Doppler energy extraction: new solar cell technology
in [Physics]
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From: Timo Nieminen on 3 Jul 2010 23:54 On Jul 3, 11:00 pm, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com> wrote: > Timo Nieminen wrote: > > On Jul 2, 1:16 am, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com> > > wrote: > >> Timo Nieminen wrote: > >>> On Jun 30, 11:26 pm, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com> > >>> wrote: > >>>> Let's suppose a Crookes type of device, so that the experiment could be > >>>> done on a lab table. Upon getting the vanes to high speed the laser > >>>> light, being red shifted substantially, should yield ever more drive to > >>>> the device. If we could get a 50% redshift, then we'd be taking 50% of > >>>> the laser energy and converting it into mechanical energy, or > >>>> acceleration of the vane. Should this work? > >>> Why shouldn't it work? But be careful with language! You write "more > >>> drive", but note well that this means "more power", not "more force". > >>> With the device as described, you'd get _less_ force as it spins > >>> faster (but still more power). Why less force? Unlike the previous > >>> analysis of the redshift/blueshift, which was done with a chnage of > >>> coordinate system, you're talking about a real motion of the vane > >>> relative to the source, so a redshift seen by the vane, so a reduction > >>> in power as seen by the vane. > >> Yet still returning to the source is 50% of what came from it (at the > >> normal incidence angle), and so we have entered a conundrum. Where did > >> the other 50% go? I do see the argument for diminishing return on the > >> vane as you state it here, and I likewise see the argument from the > >> source perspective as well. We are nearly building a Demon. > > >> I never have appreciated the random coordinate system change that you > >> constructed, and I see no reason to enter it into this conversation > >> other than to cloud the discussion. The redshift which you and I have > >> been speaking of for the last few posts has nothing to do with a > >> randomly chosen frame. > > > Not _random_. Sit down, and do some calculations. Since you don't > > believe various stuff relating to radiation momentum, just sit down > > and do some classical mechanicws calculations. Are you OK with > > Newtonian mechanics? Do some elastic collision problems, the same > > problem in some different inertial coordinate systems. If you're up to > > it, do the same for some classical "hosing" problems, like hosing a > > stream of continuous fluid onto a "reflector", with no heating, no > > loss of energy. > > > What assumptions go into this? Conservation of energy and momentum, > > and that these conservation laws hold in all inertial coordinate > > systems, and that (under Galileian relativity) the force and > > acceleration are the same in all inertial coordinate systems. > > > And what conundrum? Beam loses energy due to Doppler shift, work is > > done on the target because it's moving. The reduction in beam energy > > is equal to the work. Where is the conundrum? Choose any inertial > > coordinate system, and energy and momentum are conserved, and we see > > the same force in all such systems. What is the problem, the > > conundrum? > > I've already stated it clearly without confusion: > "Yet still returning to the source is 50% of what came from it > (at the normal incidence angle), and so we have entered a conundrum. > Where did the other 50% go?" > You've already admitted that the vane too sees less power: > "you're talking about a real motion of the vane > relative to the source, so a redshift seen by the vane, > so a reduction in power as seen by the vane." > > We have just constructed a break with conservation of energy, and in > both frames. We are seeing loss of energy in both frames. No. In the frame where the vane is moving, the difference in energy between the incident beam and the red-shifted reflected beam is equal to the work done. No loss of energy, no problem with conservation of energy. If you were to look at this in the vane's instantaneous rest frame, you'd see that the incident beam is of lower power, the reflected beam is of the same power as the incident beam, and no work is done. Again, there is no lost energy, no problem with conservation of energy. This isn't magic. It works the same way in classical mechanics. Look at an elastic collision along a line between a light ball and a heavy ball in the centre-of-mass rest frame. In this frame, the light ball starts at speed v, and finishes at speed v (the velocity is reversed). The light ball loses no KE, starting with (1/2)mv^2 and finishing with (1/2)mv^2. The heavy ball also gains no KE - it's KE before and after is (1/2)MV^2 (which is equal to (1/2)mv^2 * (m/M) since mv=MV). Choose a different reference frame. Try a frame where the light ball is initially moving at 2v. Starts with (1/2)m(2v)^2 = 2mv^2 KE, 4 times as much as before. Is this "increase" a problem for conservation of energy? No, since the KE just depends on our choice of coordinate system, just like speed and position depend on it. The light ball finishes with KE=0. The initial KE of the heavy ball is (1/2)M(V-v)^2, the final KE is (1/2)M(V+v)^2. No suprise that the difference is equal to 2mv^2. OK, what happens when we look at an elastic collision like this? The energy depends on our choice of coordinate system, and the work done depends on our choice of coordinate system. In each coordinate system, KE lost by light ball = KE gained by heavy ball; energy is conserved in each case. Same thing with beam-and-vane. Conservation of energy works fine, for any choice of (inertial) coordinate system we want to use, or for any velocity of vane relative to source. > We should have > to go to the relativistic analysis of the vanes and see that the inertia > has come up quite high at a 50% redshift. Still, having gotten there, > by the conservation of energy, those vanes must be absorbing 50% of the > energy of the beam still (source perspective). Meanwhile they are doing > so while seeing the beam as 50% of the source energy (50% redshifted), > and again, these are reflective plates we are using. Can't you see the > chicken and egg conflict here? It is absurd that you would refute the > conundrum from my perspective. > > Interestingly we've engaged a rotational instrument in order to achieve > this effect. You say there is no need to resolve this conflict. You also > have said that this effect is consistent with photon momentum. I believe > these both to be false statements. Do the calculation. It works the same way as the classical collision calculation. The "problems" that you say are problems are equally "problems" with classical mechanics. That is, they're not problems at all. (They can be surprising to students, baffling, troubling, but that doesn't mean that they're _problems_ .) If conservation of energy does really fail here, you should be able to show it. As in calculating the energy before, and the energy after, and showing they're different from each other. If it's inconsistent with photon momentum, again, you can show it. > The construction we are discusssing > is far away from photon momentum. We are claiming to extract 50% of the > light energy mechanically. I don't honestly know what to believe, but > this is much closer to holding a mirror up to the sun and being pushed > over, which we know will not happen. Why "much closer"? Sure, you can look at it in a frame where 50% of the incident power is "lost" through redshift. With Galileian relativity, the force is still the same, so not even a tiny bit closer to being pushed over. If you move the vane away from the source to achieve this, you get less force (since the vane then sees redshifted incident light). Further from being pushed over by sunlight, not closer. Sure, you might want to use special relativity instead of Galileian relativity. Go ahead, calculate the dependence of the force on the motion of the vane. Sometimes, it's useful to calculate first, and claim second. -- Timo
From: Tim Golden BandTech.com on 6 Jul 2010 11:35 Timo Nieminen wrote: > On Jul 3, 11:00 pm, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com> > wrote: >> Timo Nieminen wrote: >>> On Jul 2, 1:16 am, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com> >>> wrote: >>>> Timo Nieminen wrote: >>>>> On Jun 30, 11:26 pm, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com> >>>>> wrote: >>>>>> Let's suppose a Crookes type of device, so that the experiment could be >>>>>> done on a lab table. Upon getting the vanes to high speed the laser >>>>>> light, being red shifted substantially, should yield ever more drive to >>>>>> the device. If we could get a 50% redshift, then we'd be taking 50% of >>>>>> the laser energy and converting it into mechanical energy, or >>>>>> acceleration of the vane. Should this work? >>>>> Why shouldn't it work? But be careful with language! You write "more >>>>> drive", but note well that this means "more power", not "more force". >>>>> With the device as described, you'd get _less_ force as it spins >>>>> faster (but still more power). Why less force? Unlike the previous >>>>> analysis of the redshift/blueshift, which was done with a chnage of >>>>> coordinate system, you're talking about a real motion of the vane >>>>> relative to the source, so a redshift seen by the vane, so a reduction >>>>> in power as seen by the vane. >>>> Yet still returning to the source is 50% of what came from it (at the >>>> normal incidence angle), and so we have entered a conundrum. Where did >>>> the other 50% go? I do see the argument for diminishing return on the >>>> vane as you state it here, and I likewise see the argument from the >>>> source perspective as well. We are nearly building a Demon. >>>> I never have appreciated the random coordinate system change that you >>>> constructed, and I see no reason to enter it into this conversation >>>> other than to cloud the discussion. The redshift which you and I have >>>> been speaking of for the last few posts has nothing to do with a >>>> randomly chosen frame. >>> Not _random_. Sit down, and do some calculations. Since you don't >>> believe various stuff relating to radiation momentum, just sit down >>> and do some classical mechanicws calculations. Are you OK with >>> Newtonian mechanics? Do some elastic collision problems, the same >>> problem in some different inertial coordinate systems. If you're up to >>> it, do the same