The Emission Theory of Androcles
in [Relativity]
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From: Jonah Thomas on 23 Sep 2009 08:06 "Inertial" <relatively(a)rest.com> wrote: > "Henry Wilson, DSc" <hw@..> wrote > > So if they travel different lengths, as in rotating Sagnac, they do > > not arrive in phase. > > What has the distance have to do with it > > They leave the source with the same wavelength and speed and frequency > and phase wrt the source, and they travel for the same amount of time, > and then arrive at the destination with the same wavelength and speed > and frequency wrt the destination .. then how can they possibly be out > of phase?? I'm glad you asked that. Let's back up first and look at a more general question. We are all agreed that there is no preferred inertial frame, right? You can't tell which frame is absolute rest, one frame is as good as another, the rules are the same for all of them. But nobody should say that for rotating frames. We can tell which frame is at rest with respect to rotation, within the limits of accuracy of our particular sagnac apparatus. There is a preferred rotational frame. You know whether you are rotating or not. So let's look at the situation wrt the nonrotating frame. In the nonrotating frame, when you spin your sagnac apparatus you change the angle of the detector relative to the source. Remember that the formula for the phase shift is 4Aw/(c^2-v^2) where A is the area of the circle, w is the angular velocity and v is the speed of the loop. So for small v this is about rC*2*theta/c where r is the radius, C is the circumference, and theta is the angle the apparatus travels during the average time the light travels around it. For small v the phase shift is linear for theta. So anything that gives a linear difference in phase with theta will fit the formula and presumably fit the experimental data. Classical and SR theories do this with a constant lightspeed and a difference in arrival time. The angular distance traveled is 2pi+theta in one case and 2pi-theta in the other, allowing a phase difference of 2theta. Emission theories could do it with variable lightspeed and a constant wavelength. One case it travels a distance of 2pi+theta while the other case it travels 2pi-theta, so the phase difference is 2theta. An observer on the rotating platform might see them as traveling at the same speed with a frequency and wavelength change. But then when they leave the ring for the detector the speeds change, the wavelengths become equal but the phase shift remains.
From: Inertial on 23 Sep 2009 08:33 "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20090923080611.48ab18cf.jethomas5(a)gmail.com... > "Inertial" <relatively(a)rest.com> wrote: >> "Henry Wilson, DSc" <hw@..> wrote > >> > So if they travel different lengths, as in rotating Sagnac, they do >> > not arrive in phase. >> >> What has the distance have to do with it >> >> They leave the source with the same wavelength and speed and frequency >> and phase wrt the source, and they travel for the same amount of time, >> and then arrive at the destination with the same wavelength and speed >> and frequency wrt the destination .. then how can they possibly be out >> of phase?? > > I'm glad you asked that. So am I :) > Let's back up first and look at a more general question. We are all > agreed that there is no preferred inertial frame, right? You can't tell > which frame is absolute rest, one frame is as good as another, the rules > are the same for all of them. Yeup > But nobody should say that for rotating frames. Yeup > We can tell which frame > is at rest with respect to rotation, within the limits of accuracy of > our particular sagnac apparatus. There is a preferred rotational frame. > You know whether you are rotating or not. Yeup > So let's look at the situation wrt the nonrotating frame. Yeup > In the nonrotating frame, when you spin your sagnac apparatus you change > the angle of the detector relative to the source. And its position, by the time light gets there > Remember that the formula for the phase shift is Dt = 4Aw/(c^2-v^2) That's the difference in time between arrival of the rays in an SR or simple Aether analysis where light travels at speed c in the non-rotating frame. Which at small v is close enough to the Dt = 4Aw/c^2 > where A > is the area of the circle, w is the angular velocity and v is the speed > of the loop. Yeup > So for small v this is about rC*2*theta/c where r is the radius, C is > the circumference, and theta is the angle the apparatus travels during > the average time the light travels around it. If its going around a circle .. yeup. A = pi.r^2 C = 2.pi.r A = rC/2 4A = 2rC so Dt = 4Aw/(c^2-v^2) = 