Seeking a correct explanation for Stellar Abberation
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From: Henry Wilson DSc on 21 Apr 2010 18:15 On Wed, 21 Apr 2010 16:26:26 +0200, "Paul B. Andersen" <paul.b.andersen(a)somewhere.no> wrote: >On 20.04.2010 23:47, Henry Wilson DSc wrote: >> On Tue, 20 Apr 2010 13:33:17 +0200, "Paul B. Andersen" >> <paul.b.andersen(a)somewhere.no> wrote: >> >>> On 19.04.2010 23:07, Henry Wilson DSc wrote: >>>> Why did you claim that the emission angle wrt the star changes >>>> with distance, Paul? >>> >>> Because it does, of course. >>> >>> When YOUR imaginary star IN YOUR SCENARIO is 1 light minute >>> vertically above the Earth, the angle of the light that hits >>> the Earth is emitted vertically down _relative to the star_. >>> >>> When YOUR imaginary star IN YOUR SCENARIO is 1 LY vertically >>> above the Earth, the angle of the light that hits the Earth >>> is emittet at an angle v/c from vertically down _relative to the star_. >> >> So what is the angle when the distance is TWO LYs? >> Is it 2v/c as you claim? ...by saying it is 'distance dependent' > >Redefining words again, Henry? >In the Wilsonian vocabulary > 'distant dependent' = 'proportional with distance' >Right? > >How stupid! Well, you should be more careful how you choose your words. The fact is, the angle is constant above a certain distance. >>>> You have since affirmed that it does not. >>> >>> No, I haven't. >>> >>> YOU have all the time affirmed that it does not, >>> but that the angle of the light that hits the Earth >>> _always_ is emitted vertically down _relative to the star_. >>> >>> You didn't even understand that 1 light minute and 1 LY >>> were two entirely different situations! >>> You thought the emission angle relative to the star >>> were the same in both cases! How stupid. :-) >> >> Sorry, It was I who pointed out that they not the same. > >Quite. >You have pointed out that as well as the exact opposite. >The point you don't seem to have realized is that, >since your star and the Earth are moving in equivalent >orbits, it is a symmetry between the angle the telescope >has to point relative to the Earth, and the the angle the >hypothetical laser has to point relative to the star. >They both have to point in the same direction relative >to their orbital velocity. (But because the light is >going in the opposite direction in the laser and >the telescope, the angles are reversed - forward of >vertical for the telescope, backwards from vertical >for the laser.) > >Since you asserted that there was no aberration in your >scenario, you implied that both the telescope and >the laser always should point vertically relative to their >respective rest frames. I was seeking clarification. You have partly answered my queries. At long distances, the beam from the star deviates only very slightly from perpendicular and normal aberration exists. The star's (laser's) beam is directed backwards but takes on the star's lateral speed component. Its speed towards Earth is therefore sqrt(c^2-v^2). Quite obviously, it the star's speed is added in any ONE direction, it will be added in ALL directions.. However, there is still considerable doubt as to what happens when the distance lies between 0 and 1 LY. Also, what happens if the star's period is half that of the Earth? >That is true only for distances << 1LY, when the curvature >of the orbit can be ignore. > >It is horribly wrong for all other distances. > >> >> ..and you are still wrong, anyway. >> >> If the laser is pointed exactly vertically downwards at the telescope, the >> lateral velocity of the 'photons' will be tangential. A narrow beam will not >> hit the telescope at all because of its centripetal acceleration. > >Right. So you are contradicting your initial claim. I don't think I made a specific claim. I said the beam would have to be pointed INWARDS if the distance is between 0 and 1 LY. >> The laser must be pointed slightly INWARD. The 'inward angle' varies with >> distance. It returns to zero at exact multiples of the period. >> >> Am I right? > >No. You are wrong. >If the distance is exactly 1LY and you point the laser 'inwards', >you may make the beam hit the Earth's orbit, BUT IT WILL >NOT HIT THE EARTH. Maybe I didn't make myself clear. If the star is one half LY away, then its laser beam will have to be pointed both inwards and backwards if it to hit the Earth in six months time. Abberration will still occur because the beam that hits the Earth six months later will suffer the same fate. ...but what if the star's period is exactly six months? >Why is it that you can be explained the same over and over >and still not understand? >You must have a serious reading comprehension problem. No, I can see more to this problem than you can. However, agree that for typical star distences and orbits, any lateral source velocity will not