Doppler energy extraction: new solar cell technology
in [Physics]
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From: Tim BandTech.com on 17 Jul 2010 10:31 On Jul 16, 10:11 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > On Jul 17, 11:09 am, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > > > > > On Jul 16, 8:46 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > We have: > > > > (a) The power of an EM beam (i.e., the energy flux) is affected by > > > redshift or blueshift. > > > > (b) The cycles don't disappear. (I don't see why you think this is > > > relevant, but let me leave it here on the list.) > > > > (c) P = F.v > > > > Consider a stationary perfect reflector reflecting a monochromatic > > > collimated beam of power P0 and frequency f0. By "perfect reflector", > > > I mean that no energy is absorbed by the reflector. By "absorbed", I > > > mean that none of the incident energy is transferred to the internal > > > energy of the reflector. "Internal energy" most importantly includes > > > thermal energy, but could include other forms of energy, such as > > > electronic excitations leading to fluorescence, etc. "Internal energy" > > > does not include kinetic energy due to motion or rotation of the > > > object (i.e., the reflector, in this case). > > > > We have v = 0, and, from P = F.v, no work is done, regardless of what > > > the force F is. > > > > Since no energy is absorbed (since we are considering a perfect > > > reflector), and no work is being done, the reflected beam must be of > > > power P0. Since the reflector is stationary, there is no Doppler > > > shift, and the reflected beam is of power f0. > > > > From conservation of energy, > > > > P_out = P_in - rate of increase in internal energy - rate of increase > > > of KE > > > > P_out = P0 - 0 - F.v > > > > P_out = P0 - 0 - F.0 = P0 > > > This is an argument for no acceleration of the reflector. This is what > > I've been arguing for nearly from the start. Aren't you now > > contradicting the Nichols results? > > No. > > OK, we need to stop here; no point in going past this at the moment. > > Let DP = the change in power of the beam on reflection (D as a > substitute for delta). > > For a stationary perfect reflector, DP = 0. > > We have, above, -DP = absorbed power + rate of doing work. Since the > reflector is a perfect reflector, absorbed power = 0. This reduces to > > DP = - rate of doing work = -F.v > > Since DP = 0, and v = 0, F must satisfy > > 0*F = 0. > > You have been claiming that this means that F must be zero, I have > been claiming that F can be non-zero and still satisfy this. > > Let me give an example. Take F = 1kN. (DP = 0W, v = 0 m/s.) > > 0*1000 = 0, as required. > > Try this with any number for F (that is, any finite force, zero, > positive, or negative), and you will see that 0*F=0 is satisfied for > any value of F. > > It is not valid to argue that 0*F = 0 means that F=0. > > OK so far? If not, why isn't the numerical example showing that F=1kN > is compatible with DP = 0 sufficient? But what are you arguing for? You've essentially led us down this path of incomplete construction and landed at an incomplete result. According to your analysis F can be anything, and this is just like the static case of a person trying to push over a boulder. As hard as we may push, if the boulder does not shift, then no work was done. It does not matter how hard we pushed, until we exceed some threshold of force. This of course overlooks losses, for I imagine that a human could wear themselves out attempting to move the stone and so to have worked very hard, but all of their work would have been done on themselves and their tools, not on the boulder. Back when you were insistent upon doing a classical analysis I exposed that for the energy of a reflective collision to be returned in its entirety toward the source that the reflector would have to be infinitely massive. You are nearby with this construction, but to avoid acceleration seems to me to be a problem. Still, the mapping from the classical problem to the electromagnetic seems to be of the projectile becoming massless and the target becoming a finite mass, which we will have to admit of any reflector we construct, at least to this date. There are numerous strange phenomena taking place in cold physics and I do think it would be wise to attempt to take in more of the problem rather than keep sieving it down to the classical physics. The actual materials that we can work with are very real experimental limiting factors, and so long as the reflective loss exceeds the work done to radiation pressure then we will be doing an incomplete analysis. Suppose for instance that aluminum is not reflective at some very low light intensity. This would have to be verified at cold temperature, and what we