2rCw/(c^2-v^2) so for small v we have Dt ~= 2rC w / c^2 We have w = v/r theta = v/rc theta = w/c But the units of that are m^-1. That's not an angle So where does this theta come from? Anyway .. ignoring that, we do have that (for small v) Dt ~= 2Cv / c^2 > For small v the phase shift is linear for theta. Its linear for v. > So anything that gives a linear difference in phase with theta will fit > the formula and presumably fit the experimental data. Not necessarily > Classical and SR theories do this with a constant lightspeed and a > difference in arrival time. The phase difference you show IS the difference in arrival times > The angular distance traveled is 2pi+theta > in one case and 2pi-theta in the other, allowing a phase difference of > 2theta. Nope. The theta is different for the two rays > Emission theories could do it with variable lightspeed and a constant > wavelength. How? If the arrival times are the same Dt = 0, so no phase shift. > One case it travels a distance of 2pi+theta while the other > case it travels 2pi-theta, so the phase difference is 2theta. That is not a phase difference .. just a difference of angles subtended by the curved light path. You have not shown that distance of path from where the source was when the rays were emitted to where the ray are now is going to give a difference in phase. That's the whole point. > An observer on the rotating platform might see them as traveling at the > same speed with a frequency and wavelength change. No .. In ballistic theory they would see the same frequency and same wavelength. > But then when they leave the ring for the detector > the speeds change, No > the wavelengths > become equal but the phase shift remains. They were already equal Nope. Sorry .. but you logic is wrong there. Try again.
From: Androcles on 23 Sep 2009 09:02 "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20090923080611.48ab18cf.jethomas5(a)gmail.com... > "Inertial" <relatively(a)rest.com> wrote: >> "Henry Wilson, DSc" <hw@..> wrote > >> > So if they travel different lengths, as in rotating Sagnac, they do >> > not arrive in phase. >> >> What has the distance have to do with it >> >> They leave the source with the same wavelength and speed and frequency >> and phase wrt the source, and they travel for the same amount of time, >> and then arrive at the destination with the same wavelength and speed >> and frequency wrt the destination .. then how can they possibly be out >> of phase?? > > I'm glad you asked that. > > Let's back up first and look at a more general question. We are all > agreed that there is no preferred inertial frame, right? You can't tell > which frame is absolute rest, one frame is as good as another, the rules > are the same for all of them. > > But nobody should say that for rotating frames. We can tell which frame > is at rest with respect to rotation, within the limits of accuracy of > our particular sagnac apparatus. There is a preferred rotational frame. > You know whether you are rotating or not. I'm glad you lied that. Einstein's principle of equivalence says gravity is equivalent to constant acceleration. If you are in a space elevator that undergoes constant acceleration it feels just like gravity to you, a constant force is applied to the soles of your feet, the ball you release from your hand will accelerate toward the floor of the elevator and bounce. You are not supposed to look out the window to know whether you are accelerating or not, and even if you did it would be the universe that is accelerating the other way, you are merely experiencing gravity. Now we imagine the space elevator is whirling on the end of a rope like a boleadoras and you are in it. http://en.wikipedia.org/wiki/Bolas Without cheating by looking out of the window as a scientist would, and being a good little relativist as the Holey Lord Einstein would wish, and being better off by not discovering his flaws which causes you to "win", how the fuckin' hell can you know whether you are rotating or not? You are not supposed to look out the window to know whether you are rotating or not, and even if you did it would be the universe that is rotating the other way, you are merely experiencing gravity. Everybody knows G. Galilei was WRONG! The Earth is flat and the Sun moves around it! How do I know this? I know it because Jonah Thomas says "There is a preferred rotational frame. You know whether you are rotating or not." http://textodigital.com/P/GG/_gett.php?f=i/ssv0.jpg&r=1 Even film makers like Stanley Kubrick know more physics than you and Einstein. Do carry on "winning". Whatever it is you think you've won you are welcome to. Cuddly toy made in China, perhaps. I say you've lost before you began.
