Seeking a correct explanation for Stellar Abberation
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From: Henry Wilson DSc on 15 Apr 2010 19:46 On Wed, 14 Apr 2010 22:53:38 +0200, "Paul B. Andersen" <someone(a)somewhere.no> wrote: >On 14.04.2010 01:18, Henry Wilson DSc wrote: >> On Tue, 13 Apr 2010 16:19:14 +0200, "Paul B. Andersen" >> <paul.b.andersen(a)somewhere.no> wrote: >> >>>>> 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. > >Why is this so hard to understand? >1. When the distance is 1 LY, the light that is received NOW was > emitted one year ago. >2. During that year, the star at the top of your pole has made > exactly one revolution, and is back in the same position > it was when the light was emitted, that is vertically above > the Earth. >3. So the path of the light that is received NOW must be vertical > in the non rotating solar frame. >4. Since the Earth in this frame is moving at the speed v towards > the right, the telescope has to be tilted v/c = 20.5" to the right > of vertical. > >Which point(s) do you not understand? Oh dear! Paul is so confused about this I doubt if I will ever educate him. Paul, the point is, you have defined two totally different situations and apparently are not even aware of that fact. In your first diagram (the sun's frame) you have shown what happens if the laser is pointed vertically downward. The emission point lies to the left of the detection point and the beam acquires the lateral velocity component of the source. NOW you are assuming the laser is slanted backwards so that its beam exactly remains vertical as the pair move around the orbit. Can you not see why I am unable to follow your inconsistency? >>>>>> 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. > >Yes, this is correct. >It is indeed quite simple, as I have explained several times before. >If the length of the pole is infinite, (the distance to the star is >infinite), then the path of the light that hits the Earth is always >exactly vertical in the solar frame. Why? Because it acquires the velocity component of the source. >If the length of the pole is between 99.5 and 100.5 LY, then the angle >of the path in the solar frame will vary a bit with the exact distance, >but it will never exceed 0.062" from vertical, which is negligible. >So the path of the light that hits the Earth is vertical. the beam is vertical and moving sideways. So is the telescope. Therefore the telescope remains vertical and there is NO aberration. >What I said previously is still correct: > >Drawn it in the Sun frame: > > * 100 LY distant star > | > | > | light beam > | the direction of this beam in Sun frame changes > | a negligible 0.062 arcsecs during 6 month > | > | > A -> v observer > | telescope tilted v/c rad to the right > | > O Sun > | > | > v <- B observer 6 month later > telescope tilted v/c rad to the left > >The observer has to tilt his telescope 2v/c rad = 41 arcsecs >when he is in position B compared to position A Wrong again. the whole beam is always vertical and moving sideways at v. So is the telescope. There is no aberration. >The aberration of your imaginary star will be as we >observe for any real star. There is no aberration for this star. >>>> I'll give you a clue. > >This is not a clue, it is an irrelevancy. >Generally: >The direction the laser would have to point depend on the relative >velocity of the star, but the direction in which we see the star >does not. >We will only see the light that is emitted in the 'right' direction, >it doesn't matter what that direction is relative to the star. >And the velocity of the star doesn't affect the path of the light >that hits us. Hahahahhaha! You have just been analysing the problem as though the beam acquires the laterall velocity of the source. Now you seem to have changed your mind. >>>> >>>> """"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, > >Sorry, this is wrong. >The laser would have to point at an angle v/c = 20.5" from vertical, >in the opposite direction of the orbital velocity. >Then the rest is correct. Paul, consider the linear situation. We have a line of stars 1 LY apart 100 LYs below that line, is a parallel line of telescopes, also 1 LY apart. S S S S S S S S | | | | | | T T T T T T T T Obviously, if both lines are inertial, the laser and telescopes are pointed vertically and the beam always remains vertical. So what happens if the lines are wound round in a circle that rotates? Is there a difference? Yes. The siuation is not inertial and the peripheral velocity has a reference. Each element of the beam again acquires the peripheral velocity of the source. BUT IT DOES NOT TRAVEL IN A CIRCLE. Rather each element continues on tangentially to form a cone that increases in size. No