The Emission Theory of Androcles
in [Relativity]
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From: Henry Wilson, DSc on 4 Sep 2009 17:24 On Fri, 4 Sep 2009 11:58:46 +0100, "Androcles" <Headmaster(a)Hogwarts.physics_n> wrote: > >"Henry Wilson, DSc" <hw@..> wrote in message >news:6ok1a5hqp8tmig1g06d5gujlb7flio3390(a)4ax.com... >>>I'm not completely clear about all the assumptions for doppler in sound. > >Well, I'm sorry to hear that. Never mind, H. Just see my post to Jonah, >he said the same thing as you. >I will give you some advice, though. If you snip and attribute what >Jonah said as if I said it, I'll snip and attribute what Jonah said >as if you said it. I didn't do it. I replied to his post only. i didn't quote you at all nor did I refer to anything you said. I'm your 'friend' remember. Henry Wilson...www.users.bigpond.com/hewn/index.htm Einstein...World's greatest SciFi writer..
From: Androcles on 4 Sep 2009 17:49 "Henry Wilson, DSc" <hw@..> wrote in message news:rb13a59fcipkc76c8gu5b1h6om23jk2kvi(a)4ax.com... > On Fri, 4 Sep 2009 11:58:46 +0100, "Androcles" > <Headmaster(a)Hogwarts.physics_n> > wrote: > >> >>"Henry Wilson, DSc" <hw@..> wrote in message >>news:6ok1a5hqp8tmig1g06d5gujlb7flio3390(a)4ax.com... >>>>I'm not completely clear about all the assumptions for doppler in sound. >> >>Well, I'm sorry to hear that. Never mind, H. Just see my post to Jonah, >>he said the same thing as you. >>I will give you some advice, though. If you snip and attribute what >>Jonah said as if I said it, I'll snip and attribute what Jonah said >>as if you said it. > > I didn't do it. -- the little boy said with chocolate crumbs all over his > face. > I replied to his post only. i didn't quote you at all nor did I > refer to anything you said. I'm your 'friend' remember. Peculiar how my name was at the top where it said On Thu, 3 Sep 2009 20:09:50 -0400, Jonah Thomas <jethomas5(a)gmail.com> wrote: >"Androcles" <Headmaster(a)Hogwarts.physics_n> wrote: > >I'm not completely clear about all the assumptions for doppler in sound. --which I did NOT write. Anyway, you are more of a drinking buddy than a friend, ya daft old sheep shagger. You certainly don't know how to measure a wavelength -- only you can be in two places at once.
From: Jonah Thomas on 4 Sep 2009 19:17 hw@..(Henry Wilson, DSc) wrote: > Jonah Thomas <jethomas5(a)gmail.com> wrote: > >hw@..(Henry Wilson, DSc) wrote: > > > >> In EmT, a star that is moving sideways will be emitting light in > >all> directions. When you see it APPARENTLY moving sideways > >perpendicular> to your LOS, it will in fact be well ahead of that > >position. > > > >I don't think so, now that I've thought about it. When it appears to > >you that the star is moving sideways perpendicular to your LOS, it > >will in fact be perpendicular to your LOS. See picture. > > A point cannot be 'perpendicular'. You should think before you write. > Did you mean that 'its velocity vector IS perpendicular when it > APPEARS to be perpendicular'? > > I don't think so. > > >The photons that enter > >> your telescope barrel and move down its central axis will have in > >fact> been emitted at an angle arctan(v/c) leaning back from the > >star's> velocity vector. > >> > >> --------------------S->v > >> .->v > >> .->v > >> .->v > >> .->v > >> .->v > >> || telescope v=0 > > > >You're having the velocity go to the right here, aren't you? in that > >case your picture is backward. > > It isn't. Now I think you're right and I'm wrong. I hate it when that happens. > You see the light that goes down the centre of your telescope. If your > diagram was correct none would get to the bottom. You should think > before you write. That makes sense. I had it rationalised out with calculus, too. But now I see it your way. Clearly I don't really have this worked out yet. > >> Light moves at c wrt its source and the whole line of photons is > >> moving > >> sideways at v in your frame. It's speed down your telescope is > >> sqrt(c^2 - v^2). Therefore EmT would expect a red shift. > > > >Its speed would be sqrt(c^2-v^2) if the light reached you from the > >source after the source had passed you. Instead the light is > >traveling at (c^2+v^2) and c^2 in your direction. > > The ray travels at c in the star's frame leaning backwards as I've > shown. It has a component velocity v perpendicular to your LOS. The > resultant velocity is sqrt(c^2-v^2) in your frame along your > LOS...ie., straight down the centre of your telescope. Yes, I see that now. So the velocity has to be slow, at sqrt(c^2-v^2) and you do get a shift in the light. How much redshift do you expect? Since wavelength * frequency = speed if the speed is slow -- sqrt(c^2-v^2) -- and the frequency is the same, then the wavelength must be shorter, right? That would be a blueshift. But maybe the wavelength is the same and then the frequency must be slowed. Or somewhere inbetween. Does it matter? When the speed slows do you get a shift in interference pattern regardless what combination of frequency and wavelength change it causes? > Light speed is modified as it travels through any rare medium. That's > called extinction...not a very appropriate name I agree. Variable star > data suggests that all light traveling in a particular direction tends > toward a common speed.