for some classical "hosing" problems, like hosing a >>> stream of continuous fluid onto a "reflector", with no heating, no >>> loss of energy. >>> What assumptions go into this? Conservation of energy and momentum, >>> and that these conservation laws hold in all inertial coordinate >>> systems, and that (under Galileian relativity) the force and >>> acceleration are the same in all inertial coordinate systems. >>> And what conundrum? Beam loses energy due to Doppler shift, work is >>> done on the target because it's moving. The reduction in beam energy >>> is equal to the work. Where is the conundrum? Choose any inertial >>> coordinate system, and energy and momentum are conserved, and we see >>> the same force in all such systems. What is the problem, the >>> conundrum? >> I've already stated it clearly without confusion: >> "Yet still returning to the source is 50% of what came from it >> (at the normal incidence angle), and so we have entered a conundrum. >> Where did the other 50% go?" >> You've already admitted that the vane too sees less power: >> "you're talking about a real motion of the vane >> relative to the source, so a redshift seen by the vane, >> so a reduction in power as seen by the vane." >> >> We have just constructed a break with conservation of energy, and in >> both frames. We are seeing loss of energy in both frames. > > No. In the frame where the vane is moving, the difference in energy > between the incident beam and the red-shifted reflected beam is equal > to the work done. No loss of energy, no problem with conservation of > energy. > > If you were to look at this in the vane's instantaneous rest frame, > you'd see that the incident beam is of lower power, the reflected beam > is of the same power as the incident beam, and no work is done. Again, > there is no lost energy, no problem with conservation of energy. > > This isn't magic. It works the same way in classical mechanics. Look > at an elastic collision along a line between a light ball and a heavy > ball in the centre-of-mass rest frame. In this frame, the light ball > starts at speed v, and finishes at speed v (the velocity is reversed). > The light ball loses no KE, starting with (1/2)mv^2 and finishing with > (1/2)mv^2. The heavy ball also gains no KE - it's KE before and after > is (1/2)MV^2 (which is equal to (1/2)mv^2 * (m/M) since mv=MV). > > Choose a different reference frame. Try a frame where the light ball > is initially moving at 2v. Starts with (1/2)m(2v)^2 = 2mv^2 KE, 4 > times as much as before. Is this "increase" a problem for conservation > of energy? No, since the KE just depends on our choice of coordinate > system, just like speed and position depend on it. The light ball > finishes with KE=0. The initial KE of the heavy ball is (1/2)M(V-v)^2, > the final KE is (1/2)M(V+v)^2. No suprise that the difference is equal > to 2mv^2. > > OK, what happens when we look at an elastic collision like this? The > energy depends on our choice of coordinate system, and the work done > depends on our choice of coordinate system. In each coordinate system, > KE lost by light ball = KE gained by heavy ball; energy is conserved > in each case. > > Same thing with beam-and-vane. Conservation of energy works fine, for > any choice of (inertial) coordinate system we want to use, or for any > velocity of vane relative to source. > >> We should have >> to go to the relativistic analysis of the vanes and see that the inertia >> has come up quite high at a 50% redshift. Still, having gotten there, >> by the conservation of energy, those vanes must be absorbing 50% of the >> energy of the beam still (source perspective). Meanwhile they are doing >> so while seeing the beam as 50% of the source energy (50% redshifted), >> and again, these are reflective plates we are using. Can't you see the >> chicken and egg conflict here? It is absurd that you would refute the >> conundrum from my perspective. >> >> Interestingly we've engaged a rotational instrument in order to achieve >> this effect. You say there is no need to resolve this conflict. You also >> have said that this effect is consistent with photon momentum. I believe >> these both to be false statements. > > Do the calculation. It works the same way as the classical collision > calculation. The "problems" that you say are problems are equally > "problems" with classical mechanics. That is, they're not problems at > all. (They can be surprising to students, baffling, troubling, but > that doesn't mean that they're _problems_ .) > > If conservation of energy does really fail here, you should be able to > show it. As in calculating the energy before, and the energy after, > and showing they're different from each other. > > If it's inconsistent with photon momentum, again, you can show it. > >> The construction we are discusssing >> is far away from photon momentum. We are claiming to extract 50% of the >> light energy mechanically. I don't honestly know what to believe, but >> this is much closer to holding a mirror up to the sun and being pushed >> over, which we know will not happen. > > Why "much closer"? Sure, you can look at it in a frame where 50% of > the incident power is "lost" through redshift. Yes, this is the lab table reference frame; the one we'd be able