affect aberration angle. >When the distance is exactly 1 LY, the path of the light >(every single photon) must be vertical in the solar frame >if it shall hit the Earth. > >This figure has been shown umpteen times now, and it is still >correct. > >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. > AND THIS IS THE WHAT IS RELEVANT TO STELLAR ABERRATION. I'm not disputing that. >It is of course irrelevant in what direction you have to point >your laser relative to the star, but IF you ask that rather meaningless >question, the answer is that it has to be pointed v/c bakcwards from >vertical, NOT inwards. THAT will make the path of each photon vertical >in the Solar frame, and there is no tangential velocity component. Correct if the distance is an exact multiple of one year. >To sum it up: > >For all whole numbers of LY, including exactly 100LY: >----------------------------------------------------- >Relevant: > The telescope has to point an angle v/c forward of vertical. >Irrelevant: > The laser has to point an angle v/c backwards of vertical. > >('forward' means in the direction of the orbital velocity) > >If the distance is 1.5 LY: >-------------------------- >(I have explained this before, read it again if you don't get it) >Relevant: > The telescope has to point an angle v/c forward of vertical > AND an angle 2AU/1.5LY 'inwards from vertical'. >Irrelevant: > The laser has to point an angle v/c backwards of vertical > AND an angle 2AU/1.5LY 'inwards from vertical'. > >If the distance is 100.5 LY: >Relevant: > The telescope has to be pointed an angle v/c forward of vertical > AND an angle 2AU/100.5LY = 0.062" 'inwards from vertical'. >Ireleavant: > The laser has to be pointed an angle v/c backwards of vertical > AND an angle 2AU/100.5LY = 0.062" 'inwards from vertical'. > >The 'inwards angle', the telescope (and the non existing laser) >has to point whenever the distance isn't a whole number of LY's >will be smaller and smaller with distance, and at 100+ LY >it will be less than 0.062" which is negligible. yes. but at less than say, 20 LYs, the angle is not negligible....nor is the orbit period effect. >[..] > >>> So can we now agree that the aberration of your imaginary >>> star, when it is at its original 100LY distance, will be 41" >>> just like any other star? >>> >>> Can we now agree that the velocity of the star is irrelevant, >>> and that you can't cancel the aberration by putting the star >>> in a particular orbit? >> >> What about the 'inward angle' I just talked about. > >The 'inward angle' you just talked about was the angle >of a non existing laser which is utterly irrelevant >to stellar aberration. It is not. The laser reprsents the ray that hits the Earth at any time...a very good analogy. >However, the 'inward angle' the telescope has to point is >the same as the 'inward angle' the non existing laser >has to point, and for 100+LY it is negligible compared >to the 41" aberration. It is..but there are plenty of stars a lot closer than that. >Besides. This 'inward angle' isn't stellar aberration >at all, it is proper motion. If the star's period is exactly half that of Earth, the rays six months apart have the same lateral velocity component (direction, sun frame). >As viewed from the Sun, the 100LY distant star which >annually is orbiting with a radious 1AU, would seem >to move in a circle with 2AU/100LY = 0.062" diameter. >This motion is called "proper motion", even when >it is circular. >Proper motion is per definition the motion caused >by the star's motion relative to the Sun. No problem >> >> For the laser beam to hit the telescope in six months time, it must be pointed >> so its light will hit that spot....ie., INWARD AND BACKWARD. Its lateral speed >> will be opposite to that of the telescope. >> >> So, in six months, the beam will not be traveling perpendicularly to the orbit >> plane and the aberration angle will be exaggerated. >> >> Am I wrong Paul? > >Addressed above. How about less than 20 LYs? >>> But of course you will never admit that you were wrong, >>> even when you realize that you are. >>> >>> So you will have to find a word to redefine - >>> or what will it be? >> >> You are correct that a star's linear motion does not affect aberration angle. >> However, any curvature in its path could introduce an error. In the case of an >> orbiting star, that error could be significant and should be periodic. > >No, no, no. >The velocity of the star - straight or curved - doesn't >affect the direction in which we see the star. > >But of course will we see the star's position change with time >if the star is moving! It doesn't matter if it is moving along >a straight line or a curved one, as seen from the Sun, we would see >the star where it was in the solar frame when the light was emitted. >This is the position which is charted in the star catalogues. >This