The Emission Theory of Androcles
in [Relativity]
|
From: Androcles on 6 Sep 2009 03:40 "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20090906005619.672ec82d.jethomas5(a)gmail.com... > "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. You have to "face" where the source was when the signal was emitted, then it's the shortest distance. Sound and light signals have different speeds, a deaf man sees the source where it was, a blind man hears the source where it was, you see and hear both but they are not simultaneous. This prompts you to say you hear the sound "behind" the source and me to say I hear the source from where it came from. >> > (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. Ok, I was writing from my faulty memory of colours. > Yes, he only sees one at a time. In http://yfrog.com/02starmg he only sees the first one at one time. The trail of red squares doesn't exist, they are all history. I drew a history trail like yours in http://www.androcles01.pwp.blueyonder.co.uk/Wave/Aberration.gif but when I animated it I deleted all the cubes except one in each frame. So one image shows the light hits the edge, but it really hits the centre and travels down the axis. This is because the one cube cannot be in two places at once. > But he *could* post pickets who would see them at other times and > places. The pickets would need to be on Mars, Venus, a Jovian Moon. Perhaps you'd better work out how to get a picket to Mars before you say "could" so glibly. > 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. > The historical wavelength never changes wrt the source, the measured wavelength is impossible because the photon can't be in two places at once. Your trail of red squares is a wiggle in timespace. >> >> > 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? No. I'm not interested in comparing where the star is now with where it was last seen. Prophecies are for palmists, fortune tellers, tea leaf readers, horoscope writers and relativists. I can't see where it is now, I'll have to wait for the light to reach me. >> 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? No. I'm not interested in comparing where the star is now with where it was last seen. Prophecies are for palmists, fortune tellers, tea leaf readers, horoscope writers and relativists. I can't see where it is now, I'll have to wait for the light to reach me. > ** 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? ** They don't travel in straight lines. This ball doesn't travel in a straight line. http://ww2010.atmos.uiuc.edu/(Gh)/guides/mtr/fw/gifs/coriolis.mov (Well, it does in the playground, but not for the observer on the carousel). Since you are on the moving Earth, nor does light. http://www.androcles01.pwp.blueyonder.co.uk/Shapiro/Crapiro.htm Being god, my frame of reference is different to you mere mortals riding on the blue ball. > >> > 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. You mean the history wave? >> > 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? Yes it is. http://www.androcles01.pwp.blueyonder.co.uk/1st/Postulates.htm You are expected to know what relative motion is, Einstein can't tell you. > >> 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. It won't help that the idiot gave you a bum steer. >> >> > 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: Inertial on 6 Sep 2009 08:41 "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20090906010601.11d351bb.jethomas5(a)gmail.com... > "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. It inherits direction, not the speed > The light would not just go straight from S0 when the source > has a velocity. See the following diagram to see how that same light appears from the source. It has to have a component going backwards with speed v so that it does not move relative to the telescope. > I think your picture only represents what might happen > if EmT is not correct. No .. it is correct regardless. That is how light MUST be travelling if it goes down the telescope. It MUST be parallel to the telescope tube. >> 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 10:52 "Androcles" <Headmaster(a)Hogwarts.physics_n> wrote: > "Jonah Thomas" <jethomas5(a)gmail.com> wrote > > "Androcles" <Headmaster(a)Hogwarts.physics_n> wrote: > >> "Jonah Thomas" <jethomas5(a)gmail.com> wrote > >> > "Androcles" <Headmaster(a)Hogwarts.physics_n> wrote: > >> >> "Jonah Thomas" <jethomas5(a)gmail.com> wrote > >> >> > 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. > > You have to "face" where the source was when the signal was emitted, > then it's the shortest distance. That's what I'm running into. If you face where the source was when the signal was emitted, that's the direction I'd expect particles emitted from the source to be coming from. But the wave front is coming from a different direction, because it travels at c+v' and v' varies with direction of travel. So any individual particle of light is traveling in a straight line from the source, and it doesn't matter to the particle that there are other particles that travel at other speeds nearby. But the wave -- a Mexican wave, maybe -- has an existence that's different from that of the particles, just like a mexican wave is different