From: Jonah Thomas on 23 Sep 2009 10:04 "Inertial" <relatively(a)rest.com> wrote: > "Jonah Thomas" <jethomas5(a)gmail.com> wrote > > "Inertial" <relatively(a)rest.com> wrote: > >> "Henry Wilson, DSc" <hw@..> wrote > > > >> > So if they travel different lengths, as in rotating Sagnac, they > >do> > not arrive in phase. > >> > >> What has the distance have to do with it > >> > >> They leave the source with the same wavelength and speed and > >frequency> and phase wrt the source, and they travel for the same > >amount of time,> and then arrive at the destination with the same > >wavelength and speed> and frequency wrt the destination .. then how > >can they possibly be out> of phase?? > > > > I'm glad you asked that. > > So am I :) > > > Let's back up first and look at a more general question. We are all > > agreed that there is no preferred inertial frame, right? You can't > > tell which frame is absolute rest, one frame is as good as another, > > the rules are the same for all of them. > > Yeup > > > But nobody should say that for rotating frames. > > Yeup > > > We can tell which frame > > is at rest with respect to rotation, within the limits of accuracy > > of our particular sagnac apparatus. There is a preferred rotational > > frame. You know whether you are rotating or not. > > Yeup > > > So let's look at the situation wrt the nonrotating frame. > > Yeup > > > In the nonrotating frame, when you spin your sagnac apparatus you > > change the angle of the detector relative to the source. > > And its position, by the time light gets there > > > Remember that the formula for the phase shift is Dt = 4Aw/(c^2-v^2) > > That's the difference in time between arrival of the rays in an SR or > simple Aether analysis where light travels at speed c in the > non-rotating frame. Is the phase shift directly proportional to that time difference? > Which at small v is close enough to the Dt = 4Aw/c^2 > > > where A > > is the area of the circle, w is the angular velocity and v is the > > speed of the loop. > > Yeup > > > So for small v this is about rC*2*theta/c where r is the radius, C > > is the circumference, and theta is the angle the apparatus travels > > during the average time the light travels around it. > > If its going around a circle .. yeup. > > A = pi.r^2 > C = 2.pi.r > A = rC/2 > 4A = 2rC > > so > > Dt = 4Aw/(c^2-v^2) = 2rCw/(c^2-v^2) > > so for small v we have > > Dt ~= 2rC w / c^2 > > We have w = v/r > > theta = v/rc > theta = w/c > But the units of that are m^-1. That's not an angle Oh. Yes. You're right. w is radians/second, c is meters/second. so we get radians/meter. I don't want to divide by c but by t. > So where does this theta come from? I naively believe that the phase shift is directly proportional to the time difference in the classical and SR cases. So I want to substitute phase shift on the left side of the equation. But of course its units are not time. > Anyway .. ignoring that, we do have that (for small v) > > Dt ~= 2Cv / c^2 > > > For small v the phase shift is linear for theta. > > Its linear for v. Sure, in the classical case the difference in arrival time is directly proportional to v, as is theta when v is small. > > So anything that gives a linear difference in phase with theta will > > fit the formula and presumably fit the experimental data. > > Not necessarily Do you have a counterexample or are you asking me to correctly work the math? > > Classical and SR theories do this with a constant lightspeed and a > > difference in arrival time. > > The phase difference you show IS the difference in arrival times Yes, exactly. They create theta with their difference in arrival times. > > The angular distance traveled is 2pi+theta > > in one case and 2pi-theta in the other, allowing a phase difference > > of 2theta. > > Nope. The theta is different for the two rays I chose theta to be the average of the two. > > Emission theories could do it with variable lightspeed and a > > constant wavelength. > > How? If the arrival times are the same Dt = 0, so no phase shift. I keep telling you. If the "wavelength", the rate the "wave" shifts its "phase" is directly proportional to distance in the noninertial frame, then when they go different distances in the noninertial frame they will end with different phases. And they do go different distances in the noninertial frame in the same time. You can argue about whether the rate of phase change should be proportional to distance instead of time, but that's a different argument. > > One case it travels a distance of 2pi+theta while the other > > case it travels 2pi-theta, so the phase difference is 2theta. > > That is not a phase difference .. just a difference of angles > subtended by the curved light path. You have not shown that distance > of path from where the