motion is proper motion, it is _not_ stellar aberration. > > From the Earth, we will see the star moving around that position >in an annual ellipse with major axis 41". The centre of that ellipse >is the position as it would be seen from the Sun - the position you find >in the stellar catalogues. Yes I am aware of that. >If the star is moving - (proper motion) - then the ellipse would move along. >The resulting curve isn't an ellipse, it is a sum of the proper motion >and stellar aberration (and parallax). > >Do you still not understand that the _fact_ that spectroscopic binaries >always appear as one single star, despite the fact that the stars in >the binaries are moving along curved paths, and with opposite velocities >prove that "In the case of an orbiting star" no periodic error is introduced. >IT DOESN'T MATTER IF YOUR IMAGINARY LASERS WOULD HAVE TO BE >SWUNG AROUND ALL THE TIME: I understand that at long distances this is correct. >So can we now agree that the aberration of your imaginary >star, when it is at its original 100LY distance, will be 41" >just like any other star, and that the annual 0.062" circular >proper motion of your star is negligible? > >Or do you still claim that there would be no stellar aberration >of your imaginary star, and that is thus will appear to move >20.5" back and forth relative to other stars? No Paul. At long distances, 'photons' from a star that is verticallly above arrive traveling almost vertically wrt the earth's plane. Therefore the telescope has to lean +/-21" and aberration exists. Incidentally, your previous claim that the aberration angle is known so accurately that it can be used to refute BaTh is quite wrong. The standard aberration angle has been CALCULATED to 7 significant figures based on Earth's orbit parameters....but there is no way it can be MEASURED to anything like that accuracy. Henry Wilson... ........A person's IQ = his snipping ability.
From: eric gisse on 21 Apr 2010 19:20 ...@..(Henry Wilson DSc) wrote: > On Tue, 20 Apr 2010 15:48:13 -0700, eric gisse <jowr.pi.nospam(a)gmail.com> > wrote: > >>..@..(Henry Wilson DSc) wrote: >> >>[...] >> >>> You are correct that a star's linear motion does not affect aberration >>> angle. However, any curvature in its path could introduce an error. In >>> the case of an orbiting star, that error could be significant and should >>> be periodic. >> >>What's it like to discover all these possibly-could-may-be important >>effects from the comfort of your arm chair without having to do things >>like actually look through a telescope? >> >>Must be pretty neat. > > When are you going to say something intelligent? When are you going to explain what you were doing when you posted forged degrees? > > > Henry Wilson... > > .......A person's IQ = his snipping ability.
From: Paul B. Andersen on 24 Apr 2010 17:35 On 22.04.2010 23:58, Henry Wilson DSc wrote: > On Thu, 22 Apr 2010 16:01:02 +0200, "Paul B. Andersen" > <paul.b.andersen(a)somewhere.no> wrote: > >> On 22.04.2010 00:15, Henry Wilson DSc wrote: >>> On Wed, 21 Apr 2010 16:26:26 +0200, "Paul B. Andersen" >>> <paul.b.andersen(a)somewhere.no> wrote: >>> > > > snip > >>>> >>>> No, no, no. >>>> The velocity of the star - straight or curved - doesn't >>>> affect the direction in which we see the star. >>>> >>>> But of course will we see the star's position change with time >>>> if the star is moving! It doesn't matter if it is moving along >>>> a straight line or a curved one, as seen from the Sun, we would see >>>> the star where it was in the solar frame when the light was emitted. >>>> This is the position which is charted in the star catalogues. >>>> This motion is proper motion, it is _not_ stellar aberration. >>>> >>>> From the Earth, we will see the star moving around that position >>>> in an annual ellipse with major axis 41". The centre of that ellipse >>>> is the position as it would be seen from the Sun - the position you find >>>> in the stellar catalogues. >>> >>> Yes I am aware of that. >>> >>>> If the star is moving - (proper motion) - then the ellipse would move along. >>>> The resulting curve isn't an ellipse, it is a sum of the proper motion >>>> and stellar aberration (and parallax). >>>> >>>> Do you still not understand that the _fact_ that spectroscopic binaries >>>> always appear as one single star, despite the fact that the stars in >>>> the binaries are moving along curved paths, and with opposite velocities >>>> prove that "In the case of an orbiting star" no periodic error is introduced. >>>> IT DOESN'T MATTER IF YOUR IMAGINARY LASERS WOULD HAVE TO BE >>>> SWUNG AROUND ALL THE TIME: >>> >>> I understand that at long distances this is correct. >>> >>>> So can we now agree that the aberration of your imaginary >>>> star, when it is at its original 100LY distance, will be 41" >>>> just like any other star, and that the annual 