The Emission Theory of Androcles
in [Relativity]
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From: Jonah Thomas on 5 Sep 2009 19:04 "Androcles" <Headmaster(a)Hogwarts.physics_n> wrote: > "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message > > 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. If the movement of the particle I highlighted in red is what's seen, then the star is seen behind where it actually is. But if the movement of the waves are what's seen, then the star appears to be exactly in the direction that it actually is. (Because it hasn't deviated from that path. It could go somewhere else and you wouldn't find out until the later light arrives to tell you. But the old light moves with the old velocity, and it points to where the star is now -- if it's the waves that matter.) > > http://yfrog.com/02starmg > The star is seen behind where it actually is. If the movement of the particle I highlighted in red is what you see, then the star is seen behind where it actually is. But if the movement of the waves are what's seen, then the star appears to be in exactly the direction that it actually is. Either way. > > 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? With the source traveling at 0.5c, in the one case you get light particles traveling toward you at .877c while the wave is traveling at c. In the other case you get light particles traveling toward you at c while the wave is traveling at .877c. At least one of those cases you will get a frequency shift or a wavelength shift or both. Is there one where that won't happen? I say it ought to be the one I labeled "start" and not "starm" because it is exactly like the sound model where the source is at rest wrt the air. I don't know anybody who's actually listened much while the air was rushing past him at half the speed of sound, but the equation says to expect no doppler shift then. Yesterday I was convinced it was correct. Now I wonder whether that applies to light. And Wilson says you have to point your telescope off at an angle, and the waves will be refracting something weird that way if it works. > > 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 I see how that would apply if you were traveling at 0.5c along with the source. The problems that Einstein had to invent SR for just vanish with EmT. Your picture would fit perfectly. But right now we're looking at the case where the observer is not traveling parallel to the source, and if light travels in straight lines like particles who have additive velocities, and it isn't aimed straight down the telescope, then that light is not going to reach the bottom unreflected.
From: Androcles on 5 Sep 2009 20:08 "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20090905190405.02094869.jethomas5(a)gmail.com... > "Androcles" <Headmaster(a)Hogwarts.physics_n> wrote: >> "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message >> > 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. > > If the movement of the particle I highlighted in red is what's seen, > then the star is seen behind where it actually is. > > But if the movement of the waves are what's seen, then the star appears > to be exactly in the direction that it actually is. Nope. You see the star at 1:01 o'clock where it was at 12:00 o'clock, it took an hour and a minute for the light to reach you. Construct a table. Time of emission. theta. Time of arrival: 12:00 89 12:00 + 1-cos (89) 1:00 90 1:00 + 1- cos(90) 2:00 91 2:00 + 1- cos (91) 3:00 92 3:00 + 1- cos (92) If the next "wave" or circle is seen at a moment later it is seen in a different position. > (Because it hasn't > deviated from that path. It could go somewhere else and you wouldn't > find out until the later light arrives to tell you. But the old light > moves with the old velocity, and it points to where the star is now -- > if it's the waves that matter.) It's simple vector addition. >> > http://yfrog.com/02starmg >> The star is seen behind where it actually is. > > If the movement of the particle I highlighted in red is what you see, > then the star is seen behind where it actually is. > > But if the movement of the waves are what's seen, then the star appears > to be in exactly the direction that it actually is. > > Either way. The red observer can only see each circle when it arrives, the time of arrival is constantly changing and so is the duration of arrival between circles. You are looking at the diagram and see the circles equidistant because you are a god looking down, but the red observer does NOT see them arrive at equal time intervals, only sees one circle at a time, only sees a point on the circle. >> > 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? > > With the source traveling at 0.5c, in the one case you get light > particles traveling toward you at .877c while the wave is traveling at > c. Wrong. You see 1.25c, it is blue shifted on approach. You are still making the hidden assumption that c cannot be exceeded, that's what confuses you and why you should condemn Einstein for the charlatan he was. Wilson will tell you the same and he's a used car salesman who'll sell you a used VW camper van he wants to get rid of. f' = f. c+v.cos(60)/c = 1 + 0.5 * 0.5 / 1 = f* 1.25 As the star leaves and passes through 60 degrees, f' = f *0.75, red shifted. When the star is crossing the T, f' = f * 0.5 * cos(90) /c = 1 + 0.5 * 0.0 / 1 = f * 1 > In the other case you get light particles traveling toward you at c > while the wave is traveling at .877c. > > At least one of those cases you will get a frequency shift or a > wavelength shift or both. Is there one where that won't happen? I say it > ought to be the one I labeled "start" and not "starm" because it is > exactly like the sound model where the source is at rest wrt the air. I > don't know anybody who's actually listened much while the air was > rushing past him at half the speed of sound, but the equation says to > expect no doppler shift then. Yesterday I was convinced it was correct. > Now I wonder whether that applies to light. And Wilson says you have to > point your telescope off at an angle, and the waves will be refracting > something weird that way if it works. You do have to point it at an angle, that's aberration. http://www.androcles01.pwp.blueyonder.co.uk/Wave/Aberration.gif The Earth moves at 0.0001c in its orbit around the Sun so the angle is 0.0001 radians from the vertical, 20.6 arc minutes. In astronomy that's a lot. The only thing that is weird is your mathematics that Einstein tripped you up with. > >> > 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 > > I see how that would apply if you were traveling at 0.5c along with the > source. The problems that Einstein had to invent SR for just vanish with > EmT. Your picture would fit perfectly. But right now we're looking at > the case where the observer is not traveling parallel to the source, and > if light travels in straight lines like particles who have additive > velocities, and it isn't aimed straight down the telescope, then that > light is not going to reach the bottom unreflected. He's only got two choices, as do you. Either stand in the road and let the oncoming traffic wave hit you or stand to the side and let the traffic wave pass. If it passes you'll hear doppler shift, if it hits you, you won't hear anything except an ambulance wave and then only if you are lucky. It doesn't matter which way you are facing, either.
From: Inertial on 5 Sep 2009 20:28 "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 >> > The correct diagrams are as follows (hoping ASCII art works ok) From telescope point of view: Light emitted when star at S0 -S0------S->v . . . . . . . |.| telescope v=0 From star point of view: Light emitted from S when telescope at |0| S star v=0 . . . . . . . |.|-----|0| <-v
From: Jonah Thomas on 6 Sep 2009 00:56 "Androcles" <Headmaster(a)Hogwarts.physics_n> wrote: > "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message > > "Androcles" <Headmaster(a)Hogwarts.physics_n> wrote: > >> "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message > >> > hw@..(Henry Wilson, DSc) wrote: > >> >> 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. > > > > If the movement of the particle I highlighted in red is what's seen, > > then the star is seen behind where it actually is. > > > > But if the movement of the waves are what's seen, then the star > > appears to be exactly in the direction that it actually is. > > Nope. You see the star at 1:01 o'clock where it was at 12:00 o'clock, > it took an hour and a minute for the light to reach you. > > Construct a table. > Time of emission. theta. Time of arrival: > 12:00 89 12:00 + 1-cos > (89) 1:00 90 1:00 + 1- > cos(90) 2:00 91 2:00 + > 1- cos (91) 3:00 92 > 3:00 + 1- cos (92) > > If the next > "wave" or circle is seen at a moment later it is seen in a different > position. Sure, but it's passing through the right spot. At any moment the waves are pointing at the source, unless the source accelerates. Wherever you are, whatever your own velocity, the source's waves always face you. A result that's different from other theories except maybe SR. > > (Because it hasn't > > deviated from that path. It could go somewhere else and you wouldn't > > find out until the later light arrives to tell you. But the old > > light moves with the old velocity, and it points to where the star > > is now -- if it's the waves that matter.) > > It's simple vector addition. Yes, exactly. > >> > http://yfrog.com/02starmg > >> The star is seen behind where it actually is. > > > > If the movement of the particle I highlighted in red is what you > > see, then the star is seen behind where it actually is. > > > > But if the movement of the waves are what's seen, then the star > > appears to be in exactly the direction that it actually is. > > > > Either way. > > The red observer can only see each circle when it arrives, the time of > arrival is constantly changing and so is the duration of arrival > between circles. You are looking at the diagram and see the circles > equidistant because you are a god looking down, but the red observer > does NOT see them arrive at equal time intervals, only sees one circle > at a time, only sees a point on the circle. The observer looks blue to my browser. Yes, he only sees one at a time. But he *could* post pickets who would see them at other times and places. They could compare notes afterward. What I see from