to do measurements from most easily. I see you've put lost in quotes here so I presume you do have further analysis within this frame, otherwise, when we remove these quotes we have a conservation of energy problem. Isn't it it true that Einstein's relativity relies upon inertial reference frames? We do not then really have the option to use the vane as a reference frame, for it is a rotational (and possibly accelerated) frame. The amount of analysis at the moving reflective vane is daunting. We're going to have some sort of spectroscopic output with angle, and then a beam splitting as the next vane encounters the beam. Could it be that all of this redirecting of the beam consumes the other 50% of the energy? This would be a new effect to me, but so was scattering about a month ago. I suppose that the best we can do is declare a trap. The problem is essentially the same as with a relativistically moving reflector away from the source without rotation. Within the source frame we see less energy returning, and so the problem degrades to this case again, though with the twist of the Crookes radiometer that portion is just more hidden. We can neglect any work done on the mirror in this situation and just consider imposing a 0.5c velocity on the mirror to obtain the 50% figure. We can see that over time a pulse of the light which returns to the source from the reflector will be longer in duration. If and when the source energy extinguishes itself the reflector will still be returning energy. The quantity of wavelengths will be conserved, and so the 50% energy figure was over constant time, and had we integrated over all time we would admit that the energy is still all there. Returning to the rotational Crookes device we see that we have pulsed a continuous beam. Though this wording is loose, the redshifted beam returning to the source will be longer in duration, and will conserve the number of cycles of wave energy. To simplify, we could impose a pulsed laser pulsing a short pulse each time a vane comes normal to the source, and then even with the vanes moving at 0.5c (some fudge here for rotational complexity) then we'll see the pulses are twice as long in duration returning to the source. I remember deleting a quip in an earlier post that you would haggle over power versus energy, and here it is virtually bighting me in the back. We should constrain this method of energy extraction to nonrelativistic redshifting collectors. Orbital electrons are not unlike these Crookes devices, but they are light enough to take the kick and go somewhere with it. I find it hard to falsify this analysis above, and we overlooked this back at the inception of the effect w.r.t. photon momentum. By this analysis there never was any dE due to this effect. So, what is the proper stance on wave/energy if a reflector is accelerated by the wave? Is the reflected beam nonrelativistically redshifted? This would be the best continuous solution, but the mechanism is still cloudy. This conservation of energy will require a redshift not due to any velocity and will instead be generating an acceleration. I still prefer an electron based interpretation here, and suspect that the perfect reflector will not provide any acceleration, since there will be a conservation of energy problem. Again, the final answer to me is to argue that there will not be any perfect reflector. - Tim > With Galileian > relativity, the force is still the same, so not even a tiny bit closer > to being pushed over. If you move the vane away from the source to > achieve this, you get less force (since the vane then sees redshifted > incident light). Further from being pushed over by sunlight, not > closer. Sure, you might want to use special relativity instead of > Galileian relativity. Go ahead, calculate the dependence of the force > on the motion of the vane. > > Sometimes, it's useful to calculate first, and claim second. > > -- > Timo
From: Timo Nieminen on 6 Jul 2010 15:36 On Jul 7, 1:35 am, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com> wrote: > > Yes, this is the lab table reference frame; the one we'd be able to do > measurements from most easily. I see you've put lost in quotes here so I > presume you do have further analysis within this frame, otherwise, when > we remove these quotes we have a conservation of energy problem. Energy before = energy after, no loss, no problem. > Isn't it it true that Einstein's relativity relies upon inertial > reference frames? We do not then really have the option to use the vane > as a reference frame, for it is a rotational (and possibly accelerated) > frame. Technically, no, Einstein's relativity doesn't rely on inertial reference frames, neither special relativity nor general relativity. In SR, it is easiest to use inertial reference frames. Just use the instantaneous rest frame of the vane - works for the problem at hand. Or even easier, just use Galileian relativity. You don't seem to have tried to do this problem in the easiest possible way. So why would you want to make it harder? Just do it the easy way. > I suppose that the best we can do is declare a trap. No, the best we can do is to sit down, and do the calculations. Then we see that the change in energy in the perfect reflection of the wave