from the people who make it, and doesn't travel at the same speed they do. And if you measure interference or redshift you aren't measuring that on individual particles, you're measuring it on the wave. It's the speed of the wave that matters then. Unless the motion of the particles somehow gets in the way. > Sound and light signals have different > speeds, a deaf man sees the source where it was, a blind man hears > the source where it was, you see and hear both but they are not > simultaneous. > This prompts you to say you hear the sound "behind" the source and > me to say I hear the source from where it came from. Yes. But that's when the air you'r listening through isn't moving relative to you. When you're listening for that sound in a wind that's going 170 meters per second -- when you listen to somebody who's screaming while they're blown away -- it will sound different. It's like the case when you're traveling at half the speed of sound and the source is stationary. You get a doppler effect because you run into the waves at a different speed, but you run into each of them at the same angle you would if you were sitting still at that spot. (Or is there a calculus trick I should be applying?) > >> >> > 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 blue 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 blue > >observer> does NOT see them arrive at equal time intervals, only sees > >one circle> at a time, only sees a point on the circle. > > > > Yes, he only sees one at a time. > > In http://yfrog.com/02starmg he only sees the first one at one time. > The trail of red squares doesn't exist, they are all history. Yes. they are the history of the movement of one particle of light. Is that irrelevant? I feel like we're actually looking at the wave and not the movement of the particle through time. The movement of the particle seems completely irrelevant to me until you need it to go down a telescope. > I drew a history trail like yours in > http://www.androcles01.pwp.blueyonder.co.uk/Wave/Aberration.gif > but when I animated it I deleted all the cubes except one in each > frame. So one image shows the light hits the edge, but it really hits > the centre and travels down the axis. This is because the one cube > cannot be in two places at once. I may have missed your point. I saw the cylinder moving while the dot traveled, so that the dot could go down the center because of that movement. It doesn't matter that the wall of the cylinder hits places where the dot used to be, if the dot isn't there any more. I drew the history trails because I thought it mattered where the particles came from. But if I ignore the particles and just look at the circles then everything seems to work out. It's only when I pay attention to the particles that I start having objections that need more theory to deal with them. > > But he *could* post pickets who would see them at other times and > > places. > > The pickets would need to be on Mars, Venus, a Jovian Moon. > Perhaps you'd better work out how to get a picket to Mars before > you say "could" so glibly. Ah, the argument from practicality. I want to look at what the theory says will happen. In a second line of advance I want to look at what we can observe if the theory is true. You are quibbling about what is practical for us to observe today, when we are mostly stuck on one planet immersed in air. The distances are so short that it's mostly practical to measure speed with interferometry.... > > 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. > The historical wavelength never changes wrt the source, the measured > wavelength is impossible because the photon can't be in two places at > once. Your trail of red squares is a wiggle in timespace. If you can't actually see a wave crest then you're stuck with indirect methods. For example -- make a standing wave. Say, with masers. Put something in the way that absorbs light of that frequency and look for the bands where it does absorb some. Or is it just impossible? > >> >> > 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? > > No. I'm not interested in comparing where the star is now with where > it was last seen. Prophecies are for palmists, fortune tellers, tea > leaf readers, horoscope writers and relativists. I can't see where it > is now, I'll have to wait for the light to reach me. When the light reaches you, the particles are presumably moving in straight lines from the source. But the waves are moving in a different direction, because every individual piece of them is moving at v in direction x in addition to their other movement. What we can measure is the wave. Do you point the telescope in the direction the particles came from, or in the direction of the wavefront? > >> 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? > > No. I'm not interested in comparing where the star is now with where > it was last seen. Prophecies are for palmists, fortune tellers, tea > leaf readers, horoscope writers and relativists. I can't see where it > is now, I'll have to wait for the light to reach me. I'm interested in which direction do you point the telescope. > > ** 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? ** > > > They don't travel in straight lines. This