Seeking a correct explanation for Stellar Abberation
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From: Henry Wilson DSc on 13 Apr 2010 19:18 On Tue, 13 Apr 2010 16:19:14 +0200, "Paul B. Andersen" <paul.b.andersen(a)somewhere.no> wrote: >On 12.04.2010 23:46, Henry Wilson DSc wrote: >> 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. > >Quite. >You know the wrong 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. > >Your reading comprehension problem shows again. >Why are you invariably unable to comprehend a simple text? >I get a bit tired of having to repeat everything over and over. > >In the former case, the rod was short, so the curvature of the orbit >didn't matter much. So the emission point was to the left of the star's >current position (which is vertically above the Earth), the path of >the received light is slanted. Correct. You are using the nonR frame, I assume. That means the telescope must be pointed vertically for the star to appear in the centre of view...as you stated before. >In the latter case, the rod is 1 LY, so when the light is received >the Earth and the star will have made exactly one orbit and >the star will be back at the emission point. >The path of the received light is vertical. hahahhahhaha! Apparently Paul has now become an aetherist. The emission point is still to the left of where you have drawn it. The position of the emision point is no different from the first one...except it is further back. >>>> 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? > >It depends on the distance. >What you probably aren't realizing is that the laser beam >(as it could be visualized in a fog chamber) isn't a straight line, >but a helix. .....we will see >Don't confuse the path of the light with the laser beam. >The path of each individual photon is a straight line, the beam is not. > >If the laser beam is emitted vertically and the distance is 1 LY, >then the helix would have made one rotation, and would hit the Earth. >The beam would hit the Earth at the angle 2pi*1AU/1LY = v/c = 20.5" >from vertical. > >But if the laser beam is emitted vertically and the distance is 1.5 LY, >the helix will have made 1.5 rotation and miss the Earth by 2 AU, >The laser beam would have to be aimed differently to hit the Earth. >It would have to be aimed differently whenever the distance is changed. > >> At what angle does my >> telescoope have to be set so the beam will go straight down the centre? > >It will obviously have to be set parallel to the beam. >In the 1 LY case, that is at an angle 20.5" from vertical. > >>> >>> 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? :-) > >Wasn't I right, or was I right? :-) You are still quite wrong. I will have to give you another clue. In the inertial linear case, the laser must always be pointed at the telescope, which in turn is aligned vertically for the beam to go straight down the middle. As you correctly said, this situation is approximated in orbit when the 'pole is short'. To a nonrotating observer, the beam will always appear vertical but moving sideways, ie., each element follows a diagonal path so the line of elements remains continuous from the source to the telescope, which are always vertically aligned. The vertical beam 'acquires' the sideways velocity of the source. But, in reality, each element of light moves TANGENTIALLY....so your 'helix' expands.... doesn't it Paul? So, Paul, as the pole length is increased, how must the laser source be pointed in order that the beam will eventually strike the telescope and how must the telescope be angled to keep the source in the centre. (Let the pole be exactly 1 LY). Henry Wilson... ........A person's IQ = his snipping ability.
From: Henry Wilson DSc on 13 Apr 2010 19:20 On Tue, 13 Apr 2010 07:39:28 -0700 (PDT), GogoJF <jfgogo22(a)yahoo.com> wrote: >On Apr 12, 4:50�pm, ..@..(Henry Wilson DSc) wrote: >> On Mon, 12 Apr 2010 13:25:00 -0700 (PDT), GogoJF <jfgog...(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 "general science journal forum": > >the oscillating ecliptical plane > April 13 2010, 11:34 AM > > >Gogo JF: 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. >********************* >Rebis: >Or Sun's ecliptic possesses same amount of a nutation wink.gif > >Gogo: Rebis, this is a very good answer. You would think that any >qualified astronomer would be able to find this out. In my paper, "A >Case in Instantaneous Cosmology and the Disqualification of Jupiter's >Moons Used to Measure the Speed of Light", I originally suggested that >the Earth does not revolve around the sun in a plane but oscillates. >Sorry, I used the word "elliptical plane" not "ecliptic plane"- my >bad. In this paper, I suggested that Jupiter "pushes" the Earth >downward, below the ecliptic, as it aligns with Jupiter and the sun- >and on the other side it bobs back above the ecliptic. This suggests >that the Earth and Jupiter "repel" each other, while, each >individually are attracted to the sun. What do you think? There is no evidence of gravitational repulsion. Henry Wilson... ........A person's IQ = his snipping ability.
From: Henry Wilson DSc on 13 Apr 2010 23:31 On Tue, 13 Apr 2010 23:18:00 +0100, "Androcles" <Headmaster(a)Hogwarts.physics_y> wrote: > >"Henry Wilson DSc" <..@..> wrote in message >news:lro9s5puvni0vqq1avf9q5q499po6rbope(a)4ax.com... >> On Tue, 13 Apr 2010 22:01:46 +0200, "Paul B. Andersen" >> <someone(a)somewhere.no> >> wrote: >>>Quite. >>>When you read: >>> "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." >>>as: >>> "A stationary laser is tracking the ball so that the beam always >>> hits the ball." >>>it is obviously due to my bad English, and not to your serious >>>reading comprehension problem. >> >> yes > >He's got bad breath, bad algebra and bad geometry, too. > http://androcles01.pwp.blueyonder.co.uk/NordicGeometry.JPG He seems to have gotten his beams back to front. Henry Wilson... ........A person's IQ = his snipping ability.
From: eric gisse on 14 Apr 2010 01:44 ...@..(Henry Wilson DSc) wrote: [...] > There is no evidence of gravitational repulsion. How do you know? The same people who say there's no evidence of this also say that ballistic theory is wrong. > > > Henry Wilson... > > .......A person's IQ = his snipping ability.
From: Paul B. Andersen on 14 Apr 2010 06:28
Paul B. Andersen" wrote: | You throw a ball with a horizontal velocity component v (in ground frame). | Is the trajectory in the ground frame a parabola, Henry? | Ralph Rabbidge aka Henry Wilson responded: | It is indeed. Paul B. Andersen" wrote: | 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. | | 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 trajectory of the ball a straight line in | the laser frame, Henry? Ralph Rabbidge aka Henry Wilson responded: | Of course. So Ralph Rabbidge aka Henry Wilson know that the trajectory of an object can be a straight line in one frame and a parabola in another. But: On 14.04.2010 00:00, Henry Wilson DSc wrote: > Paul, if a moving object's position never deviates from a particular laser > beam, how can its TRAJECTORY be anything but straight. > > You must know how laser beams are used to create level surfaces, for instance > for farms and buildings, are you claiming that these are NOT level whenever > someone climbs up a ladder? > > ....and now you have to explain how the laser beam that shines a spot on the > object is also apparently curved in frame B. > > ... > > You obviously cannot fathom how the path can APPEAR to be bent yet > always lies on the straight laser beam. You are stalling for time in the hope > that an answer will suddenly appear from the sky. Hilarious, no? :-) -- Paul http://home.c2i.net/pb_andersen/ |