from a moving mirror works in just the same way as the elastic bouncing off of a ball from a moving wall. > Within the source frame we see less > energy returning, and so the problem degrades to this case again, though > with the twist of the Crookes radiometer that portion is just more > hidden. We can neglect any work done on the mirror in this situation and > just consider imposing a 0.5c velocity on the mirror to obtain the 50% > figure. Why would you ignore the work done on the mirror? If you're worried about conservation of energy, why would you ignore the work being done? The rate of doing work on a moving object by a force is Fv; include this, and you might just find that your problems with conservation of energy go away. > I remember deleting a quip in an earlier post that you would haggle over > power versus energy, and here it is virtually bighting me in the back. No, power vs force. > I find it hard to falsify this analysis above, and we overlooked this > back at the inception of the effect w.r.t. photon momentum. By this > analysis there never was any dE due to this effect. Go back, and re-read the original moving mirror analysis. It's there, it isn't overlooked. > So, what is the > proper stance on wave/energy if a reflector is accelerated by the wave? > Is the reflected beam nonrelativistically redshifted? Yes. Or relativistically redshifted. Depending on what you mean by "relativistically". It's just the same as in classical mechanics. With a lossless collision (i.e., an elastic collision), if the reflector gains energy, the reflected object must lose energy. Choose a reference frame where the reflector doesn't gain energy, and the reflected object doesn't lose (or gain) energy. > This would be the > best continuous solution, but the mechanism is still cloudy. This > conservation of energy will require a redshift not due to any velocity > and will instead be generating an acceleration. I still prefer an > electron based interpretation here, and suspect that the perfect > reflector will not provide any acceleration, since there will be a > conservation of energy problem. "Suspect" doesn't go very far in physics. "Experiment" does, and, in its place, derivation from theory does too. > Again, the final answer to me is to > argue that there will not be any perfect reflector. Fine. By the same argument, there can't be an elastic collision either. It seems premature to me to decide on a final answer based on "suspect", especially when it is contradicted by experiment. YMMV.
From: NoEinstein on 6 Jul 2010 23:44 On Jul 6, 11:35 am, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com> wrote: > Dear Tim: Talk is cheap. You, and-or Timo keep mentioning "waves" impacting vanes. Ocean waves have MASS, photons do not. Like I've explained, but you two won't listen and learn, light in the Universe can travel perfectly well across the Swiss Cheese voids between galaxies. Those voids were where the ENERGY was scavenged to create the galaxies. That energy is the ether, which stays close to where the masses are. If there was a "wave" associated with light, the ether would be it. But like photons, the ether is massless, and thus incapable of imparting a thrust to any object, including: the vanes of the Crookes Radiometer; sails in space; nor to the planets of the solar system which would already have been pushed into the space beyond, IF light has thrust. I've explained why the tails of comets point away from the sun. It's not due to the thrust of light, but due to the diminishment of light because of the SHADING provided by dust particles closer to the Sun. Solar radiation creates the ether push on the opposing sides of masses that makes the planets stay in orbit. Photon exchange is part of the mechanism of gravity. It has absolutely nothing to do with... space- time nor... "relativistic" anything. Apparently, you and Timo are more interested in playing pedants with the errant status quo physics than you are in learning the true mechanism of gravity, and the true reason the Crookes Radiometer rotates black squares trailing. I'm perfect OK with have you two continue your solitary exchanges. But I would ask this: Please don't "change the subject" on my post. You and Timo are perfectly at liberty to do your shallow talking, elsewhere. NoEinstein 6-05-10 Dear Tim: My reading speed can't take in all that you are saying. But I don't disagree. The 'missing link' of your Black Body rationale is that the white light coming in has more energy per 'ray'. To have the same infrared radiation, there have to be a higher number of rays. It's likely that it's the number of rays that account for the amount of ether ejected along with the photons. In the Crookes, light reflecting from the white squares, and the ether ejected in the same direction, push some of the argon atoms toward the adjacent (opposed) black squares. It is the mass of those argon atoms which causes the vanes to rotate, not heat "rocketing" from the porous edges of the vanes! NoEinstein Taking a Fresh Look at the Physics of Radiometers. http://groups.google.com/group/sci.physics/browse_thread/thread/3ebe85495d1929b0/ba1163422440ffd9?hl=en#ba1163422440ffd9 A Proposed Gravity-Propelled Swing