would then observe is a threshold energy upon which the aluminum begins to reflect; the threshold being that which essentially primes the aluminum into being reflective. Such a feature could tie back to the Nichols experiment and is not unlike the charge pumping that I've suggested, though it would be of a more complex geometry. You yourself have stated that these do not accomodate the light trap, which is supposed to be a cold environment. You claim that the metallic particles are driven out of the trap right? Or are they merely drifting out? The metals which provide the most generic reflectance (overlooking TIR) seem to be soft and conductive. Why is this? The standard atomic theory seems to answer much here, but as to the Fourier analysis, well, here I do think there are some weaknesses, for we are taught that the only interaction that photons have with atoms is via electrons jumping or descending in discrete shells. These material analyses become an assay of data that have some detail. One site I was reading attributes the color of gold to a relativistic phenomenon, though I don't actually remember how they got there. I guess, since we are attempting an investigation and both seem willing to stray and follow lines of thought for their own sake, which I am supportive of, we could try to go down there. As I search for 'mechanism of reflection' and such the closest I find is http://physics-edu.org/reflection.htm which is quite lame. Better is http://en.wikipedia.org/wiki/Reflection_(physics)#Mechanisms_of_reflection but still, there is no analysis of the permittivity and so forth, which within electrical engineering will expose two types of reflectors: the short ciruit reflector( the one the wiki talks about with pi phase change) and the open circuit reflector; with no phase change. Trying to delve into the optical material world is daunting, and it does seem easier to go from EE stuff, which is to say RF; lower frequencies, where a coax cable can carry a steady kilowatt, though with some loss. - Tim > > > > (We could write the rate of increase of internal energy in terms of an > > > absorption coefficient a as a*P0. For a perfect reflector, a=0.) > > > > Note well that the reflected beam is of power P0 and frequency f0 for > > > _any_ force F, zero or non-zero, that the beam might exert on the > > > reflector. > > > > The perfect reflector, with no absorption, is an idealisation. It > > > doesn't matter that it can't be achieved with a real reflector; we are > > > considering an idealised thought experiment here, just as we consider > > > frictionless surfaces, rigid bodies, inextensible strings, light > > > strings, monochromatic collimated beams, and so on. We can get > > > usefully close to the idealisation with a real reflector, so a > > > reasonable approximation of the thought experiment can be realised. If > > > you feel that we come to some wrong result due to these idealisations, > > > do say so, but preferably with some concrete details of why. > > > > The usage of "perfect reflector" and "absorbed" above are consistent > > > with standard usage in physics. These are technical terms, and do not > > > necessarily mean the same thing as the same terms in everyday English. > > > It will be important to use these terms consistently, and not change > > > their meanings in the middle of the discussion. > > > > Do these terms need further clarification? (Yes/no? If yes, what do > > > you suggest?) > > > > For the next step, it will be necessary to consider relatively moving > > > reference frames. It will be sufficient to use Galileian relativity. > > > In principle, we could also use Lorentz/Einstein relativity (i.e., > > > SR), taking limits to first order in v/c if we wish. > > > > Under Galileian relativity, mass, force, acceleration, and internal > > > energy are invariant under change of inertial reference frame (i.e., > > > they are the same in all inertial reference frames). Velocity of an > > > object, and its KE, are not invariant under such transformations. > > > (Also true to first order in v/c in SR.) Agreed? > > > I guess so. You are not yielding a complete argument, but I am willing > > to go along. So far you seem to be supporting the idea that there is > > no theoretical support for the radiation pressure acceleration effect, > > from energy conservation of a perfect reflector.
