Doppler energy extraction: new solar cell technology
in [Physics]
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From: Tim Golden BandTech.com on 11 Jul 2010 08:05 Timo Nieminen wrote: > On Jul 10, 10:32 pm, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com> > wrote: >> Timo Nieminen wrote: >>> They take longer by a longer and longer amount of time. It isn't a >>> constant delay. >> For a constant velocity this is a linear problem. There is nothing wrong >> here. > > It still isn't a constant delay. The energy in during time T equals > the energy out during time 2T. Whether or not the cycles eventually > get back doesn't matter - the power matters, not eventual conservation > of energy. > > Deficit spending at twice one's income is still deficit spending, > getting more and more in debt, even if every borrowed dollar is paid > back (with an every increasing delay). I have no idea why a constant delay is important to you. There has been no requirement like this stated before in the discussion. It is irrelevant. For any accelerated reflector the delay that the source will see in returning light will vary, so this variable delay is a general observation and is not a falsification at all. > >>> Something that is the same in all the reference frames we might >>> choose. Maybe not what we would call a conservation law. This is >>> invariance of phase, not conservation of energy. >> Actually, it pretty directly transposes to conservation of energy via >> e = h f . > > OK, use E=hf if you want. The photons are countable, the count must > stay the same despite the change in reference frame, the energy of the > photons depends on the choice of reference frame. The redshifted wave > has less power. End of story, yes? > >> 10 cycles at one frequency are equivalent in energy to ten cycles at any >> other frequency. > > Does it matter? For the 50% redshift case, our wavelength is doubled. > If the total energy in the cycle is the same, then we have E', the E > field in the redshift frame, given by E' = E/sqrt(2). If the total > energy of the cycle is the same, and the cycle now occupies twice the > spatial extent, the energy density must be halved, and thus E^2 must > be halved. Likewise, H'=H/sqrt(2). The energy flux, the Poynting > vector, ExH, is halved as well. We need half the power since the duration of the wave has doubled. Otherwise we will not have conservation of energy. You are only supporting my claim. We know that power is instantaneous and only when we integrate that power over time do we get energy. Send a blip of light, then redshift it 50%: it doubles in duration or length. > > Since the EM wave travels at the same speed in both frames, the energy > flux is proportional to the energy density. The energy flux, the > power, is what matters. > > One can do the full SR calculation, too. For a plane EM wave, when > transforming to a reference frame moving along the direction of > propagation such that we have a redshift, we have (ref., e.g., > http://scienceworld.wolfram.com/physics/ElectromagneticFieldTensor.html) > > E = gamma (E - vB) > B = gamma (B - v/c^2 E) > >> This has nothing to do with phase until some strict >> experiment is setup. > > No, "phase" tells us where the cycles begin and end. N cycles in > meaning N cycles out is invariance of phase. But irrelevant. > >> This is a plane wave of monochromatic light, and >> matches the Maxwell wave interpretation, though the energy can be shown >> through the photon energy equation. > > E = hf tells us the power is reduced. Maxwell tells us the power is > reduced. Surely this is enough? I tell you the power is reduced too. Yes, the power of the wave is reduced, and so to conserve energy we should consider that the duration of a 50% redshifted 1 ns source pulse is 2 ns. Until you falsify this then we have no disagreement. You argue in support of zero acceleration to the reflector. - Tim > >> I understand that this is a sidetrack, but still, since you've decided >> to critique it, the photon's reference frame travels at c [...] >> This is not standard relativity >> theory, but the math does account for it. You say that they do not have >> any inertial rest frame, but I just gave them one without conflict, >> other than a division by zero for the spatial dimension. > > Invalid mathematics is not "without conflict". The math _doesn't_ > account for it. The equivalent choice of reference frame in Galileian > relativity is one at infinite velocity. Doesn't work either, > mathematically. > >>>> You'd have to read the article. They present redshift as an energy >>>> conservation problem. >>> Cosmological redshift. A "problem" for conservation of energy of the >>> universe. As the article concludes, "So the universe does not violate >>> the conservation of energy; rather it lies outside that law�s >>> jurisdiction." >> They failed to assess the energy delay of redshifted light. Who else is >> failing to do this? > > This is your conclusion from the article? Why so? I see that we have overlooked this argument. I see that they are overlooking this simple argument. Who else is overlooking this simple argument? This is my question which you answer with a question. Why so? > > How is this at all relevant to radiation pressure? None of this is relevant to radiation pressure so far as I can tell. Redshift has nothing to do with radiation pressure. They are two different things, and my understanding of radiation pressure is not coming along so well. Your own attachment to a redshift argument on radiation pressure is not going so well any more. - Tim