ball doesn't travel in a > straight line. > http://ww2010.atmos.uiuc.edu/(Gh)/guides/mtr/fw/gifs/coriolis.mov > > (Well, it does in the playground, but not for the observer on the > carousel). > > Since you are on the moving Earth, nor does light. > http://www.androcles01.pwp.blueyonder.co.uk/Shapiro/Crapiro.htm > Being god, my frame of reference is different to you mere mortals > riding on the blue ball. I don't see where coriolis applies to a source moving in a straight line in direction x at velocity v, and a stationary observer. > >> > 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. > > You mean the history wave? No, I mean the wave crest that the observer gets right now. The last wave crest came at a slightly different direction and speed, and the next one will be different too, but right now the wave is at 90 degrees and traveling 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? > > Yes it is. > http://www.androcles01.pwp.blueyonder.co.uk/1st/Postulates.htm > You are expected to know what relative motion is, Einstein can't tell > you. This is important. With other theories, the direction of the waves is always the direction of the light. That's what the direction of the light *means* to them. With EmT you can have light that's actually traveling in a different direction and speed from its waves. This complication is bound to have a whole lot of implications. For one, you will get kinds of interference etc that are forbidden to other theories, unless it turns out somehow that they cancel out. This is exciting! > >> >> > 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: Androcles on 6 Sep 2009 15:47 "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20090906105221.47149fe1.jethomas5(a)gmail.com... > "Androcles" <Headmaster(a)Hogwarts.physics_n> wrote: >> "Jonah Thomas" <jethomas5(a)gmail.com> wrote >> > "Androcles" <Headmaster(a)Hogwarts.physics_n> wrote: >> >> "Jonah Thomas" <jethomas5(a)gmail.com> wrote >> >> > "Androcles" <Headmaster(a)Hogwarts.physics_n> wrote: >> >> >> "Jonah Thomas" <jethomas5(a)gmail.com> wrote > >> >> >> > 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. >> >> You have to "face" where the source was when the signal was emitted, >> then it's the shortest distance. > > That's what I'm running into. If you face where the source was when the > signal was emitted, that's the direction I'd expect particles emitted > from the source to be coming from. > > But the wave front is coming from a different direction, because it > travels at c+v' and v' varies with direction of travel. You continue talking about waves. Waves are history. You standing on a beach and a 6' high roller comes in, you get very wet. Game over. It doesn't matter what the next roller does or whether you are facing the shore or facing out to sea. Put the shoe on the other foot. You are approaching a point just to the side of me whilst following a straight line. You brush past me and continue on in your straight line. It doesn't matter which way I'm facing, you come past me. > So any individual particle of light is traveling in a straight line from > the source, and it doesn't matter to the particle that there are other > particles that travel at other speeds nearby. But the wave -- a Mexican > wave, maybe -- has an existence that's different from that of the > particles, just like a mexican wave is different from the people who > make it, and doesn't travel at the same speed they do. And if you > measure interference or redshift you aren't measuring that on individual > particles, you're measuring it on the wave. It's the speed of the wave > that matters then. Unless the motion of the particles somehow gets in > the way. > >> Sound and light signals have different >> speeds, a deaf man sees the source where it was, a blind man hears >> the source where it was, you see and hear both but they are not >> simultaneous. >> This prompts you to say you hear the sound "behind" the source and >> me to say I hear the source from where it came from. > > Yes. But that's when the air you'r listening through isn't moving > relative to you. When you're listening for that sound in a wind that's > going 170 meters per second -- when you listen to somebody who's > screaming while they're blown away -- it will sound different. It's like > the case when you're traveling at half the speed of sound and the source > is stationary. You get a doppler effect because you run into the waves > at a different speed, but you run into each of them at the same angle > you would if you were sitting still at that spot. (Or is there a > calculus trick I should be applying?) No. All you need is f' = f. (c-v.cos(phi))/c >> >> >> > 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 blue 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 blue >> >observer> does NOT see them arrive at equal time intervals, only sees >> >one circle> at a time, only sees a point on the circle. >> > >> > Yes, he only sees one at a time. >> >> In http://yfrog.com/02starmg he only sees the first one at one time. >> The trail of red squares doesn't exist, they are all history. > > Yes. they are the history of the movement of one particle of light. Is > that irrelevant? Yes, that's your wave. You only get wet