From: Timo Nieminen on 17 Jul 2010 20:27 On Jul 18, 12:31 am, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > On Jul 16, 10:11 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > > On Jul 17, 11:09 am, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > > > > On Jul 16, 8:46 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > > We have: > > > > > (a) The power of an EM beam (i.e., the energy flux) is affected by > > > > redshift or blueshift. > > > > > (b) The cycles don't disappear. (I don't see why you think this is > > > > relevant, but let me leave it here on the list.) > > > > > (c) P = F.v > > > > > Consider a stationary perfect reflector reflecting a monochromatic > > > > collimated beam of power P0 and frequency f0. By "perfect reflector", > > > > I mean that no energy is absorbed by the reflector. By "absorbed", I > > > > mean that none of the incident energy is transferred to the internal > > > > energy of the reflector. "Internal energy" most importantly includes > > > > thermal energy, but could include other forms of energy, such as > > > > electronic excitations leading to fluorescence, etc. "Internal energy" > > > > does not include kinetic energy due to motion or rotation of the > > > > object (i.e., the reflector, in this case). > > > > > We have v = 0, and, from P = F.v, no work is done, regardless of what > > > > the force F is. > > > > > Since no energy is absorbed (since we are considering a perfect > > > > reflector), and no work is being done, the reflected beam must be of > > > > power P0. Since the reflector is stationary, there is no Doppler > > > > shift, and the reflected beam is of power f0. > > > > > From conservation of energy, > > > > > P_out = P_in - rate of increase in internal energy - rate of increase > > > > of KE > > > > > P_out = P0 - 0 - F.v > > > > > P_out = P0 - 0 - F.0 = P0 > > > > This is an argument for no acceleration of the reflector. This is what > > > I've been arguing for nearly from the start. Aren't you now > > > contradicting the Nichols results? > > > No. > > > OK, we need to stop here; no point in going past this at the moment. > > > Let DP = the change in power of the beam on reflection (D as a > > substitute for delta). > > > For a stationary perfect reflector, DP = 0. > > > We have, above, -DP = absorbed power + rate of doing work. Since the > > reflector is a perfect reflector, absorbed power = 0. This reduces to > > > DP = - rate of doing work = -F.v > > > Since DP = 0, and v = 0, F must satisfy > > > 0*F = 0. > > > You have been claiming that this means that F must be zero, I have > > been claiming that F can be non-zero and still satisfy this. > > > Let me give an example. Take F = 1kN. (DP = 0W, v = 0 m/s.) > > > 0*1000 = 0, as required. > > > Try this with any number for F (that is, any finite force, zero, > > positive, or negative), and you will see that 0*F=0 is satisfied for > > any value of F. > > > It is not valid to argue that 0*F = 0 means that F=0. > > > OK so far? If not, why isn't the numerical example showing that F=1kN > > is compatible with DP = 0 sufficient? > > But what are you arguing for? You've essentially led us down this path > of incomplete construction and landed at an incomplete result. You have repeated claimed that DP = 0 means that F = 0. You have made these claims for a stationary reflector, that the lack of absorption and the lack of Doppler shift means that there can't be any force. It comes down to DP = -F.v, in the absence of absorption. Your claim has, repeatedly, been that DP=0 requires F=0. This is an incorrect claim, and the above is an attempt to show that it is incorrect. Since we have returned to this claim for the 3rd or 4th time (Why? Isn't it obvious that 0=0*F allows non-zero F?), it needs to be dealt with before proceeding. > According to your analysis F can be anything, and this is just like > the static case of a person trying to push over a boulder. As hard as > we may push, if the boulder does not shift, then no work was done. No, you are overgeneralising. P = F.v works even if the boulder is accelerating. It does not need to be fixed in place. It is sufficient that v(t=t1)=0 for P(t1)=0. If we also have DP(t1)=0, then DP = -F.v is still satisfied, for any F. Yes, that is the point, that F _can_ be anything. You have been arguing that, under the same conditions, you _must_ have F=0. > It > does not matter how hard we pushed, until we exceed some threshold of > force. This of course overlooks losses, for I imagine that a human > could wear themselves out attempting to move the stone and so to have > worked very hard, but all of their work would have been done on > themselves and their tools, not on the boulder. Irrelevant if the human expends energy exerting a force. What matters is the work done by the force acting on the boulder. > Back when you were insistent upon doing a classical analysis I exposed > that for the energy of a reflective collision to be returned in its > entirety toward the source that the reflector would have to be > infinitely massive. No, this is wrong, too. (Surely you mean _power_ here, not energy? Or are you ignoring your Doppler shift "delay"?) For the instantaneous power to be returned entirely, it is sufficient to have v=0. We have 