From: Timo Nieminen on 11 Jul 2010 17:17 On Jul 11, 10:05 pm, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com> wrote: > >> They failed to assess the energy delay of redshifted light. Who else is > >> failing to do this? > > > This is your conclusion from the article? Why so? > > I see that we have overlooked this argument. I see that they are > overlooking this simple argument. Who else is overlooking this simple > argument? This is my question which you answer with a question. Why so? Tamara doesn't overlook this in the article. At least, not for my reading of "delay". I was wondering why you thought she did. > > How is this at all relevant to radiation pressure? > > None of this is relevant to radiation pressure so far as I can tell. > Redshift has nothing to do with radiation pressure. They are two > different things, and my understanding of radiation pressure is not > coming along so well. Your own attachment to a redshift argument on > radiation pressure is not going so well any more. OK, if you think it isn't relevant, let us cut it. If you think that deficit spending is "conservation of money", fine. Perhaps it is, for some definition of conservation of money, but the bank account keeps going further and further into the red. What matters for the current point is that the _power_, the energy per unit time, is reduced, and we agree on that. Let me instead return to a simpler point that you haven't answered yet, but which is very relevant. (Why is it relevant? Your argument has centered on "a = 0 means P = 0" for many iterations.) What is the dependence of the rate of work done on an object by a force on the velocity and acceleration of the object? What is the explicit formula for P(v,a)? Classical mechanics will do fine. -- Timo
From: spudnik on 11 Jul 2010 22:53 what is the problemma with "doppler shift can only be used, where the redshift can be assumed to be due to velocity-away-from-the-detector; for example, I' m singing in the key of F# (where middle C is 256 cycles per second): how fast do I have to go away from you, or you away from me, for it to sound like C (either higher or lower or C=256cps) ??... assume equal temperament, steps in powers of the second root of 12 -- integer powers. > What is the explicit formula for P(v,a)? <endquote> <EOF(no_spaces)> thus&so: actually, they were using Vedic astrology, with a Ptolemaic back-up from JPL ... and I don't mean, Dianetics and/ or Scientology, TM. > LAtribcoTimes: Some Scientists Surprised; Admin "Politics <verbing> Science" thus&so: if you let time be a spatial dimension, the morons win ... if you can't convert it, back. > "Ordinarily he is insane. But he has lucid moments when he is > -- Heinrich Heine thus&so; very funny, mister President -- and, I read a book like that! > > travel agent and say "get me door-to-door to the Library of Congress". thus&so: well, what did he do, when he got to n=67? (sumorial ?-) > Fermat proved all successes of the exponent 2 are non-successes for exp.4.. I would speculate that Fermat squared the square of the Pythagorean sides to get the fourth power he used in his proof, and that he used fractions in his own explorations. The ancients, I have read, used fractions. Number theory also simplifies overwhelming information. thus&so: so, how about for base-3? -- not "sumorial," if that's not a pun. > "Generalization to digits beyond the first". --les ducs d'oil! http://wlym.com
From: Tim BandTech.com on 12 Jul 2010 23:51 On Jul 11, 5:17 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > On Jul 11, 10:05 pm, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com> > wrote: > > > >> They failed to assess the energy delay of redshifted light. Who else is > > >> failing to do this? > > > > This is your conclusion from the article? Why so? > > > I see that we have overlooked this argument. I see that they are > > overlooking this simple argument. Who else is overlooking this simple > > argument? This is my question which you answer with a question. Why so? > > Tamara doesn't overlook this in the article. At least, not for my > reading of "delay". I was wondering why you thought she did. You are substantiating a claim without any evidence. Furthermore, you are stating support for the delay paradigm. Care to form a quote here? I only read the article in a local library, and in light of our discussion see that the argument is flawed by overlooking the redshift as energy delay. I seriously doubt that Tamara, whoever that is, has cited the delay effect. By stonewalling you are not furthering any argument. I ask you now what are the stones in your wall? - Tim > > > > How is this at all relevant to radiation pressure? > > > None of this is relevant to radiation pressure so far as I can tell. > > Redshift has nothing to do with radiation pressure. They are two > > different things, and my understanding of radiation pressure is not > > coming along so well. Your own attachment to a redshift argument on > > radiation pressure is not going so well any more. > > OK, if you think it isn't relevant, let us cut it. If you think that > deficit spending is "conservation of money", fine. Perhaps it is, for > some definition of conservation of money, but the bank account keeps > going further and further into the red. What matters for the current > point is that the _power_, the energy per unit time, is reduced, and > we agree on that. > > Let me