once. Another wave comes from a different direction, so it's irrelevant. Stand on a bridge over a flowing river, and drop pebbles in. You'll see circular ripples, but you'll also see the circles moving downstream with the flow of the river. > I feel like we're actually looking at the wave and not > the movement of the particle through time. The movement of the particle > seems completely irrelevant to me until you need it to go down a > telescope. > >> I drew a history trail like yours in >> http://www.androcles01.pwp.blueyonder.co.uk/Wave/Aberration.gif >> but when I animated it I deleted all the cubes except one in each >> frame. So one image shows the light hits the edge, but it really hits >> the centre and travels down the axis. This is because the one cube >> cannot be in two places at once. > > I may have missed your point. I saw the cylinder moving while the dot > traveled, so that the dot could go down the center because of that > movement. It doesn't matter that the wall of the cylinder hits places > where the dot used to be, if the dot isn't there any more. > > I drew the history trails because I thought it mattered where the > particles came from. But if I ignore the particles and just look at the > circles then everything seems to work out. It's only when I pay > attention to the particles that I start having objections that need more > theory to deal with them. > Make your theory fit the known facts. >> > But he *could* post pickets who would see them at other times and >> > places. >> >> The pickets would need to be on Mars, Venus, a Jovian Moon. >> Perhaps you'd better work out how to get a picket to Mars before >> you say "could" so glibly. > > Ah, the argument from practicality. I want to look at what the theory > says will happen. In a second line of advance I want to look at what we > can observe if the theory is true. You are quibbling about what is > practical for us to observe today, when we are mostly stuck on one > planet immersed in air. The distances are so short that it's mostly > practical to measure speed with interferometry.... Engineers do that. We find theories seldom work. >> > 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. > >> The historical wavelength never changes wrt the source, the measured >> wavelength is impossible because the photon can't be in two places at >> once. Your trail of red squares is a wiggle in timespace. > > If you can't actually see a wave crest then you're stuck with indirect > methods. > > For example -- make a standing wave. Say, with masers. Put something in > the way that absorbs light of that frequency and look for the bands > where it does absorb some. > > Or is it just impossible? Waves are history. http://paws.kettering.edu/~drussell/Demos/SHO/damp.html The mass is only travelling vertically, the wave is a historical document, recording where the mass was in the past. If you move the mass and spring sideways then you have a waveLENGTH, but it is INDEPENDENT of waveTIME. The mass could be a car and the car has springs and dampers that are misnamed "shock absorbers". The real shock absorbers are the springs and the tyres. >> >> >> > 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? >> >> No. I'm not interested in comparing where the star is now with where >> it was last seen. Prophecies are for palmists, fortune tellers, tea >> leaf readers, horoscope writers and relativists. I can't see where it >> is now, I'll have to wait for the light to reach me. > > When the light reaches you, the particles are presumably moving in > straight lines from the source. But the waves are moving in a different > direction, because every individual piece of them is moving at v in > direction x in addition to their other movement. What we can measure is > the wave. > > Do you point the telescope in the direction the particles came from, or > in the direction of the wavefront? http://www.androcles01.pwp.blueyonder.co.uk/JT.GIF Your red squares are the history of ONE photon. My green squares are the actual positions of a stream of photons, one after the other, all of which collide with the observer. > >> >> 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? >> >> No. I'm not interested in comparing where the star is now with where >> it was last seen. Prophecies are for palmists, fortune tellers, tea >> leaf readers, horoscope writers and relativists. I can't see where it >> is now, I'll have to wait for the light to reach me. > > I'm interested in which direction do you point the telescope. > >> > ** 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? ** >> >> >> They don't travel in straight lines. This ball doesn't travel in a >> straight line. >> http://ww2010.atmos.uiuc.edu/(Gh)/guides/mtr/fw/gifs/coriolis.mov >> >> (Well, it does in the playground, but not for the observer on the >> carousel). >> >> Since you are on the moving Earth, nor does light. >> http://www.androcles01.pwp.blueyonder.co.uk/Shapiro/Crapiro.htm >> Being god, my frame of reference is different to you mere mortals >> riding on the blue ball. > > I don't see where coriolis applies to a source moving in a straight line > in direction x at velocity v, and a stationary observer. http://www.androcles01.pwp.blueyonder.co.uk/JT.GIF I