0.062" circular >>>> proper motion of your star is negligible? >>>> >>>> Or do you still claim that there would be no stellar aberration >>>> of your imaginary star, and that is thus will appear to move >>>> 20.5" back and forth relative to other stars? >>> >>> No Paul. At long distances, 'photons' from a star that is verticallly above >>> arrive traveling almost vertically wrt the earth's plane. Therefore the >>> telescope has to lean +/-21" and aberration exists. >> >> So now you agree that the aberration of your imaginary star is >> just like the aberration of any other star. >> >> Remarkable! >> >> So now you will probably claim that you never disagreed with me in >> the first place, so we have had this long discussion about nothing. Wasn't I right, or was I right? :-) > Well, I did know the answer all along...but it hasn't been for nothing. > > The real motive for this thread was to point out the folly of using rotating > frames. > I said originally that since the star and earth were always MAR, it stands to > reason that the telescope should always point directly at the star and there > should be no aberration. So you did. But admitting having done a blunder is no option for Doctor Ralph Rabbidge aka Henry Wilson; better then to claim that you never meant what you said. :-) > That is what one might conclude, using the rotating frame....(That of the > orbit). > > Your Sagnac argument is identical to this. You say, if you regard a ring gyro > in its rotating frame, it never rotates..... and there can be no fringe > displacement according to BaTh. > > Do you now see how stupid your Sagnac 'BaTh refutation' is? Frustrating not to be able to find any errors in it, so you have to resort to stupidities like the above, eh? :-) See the interferometers that never rotates: http://home.c2i.net/pb_andersen/pdf/sagnac_ring.pdf http://home.c2i.net/pb_andersen/pdf/four_mirror_sagnac.pdf http://home.c2i.net/pb_andersen/FourMirrorSagnac.html >>> Incidentally, your previous claim that the aberration angle is known so >>> accurately that it can be used to refute BaTh is quite wrong. >>> >>> The standard aberration angle has been CALCULATED to 7 significant figures >>> based on Earth's orbit parameters....but there is no way it can be MEASURED to >>> anything like that accuracy. >> >> The radial velocity of some close binaries is varying with >> an amplitude of 300 km/s. So according to the emission theory, >> the speed of light from those stars should vary by c(1 +/- 10^-3). >> That means that the aberration angle v/c should change by +/-10^-7 radians. >> These binaries have a short period (days). 10^-7 radians = 0.02" = 20mas. >> That means that these stars should move 20mas back and forth relative to >> other stars in the region, with a period in the order of days. >> Hipparcos could measure such a motion with a precision of 1.5 mas. > > All quoted stellar orbit periods and velocities are likely to be wrong. > > Rotation frequencies can be exaggerated by factors of one hundred or more due > to time compression. No, it can not, not even according to the emission theory. 'Time compression' is according to the emission theory caused by acceleration of the source. If the orbital period was to be "exaggerated by factors of one hundred times or more" the whole orbiting system would have to be accelerated much more than the acceleration of the orbiting star. Which of course is ridiculous. We have discussed this before, you obviously still don't understand what the emission theory predicts. > Calculated velocities, using spectral line shifts, can also be out by large > amounts because wavelength shifts are affected far more by ADoppler than > VDoppler...and nobody has even heard of the former. > >> So yes, if such an aberration anomaly had existed, Hipparcos would >> have detected it. >> It didn't. > > Since I am the only person on this Earth who understands ADoppler, it is not > surprising that almost ALL astronomy is completely wrong. Your "Adoppler" is the same as your "time compression", namely the Doppler shift due to source acceleration predicted by the emission theory. And of course you are not the only person that knows this, it is a rather trivial and obvious consequence of the emission theory. But all astronomers know of course that the emission theory is falsified, so there is no "Adoppler" in the real world. On Fri, 07 Apr 2006 "Paul B. Andersen" wrote: | There is but one type of what you call "time compression", | and it is basically the same phenomenon as the Doppler shift. | And it isn't hard to calculate at all. | If an event happens at the source at the time t, | and this event is observed by the observer at the time to, | then the "time compression" simply is : dto/dt. | | Let me first demonstrate this "time compression" | when we assume that the speed of light is c in | the observer's rest