Taking a Fresh Look at the Physics of Radiometers.
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From: Timo Nieminen on 5 Jun 2010 01:19 On Jun 5, 1:53 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > Dear Timo: You had said, earlier: "This isn't true. The force on the > vanes has been measured in vacuum, and the force is in the opposite > direction to the usual Crookes radiometer thermal force. If not for > friction, the radiometer in vacuum would rotate "backwards"." Since > my New Science has the direction of rotation identical to the Crookes > (black squares trailing), your statement seemed to be saying that my > New Science is wrongwhich it of course, isn't. The wrongly assumed > 'forward' rotation is for the white squares to trail. Friction can > STOP or prevent a rotation, but never change its direction. If you > are wishing to change the subject to "momentum", you are way over your > head. I wrote the book on momentum and KE. PD has fought on those > subjects for three years, and has lost (to me). The world doesn't > need any more PDs! NoEinstein Job 8:2
From: NoEinstein on 5 Jun 2010 12:55 On Jun 5, 1:19 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > Dear Timo: Have I turned you... religious? "How long wilt (I) speak these things? And how long will the words of (my) mouth be like a strong wind?" Answers: I'll keep speaking the truth until the largely dull 'scientific' community wises up. Since my New Science fits the observations of the entire Universe, I can speak, with authority, that no "detail" of my science will change the whole of my discoveries. Note: Throughout the Universe, "the wind" is most often ether flow caused by pressure differentials. The maximum pressure is always closest to where the mass(es) is most concentrated. Photon emission (including infrared) depletes part of the ether INSIDE matter. That causes a vacuum that keeps drawing in new ether from outside. Matter that receives photons from other matter isn't as 'deficient' in ether on the facing sides. So, the net 'replacement' energy will flow in on the opposing sides. It is the DRAG of the flowing ether on the matter on the opposing sides of the masses that PUSHES the masses together. I. e.: the moon and the Earth are orbitally bound in this way. Thats how gravity works! NoEinstein > > On Jun 5, 1:53 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > Dear Timo: You had said, earlier: "This isn't true. The force on the > > vanes has been measured in vacuum, and the force is in the opposite > > direction to the usual Crookes radiometer thermal force. If not for > > friction, the radiometer in vacuum would rotate "backwards"." Since > > my New Science has the direction of rotation identical to the Crookes > > (black squares trailing), your statement seemed to be saying that my > > New Science is wrongwhich it of course, isn't. The wrongly assumed > > 'forward' rotation is for the white squares to trail. Friction can > > STOP or prevent a rotation, but never change its direction. If you > > are wishing to change the subject to "momentum", you are way over your > > head. I wrote the book on momentum and KE. PD has fought on those > > subjects for three years, and has lost (to me). The world doesn't > > need any more PDs! NoEinstein > > Job 8:2
From: Tim BandTech.com on 5 Jun 2010 14:48 On Jun 4, 5:40 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > On Jun 4, 9:59 pm, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > > On Jun 3, 5:59 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: <snip> > > > Consider a steady beam of monchromatic light incident on a stationary > > > reflector. Each optical cycle has energy E (i.e., the total energy of > > > a length of the beam equal to 1 wavelength). The incident power P is > > > thus P = Ef, where f is the optical frequency.The output power is the > > > same, since the reflected beam is identical, except for the reversed > > > direction. Since the reflector is stationary, no work is being done on > > > it. > > > > Now change to a coordinate system where the reflector is in motion, at > > > a speed v away from the source of the incident beam. (For simplicity, > > > assume v << c.) The incident beam is now blue-shifted, and the > > > frequency of the incident beam in this coordinate system is now f(1+v/ > > > c), and the power is P_in = Ef(1+v/c). Similarly, the reflected beam > > > now has power P_out = Ef(1-v/c). > > > In the frame of the reflector the light will appear redder, since the > > reflector is moving away from the source by v, so I disagree with your > > blue-shifted claim of the incident beam. > > We only changed the coordinate system; we didn't introduce any > relative motion between the source and the reflector. All we do is > choose a new coordinate system with origin moving in the direction > from reflector to source. > > In the frame of the reflector, everything should (must!) remain > identical. The question is, what does it look like in _another_ > reference frame. <snip> You've contradicted yourself Timo. In your previous post you state(above): "Now change to a coordinate system where the reflector is in motion, at a speed v away from the source of the incident beam. (For simplicity, assume v << c.) The incident beam is now blue-shifted..." which is clearly in error, and now to double cover your error you've misinterpreted your own post: "We only changed the coordinate system; we didn't introduce any relative motion between the source and the