see a curved path. > >> >> > 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. >> >> You mean the history wave? > > No, I mean the wave crest that the observer gets right now. The last > wave crest came at a slightly different direction and speed, and the > next one will be different too, but right now the wave is at 90 degrees > and traveling 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? >> >> Yes it is. >> http://www.androcles01.pwp.blueyonder.co.uk/1st/Postulates.htm >> You are expected to know what relative motion is, Einstein can't tell >> you. > > This is important. With other theories, the direction of the waves is > always the direction of the light. That's what the direction of the > light *means* to them. With EmT you can have light that's actually > traveling in a different direction and speed from its waves. Waves are history, a trace of where a photon has been. Light is stream of photons, like bullets from a minigun. http://www.youtube.com/watch?v=tyF0G7g4KfY http://www.youtube.com/watch?v=9C5FYsjluJc > This > complication is bound to have a whole lot of implications. For one, you > will get kinds of interference etc that are forbidden to other theories, > unless it turns out somehow that they cancel out. > > This is exciting! > >> >> >> > 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 20:21
"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 > >> > "Androcles" <Headmaster(a)Hogwarts.physics_n> wrote: > >> >> "Jonah Thomas" <jethomas5(a)gmail.com> wrote > >> >> > "Androcles" <Headmaster(a)Hogwarts.physics_n> wrote: > >> >> >> "Jonah Thomas" <jethomas5(a)gmail.com> wrote > > > >> >> >> > 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. > >> > >> You have to "face" where the source was when the signal was > >emitted,> then it's the shortest distance. > > > > That's what I'm running into. If you face where the source was when > > the signal was emitted, that's the direction I'd expect particles > > emitted from the source to be coming from. > > > > But the wave front is coming from a different direction, because it > > travels at c+v' and v' varies with direction of travel. > > You continue talking about waves. Waves are history. You standing > on a beach and a 6' high roller comes in, you get very wet. Game over. > It doesn't matter what the next roller does or whether you are facing > the shore or facing out to sea. > Put the shoe on the other foot. You are approaching a point just to > the side of me whilst following a straight line. You brush past me > and continue on in your straight line. It doesn't matter which way I'm > facing, you come past me. I'm real unclear about all this. People talk about waves to explain interference. They figure that if you have a wave, and you split it into two parts and then let them rejoin, and one of them is one half cycle slower, or 1.5 cycles slower, or 2.5 cycles slower etc, then they will cancel and that's interference. To do this they assume that the wave is many cycles long because otherwise how could you get the part that arrives 2.5 or 3.5 or 4.5 cycles later to cancel out stuff that has already come and gone? Wave theory works for a variety of things. The doppler equations assume waves, don't they? It's a frequency shift you find with them. You don't just get hit by the first wave, game over. You get a series of sound waves and you get a frequency for them. I'm going to use waves until I see how to get those results without them. > > So any individual particle of light is traveling in a straight line > > from the source, and it doesn't matter to the particle that there > > are other particles that travel at other speeds nearby. But the wave > > -- a Mexican wave, maybe -- has an existence that's different from > > that of the particles, just like a mexican wave is different from > > the people who make it, and doesn't travel at the same speed they > > do. And if you measure interference or redshift you aren't measuring > > that on individual particles, you're measuring it on the wave. It's > > the speed of the wave that matters then. Unless the motion of the > > particles somehow gets in the way. > > > >> Sound and light signals have different > >> speeds, a deaf man sees the source where it was, a blind man hears > >> the source where it was, you see and hear both but they are not > >> simultaneous. > >> This prompts you to say you hear the sound "behind" the source and > >> me to say I hear the source from where it came from. > > > > Yes. But that's when the air you'r listening through isn't moving > > relative to you. When you're listening for that sound in a wind > > that's going 170 meters per second -- when you listen to somebody > > who's screaming while they're blown away -- it will sound different. > > It's like the case when you're traveling at half the speed of sound > > and the source is stationary. You get a doppler effect because you > > run into the waves at a different speed, but you run into each of > > them at the same angle you would if you were sitting still at that > > spot. (Or is there a calculus trick I should be applying?) > > No. All you need is f' = f. (c-v.cos(phi))/c Let's consider the different effects. You can estimate where sounds come from because you