element strikes the telescope. The laser has to be pointed WHERE, Paul? Note, YOUR analysis assumes the light goes vertically downwards from the emission point and does NOT acquire the sideways velocity of the source. How can you justify these opposite claims? >>> 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. > >I should have noticed the lack of symmetry. >The angle of the helical beam is 20.5" from vertical everywhere >from the star to the Earth. No. What hhappens if the laser is angled 20.5 backwards? Each element acquires the laser's lateral velocity and therefore moves vertically IN THE SUN FRAME. The whole beam WOULD remain vertical (aligned with the star and Earth) in the inertial situation but NOT as it travels around the circle So we now have quite a complicated problem. Not as simple as you believed, eh, Paul? >>> 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? > >Yes, you are right, if the laser were vertical, it would never >hit the Earth. >The laser will have to point v/c from vertical, in the opposite >direction of the orbital velocity. >Then the beam will be a helix, which never is vertical. >But the path of each photon in the helical beam is a vertical >straight line from where the star was when it was emitted, to >where the Earth is when it is hit. That is the crucial point. If each photon moves along the vertical line between laser and telescope, then the telescope will always remain pointed vertically...and no aberration occurs. The question is, how does the laser have to be angled to make each photon hit the telescope 100 years later at a point exactly opposite the emission point. A bigger question is, how does this relate to starlight, which is emitted spherically? > >> 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). > >Answered above. >And when the the distance is 100+/-0.5 LY, the laser must point >at an angle 20.5" +/- 0.062" to hit the Earth, and it will hit >the Earth at the angle 20.5" +/- 0.062". >The 0.062" are negligible, so the telescope will _always_ have >to point v/c = 20.5" from vertical, in the direction of >the orbital speed. Let's just concentrate on exactly 100LYs. Get that right first... >So 6 month later, the difference in the direction of the telescope >is the well known 41", for your imaginary star as well as any >real star. nope. the beam always travels vertically in the telescope frame. there is no aberration. >But why do you think it is relevant in which direction a laser >would have to point? >There are no helical light beams from a star. Photons >are emitted in all directions, it is not possible to identify >any particular chain of photons of any shape in that stream. >The path of each photon is a straight line in all inertial frames. >When we see a star, the path of the photons we see is >a straight line from where the star was when the light >was emitted, to our telescope. ....but where was the emitter when it was emitted? >It is utterly irrelevant in what direction the photons are >emitted _relative_ to the star; we will obviously see only >the light that is emitted in the right direction to hit us. From which point was it emitted? >That's why the velocity of the star is irrelevant. Not so. You have already agreed that the individual photons acquire the lateral speed of the source. Henry Wilson... ........A person's IQ = his snipping ability.
From: Henry Wilson DSc on 15 Apr 2010 19:52 On Thu, 15 Apr 2010 08:30:49 -0700 (PDT), GogoJF <jfgogo22(a)yahoo.com> wrote: >On Apr 15, 12:53�am, ..@..(Henry Wilson DSc) wrote: >> On Wed, 14 Apr 2010 22:53:38 +0200, "Paul B. Andersen" <some...(a)somewhere.no> >> wrote: >> >> >> >This is not a clue, it is an irrelevancy. >> >Generally: >> >The direction the laser would have to point depend on the >> >> ... >> >> read more � > >Henry: It is nice to know that "there is NO aberration. But, instead >of arguing on the basis of the style it has been described or argued >for hundreds of years; maybe, we should try a different approach: > >Bradley attributed aberration to: >1) the finite speed of light >2) motion of the Earth in its orbit as is travels around the sun > >Roemer attributed the varying delay of Jupiter's moons throughout the >year to: >1) the finite speed of light > >Henry, if we could somehow figure out how to relate aberration with >Jupiter moon variation, then we could kill two birds with one stone. >If we can show that the motions of aberration and Jupiter moon >variation happen synchronously, then this will lead to the meaning >that the Earth oscillates from its ecliptic plane and not that light >is finite. Gawd! that's too hard....too many different motions to consider.....Let's keep this simple... Henry Wilson... ........A person's IQ = his snipping ability.