(Andro strongly disagrees) I can certainly imagine that. I was imagining what happens when light at c+v enters the earth's atmosphere, where particles do cerenkov radiation because lightspeed has turned so slow. You can't have superfast light in atmosphere, and interstellar space isn't completely empty either. > If you want to see how the brightness of orbiting stars should vary > due to the bunching and separation of c+v light you can spend some > time running mty very comprehensive program that does all the > calculations for you. www.users.bigpond.com/hewn/variables.exe > It is not a virus. I do better with source code, but there's no hurry. I have more than enough to occupy my limited spare time for awhile. > >Clearly you guys have put a lot of thought into this. And it looks > >like it takes a lot of thought to design an experiment that tells the > >difference. I find it surprising that systems which at first sight > >look so radically different would be so hard to distinguish. > > It boils down to the difficulty in measuring OWLS from a moving > source. Only very recently have the means become available...but of > course no one will get the funding because the physics establishment > is dominated by diehard Einsteinians. Any chance that the data might become available as a side result from something else? Worth watching for....
From: Jonah Thomas on 5 Sep 2009 16:51 hw@..(Henry Wilson, DSc) wrote: > Jonah Thomas <jethomas5(a)gmail.com> wrote: > > hw@..(Henry Wilson, DSc) wrote: > >The photons that enter > >> your telescope barrel and move down its central axis will have in > >fact> been emitted at an angle arctan(v/c) leaning back from the > >star's> velocity vector. > >> > >> --------------------S->v > >> .->v > >> .->v > >> .->v > >> .->v > >> .->v > >> || telescope v=0 > > > >You're having the velocity go to the right here, aren't you? in that > >case your picture is backward. > > It isn't. > > > > > ----------S->v....S > > .->v > > .->v > > .->v > > .->v > > .->v > > .->v > > .->v > > .->v > > || telescope v=0 > > > >This is light that moves at c toward you, and at v to the side. In > >the time that the light reaches you at an angle, the source moves > >just enough that it is lined up too. > > You see the light that goes down the centre of your telescope. If your > diagram was correct none would get to the bottom. You should think > before you write. I am still very confused, but at least I'm getting it clearer what it is I'm confused about. http://yfrog.com/5ystartg http://yfrog.com/02starmg Here are two different cases. The yellow source is traveling at about 0.5c compared to the stationary observer, who is 1 distance unit away at the closest approach. It takes 1 time unit for light to get from a stationary source at that closest spot to the observer. In one case the light leaves at a 60 degree angle when the source is still 0.5 distance unit from closest approach. Since the light travels at c+v, in one time unit it goes 1 distance unit sideways and 0.5 distance unit forward, and reaches the observe after 1 time unit, when the source is at the closest approach. The point on the wavefront marked in red has traveled at this 60 degree angle the whole distance and presumably will keep traveling at that angle. The wavefront at this moment is at 90 degrees, precisely facing the observer. In the other case the light leaves at a 90 degree angle when the source is at closest approach. It is leaving at a -60 degree angle relative to the source, but because of the source's forward motion which carries over to the light, it actually travels at a 90 degree angle and will be traveling in that direction when it reaches the observer. The wavefront, though, is moving at a 60 degree angle at that time relative to the observer. The light traveles at sqrt(3)/2 or about 0.87 c. So, what direction do you think the light is coming from, when the individual elements are coming at one angle but the wavefront comes from another angle? I know the answer for sound -- it sounds like the sound is coming perpendicular to the wavefront. Sound is a compression wave, and you get the direction by the delay in compression for one ear compared to the other. If there's a high wind and the whole compression wave is going sideways compared to the direction of the wavefront you mostly don't notice -- the new air is as compressed as the old air. Does that work for light too? I'm confused. At first thought it seems like it ought to, but as you point out that would require light that's been going in one direction to make a sudden 30 degree turn when it enters your telescope. The other picture shows the on-the-other-hand. These photons have been traveling straight at .87c, and the ones a little bit to either side have been traveling faster and slower. The wavefront is at 60 degrees, not 90 degrees. Interference and redshifts are related problems. In the first case if you point your telescope at a 60 degree angle (or whatever works) you should get light that is traveling toward you at sqrt(c^2+v^2). Its wavelength in the direction of the waves' apparent source is the original wavelength -- if you look at the picture at any one time you can't tell it from a stationary source. The frequency at this time is the original frequency, though it has been higher and is falling. Do you get a blueshift? The second picture gives you waves traveling at c at -60 degrees, but individual particles in those waves are traveling at 90 degrees at 0.87 c. Very confusing. My intuition says to pay attention to the waves and not the particles. But that could easily be wrong.