Experiment. http://groups.google.com/group/sci.physics/browse_thread/thread/3052e7f7b228a800/aef3ee7dc59b6e2f?hl=en&q=gravity+swing Shedding New Light on Comet Tails http://groups.google.com/g/d8e7fef4/t/fbb6a213b8c465b3/.../187797453b40de4f?... > > Timo Nieminen wrote: > > On Jul 3, 11:00 pm, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com> > > wrote: > >> Timo Nieminen wrote: > >>> On Jul 2, 1:16 am, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com> > >>> wrote: > >>>> Timo Nieminen wrote: > >>>>> On Jun 30, 11:26 pm, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com> > >>>>> wrote: > >>>>>> Let's suppose a Crookes type of device, so that the experiment could be > >>>>>> done on a lab table. Upon getting the vanes to high speed the laser > >>>>>> light, being red shifted substantially, should yield ever more drive to > >>>>>> the device. If we could get a 50% redshift, then we'd be taking 50% of > >>>>>> the laser energy and converting it into mechanical energy, or > >>>>>> acceleration of the vane. Should this work? > >>>>> Why shouldn't it work? But be careful with language! You write "more > >>>>> drive", but note well that this means "more power", not "more force". > >>>>> With the device as described, you'd get _less_ force as it spins > >>>>> faster (but still more power). Why less force? Unlike the previous > >>>>> analysis of the redshift/blueshift, which was done with a chnage of > >>>>> coordinate system, you're talking about a real motion of the vane > >>>>> relative to the source, so a redshift seen by the vane, so a reduction > >>>>> in power as seen by the vane. > >>>> Yet still returning to the source is 50% of what came from it (at the > >>>> normal incidence angle), and so we have entered a conundrum. Where did > >>>> the other 50% go? I do see the argument for diminishing return on the > >>>> vane as you state it here, and I likewise see the argument from the > >>>> source perspective as well. We are nearly building a Demon. > >>>> I never have appreciated the random coordinate system change that you > >>>> constructed, and I see no reason to enter it into this conversation > >>>> other than to cloud the discussion. The redshift which you and I have > >>>> been speaking of for the last few posts has nothing to do with a > >>>> randomly chosen frame. > >>> Not _random_. Sit down, and do some calculations. Since you don't > >>> believe various stuff relating to radiation momentum, just sit down > >>> and do some classical mechanicws calculations. Are you OK with > >>> Newtonian mechanics? Do some elastic collision problems, the same > >>> problem in some different inertial coordinate systems. If you're up to > >>> it, do the same for some classical "hosing" problems, like hosing a > >>> stream of continuous fluid onto a "reflector", with no heating, no > >>> loss of energy. > >>> What assumptions go into this? Conservation of energy and momentum, > >>> and that these conservation laws hold in all inertial coordinate > >>> systems, and that (under Galileian relativity) the force and > >>> acceleration are the same in all inertial coordinate systems. > >>> And what conundrum? Beam loses energy due to Doppler shift, work is > >>> done on the target because it's moving. The reduction in beam energy > >>> is equal to the work. Where is the conundrum? Choose any inertial > >>> coordinate system, and energy and momentum are conserved, and we see > >>> the same force in all such systems. What is the problem, the > >>> conundrum? > >> I've already stated it clearly without confusion: > >> "Yet still returning to the source is 50% of what came from it > >> (at the normal incidence angle), and so we have entered a conundrum. > >> Where did the other 50% go?" > >> You've already admitted that the vane too sees less power: > >> "you're talking about a real motion of the vane > >> relative to the source, so a redshift seen by the vane, > >> so a reduction in power as seen by the vane." > > >> We have just constructed a break with conservation of energy, and in > >> both frames. We are seeing loss of energy in both frames. > > > No. In the frame where the vane is moving, the difference in energy > > between the incident beam and the red-shifted reflected beam is equal > > to the work done. No loss of energy, no problem with conservation of > > energy. > > > If you were to look at this in the vane's instantaneous rest frame, > > you'd see that the incident beam is of lower power, the reflected beam > > is of the same power as the incident beam, and no work is done. Again, > > there is no lost energy, no problem with conservation of energy. > > > This isn't magic. It works the same way in classical mechanics. Look > > at an elastic collision along a line between a light ball and a heavy > > ball in the centre-of-mass rest frame. In this frame, the light ball > > starts at speed v, and finishes at speed v (the velocity is reversed). > > The light ball loses no KE, starting with (1/2)mv^2 and finishing with > > (1/2)mv^2. The heavy ball also gains no KE - it's KE before and after > > is (1/2)MV^2 (which is equal to (1/2)mv^2 * (m/M) since mv=MV). > > > Choose a different reference frame. Try a frame where the light ball > > is initially moving at 2v. Starts with (1/2)m(2v)^2 = 2mv^2 KE, 4 > > times as much as before. Is this "increase" a