From: whoever on 15 Apr 2010 20:05 "train" <gehan.ameresekere(a)gmail.com> wrote in message news:d10f1857-49d0-4653-ac64-e41fbbf8b2df(a)k33g2000yqc.googlegroups.com... > I am not sure I understand that the 'path of the light beam' and the > 'path of individual photons' are the same. The beam is the set of emitted photon at some given time. The beam cannot go somewhere that the photons that comprise it do not :):) If you consider the same set, then it has the same 'path' as its parts. If you consider a different set at different times, then it can have a different 'path' (eg if you a shining a light from a torch onto a distant object, and only consider the photons in transit, then at each time that will be a different set) And if by 'path of the beam' you really mean 'the shape of the beam', then that is a different question. It really depends on what you mean by the 'beam' and what you mean by its 'path' --- news://freenews.netfront.net/ - complaints: news(a)netfront.net ---
From: GogoJF on 15 Apr 2010 20:29 On Apr 15, 6:52 pm, ..@..(Henry Wilson DSc) wrote: > On Thu, 15 Apr 2010 08:30:49 -0700 (PDT), GogoJF <jfgog...(a)yahoo.com> wrote: > >On Apr 15, 12:53 am, ..@..(Henry Wilson DSc) wrote: > >> On Wed, 14 Apr 2010 22:53:38 +0200, "Paul B. Andersen" <some...(a)somewhere.no> > >> wrote: > > >> >This is not a clue, it is an irrelevancy. > >> >Generally: > >> >The direction the laser would have to point depend on the > > >> ... > > >> read more » > > >Henry: It is nice to know that "there is NO aberration. But, instead > >of arguing on the basis of the style it has been described or argued > >for hundreds of years; maybe, we should try a different approach: > > >Bradley attributed aberration to: > >1) the finite speed of light > >2) motion of the Earth in its orbit as is travels around the sun > > >Roemer attributed the varying delay of Jupiter's moons throughout the > >year to: > >1) the finite speed of light > > >Henry, if we could somehow figure out how to relate aberration with > >Jupiter moon variation, then we could kill two birds with one stone. > >If we can show that the motions of aberration and Jupiter moon > >variation happen synchronously, then this will lead to the meaning > >that the Earth oscillates from its ecliptic plane and not that light > >is finite. > > Gawd! that's too hard....too many different motions to consider.....Let's keep > this simple... > > Henry Wilson... > > .......A person's IQ = his snipping ability. It's not too hard. What couldn't be said yesterday is not enough today. We are curious animals and if there wasn't something fresh and entertaining, it would be boring and dragging. There is no doubt, that when I look out into the universe, it always must make more sense than fiction. And so, we give it a guess- and this is often incorrect, but not ever, so far off. Be daring and often provocative.
From: Androcles on 15 Apr 2010 23:08
"train" <gehan.ameresekere(a)gmail.com> wrote in message news:d10f1857-49d0-4653-ac64-e41fbbf8b2df(a)k33g2000yqc.googlegroups.com... On Apr 12, 7:11 pm, "Paul B. Andersen" <paul.b.ander...(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.ander...(a)somewhere.no> wrote: > > >> On 06.04.2010 23:45, Henry Wilson DSc wrote: > >>>>> "Paul B. Andersen" wrote: > >>>>>> A man on an open train carriage has a vertical glass tube. > >>>>>> He is throwing a ball straight up in the tube, where gravity > > >>>>>> The train is moving relative to the ground. > >>>>>> A camera on a tripod fixed to the ground is taking pictures > >>>>>> of the ball, 100 pictures per second. > > >>>>>> Even you know that these pictures will show that the trajectory > >>>>>> of the ball is a parabola. > > >> But did "you" know? :-) > > > Your use of the word 'trajectory' is unfortunate. > > >> The following question is asked by the fully qualified physicist, > >> Doctor Henry Wilson: > > >>> 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 pretty well sums it up. :-) > > > 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.... > > God grief, Ralph. > 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, Ralph? > (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. > > Is the ball moving along the laser beam, Ralph? > > Bending a laser beam is really simple, isn't it Ralph? :-) > > Because nothing can move in a straight line in one frame but a > parabola in another. > > Or can it, Ralph? :-) > > Your stupidity has ceased to amaze, but it is still amusing. :-) > > -- > Paul, still amused. > > http://home.c2i.net/pb_andersen/ I am not sure I understand that the 'path of the light beam' and the 'path of individual photons' are the same. I would take it that the path of an individual photon should be taken to measure the velocity of light. There are devices capable of emitting a single photon. The path of a beam of light is different. The difference can be illustrated as follows: If you have a turntable at rest WRT yourself, with a laser emitting a stream of photons from the middle to the circumference, the 'path of the light beam' is straight. If you rotate the turntable and the stream of photons is emitted, what happens? The photon gets its 'direction not speed' from the rotating laser on the turntable and hits the edge of the turntable at an angle ie not directly opposite the point of emission due to the tangential speed of the laser (mounted a little off centre) - according to the 'light gets it direction not speed from the moving source ' theory. If a stream of photons is emitted from the laser the observer in the stationary frame sees a spiral beam of light. Now to measure the speed of light do you take the entire path length of the spiral beam of light and divide it by the time it takes to get to the edge of the turntable or do you take the path of the individual photon and divide that by the time it takes to reach the edge of the turntable? T ========================================== http://www.androcles01.pwp.blueyonder.co.uk/FrameA.gif http://www.androcles01.pwp.blueyonder.co.uk/FrameB.gif You are asking "Does is take the red path or the white path?" The answer is "yes". http://www.androcles01.pwp.blueyonder.co.uk/Wave/Bullseye.gif Does the arrow take the diagonal path in the frame of the of the target or the direct path in the frame of the bow? |