From: Androcles on 5 Sep 2009 17:33
"Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20090905165157.5d62db7b.jethomas5(a)gmail.com... > hw@..(Henry Wilson, DSc) wrote: >> Jonah Thomas <jethomas5(a)gmail.com> wrote: >> > hw@..(Henry Wilson, DSc) wrote: > >> >The photons that enter >> >> your telescope barrel and move down its central axis will have in >> >fact> been emitted at an angle arctan(v/c) leaning back from the >> >star's> velocity vector. >> >> >> >> --------------------S->v >> >> .->v >> >> .->v >> >> .->v >> >> .->v >> >> .->v >> >> || telescope v=0 >> > >> >You're having the velocity go to the right here, aren't you? in that >> >case your picture is backward. >> >> It isn't. >> >> > >> > ----------S->v....S >> > .->v >> > .->v >> > .->v >> > .->v >> > .->v >> > .->v >> > .->v >> > .->v >> > || telescope v=0 >> > >> >This is light that moves at c toward you, and at v to the side. In >> >the time that the light reaches you at an angle, the source moves >> >just enough that it is lined up too. >> >> You see the light that goes down the centre of your telescope. If your >> diagram was correct none would get to the bottom. You should think >> before you write. > > I am still very confused, but at least I'm getting it clearer what it is > I'm confused about. > > http://yfrog.com/5ystartg The star is seen behind where it actually is. > http://yfrog.com/02starmg The star is seen behind where it actually is. > > Here are two different cases. The yellow source is traveling at about > 0.5c compared to the stationary observer, who is 1 distance unit away at > the closest approach. It takes 1 time unit for light to get from a > stationary source at that closest spot to the observer. > > In one case the light leaves at a 60 degree angle when the source is > still 0.5 distance unit from closest approach. Since the light travels > at c+v, in one time unit it goes 1 distance unit sideways and 0.5 > distance unit forward, and reaches the observe after 1 time unit, when > the source is at the closest approach. The point on the wavefront marked > in red has traveled at this 60 degree angle the whole distance and > presumably will keep traveling at that angle. The wavefront at this > moment is at 90 degrees, precisely facing the observer. > > In the other case the light leaves at a 90 degree angle when the source > is at closest approach. It is leaving at a -60 degree angle relative to > the source, but because of the source's forward motion which carries > over to the light, it actually travels at a 90 degree angle and will be > traveling in that direction when it reaches the observer. The wavefront, > though, is moving at a 60 degree angle at that time relative to the > observer. The light traveles at sqrt(3)/2 or about 0.87 c. > > So, what direction do you think the light is coming from, when the > individual elements are coming at one angle but the wavefront comes from > another angle? I know the answer for sound -- it sounds like the sound > is coming perpendicular to the wavefront. Sound is a compression wave, > and you get the direction by the delay in compression for one ear > compared to the other. If there's a high wind and the whole compression > wave is going sideways compared to the direction of the wavefront you > mostly don't notice -- the new air is as compressed as the old air. > > Does that work for light too? I'm confused. At first thought it seems > like it ought to, but as you point out that would require light that's > been going in one direction to make a sudden 30 degree turn when it > enters your telescope. > > The other picture shows the on-the-other-hand. These photons have been > traveling straight at .87c, and the ones a little bit to either side > have been traveling faster and slower. The wavefront is at 60 degrees, > not 90 degrees. The star is seen behind where it actually is in both images, and in the direction where it was. What other hand? > Interference and redshifts are related problems. In the first case if > you point your telescope at a 60 degree angle (or whatever works) you > should get light that is traveling toward you at sqrt(c^2+v^2). Its > wavelength in the direction of the waves' apparent source is the > original wavelength -- if you look at the picture at any one time you > can't tell it from a stationary source. The frequency at this time is > the original frequency, though it has been higher and is falling. Do you > get a blueshift? > > The second picture gives you waves traveling at c at -60 degrees, but > individual particles in those waves are traveling at 90 degrees at 0.87 > c. Very confusing. > > My intuition says to pay attention to the waves and not the particles. > But that could easily be wrong. You don't have any waves, you have expanding circles. Aberration of light: http://www.androcles01.pwp.blueyonder.co.uk/Wave/Aberration.gif |