reflector." This is a twist on the old dodge tactic that usenet suffers from. Come Timo, let's please admit our mistakes in order to seek the truth. The self contradiction above I can accept as overlooked by you, but we are tendrilling away from the core argument within this context, as I've already argued. Therefor I isolate this information here in order to see if we can come to some agreement. Without credibility what is the value of discussion? I will pursue the momentum conservation argument on the reflective surface in another post later, upon which you argue that my own credibility is suffering. Here please, lets at least clean up and move onward. Redshifting of light has nothing to do with the claimed existence of radiation pressure. Still, it is nearby so I will not dismiss this topic completely, and the definition of a 'perfect reflector' does come under scrutiny under these conditions, which could be beneficial in the provision of a new theory. There are fundamental problems with humans doing science and the assumption that the scientist makes of getting beyond his human behavior cannot be granted unconditionally. This is cause to consider all problems as open problems, for assumptions that we make and observations that we fail to make can go invisible. This will be proven in time, and already has been proven many times. The ability to practice false belief systems is beyond no man, including myself. Falsification is a worthy method, which is a skeptical approach, but the ideal outcome is a replacement theory. It is not quite enough to declare a belief system wrong. - Tim
From: Tim BandTech.com on 5 Jun 2010 15:25 On Jun 5, 12:55 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > On Jun 5, 1:19 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > Dear Timo: Have I turned you... religious? "How long wilt (I) speak > these things? And how long will the words of (my) mouth be like a > strong wind?" Answers: I'll keep speaking the truth until the > largely dull 'scientific' community wises up. Since my New Science > fits the observations of the entire Universe, I can speak, with > authority, that no "detail" of my science will change the whole of my > discoveries. Note: Throughout the Universe, "the wind" is most often > ether flow caused by pressure differentials. The maximum pressure is > always closest to where the mass(es) is most concentrated. Photon > emission (including infrared) depletes part of the ether INSIDE > matter. That causes a vacuum that keeps drawing in new ether from > outside. Matter that receives photons from other matter isn't as > 'deficient' in ether on the facing sides. So, the net 'replacement' > energy will flow in on the opposing sides. It is the DRAG of the > flowing ether on the matter on the opposing sides of the masses that > PUSHES the masses together. I. e.: the moon and the Earth are > orbitally bound in this way. Thats how gravity works! NoEinstein > > > > > > On Jun 5, 1:53 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > Dear Timo: You had said, earlier: "This isn't true. The force on the > > > vanes has been measured in vacuum, and the force is in the opposite > > > direction to the usual Crookes radiometer thermal force. If not for > > > friction, the radiometer in vacuum would rotate "backwards"." Since > > > my New Science has the direction of rotation identical to the Crookes > > > (black squares trailing), your statement seemed to be saying that my > > > New Science is wrongwhich it of course, isn't. The wrongly assumed > > > 'forward' rotation is for the white squares to trail. Friction can > > > STOP or prevent a rotation, but never change its direction. If you > > > are wishing to change the subject to "momentum", you are way over your > > > head. I wrote the book on momentum and KE. PD has fought on those > > > subjects for three years, and has lost (to me). The world doesn't > > > need any more PDs! NoEinstein > > > Job 8:2 Hi NoEinstein. I honestly haven't carefully considered your argument, but I do believe that it will be consistent with gravitational shadowing, and perhaps a means of describing some of what modern jibe calls 'dark matter'. This is a bit of a scary subject, for it would bring huge interest from astrologers who concern themselves with the dynamics of such situations. Will we be forced to admit under this theory that when the moon passes between the sun and the earth that some of the gravitational force has disappeared? This is what I mean by gravitational shadowing. It is an interesting concept, especially when tied into electromagnetic radiation. I don't feel it is a complete theory, and feel spread pretty thin trying to consider it within the radiation pressure claim. Still, they are nearby to each other. There are so many possible variations on the Crooks and Nichols radiometers. There is not necessarily any need to expose more than one vane of the Crooks device to light in order to experiment, so that the black/silver conundrum could be deleted. Further I have not found any analysis on a spinning bearing versus a quartz fiber's friction, but I presume that a fine enough pin style bearing on a hard jewel concavity could have a very slight amount of friction relative to the torque of a fairly distant vane, not to mention more advanced options such as a magnetic bearing. I've attempted some research on this but found nothing yet. Any links on this bearing friction analysis are welcome. I have made some simple bearing out of copper arms resting on a nail and it is such a nice simple thing that a child can do it. Even wood on wood bearings are nicely behaved, and can make pretty windvanes, a few of which I have around my garden. - Tim