above is that the waves are equidistant and they look equidistant no matter what inertial frame you're in. The actual wavelength never changes, but if you measure wavelength by seeing how fast waves roll over you at one point then your measurements can change. > >> > 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? > > > > With the source traveling at 0.5c, in the one case you get light > > particles traveling toward you at .877c while the wave is traveling > > at c. > > Wrong. You see 1.25c, it is blue shifted on approach. You are > still making the hidden assumption that c cannot be exceeded, that's > what confuses you and why you should condemn Einstein for the > charlatan he was. Wilson will tell you the same and he's a used car > salesman who'll sell you a used VW camper van he wants to get rid of. I think you must be looking at the other case. I'm talking about starm, where the wave appears to be traveling at c while the particle I marked travels at sqrt(3)/2. In the "start" case, the particle of light is going at c to the side and at .5c forward, so the particle of light travels at sqrt(5)/2 and the wave front is traveling at c when the particle arrives. The particle is traveling faster than c, but the wave is only traveling straight toward the observer at c. And it's waves that get red-shifted or dopplered, which I haven't yet become certain are the same thing. > f' = f. c+v.cos(60)/c = 1 + 0.5 * 0.5 / 1 = f* 1.25 Is that when the waves appear to be coming in at 60 degrees? The light that left the source much earlier? > As the star leaves and passes through 60 degrees, > f' = f *0.75, red shifted. And this is the light that left the source at the T? > When the star is crossing the T, > f' = f * 0.5 * cos(90) /c = 1 + 0.5 * 0.0 / 1 = f * 1 This is the light that arrives at the observer when the star is crossing the T, that left the star when it was at -60 degrees? ** So anyway, is there any value in thinking about light particles that travel in straight lines, or is it better to just look at how the waves travel? ** > > In the other case you get light particles traveling toward you at c > > while the wave is traveling at .877c. I got that wrong. It should have been light at c*sqrt(5)/2 and the wave at c. > > At least one of those cases you will get a frequency shift or a > > wavelength shift or both. Is there one where that won't happen? I > > say it ought to be the one I labeled "start" and not "starm" because > > it is exactly like the sound model where the source is at rest wrt > > the air. I don't know anybody who's actually listened much while the > > air was rushing past him at half the speed of sound, but the > > equation says to expect no doppler shift then. Yesterday I was > > convinced it was correct. Now I wonder whether that applies to > > light. And Wilson says you have to point your telescope off at an > > angle, and the waves will be refracting something weird that way if > > it works. > > You do have to point it at an angle, that's aberration. > http://www.androcles01.pwp.blueyonder.co.uk/Wave/Aberration.gif But that isn't what's happening in this case, right? > The Earth moves at 0.0001c in its orbit around the Sun so the angle > is 0.0001 radians from the vertical, 20.6 arc minutes. In astronomy > that's a lot. The only thing that is weird is your mathematics that > Einstein tripped you up with. My current problems are probably pre-Einstein. > >> > 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 > > I see how that would apply if you were traveling at 0.5c along with > > the source. The problems that Einstein had to invent SR for just > > vanish with EmT. Your picture would fit perfectly. But right now > > we're looking at the case where the observer is not traveling > > parallel to the source, and if light travels in straight lines like > > particles who have additive velocities, and it isn't aimed straight > > down the telescope, then that light is not going to reach the bottom > > unreflected. > He's only got two choices, as do you. Either stand in the road and > let the oncoming traffic wave hit you or stand to the side and let > the traffic wave pass. If it passes you'll hear doppler shift, if it > hits you, you won't hear anything except an ambulance wave and then > only if you are lucky. It doesn't matter which way you are facing, > either. I don't see how that relates.
From: Jonah Thomas on 6 Sep 2009 01:06
"Inertial" <relatively(a)rest.com> wrote: > "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message > > 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 > >> > > > The correct diagrams are as follows (hoping ASCII art works ok) > > From telescope point of view: > > Light emitted when star at S0 > > -S0------S->v > . > . > . > . > . > . > . > |.| telescope v=0 I don't understand. If EmT is correct, then everybody would agree that when the source travels at v then the light it emits will inherit that velocity. The light would not just go straight from S0 when the source has a velocity. I think your picture only represents what might happen if EmT is not correct. > From star point of view: > > Light emitted from S when telescope at |0| > > S star v=0 > . > . > . > . > . > . > . > |.|-----|0| <-v |