3 cases: (a) We have v=0 instantaneously only. The object is accelerating, but v(t) = 0 for some t=t1. At t=t1, we have DP(t1)=0. (b) The object is not accelerating. This will occur if (i) F_reflection = 0, and F_total = F_reflection, (ii) mass is infinite, or (iii) F_total = 0 and F_reflection is non-zero, which requires F_other = -F_reflection, some restraining force preventing the acceleration. Note well that infinite mass only means a=0, not v=0. In particular, infinite mass can be accompanied by non-zero v, and non-zero DP. That is, infinite mass _does not_ mean that all incident power will be returned. > You are nearby with this construction, but to > avoid acceleration seems to me to be a problem. Still, the mapping > from the classical problem to the electromagnetic seems to be of the > projectile becoming massless and the target becoming a finite mass, > which we will have to admit of any reflector we construct, at least to > this date. There are numerous strange phenomena taking place in cold > physics and I do think it would be wise to attempt to take in more of > the problem rather than keep sieving it down to the classical physics. Deal with the basics first. The basics - conservation of energy and momentum - still work with all the complicated details added. (Cut the rest for now. There isn't any big secret in the classical EM picture of reflection, either by a dielectric or a conductor - there is a change in wave impedance, there is reflection. In either case, direct calculation of the EM force acting on the reflector, via the Lorentz force acting on induced currents and polarisation, yields the same result as the conservation of energy/momentum radiation pressure calculation. Let us get past P=-F.v first before we try to deal with some serious EM. If we get past the basics, then we can return to this.)
From: Tim BandTech.com on 18 Jul 2010 10:46 On Jul 17, 8:27 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > On Jul 18, 12:31 am, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > > > > > On Jul 16, 10:11 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > On Jul 17, 11:09 am, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > > > > > On Jul 16, 8:46 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > > > We have: > > > > > > (a) The power of an EM beam (i.e., the energy flux) is affected by > > > > > redshift or blueshift. > > > > > > (b) The cycles don't disappear. (I don't see why you think this is > > > > > relevant, but let me leave it here on the list.) > > > > > > (c) P = F.v > > > > > > Consider a stationary perfect reflector reflecting a monochromatic > > > > > collimated beam of power P0 and frequency f0. By "perfect reflector", > > > > > I mean that no energy is absorbed by the reflector. By "absorbed", I > > > > > mean that none of the incident energy is transferred to the internal > > > > > energy of the reflector. "Internal energy" most importantly includes > > > > > thermal energy, but could include other forms of energy, such as > > > > > electronic excitations leading to fluorescence, etc. "Internal energy" > > > > > does not include kinetic energy due to motion or rotation of the > > > > > object (i.e., the reflector, in this case). > > > > > > We have v = 0, and, from P = F.v, no work is done, regardless of what > > > > > the force F is. > > > > > > Since no energy is absorbed (since we are considering a perfect > > > > > reflector), and no work is being done, the reflected beam must be of > > > > > power P0. Since the reflector is stationary, there is no Doppler > > > > > shift, and the reflected beam is of power f0. > > > > > > From conservation of energy, > > > > > > P_out = P_in - rate of increase in internal energy - rate of increase > > > > > of KE > > > > > > P_out = P0 - 0 - F.v > > > > > > P_out = P0 - 0 - F.0 = P0 > > > > > This is an argument for no acceleration of the reflector. This is what > > > > I've been arguing for nearly from the start. Aren't you now > > > > contradicting the Nichols results? > > > > No. > > > > OK, we need to stop here; no point in going past this at the moment. > > > > Let DP = the change in power of the beam on reflection (D as a > > > substitute for delta). > > > > For a stationary perfect reflector, DP = 0. > > > > We have, above, -DP = absorbed power + rate of doing work. Since the > > > reflector is a perfect reflector, absorbed power = 0. This reduces to > > > > DP = - rate of doing work = -F.v > > > > Since DP = 0, and v = 0, F must satisfy > > > > 0*F = 0. > > > > You have been claiming that this means that F must be zero, I have > > > been claiming that F can be non-zero and still satisfy this. > > > > Let me give an example. Take F = 1kN. (DP = 0W, v = 0 m/s.) > > > > 0*1000 = 0, as required. > > > > Try this with any number for F (that is, any