instead return to a simpler point that you haven't answered > yet, but which is very relevant. (Why is it relevant? Your argument > has centered on "a = 0 means P = 0" for many iterations.) > > What is the dependence of the rate of work done on an object by a > force on the velocity and acceleration of the object? > > What is the explicit formula for P(v,a)? I don't honestly see this as a valid question. Why are you attempting to put velocity and acceleration both in the equation? Is this at the reflector? Whatever the velocity of the reflector there is no power due to velocity, so this portion d P(v,a) / dv = 0. I guess this is a partial derivative, but I don't know a text format for it so am using 'd'. Should we just consider P(x), with a position x of the reflector? What I am stating is that the P(x) is zero, and so the acceleration is zero, due to light hitting the reflector, because all of the light is returned toward the source. But this whole nomenclature is new, so I am not probably stating this clearly. What is P()? > > Classical mechanics will do fine. > > -- > Timo
From: Timo Nieminen on 13 Jul 2010 00:39
On Mon, 12 Jul 2010, Tim BandTech.com wrote: > On Jul 11, 5:17 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > On Jul 11, 10:05 pm, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com> > > wrote: > > > > > >> They failed to assess the energy delay of redshifted light. Who else is > > > >> failing to do this? > > > > > > This is your conclusion from the article? Why so? > > > > > I see that we have overlooked this argument. I see that they are > > > overlooking this simple argument. Who else is overlooking this simple > > > argument? This is my question which you answer with a question. Why so? > > > > Tamara doesn't overlook this in the article. At least, not for my > > reading of "delay". I was wondering why you thought she did. > > You are substantiating a claim without any evidence. Furthermore, you > are stating support for the delay paradigm. Care to form a quote here? > I only read the article in a local library, and in light of our > discussion see that the argument is flawed by overlooking the redshift > as energy delay. I seriously doubt that Tamara, whoever that is, has > cited the delay effect. (a) It's irrelevant. From the beginning when you introduced this paper into the discussion, I said it was irrelevant. In your previous message, you agreed that it was irrelevant. But since you insist, note how she wrote "the drop in energy is just a matter of perspective and relative motion". (b) She writes in terms of photons, so your "delay" isn't explicitly mentioned. What is mentioned is that the photon doesn't "lose" energy, which is photon-speak for essentially the same thing. (c) Tamara is the author, as you would note if you looked at the first page of the article. (d) Why did you think that she "failed to assess the energy delay of redshifted light" in any way that counts as "failing"? (e) Why do I answer your question ("Who else is failing to do this?") with a question? For clarification. You surely don't want the literal answer, a list of the people who haven't assessed it - a list which would be about 6 billion names long, to just include living humans. You appear to think it's a Big Problem. Just what kind of problem? Anyway, it isn't relevant to radiation force. What is relevant is that the power is affected by redshift or blueshift. As of the previous pair of posts, we were agreed on that. > By stonewalling you are not furthering any argument. I ask you now > what are the stones in your wall? You're the one who's stonewalling: > > Let me instead return to a simpler point that you haven't answered > > yet, but which is very relevant. (Why is it relevant? Your argument > > has centered on "a = 0 means P = 0" for many iterations.) > > > > What is the dependence of the rate of work done on an object by a > > force on the velocity and acceleration of the object? > > > > What is the explicit formula for P(v,a)? > > I don't honestly see this as a valid question. Why are you attempting > to put velocity and acceleration both in the equation? Is this at the > reflector? Whatever the velocity of the reflector there is no power > due to velocity, so this portion > d P(v,a) / dv = 0. > I guess this is a partial derivative, but I don't know a text format > for it so am using 'd'. Should we just consider P(x), with a position > x of the reflector? What I am stating is that the P(x) is zero, and so > the acceleration is zero, due to light hitting the reflector, because > all of the light is returned toward the source. But this whole > nomenclature is new, so I am not probably stating this clearly. What > is P()? P = rate of doing work. If a force is acting on some object, it might be doing work on the object. I've asked this a few times already, and you haven't answered. So, let me give you the answer: P = Fv. It doesn't depend on the acceleration. It depends on the velocity. Note well that when v = 0, P = 0. Yes, I know that you're saying that P=0 means that a=0. This is wrong. What this basic result from classical mechanics tells us is that when v=0, no work is being done on the object. Doesn't matter what the force is, what causes the force, or what the object is; the rate of doing work on a stationary object is zero. If the object is only instantaneously stationary, the rate is still zero at the instant when v=0. What this means is that P=0 doesn't mean that you must have F=0. -- Timo |