The Emission Theory of Androcles
in [Relativity]
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From: Jonah Thomas on 3 Sep 2009 11:09 Androcles, I worked out my problem about doppler. "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 > > First, did I get the directions and speeds right? > > > > Say it's source is traveling in direction V at speed v. The particle > > is emitted in direction P at speed c. The actual direction S and > > speed s is the vector sum of vV + cP. > > > > Did I get the quickest path right? Yes, I got the quickest path right, and that has no effect on doppler. > If it crosses a T then Doppler says > http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img107.gif > (without the division by the silly sqrt(1-v^2/c^2) term in the > divisor), so Einstein agrees with Doppler and then blue-shifts the > result. If it was sound, you'd apply the doppler formula that's appropriate for sound when you are moving relative to the air and the source is not. If thinking in terms of waves is appropriate at all, then that is the right model. The situation when you move past a stationary source of sound at 50 mph is just like the one where you stand still and a source of sound moves past you at 50 mph with a 50 mph tailwind. Either way, the distortion is minimised when the direction of the source is orthogonal to the direction of the motion. The actual sideways velocity then does not matter -- it will have its minimal effect when you are traveling along the compression wave, and all the wave crests are traveling sideways at the same speed. If you could take a snapshot of the waves at that time you couldn't tell them from a stationary source and a stationary observer. You can detect the sideways velocity by the doppler effect -- by the speed that the pitch changes -- but still the time the difference passes from positive to negative comes right then. My intuition still isn't tuned to the whole wave moving. I think that when the already-emitted waves don't move with the source -- when they expand at c and you get that off-center picture of them that would turn into a shock wave if the speed ever got up to c -- that then the doppler effect would cancel out at a different place, at a noticeable angle away from right angles. When the SR guys say that relativity predicts a doppler effect at 90 degrees but nothing else does I think they are mistaken. But EmT *does* predict the doppler effect will cancel at precisely 90 degrees. So this is a place where SR and EmT give different results, that might be somehow testable. > > What is it we really observe when we see interference > > patterns? > > Photons that enter our eyes from different locations. Sure. One spot is lit up and another spot is not. We can measure the distance between them. When the light comes through a diffraction grating, only light in equally-spaced bands gets through, and the accepted theory is that this light spreads out in all directions, To reach any particular point, light from different bands will travel different distances, and the light that is in phase will reinforce itself. The formula for where to expect reinforcement is d sin(theta) = ml where d -- the distance between lines in the grating l = wavelength m = some integer theta = some angle that makes the rest true. But we could easily replace l by c/f, lightspeed/frequency and get the same result. > > Are we really measuring wavelength or frequency or some > > combination or something else? > > It's impossible to observe a wavelength or a frequency. > A wavelength is a distance from where something was to where it is > now. The red pointer is attempting to measure a wavelength, and > gets it wrong. > http://www.androcles01.pwp.blueyonder.co.uk/Wave/Relative.gif Sure. Ideally you would measure a wavelength by measuring a distance, all at once. If you can't actually measure the distance then you have to infer the distance from the known speed and the known frequency. Or with the diffraction pattern you can measure a length, and since you know that c is constant that gets you both the wavelength and the frequency. But if you don't know that c is a constant after all, then it doesn't work. It's all assumed based on that.
From: Androcles on 4 Sep 2009 06:58 "Henry Wilson, DSc" <hw@..> wrote in message news:6ok1a5hqp8tmig1g06d5gujlb7flio3390(a)4ax.com... >>I'm not completely clear about all the assumptions for doppler in sound. Well, I'm sorry to hear that. Never mind, H. Just see my post to Jonah, he said the same thing as you. I will give you some advice, though. If you snip and attribute what Jonah said as if I said it, I'll snip and attribute what Jonah said as if you said it.
From: Jonah Thomas on 4 Sep 2009 13:30 "Androcles" <Headmaster(a)Hogwarts.physics_n> wrote: > "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message > > Androcles, I worked out my problem about doppler. Here is my detailed reasoning why the formula works. It will be obvious to anybody who's done it before so just skip over it unless you think there might be a flaw you can point out. I just wanted to record it while it's fresh in my mind. I'll give an example with numbers on it because that makes it more concrete. Say EmT is true, and you are approaching an observer at 0.5c, and at the moment you are 10 light-seconds away. It will take you 20 seconds to get there. You have been sending a 1 hertz signal all along. This could be a laser that you pulse once a second or any other periodic thing that the observer can observe from your signal. So over the last ten seconds before now, you sent out 10 pulses. And none of them have reached the observer yet, they're still in transit. Ten seconds ago you were 15 light seconds away, and since the light travels at 1.5c toward the observer, the first one is reaching him right now. That's ten pulses that are in transit. Over the next 20 seconds traveling at 0.5c, you will reach the observer. And during that time you will put out 20 more pulses. All of them will reach the observer within the next 20 seconds, and no others will except the 10 pulses you have already sent. Within those 20 seconds the observer will receive 30 pulses. The total number of pulses the observer receives is f0*d/v the number of pulses generated while you travel plus f0*d/c The number of pulses already on their way #cycles = f0*d(1/v + 1/c) #cycles/second = f0*d(v+c)/cv * v/d f' = f*(v+c)/c Similarly, after you pass then it becomes f' = f(c-v)/c At 0.5c away, in a unit time half the pulses don't arrive yet, they go into the lengthening pipeline to arrive later. And once it works for the forward and backward case, the sideways cases are just trig. Given something that's x units forward and y units sideways, the total lightspeed in that direction is c*sqrt (x^2 + y^2)/[(xc/(c+v))^2 + y^2]. In english, the time it takes to get from (0,0) to (x,y) is reduced to the time it would take to get to (xc/(c+v),y) It's much easier to see that it's the appropriate formula after I see how it works.
From: Androcles on 4 Sep 2009 14:59 "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20090904133050.73f8090f.jethomas5(a)gmail.com... > "Androcles" <Headmaster(a)Hogwarts.physics_n> wrote: >> "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message > >> > Androcles, I worked out my problem about doppler. > > Here is my detailed reasoning why the formula works. Good for you. I'm having a snipfest, so I won't bother to read it.
From: Androcles on 4 Sep 2009 15:04
"Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20090904134702.356255b8.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: > >> > 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. >> > >> > >> > ----------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. >> > >> > If you could look at all the waves at the same time, they would look >> > like the source is directly to your side and it has as lot of >> > concentric circles around it, and one of them is touching you -- it >> > would look exactly like the picture when nothing is moving -- at >> > that instant. >> > >> > Even though the wave as a whole is traveling to the side at v, the >> > wave front at that moment is traveling in precisely your direction >> > at c. > > Hey, Androcles, did I draw the picture right or did Wilson? Dunno, today is snipfest day. Seems to me that Henri is right, but I haven't checked. Give up, anything you write today will be snipped and ignored. |