finite force, zero, > > > positive, or negative), and you will see that 0*F=0 is satisfied for > > > any value of F. > > > > It is not valid to argue that 0*F = 0 means that F=0. > > > > OK so far? If not, why isn't the numerical example showing that F=1kN > > > is compatible with DP = 0 sufficient? > > > But what are you arguing for? You've essentially led us down this path > > of incomplete construction and landed at an incomplete result. > > You have repeated claimed that DP = 0 means that F = 0. You have made > these claims for a stationary reflector, that the lack of absorption > and the lack of Doppler shift means that there can't be any force. > > It comes down to DP = -F.v, in the absence of absorption. Your claim > has, repeatedly, been that DP=0 requires F=0. This is an incorrect > claim, and the above is an attempt to show that it is incorrect. Since > we have returned to this claim for the 3rd or 4th time (Why? Isn't it > obvious that 0=0*F allows non-zero F?), it needs to be dealt with > before proceeding. > > > According to your analysis F can be anything, and this is just like > > the static case of a person trying to push over a boulder. As hard as > > we may push, if the boulder does not shift, then no work was done. > > No, you are overgeneralising. P = F.v works even if the boulder is > accelerating. It does not need to be fixed in place. It is sufficient > that v(t=t1)=0 for P(t1)=0. If we also have DP(t1)=0, then DP = -F.v > is still satisfied, for any F. > > Yes, that is the point, that F _can_ be anything. You have been > arguing that, under the same conditions, you _must_ have F=0. Ahh. If the boulder is in outer space, then yes, the force must be zero. If the force is even slight then the boulder accelerates. Likewise for a free reflector at steady zero velocity the force must be zero. I don't understand how the discussion has degenerated this badly. I have presented you with the condition where a variable force does no work, but they are not the conditions of a reflector floating in space, or tethered on a quartz fiber as a torsion instrument(the Nichols experiment). You have provided constraints without providing the experiment. Please, if you are going to make a clear argument, don't just plop down an equation with artificial constraints. Context, please. Beyond this, you've chopped up your argument in such a way that we still do not see any complete presentation of your argument. My own argument fits in one paragraph, and I see no need to stretch a contradiction out to a decapost or so. The stone wall effect is yours, not mine. > > > It > > does not matter how hard we pushed, until we exceed some threshold of > > force. This of course overlooks losses, for I imagine that a human > > could wear themselves out attempting to move the stone and so to have > > worked very hard, but all of their work would have been done on > > themselves and their tools, not on the boulder. > > Irrelevant if the human expends energy exerting a force. What matters > is the work done by the force acting on the boulder. > > > Back when you were insistent upon doing a classical analysis I exposed > > that for the energy of a reflective collision to be returned in its > > entirety toward the source that the reflector would have to be > > infinitely massive. > > No, this is wrong, too. (Surely you mean _power_ here, not energy? Or > are you ignoring your Doppler shift "delay"?) Well, I am discussing from a stationary reference frame. When the target accelerates away from the ball, the return velocity of the ball will be diminished. This same effect I argue on the light case; that in order for the target to accelerate there must have been some energy absorbed, and thus the energy cannot have returned to the source. Yet we have no mechanism for this with a perfect reflector. Recently I've disproven this even with redshift. How is it that you feel comfortable dodging acceleration in your new argument, and so happy to have derived a force that can be anything? This indicates something and I suspect it is denial of the accuracy of my argument, which to be maintained requires your level of argumentation. I am still open to being falsified, but in no way do I see the raw P=fv as doing that. You've already used it to argue that the Nichols experiment will fail(v=0), and somehow go on now to attempt to falsify my claim, which connects directly. These are your own contradictions which I am interpreting, and which I have attempted to confront you on in my prior post, and which you continue to ignore. Likewise there was denial of my redshift argument, which you then go on to support in a later post. You see, without admission of the truth, you cannot argue for the truth. This does not bode well, and I do now question the validity of this discussion, as