problem for conservation > > of energy? No, since the KE just depends on our choice of coordinate > > system, just like speed and position depend on it. The light ball > > finishes with KE=0. The initial KE of the heavy ball is (1/2)M(V-v)^2, > > the final KE is (1/2)M(V+v)^2. No suprise that the difference is equal > > to 2mv^2. > > > OK, what happens when we look at an elastic collision like this? The > > energy depends on our choice of coordinate system, and the work done > > depends on our choice of coordinate system. In each coordinate system, > > KE lost by light ball = KE gained by heavy ball; energy is conserved > > in each case. > > > Same thing with beam-and-vane. Conservation of energy works fine, for > > any choice of (inertial) coordinate system we want to use, or for any > > velocity of vane relative to source. > > >> We should have > >> to go to the relativistic analysis of the vanes and see that the inertia > >> has come up quite high at a 50% redshift. Still, having gotten there, > >> by the conservation of energy, those vanes must be absorbing 50% of the > >> energy of the beam still (source perspective). Meanwhile they are doing > >> so while seeing the beam as 50% of the source energy (50% redshifted), > >> and again, these are reflective plates we are using. Can't you see the > >> chicken and egg conflict here? It is absurd that you would refute the > >> conundrum from my perspective. > > >> Interestingly we've engaged a rotational instrument in order to achieve > >> this effect. You say there is no need to resolve this conflict. You also > >> have said that this effect is consistent with photon momentum. I believe > >> these both to be false statements. > > > Do the calculation. It works the same way as the classical collision > > calculation. The "problems" that you say are problems are equally > > "problems" with classical mechanics. That is, they're not problems at > > all. (They can be surprising to students, baffling, troubling, but > > that doesn't mean that they're _problems_ .) > > > If conservation of energy does really fail here, you should be able to > > show it. As in calculating the energy before, and the energy after, > > and showing they're different from each other. > > > If it's inconsistent with photon momentum, again, you can show it. > > >> The construction we are discusssing > >> is far away from photon momentum. We are claiming to extract 50% of the > >> light energy mechanically. I don't honestly know what to believe, but > >> this is much closer to holding a mirror up to the sun and being pushed > >> over, which we know will not happen. > > > Why "much closer"? Sure, you can look at it in a frame where 50% of > > the incident power is "lost" through redshift. > > Yes, this is the lab table reference frame; the one we'd be able to do > measurements from most easily. I see you've put lost in quotes here so I > presume you do have further analysis within this frame, otherwise, when > we remove these quotes we have a conservation of energy problem. > > Isn't it it true that Einstein's relativity relies upon inertial > reference frames? We do not then really have the option to use the vane > as a reference frame, for it is a rotational (and possibly accelerated) > frame. > > The amount of analysis at the moving reflective vane is daunting. We're > going to have some sort of spectroscopic output with angle, and then a > beam splitting as the next vane encounters the beam. Could it be that > all of this redirecting of the beam consumes the other 50% of the > energy? This would be a new effect to me, but so was scattering about a > month ago. > > I suppose that the best we can do is declare a trap. The problem is > essentially the same as with a relativistically moving reflector away > from the source without rotation. Within the source frame we see less > energy returning, and so the problem degrades to this case again, though > with the twist of the Crookes radiometer that portion is just more > hidden. We can neglect any work done on the mirror in this situation and > just consider imposing a 0.5c velocity on the mirror to obtain the 50% > figure. We can see that over time a pulse of the light which returns to > the source from the reflector will be longer in duration. If and when > the source energy extinguishes itself the reflector will still be > returning energy. The quantity of wavelengths will be conserved, and so > the 50% energy figure was over constant time, and had we integrated over > all time we would admit that the energy is still all there. > > Returning to the rotational Crookes device we see that we have pulsed a > continuous beam. Though this wording is loose, the redshifted beam > returning to the source will be longer in duration, and will conserve > the number of cycles of wave energy. To simplify, we could impose a > pulsed laser pulsing a short pulse each time a vane comes normal to the > source, and then even with the vanes moving at 0.5c (some fudge here for > rotational complexity) then we'll see the pulses are twice as long in > duration returning to the source. > > I remember deleting a quip in an ... > > read more »- Hide quoted text - > > - Show quoted text -
From: Tim Golden BandTech.com on 8 Jul 2010 17:10