don't just have one ear at one place, you have two. If one of them gets the signal a little later than the other, tha gives you an estimate of left to right, though it does not tell you whether it's ahead or behind or above. (Subtle differences in sound caused by the shape of your ears help some with that.) Similarly, you have more than one sensor for light, your eye has a whole array of them. You focus the light with lenses and notice which spots on your retina get the light, and there's your direction. If you aim a telescope somewhere other than the wave front then you won't see that star. Now we have a picture of it where the light starts out traveling in all directions from a particular spot. But because the source was moving, had a velocity dx/dt, the wave front isn't coming from that direction when it reaches you. It's coming from a completely different direction. So we have to either give up the wave models that worked and replace them with something new, because you can't point a telescope where the wave model says to point it and have the wave actually go into the telescope. Or we keep the wave model working and something else changes. I'm ready to try it both ways. > >> >> >> > 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 blue 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 blue> >observer> does NOT see them arrive at equal time > >intervals, only sees> >one circle> at a time, only sees a point on > >the circle.> > > >> > Yes, he only sees one at a time. > >> > >> In http://yfrog.com/02starmg he only sees the first one at one > >time.> The trail of red squares doesn't exist, they are all history. > > > > Yes. they are the history of the movement of one particle of light. > > Is that irrelevant? > > Yes, that's your wave. You only get wet once. Another wave comes > from a different direction, so it's irrelevant. Stand on a bridge over > a flowing river, and drop pebbles in. You'll see circular ripples, but > you'll also see the circles moving downstream with the flow of the > river. Good thought. The ripples are in some ways not real. A cork would go up and down and sideways a little bit and end up in about the same place. There's something happening, it is real, but not in the same way as the current that moves all the water downstream along with the corks and the ripples. The current is like, like, like -- movement of an aether! Only in your model it's like every moving source has its own aether that travels with it.... > > I feel like we're actually looking at the wave and not > > the movement of the particle through time. The movement of the > > particle seems completely irrelevant to me until you need it to go > > down a telescope. > > > >> I drew a history trail like yours in > >> http://www.androcles01.pwp.blueyonder.co.uk/Wave/Aberration.gif > >> but when I animated it I deleted all the cubes except one in each > >> frame. So one image shows the light hits the edge, but it really > >hits> the centre and travels down the axis. This is because the one > >cube> cannot be in two places at once. > > > > I may have missed your point. I saw the cylinder moving while the > > dot traveled, so that the dot could go down the center because of > > that movement. It doesn't matter that the wall of the cylinder hits > > places where the dot used to be, if the dot isn't there any more. > > > > I drew the history trails because I thought it mattered where the > > particles came from. But if I ignore the particles and just look at > > the circles then everything seems to work out. It's only when I pay > > attention to the particles that I start having objections that need > > more theory to deal with them. > > > Make your theory fit the known facts. Exactly. If I do that, I get circular waves that spread out in all directions while they move sideways. And now I see one possible way your aberration picture could fit in. Since you know the source is moving, and the image of it is moving across the sky, you set up the telescope to move with it. So at any given moment the telescope is aimed at the direction that makes the light actually reach the bottom of it, at that moment. So, which direction is that? Is it where the source was when it actually emitted the light, or is it some other direction? I don't see a better choice than the direction the wavefront is pointing, but I'm still real unclear about it all. > >> > But he *could* post pickets who would see them at other times and > >> > places. > >> > >> The pickets would need to be on Mars, Venus, a Jovian Moon. > >> Perhaps you'd better work out how to get a picket to Mars before > >> you say "could" so glibly. > > > > Ah, the argument from practicality. I want to look at what the > > theory says will happen. In a second line of advance I want to look > > at what we can observe if the theory is true. You are quibbling > > about what is practical for us to observe today, when we are mostly > > stuck on one planet immersed in air. The distances are so short that > > it's mostly practical to measure speed with interferometry.... > > Engineers do that. We find theories seldom work. All false theories will have areas that don't work. Theories seldom work because people don't bother to get true theories.... > >> > 