well as your own credibility. > > For the instantaneous power to be returned entirely, it is sufficient > to have v=0. We have 3 cases: Sufficient, but more importantly the acceleration due to that power must be zero, and v can actually be anything. Why you would care to construct the special case rather than the general case is puzzling, but acceptable. Why do we have three cases? Is it because you wish to confuse the subject? > > (a) We have v=0 instantaneously only. The object is accelerating, but > v(t) = 0 for some t=t1. At t=t1, we have DP(t1)=0. No. There is no acceleration. > > (b) The object is not accelerating. This will occur if (i) > F_reflection = 0, and F_total = F_reflection, (ii) mass is infinite, > or (iii) F_total = 0 and F_reflection is non-zero, which requires > F_other = -F_reflection, some restraining force preventing the > acceleration. > > Note well that infinite mass only means a=0, not v=0. In particular, > infinite mass can be accompanied by non-zero v, and non-zero DP. That > is, infinite mass _does not_ mean that all incident power will be > returned. Well, now you are trying to unhinge your own v=0, and that is not healthy. The a=0 that I mention should have been the crux up front. I have only said this perhaps a hundred times here, on the perfect reflector. What is this lameness that comes over you? Why do you dodge all trails leading back to Nichols? It is as if you will come out with some small fact in a decapost or two that will provide an answer. I'd like some straight talk, and I'd like it in your next post, please. > > > You are nearby with this construction, but to > > avoid acceleration seems to me to be a problem. Still, the mapping > > from the classical problem to the electromagnetic seems to be of the > > projectile becoming massless and the target becoming a finite mass, > > which we will have to admit of any reflector we construct, at least to > > this date. There are numerous strange phenomena taking place in cold > > physics and I do think it would be wise to attempt to take in more of > > the problem rather than keep sieving it down to the classical physics. > > Deal with the basics first. The basics - conservation of energy and > momentum - still work with all the complicated details added. I find this laughable. I don't care to smear you, yet I am nearly forced to by the weakness of your argument. You've derived a force that can be anything. We may substitute 1E10 Newtons, or 1E-10 Newtons, and you somehow are happy to rely upon this. Was it not true that to turn a Nichols vane the force would be more like the 1E-10 Newtons? Shouldn't we admit that there is need of a finite force? Let's consider a small reflective plate in free space, as in outer space. Beyond your P = f . v we also have the familiar f = m a and so we may as well write P = m a . v and so to overlook the acceleration will be a mistake. Here at least I have put the equations in some context, and without any context to work from there is little convincing beyond mimicing textbook equations that can take place. Even with the context, I'll have to admit that these are largely mimicry with but minor substitution. > > (Cut the rest for now. There isn't any big secret in the classical EM > picture of reflection, either by a dielectric or a conductor - there > is a change in wave impedance, there is reflection. In either case, > direct calculation of the EM force acting on the reflector, via the > Lorentz force acting on induced currents and polarisation, yields the Ahh... induced currents. This is somewhat what I was arguing for. Somehow Maxwell's original model had an electrokinetic energy term. The reflector will generally be conductive to electricity. I still don't see why a photon drives an electron in the same direction as the photon was travelling. I understand that the energy is directed, but somehow I still fail to make the connection clearly. This is where I'd rather be spending the discussion time. I don't see why the same effect should not be appropriate to the blackbody computation as well. - Tim > same result as the conservation of energy/momentum radiation pressure > calculation. Let us get past P=-F.v first before we try to deal with > some serious EM. If we get past the basics, then we can return to > this.)