From: Timo Nieminen on 5 Jun 2010 16:24
On Jun 6, 4:48 am, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > On Jun 4, 5:40 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > On Jun 4, 9:59 pm, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > > > On Jun 3, 5:59 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > <snip> > > > > Consider a steady beam of monchromatic light incident on a stationary > > > > reflector. Each optical cycle has energy E (i.e., the total energy of > > > > a length of the beam equal to 1 wavelength). The incident power P is > > > > thus P = Ef, where f is the optical frequency.The output power is the > > > > same, since the reflected beam is identical, except for the reversed > > > > direction. Since the reflector is stationary, no work is being done on > > > > it. > > > > > Now change to a coordinate system where the reflector is in motion, at > > > > a speed v away from the source of the incident beam. (For simplicity, > > > > assume v << c.) The incident beam is now blue-shifted, and the > > > > frequency of the incident beam in this coordinate system is now f(1+v/ > > > > c), and the power is P_in = Ef(1+v/c). Similarly, the reflected beam > > > > now has power P_out = Ef(1-v/c). > > > > In the frame of the reflector the light will appear redder, since the > > > reflector is moving away from the source by v, so I disagree with your > > > blue-shifted claim of the incident beam. > > > We only changed the coordinate system; we didn't introduce any > > relative motion between the source and the reflector. All we do is > > choose a new coordinate system with origin moving in the direction > > from reflector to source. > > > In the frame of the reflector, everything should (must!) remain > > identical. The question is, what does it look like in _another_ > > reference frame. > > <snip> > > You've contradicted yourself Timo. In your previous post you > state(above): > > "Now change to a coordinate system where the reflector is in > motion, at > a speed v away from the source of the incident beam. (For > simplicity, > assume v << c.) The incident beam is now blue-shifted..." > > which is clearly in error, and now to double cover your error you've > misinterpreted your own post: > > "We only changed the coordinate system; we didn't introduce any > relative motion between the source and the reflector." > > This is a twist on the old dodge tactic that usenet suffers from. Come > Timo, let's please admit our mistakes in order to seek the truth. As I said (and you cut), I was not clear. The key phrase is "now change to a coordinate system". How can changing to a different coordinate system change the relative speed of the reflector and source. Of course, the reflector is still moving in some direction, and the direction of motion is away from the source. The source is also moving, and the direction is towards the reflector. As I said already, this wasn't clear enough. * Change of coordinate system, no change in relative motion. * > The > self contradiction above I can accept as overlooked by you, but we are > tendrilling away from the core argument within this context, as I've > already argued. Therefor I isolate this information here in order to > see if we can come to some agreement. > Without credibility what is the value of discussion? I will pursue the > momentum conservation argument on the reflective surface in another > post later, upon which you argue that my own credibility is suffering. > Here please, lets at least clean up and move onward. Redshifting of > light has nothing to do with the claimed existence of radiation > pressure. Still, it is nearby so I will not dismiss this topic > completely, and the definition of a 'perfect reflector' does come > under scrutiny under these conditions, which could be beneficial in > the provision of a new theory. If light carries energy, redshift/blueshift and conservation of energy tells you everything you need to know about radiation pressure. Why avoid discussing it? Is there anything wrong with what is below? Consider a steady beam of monchromatic light incident on a stationary reflector. Each optical cycle has energy E (i.e., the total energy of a length of the beam equal to 1 wavelength). The incident power P is thus P = Ef, where f is the optical frequency.The output power is the same, since the reflected beam is identical, except for the reversed direction. Since the reflector is stationary, no work is being done on it. Now change to a coordinate system where the reflector and source are in mutual motion, at a speed v, in a direction along a line joining the source to the reflector. (For simplicity, assume v << c.) The incident beam is now blue-shifted, and the frequency of the incident beam in this coordinate system is now f(1+v/c), and the power is P_in = Ef(1+v/c). Similarly, the reflected beam now has power P_out = Ef(1-v/c). Since energy is conserved, the difference in power must be the rate of doing work on the reflector. Since work is being done on the reflector, there must be a force acting on the reflector due to the reflection. P_reflector = P_in - P_out = 2Efv/c = 2vP/c. Since P_reflector = F_reflector * v, we have: F_reflector = 2P/c, which is the radiation pressure due to complete reflection at normal incidence. |