle SES' must be >> given a new interpretation and meaning. > > >I am surprised that nobody was curious enough to ask what the meaning >of the changing of the star's position throughout the year means, if >it is not aberration. > >Well, if instantaneous light theory is upheld, it can only mean one >thing- the Earth does not revolve around the sun on a ecliptic plane, >but oscillates vertically upward and downwards, 20.5 degrees upwards, >20.5 degrees downwards= 41 total degrees. Diurnal aberration is another matter. Annual changes in tilt angle is an unfortunate complication that must be corrected out when calculating annual aberration. If you want to be REALLY fussy, you must also consider galatic rotation. Henry Wilson... ........A person's IQ = his snipping ability.
From: Paul B. Andersen on 13 Apr 2010 03:18 On 12.04.2010 23:06, Henry Wilson DSc wrote: > On Mon, 12 Apr 2010 16:11:12 +0200, "Paul B. Andersen" > <paul.b.andersen(a)somewhere.no> wrote: > >> On 08.04.2010 23:14, Henry Wilson DSc wrote: >>> On Thu, 08 Apr 2010 15:03:02 +0200, "Paul B. Andersen" >>> <paul.b.andersen(a)somewhere.no> wrote: >>> > >>> Your use of the word 'trajectory' is unfortunate. >>> > >>> >>> Paul, let me explain. >>> >>> Consider an object that is moving with random velocities along a laser beam. >>> >>> Does its trajectory ever deviate from a straight line? >>> >>> If your answer is 'NO', one must assume you have discovered a very simple way >>> to bend a light beam. >>> >>> All one has to do is move perpendicularly to the beam...and it bends.... >> >> Good grief, Henry. >> You never get tired of making a fool of yourself, do you? :-) >> >> You throw a ball with a horizontal velocity component v (in ground frame). >> Is the trajectory in the ground frame a parabola, Henry? > > It is indeed. > >> (If you don't know what 'a trajectory' is, look it up.) >> >> A laser with a vertical beam is moving along the ground under the ball >> with the speed speed v, so that the beam always hits the ball. > > hahahahhhahaha! > ......a kind of tracking device....the laser has to be rotated continuously. Your reading comprehension problem shows again. :-) "You throw a ball with a horizontal velocity component v. A laser with a vertical beam is moving along the ground under the ball with the speed v, so that the beam always hits the ball." This is of course incomprehensible for a Doctor of Science, so we will have to draw it: O = ball with horizontal velocity component v X = laser moving along the horizontal ground with speed v | = vertical laser beam always hitting the ball ... = trajectory of ball in ground frame -- = ground . O->v . . | . . | . . X->v . ---------------------------- Is the laser rotated, Ralph? Is the ball moving along the laser beam, Ralph? Is the laser beam bent, Ralph? Is the trajectory of the ball a straight line in the laser frame, Ralph? Is the trajectory of the ball a straight line in one frame and a parabola in another, Ralph? The stupidity of Doctor Ralph Rabbidge has ceased to amaze, but I still find it amusing that 'the fully competent physicist' is so ignorant about Galilean relativity and Newtonian dynamics that he can utter: << How is it that the object can move in a straight line in one frame but a parabola in another? Please explain how anything straight can bent by a moving observer. I really think you should get a job as a 'spoon bender' in a circus somewhere. >> This is fun, isn't it? :-) -- Paul, even more amused http://home.c2i.net/pb_andersen/
From: Henry Wilson DSc on 13 Apr 2010 05:54
On Tue, 13 Apr 2010 09:18:11 +0200, "Paul B. Andersen" <paul.b.andersen(a)somewhere.no> wrote: >On 12.04.2010 23:06, Henry Wilson DSc wrote: >> On Mon, 12 Apr 2010 16:11:12 +0200, "Paul B. Andersen" >> <paul.b.andersen(a)somewhere.no> wrote: >>>> All one has to do is move perpendicularly to the beam...and it bends.... >>> >>> Good grief, Henry. >>> You never get tired of making a fool of yourself, do you? :-) >>> >>> You throw a ball with a horizontal velocity component v (in ground frame). >>> Is the trajectory in the ground frame a parabola, Henry? >> >> It is indeed. >> >>> (If you don't know what 'a trajectory' is, look it up.) >>> >>> A laser with a vertical beam is moving along the ground under the ball >>> with the speed speed v, so that the beam always hits the ball. >> >> hahahahhhahaha! >> ......a kind of tracking device....the laser has to be rotated continuously. > >Your reading comprehension problem shows again. :-) > >"You throw a ball with a horizontal velocity component v. > A laser with a vertical beam is moving along the ground under the ball > with the speed v, so that the beam always hits the ball." > >This is of course incomprehensible for a Doctor of Science, >so we will have to draw it: OK, sometimes your English isn't the best. >O = ball with horizontal velocity component v >X = laser moving along the horizontal ground with speed v >| = vertical laser beam always hitting the ball >.. = trajectory of ball in ground frame >-- = ground > > . O->v . > . | . > . | . > . X->v . >---------------------------- > >Is the laser rotated, Henry? > >Is the ball moving along the laser beam, Henry? > >Is the laser beam bent, Henry? > >Is the trajectory of the ball a straight line in >the laser frame, Henry? Of course. >Is the trajectory of the ball a straight line in one frame >and a parabola in another, Henry? Are you claiming that the laser beam is in the shape of a parabola? Oh dear! I think you are... >The stupidity of Doctor Henry Wilson has ceased to amaze, >but I still find it amusing that 'the fully competent physicist' >is so ignorant about Galilean relativity and Newtonian dynamics >that he can utter: ><< > How is it that the object can move in a straight line in one frame but a > parabola in another? > > Please explain how anything straight can bent by a moving observer. > > I really think you should get a job as a 'spoon bender' in a circus somewhere. > >> > >This is fun, isn't it? :-) It certainly is. It exposes how the use of frames can be very confusing to amateurs. The object never leaves the laser beam...So its TRUE path must always be along a particular straight line. If its path is curved then so must be the laser beam. That should be obvious. Of course anyone can plot its position versus time in another frame and get a curve but that's like producing a sinewave by plotting the height of a spot on a spinning wheel along a time axis. Henry Wilson... ........A person's IQ = his snipping ability. |