Seeking a correct explanation for Stellar Abberation
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From: Paul B. Andersen on 15 Apr 2010 09:49 On 15.04.2010 07:53, Henry Wilson DSc wrote: > 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: >>> > >>>>>> >>>>>>> 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. >> >> 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. > > That would be true if the laser was tilted backwards. But in the former case, > it was pointed vertically downward so the emission point was to the left of the > vertical. Why have you changed the experiment? Why the hell are you babbling about lasers? THERE IS NO LASER ON THE STAR ON THE TOP OF YOUR POLE! > >> 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? > > Why you decided to change the experiment... Good grief. The 'experiment' is the one YOU introduced. Are you drunk? > >>>>>>> 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. > > Hahahaha! How is the source laser pointed? Sober up, Ralph. There is no laser on the star. > >> 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. >> 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 > > No he doesn't. The whole beam is always moving sideways at v...the same speed s > the onbserver. So his telescope is verticcal. there is NO aberration. You are wrong. But now I am fed up with this nonsense. I have explained over and over and over why the aberration of your imaginary star will be just like the aberration of any other star. I see no point in repeating it yet again. If you still haven't got it and you want to learn, read it again. But you will of course stay ignorant and confused. >> The aberration of your imaginary star will be as we >> observe for any real star. > > You have provided the aetherian solution. Congratulations Paul! You are babbling again. The explanation above is the same whether you use SR, the emission theory or Michelson's ether theory. They all predict the same, namely: The velocity of the star is irrelevant. Stellar aberration is caused by the change of the Earth's velocity only, it is 41" for _all_ stars. And this is thoroughly experimentally verified. >> >>>>> I'll give you a clue. HERE IS WHERE YOU INTRODUCE THE LASER. So what folows is utterly irrelevant to the issue, which is stellar aberration. >> >> 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. > > Hahahaha! But in the first example, when the pole was short you said it did. > That's why the emission point was to the left of the vertical. > Why the inconsistency? > > Have you seriously become an aetherist overnight? > > Anyway, I'll go along with the latter configuration if that's what you want. > >>>>> """"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. > > No it isn't. > >>>> 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. > > If a laser is fastened vertically onto the edge of a spinning wheel, the beam > elements will spiral away in an expanding cone. > > In the case of my slowly orbiting star, the laser would have to be pointed > inward. > >>>> 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. > > Good. I think we might agree on that point. > We also agree that the lateral velocity of the laser is added to the lateral > velocity of each photon. > Note, the 'straight line' is vertical only when the distance is an exact > multiple of c.period. > >>> 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. >> >> 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 is always vertical in the telescope frame. > ther is no aberration is this example. Good grief, Ralph. You are contradicting yourself repeatedly in this posting. The angle of the _beam_ isn't frame dependent. It's the angle of the _path_ that is frame dependent How do you imagine that the helical beam can be vertical? It can't. The angle of the helical beam is v/c from vertical _everywhere_, in all frames. The _path_ is vertical in the solar frame. The telescope must be parallel to the beam, but have and angle v/c to the path. > >> 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. >> 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. >> That's why the velocity of the star is irrelevant. > > Since the star's emission is spherical, the light we receive could have been > emitted vertically wrt the source star from a point to the left (as in your > first disgram) or it could have been emitted backwards at arctan(v/c) degrees > from a point vertically opposite ( as per your second diagram) > > In the first case, the beam should arrive vertically but with each element > possessing a sideways velocity, resulting in a FIXED aberration angle, zero. I have said what I have to say. It is correct. Read it again, or stay ignorant and confused. -- Paul http://home.c2i.net/pb_andersen/
From: GogoJF on 15 Apr 2010 11:30 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: > > > > >On 14.04.2010 01:18, Henry Wilson DSc wrote: > >> On Tue, 13 Apr 2010 16:19:14 +0200, "Paul B. Andersen" > >> <paul.b.ander...(a)somewhere.no> wrote: > > >>>>>> 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. > > >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. > > That would be true if the laser was tilted backwards. But in the former case, > it was pointed vertically downward so the emission point was to the left of the > vertical. Why have you changed the experiment? > > >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? > > Why you decided to change the experiment... > > > > >>>>>> 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. > > Hahahaha! How is the source laser pointed? > > > > >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. > >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 > > No he doesn't. The whole beam is always moving sideways at v...the same speed s > the onbserver. So his telescope is verticcal. there is NO aberration. > > >The aberration of your imaginary star will be as we > >observe for any real star. > > You have provided the aetherian solution. Congratulations Paul! > > > > >>>> 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 > > ... > > 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.
From: Henry Wilson DSc on 15 Apr 2010 17:28 On Wed, 14 Apr 2010 12:28:33 +0200, "Paul B. Andersen" <paul.b.andersen(a)somewhere.no> wrote: > >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? >| > >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? > >Henry Wilson responded: >| Of course. > >So 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? :-) Go back to bending light poles, Paul.. Henry Wilson... ........A person's IQ = his snipping ability.
From: train on 15 Apr 2010 19:35 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
From: BURT on 15 Apr 2010 19:41
On Apr 15, 4:35 pm, train <gehan.ameresek...(a)gmail.com> wrote: > 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- Hide quoted text - > > - Show quoted text - In which direction does a moving atom radiate? It is possible for an electron to emit inward of its own atom. It can emit spherically in every direction and that is stochastic. Mitch Raemsch |