From: Timo Nieminen on 18 Jul 2010 18:56 On Sun, 18 Jul 2010, Tim BandTech.com wrote: > On Jul 17, 8:27 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > On Jul 18, 12:31 am, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > > > On Jul 16, 10:11 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > On Jul 17, 11:09 am, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > > > > > On Jul 16, 8:46 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > > > > > We have: > > > > > > > > (a) The power of an EM beam (i.e., the energy flux) is affected by > > > > > > redshift or blueshift. > > > > > > > > (b) The cycles don't disappear. (I don't see why you think this is > > > > > > relevant, but let me leave it here on the list.) > > > > > > > > (c) P = F.v > > > > > > > > Consider a stationary perfect reflector reflecting a monochromatic > > > > > > collimated beam of power P0 and frequency f0. By "perfect reflector", > > > > > > I mean that no energy is absorbed by the reflector. By "absorbed", I > > > > > > mean that none of the incident energy is transferred to the internal > > > > > > energy of the reflector. "Internal energy" most importantly includes > > > > > > thermal energy, but could include other forms of energy, such as > > > > > > electronic excitations leading to fluorescence, etc. "Internal energy" > > > > > > does not include kinetic energy due to motion or rotation of the > > > > > > object (i.e., the reflector, in this case). > > > > > > > > We have v = 0, and, from P = F.v, no work is done, regardless of what > > > > > > the force F is. > > > > > > > > Since no energy is absorbed (since we are considering a perfect > > > > > > reflector), and no work is being done, the reflected beam must be of > > > > > > power P0. Since the reflector is stationary, there is no Doppler > > > > > > shift, and the reflected beam is of power f0. > > > > > > > > From conservation of energy, > > > > > > > > P_out = P_in - rate of increase in internal energy - rate of increase > > > > > > of KE > > > > > > > > P_out = P0 - 0 - F.v > > > > > > > > P_out = P0 - 0 - F.0 = P0 > > > > > > > This is an argument for no acceleration of the reflector. This is what > > > > > I've been arguing for nearly from the start. Aren't you now > > > > > contradicting the Nichols results? > > > > > > No. > > > > > > OK, we need to stop here; no point in going past this at the moment. > > > > > > Let DP = the change in power of the beam on reflection (D as a > > > > substitute for delta). > > > > > > For a stationary perfect reflector, DP = 0. > > > > > > We have, above, -DP = absorbed power + rate of doing work. Since the > > > > reflector is a perfect reflector, absorbed power = 0. This reduces to > > > > > > DP = - rate of doing work = -F.v > > > > > > Since DP = 0, and v = 0, F must satisfy > > > > > > 0*F = 0. > > > > > > You have been claiming that this means that F must be zero, I have > > > > been claiming that F can be non-zero and still satisfy this. > > > > > > Let me give an example. Take F = 1kN. (DP = 0W, v = 0 m/s.) > > > > > > 0*1000 = 0, as required. > > > > > > Try this with any number for F (that is, any finite force, zero, > > > > positive, or negative), and you will see that 0*F=0 is satisfied for > > > > any value of F. > > > > > > It is not valid to argue that 0*F = 0 means that F=0. > > > > > > OK so far? If not, why isn't the numerical example showing that F=1kN > > > > is compatible with DP = 0 sufficient? > > > > > But what are you arguing for? You've essentially led us down this path > > > of incomplete construction and landed at an incomplete result. > > > > You have repeated claimed that DP = 0 means that F = 0. You have made > > these claims for a stationary reflector, that the lack of absorption > > and the lack of Doppler shift means that there can't be any force. > > > > It comes down to DP = -F.v, in the absence of absorption. Your claim > > has, repeatedly, been that DP=0 requires F=0. This is an incorrect > > claim, and the above is an attempt to show that it is incorrect. Since > > we have returned to this claim for the 3rd or 4th time (Why? Isn't it > > obvious that 0=0*F allows non-zero F?), it needs to be dealt with > > before proceeding. > > > > > According to your analysis F can be anything, and this is just like > > > the static case of a person trying to push over a boulder. As hard as > > > we may push, if the boulder does not shift, then no work was done. > > > > No, you are overgeneralising. P = F.v works even if the boulder is > > accelerating. It does not need to be fixed in place. It is sufficient > > that v(t=t1)=0 for P(t1)=0. If we also have DP(t1)=0, then DP = -F.v > > is still satisfied, for any F. > > > > Yes, that is the point, that F _can_ be anything. You have been > > arguing that, under the same conditions, you _must_ have F=0. > > Ahh. If the boulder is in outer space, then yes, the force must be > zero. If the force is even slight then the boulder accelerates. > Likewise for a free reflector at steady zero velocity the force must > be zero. Since the rate of doing work by a force is P=F.v, your claim is equivalent to claiming that 0*F=0 requires F=0. Counterexample: 0*1000=0. > I don't understand how the discussion has degenerated this > badly. I don't know either. P=F.v is simple mathematics, 0*F=0 for