source was when the rays were emitted to where > the ray are now is going to give a difference in phase. That's the > whole point. What do I have to do? I told you that the difference in phase is directly proportional to distance traveled. I can't very well prove that to you without producing light that travels at different speeds whose difference in phase is proportional to distance traveled. What more do you want than for me to just tell you that the difference in phase is proportional to distance traveled? > > An observer on the rotating platform might see them as traveling at > > the same speed with a frequency and wavelength change. > > No .. In ballistic theory they would see the same frequency and same > wavelength. They won't see that if the light actually does have its phase shift proportional to distance in the inertial frame. If that's what happens they'll see a frequency and wavelength change because the phase shift won't be proportional to distance in their frame. > > But then when they leave the ring for the detector > > the speeds change, > > No If they saw the speeds as the same because the speeds were c+v in one direction and c-v in the other and those were their speeds too, then they won't see the speeds as the same when the light is no longer revolving with them but travels straight, still at c+v and c-v. Of course, if changing the direction of the light causes it to change speed to that of the mirror, then something else will happen. > > the wavelengths > > become equal but the phase shift remains. > > They were already equal No, they weren't. The model said they weren't. You don't get to tell me what my model says, you only get to tell me that it contradicts itself and that it's contrary to experimental evidence.
From: Jonah Thomas on 23 Sep 2009 10:09
"Androcles" <Headmaster(a)Hogwarts.physics_o> wrote: > "Jonah Thomas" <jethomas5(a)gmail.com> wrote > > Let's back up first and look at a more general question. We are all > > agreed that there is no preferred inertial frame, right? You can't > > tell which frame is absolute rest, one frame is as good as another, > > the rules are the same for all of them. > > > > But nobody should say that for rotating frames. We can tell which > > frame is at rest with respect to rotation, within the limits of > > accuracy of our particular sagnac apparatus. There is a preferred > > rotational frame. You know whether you are rotating or not. > > I'm glad you lied that. > > Einstein's principle of equivalence says gravity is equivalent to > constant acceleration. If you are in a space elevator that undergoes > constant acceleration it feels just like gravity to you, a constant > force is applied to the soles of your feet, the ball you release from > your hand will accelerate toward the floor of the elevator and bounce. > You are not supposed to look out the window to know whether you > are accelerating or not, and even if you did it would be the universe > that is accelerating the other way, you are merely experiencing > gravity. > > Now we imagine the space elevator is whirling on the end of a rope > like a boleadoras and you are in it. > > http://en.wikipedia.org/wiki/Bolas > > Without cheating by looking out of the window as a scientist would, > and being a good little relativist as the Holey Lord Einstein would > wish, and being better off by not discovering his flaws which causes > you to "win", how the fuckin' hell can you know whether you are > rotating or not? You can look at your sagnac interferometer and see that you are rotating. You don't have to look out the window. They now have something almost like a sagnac machine that can detect magnetic fields. They use light that is circularly polarised going in the two different directions, and the magnetic field affects them differently to create an interference pattern. Apparently that has almost no effect on a regular sagnac interferometer but does if you polarise the light just right. There is talk about making something like that to detect gravity waves but I haven't heard of any successes. > You are not supposed to look out the window to know whether you > are rotating or not, and even if you did it would be the universe > that is rotating the other way, you are merely experiencing gravity. > Everybody knows G. Galilei was WRONG! The Earth is flat and the > Sun moves around it! > How do I know this? I know it because Jonah Thomas says > "There is a preferred rotational frame. You know whether you are > rotating or not." > http://textodigital.com/P/GG/_gett.php?f=i/ssv0.jpg&r=1 > Even film makers like Stanley Kubrick know more physics than you > and Einstein. Do carry on "winning". Whatever it is you think you've > won you are welcome to. Cuddly toy made in China, perhaps. > I say you've lost before you began. |