frame. (In a Galilean world.) | The observed object is moving with the radial speed v. | We define positive v as approaching. | At t = 0, the distance to the object is D. | At time t, the distance to the object is | D - (integral from 0 to t of) v(t)*dt | I will write the latter term as I(vdt). | So an event happening at the time t will be observed | by the observer at the time: | to = t + (D - I(vdt))/c | dto/dt = 1 - v/c = (c - v)/c | This is observed "time compression". | Since a period of the emitted light is "compressed" | by the factor dto/dt, the Doppler shift is the inverse | of this: DS = c/(c - v) | | Now, let us do the same using the ballistic theory. | The difference is only that now the speed of light | in the observer's rest frame is c+v. | Thus: | to = t + (D - I(vdt))/(c+v) | If we assume that v/c << 1, we can write the approximation: | to = t + (D - I(vdt))*(1 - v/c)/c | dto/dt = 1 - v/c + (v/c)^2 - ((D + I(vdt))/c^2)*dv/dt | We can ignore the (v/c)^2 term. | And if we assume that that the object is in orbit, | the displacement I(vdt) can be ignored compared to D. | (The radius of the orbit is small compared to the distance | to the star.) | So we get: | dto/dt = 1 - v/c - (D/c^2)*dv/dt | | Note one very important point, though. | This equation says that the light that is | emitted at the time t will be observed | at the rate dto/dt, it says nothing about | _when_ the observation is done. | (Meaning that dto/dt doesn't have to vary | sinusoidally with to even if v(t)is sinusoidal.) | | The Doppler shift is the inverse of this: | DS = 1/(1 - v/c - (D/c2)*dv/dt) | | If we can assume that (D/c^2)*dv/dt << 1, | then this can be written: | DS = 1 + v/c + (D/c^2)*dv/dt | | Note however that according to the ballistic | theory, this assumption can NOT generally be done. | (Se remark below.) | | If you analyse under which conditions the ballistic | theory predicts that the intensity will be infinite, | you will find that it is when it predicts that | the Doppler shift also is infinite. | It is a good reason for that. When light emitted | at the time t is received at the same time as | the light emitted a time dt later, the "time compression" | dto/dt = 0, which means that a period is compressed to nothing, | the observed frequency is infinite, and the intensity | is infinite. | If the light emitted at the time t is received _after_ | the light emitted dt later, the "time compression" | dto/dt is negative (time is reversed), and the Doppler | shift is negative. | If this happens, we will at the same time observe | light emitted at another time (multiple stars), | with a different Doppler shift. | | But the bottom line is that it is a one to one | relationship between the predicted brightening of | the star and the Doppler shift. > > Tell me Paul, how is it that astronomers cannot explain why cepheid velocity > curves are a virtual mirror image of their brightness curves? But they can. > > I know the answer, Paul. ......ADoppler...or 'WaSh' (the Wilson > acceleration Shift) You are dead wrong. Yet another demonstration of your failure to understand what the emission theory predicts. Remember that 'the velocity curve' is calculated from the assumption that the Doppler shift is f = fo(1 + v/c). It the Doppler shift that is observed. -------------------------------------- For delta Cep the observed Doppler shift is about f = fo(1 +/- 0.7E-4). If we assume that the Cepheid really is an orbiting star with Doppler shift as observed, then the emission theory predicts that the brightness variation should be 1.4E-4, and the brightness should be maximum when the Doppler shift is maximum. The observed brightness variation is 2 (0.8 magnitude variation), and the brightness is minimum when the Doppler shift is maximum. So the emission theory gets the brightness variation wrong by more than 4 orders of magnitude, and the phase wrong by 180 degrees. -- Paul http://home.c2i.net/pb_andersen/
From: Paul B. Andersen on 24 Apr 2010 17:58 On 20.04.2010 23:57, Henry Wilson DSc wrote: > > I maintain that 'true trajectory' is y versus t for constant x....and that can > only be meaningful in the object's own frame. A true gem. :-) The only 'true trajectory' is a point in the object's own rest frame. A true keeper! :-) -- Paul http://home.c2i.net/pb_andersen/
From: Henry Wilson DSc on 24 Apr 2010 18:42
On Sat, 24 Apr 2010 23:58:07 +0200, "Paul B. Andersen" <someone(a)somewhere.no> wrote: >On 20.04.2010 23:57, Henry Wilson DSc wrote: >> >> I maintain that 'true trajectory' is y versus t for constant x....and that can >> only be meaningful in the object's own frame. > >A true gem. :-) > >The only 'true trajectory' is a point in the object's own rest frame. > >A true keeper! :-) Is you definition of 'trajectory' a plot of y-t or y-x? Come on! Answer! Henry Wilson... ........A person's IQ = his snipping ability. |