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. > > > >> The historical wavelength never changes wrt the source, the > >measured> wavelength is impossible because the photon can't be in two > >places at> once. Your trail of red squares is a wiggle in timespace. > > > > If you can't actually see a wave crest then you're stuck with > > indirect methods. > > > > For example -- make a standing wave. Say, with masers. Put something > > in the way that absorbs light of that frequency and look for the > > bands where it does absorb some. > > > > Or is it just impossible? > > Waves are history. > http://paws.kettering.edu/~drussell/Demos/SHO/damp.html > The mass is only travelling vertically, the wave is a historical > document, recording where the mass was in the past. > If you move the mass and spring sideways then you have a > waveLENGTH, but it is INDEPENDENT of waveTIME. Wouldn't that be more like an amplitude? > The mass could be a car and the car has springs and dampers > that are misnamed "shock absorbers". The real shock absorbers > are the springs and the tyres. Yes. > >> >> >> > 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? > >> > >> No. I'm not interested in comparing where the star is now with > >where> it was last seen. Prophecies are for palmists, fortune > >tellers, tea> leaf readers, horoscope writers and relativists. I > >can't see where it> is now, I'll have to wait for the light to reach > >me. > > > > When the light reaches you, the particles are presumably moving in > > straight lines from the source. But the waves are moving in a > > different direction, because every individual piece of them is > > moving at v in direction x in addition to their other movement. What > > we can measure is the wave. > > Do you point the telescope in the direction the particles came from, > > or in the direction of the wavefront? > > http://www.androcles01.pwp.blueyonder.co.uk/JT.GIF > Your red squares are the history of ONE photon. > My green squares are the actual positions of a stream of photons, > one after the other, all of which collide with the observer. Yes. Yes? And their speed keeps slowing, more than the distance decreases, right? By the time you get to the end, your vector sum is cancelling out the entire v side. So at the earliest time on our diagram it takes 1 time unit to get to the observer. But by the end, when the source is at the T, it takes 2/sqrt(3) time units to get there. > >> >> 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? > >> > >> No. I'm not interested in comparing where the star is now with > >where> it was last seen. Prophecies are for palmists, fortune > >tellers, tea> leaf readers, horoscope writers and relativists. I > >can't see where it> is now, I'll have to wait for the light to reach > >me. > > > > I'm interested in which direction do you point the telescope. > > > >> > ** 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? ** > >> > >> > >> They don't travel in straight lines. This ball doesn't travel in a > >> straight line. > >> http://ww2010.atmos.uiuc.edu/(Gh)/guides/mtr/fw/gifs/coriolis.mov > >> > >> (Well, it does in the playground, but not for the observer on the > >> carousel). > >> > >> Since you are on the moving Earth, nor does light. > >> http://www.androcles01.pwp.blueyonder.co.uk/Shapiro/Crapiro.htm > >> Being god, my frame of reference is different to you mere mortals > >> riding on the blue ball. > > > > I don't see where coriolis applies to a source moving in a straight > > line in direction x at velocity v, and a stationary observer. > > http://www.androcles01.pwp.blueyonder.co.uk/JT.GIF > I see a curved path. I don't see why you're interested in that particular convolution. > >> >> > 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. > >> > >> You mean the history wave? > > > > No, I mean the wave crest that the observer gets right now. The last > > wave crest came at a slightly different direction and speed, and the > > next one will be different too, but right now the wave is at 90 > > degrees and traveling 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? > >> > >> Yes it is. > >> http://www.androcles01.pwp.blueyonder.co.uk/1st/Postulates.htm > >> You are expected to know what relative motion is, Einstein can't > >tell> you. > > > > This is important. With other theories, the direction of the waves > > is always the direction of the light. That's what the direction of > > the light *means* to them. With EmT you can have light that's > > actually traveling in a different direction and speed from its > > waves. > > Waves are history, a trace of where a photon has been. Light is > stream of photons, like bullets from a minigun. > http://www.youtube.com/watch?v=tyF0G7g4KfY > http://www.youtube.com/watch?v=9C5FYsjluJc They do constructive and destructive interference. That could happen if waves were more than history. Maybe it could happen some other way if waves are history. > > This > > complication is bound to have a whole lot of implications. For one, > > you will get kinds of interference etc that are forbidden to other > > theories, unless it turns out somehow that they cancel out. > > > > This is exciting! > >> >> >> > 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. |