Timo Nieminen wrote: > On Jul 7, 1:35 am, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com> > wrote: >> Yes, this is the lab table reference frame; the one we'd be able to do >> measurements from most easily. I see you've put lost in quotes here so I >> presume you do have further analysis within this frame, otherwise, when >> we remove these quotes we have a conservation of energy problem. > > Energy before = energy after, no loss, no problem. You've deleted the most important part of my prior post. The redshift has been misinterpreted by both of us. It is merely an energy delay, for so long as the number of cycles is preserved then the energy is conserved. I've been discussing a 50% redshift, and a pulse 1 ns long from the source will come back 2 ns long at 50% redshift. The pulse will have the same number of cycles and so this conservation of cycles is a pretty good paradigm. I was just reading Scientific American, and they pose a conservation of energy problem in the cosmological redshift, but they never came up with this answer. This is a general answer and applies there too. Now, returning to the reflector, if a perfect reflector returns all of the energy from the source back to the source then by conservation of energy no work can have been done on the reflector. There is no redshift energy loss. There is no acceleration, theoretically speaking. So I can still claim that there is a conflict between theory and experiment. Experiment shows a deflection, and supposedly even a doubling effect over an absorber, within photon momentum theory. Conservation of energy requires that the perfect reflector cannot accelerate. Energy before = energy after, no loss, no acceleration. I suspect you'll address this in your next post. - Tim > >> Isn't it it true that Einstein's relativity relies upon inertial >> reference frames? We do not then really have the option to use the vane >> as a reference frame, for it is a rotational (and possibly accelerated) >> frame. > > Technically, no, Einstein's relativity doesn't rely on inertial > reference frames, neither special relativity nor general relativity. > In SR, it is easiest to use inertial reference frames. Just use the > instantaneous rest frame of the vane - works for the problem at hand. > Or even easier, just use Galileian relativity. > > You don't seem to have tried to do this problem in the easiest > possible way. So why would you want to make it harder? Just do it the > easy way. > >> I suppose that the best we can do is declare a trap. > > No, the best we can do is to sit down, and do the calculations. Then > we see that the change in energy in the perfect reflection of the wave > from a moving mirror works in just the same way as the elastic > bouncing off of a ball from a moving wall. > >> Within the source frame we see less >> energy returning, and so the problem degrades to this case again, though >> with the twist of the Crookes radiometer that portion is just more >> hidden. We can neglect any work done on the mirror in this situation and >> just consider imposing a 0.5c velocity on the mirror to obtain the 50% >> figure. > > Why would you ignore the work done on the mirror? If you're worried > about conservation of energy, why would you ignore the work being > done? > > The rate of doing work on a moving object by a force is Fv; include > this, and you might just find that your problems with conservation of > energy go away. > >> I remember deleting a quip in an earlier post that you would haggle over >> power versus energy, and here it is virtually bighting me in the back. > > No, power vs force. > >> I find it hard to falsify this analysis above, and we overlooked this >> back at the inception of the effect w.r.t. photon momentum. By this >> analysis there never was any dE due to this effect. > > Go back, and re-read the original moving mirror analysis. It's there, > it isn't overlooked. > >> So, what is the >> proper stance on wave/energy if a reflector is accelerated by the wave? >> Is the reflected beam nonrelativistically redshifted? > > Yes. Or relativistically redshifted. Depending on what you mean by > "relativistically". > > It's just the same as in classical mechanics. With a lossless > collision (i.e., an elastic collision), if the reflector gains energy, > the reflected object must lose energy. Choose a reference frame where > the reflector doesn't gain energy, and the reflected object doesn't > lose (or gain) energy. > >> This would be the >> best continuous solution, but the mechanism is still cloudy. This >> conservation of energy will require a redshift not due to any velocity >> and will instead be generating an acceleration. I still prefer an >> electron based interpretation here, and suspect that the perfect >> reflector will not provide any acceleration, since there will be a >> conservation of energy problem. > > "Suspect" doesn't go very far in physics. "Experiment" does, and, in > its place, derivation from theory does too. > >> Again, the final answer to me is to >> argue that there will not be any perfect reflector. > > Fine. By the same argument, there can't be an elastic collision > either. > > It seems premature to me to decide on a final answer based on > "suspect", especially when it is contradicted by experiment. YMMV. |