any finite F is elementary mathematics. You even agree(d) with P=F.v. If you are still claiming that F must be zero when P=0, even if v=0, please give some support for the claim, rather than simply repeating it. > > For the instantaneous power to be returned entirely, it is sufficient > > to have v=0. We have 3 cases: > > Sufficient, but more importantly the acceleration due to that power > must be zero, and v can actually be anything. Why you would care to > construct the special case rather than the general case is puzzling, > but acceptable. Why do we have three cases? Is it because you wish to > confuse the subject? > > > > > (a) We have v=0 instantaneously only. The object is accelerating, but > > v(t) = 0 for some t=t1. At t=t1, we have DP(t1)=0. > > No. There is no acceleration. Prove it. It is possible to have v=0 and a=nonzero at some time t1. Do you deny this? > > (b) The object is not accelerating. This will occur if (i) > > F_reflection = 0, and F_total = F_reflection, (ii) mass is infinite, > > or (iii) F_total = 0 and F_reflection is non-zero, which requires > > F_other = -F_reflection, some restraining force preventing the > > acceleration. > > > > Note well that infinite mass only means a=0, not v=0. In particular, > > infinite mass can be accompanied by non-zero v, and non-zero DP. That > > is, infinite mass _does not_ mean that all incident power will be > > returned. > > Well, now you are trying to unhinge your own v=0, and that is not > healthy. > The a=0 that I mention should have been the crux up front. I > have only said this perhaps a hundred times here, on the perfect > reflector. What is this lameness that comes over you? Why do you dodge > all trails leading back to Nichols? It is as if you will come out with > some small fact in a decapost or two that will provide an answer. I'd > like some straight talk, and I'd like it in your next post, please. Why do you dodge elementary mathematics? You are continuing to claim that 0*F=0 requires F=0! You agreed that the rate of doing work on an object is P=F.v. Do you disagree with that now? You didn't disagree with P_out = P_in - rate of increase in internal energy - rate of increase of KE which gives us DP = -F.v before. Do you disagree with this now? DP = 0, no change in power due to reflection, gives us F.v=0. For a stationary reflector, v=0. So, we have 0*F=0. You disagree, and claim that F must be zero. Can you show that non-zero F leads to a contradiction? That non-zero F doesn't satisfy 0*F=0? ********** THIS IS THE KEY POINT HERE! **************** To summarise: At some time t, we have no change in power on reflection (DP(t)=0), no absorption (rate of increase in internal energy = 0), v(t)=0. F(t) can be any finite force, and conservation of energy is still satisfied at time t. Yes or no? If no, support your position. ************************** > > Deal with the basics first. The basics - conservation of energy and > > momentum - still work with all the complicated details added. > > I find this laughable. I don't care to smear you, yet I am nearly > forced to by the weakness of your argument. You've derived a force > that can be anything. No, I haven't derived a force. This isn't a derivation. It's a refutation of your repeated claim that DP=0 requires F=0. If we can't get past this point - just relying on conservation of energy, basic classical mechanics, and elementary mathematics - what chance do we have with something more complicated? From past experience, you would just repeat your "but the force must be zero", and we'd return to here anyway. Deal with it now, not in 26 more posts. > Shouldn't we admit that there is need of a finite force? That F must be finite has already been stated explicitly. After all, we have 0*F=0, which is satisfied for any number F. What is 0*infinity? > Let's consider a small reflective plate in free space, as in outer > space. Beyond your > P = f . v > we also have the familiar > f = m a > and so we may as well write > P = m a . v > and so to overlook the acceleration will be a mistake. So, if P=0, and v=0, we have 0*a = 0. Are you claiming that this means that a must be zero? Btw, you have restricted this to a special case here, where the force F being considered is the total force acting on the object. It isn't necessary to make this assumption, and it isn't possible to make this assumption when modelling some of the relevant experiments. > > (Cut the rest for now. There isn't any big secret in the classical EM > > picture of reflection, either by a dielectric or a conductor - there > > is a change in wave impedance, there is reflection. In either case, > > direct calculation of the EM force acting on the reflector, via the > > Lorentz force acting on induced currents and polarisation, yields the > > Ahh... induced currents. This is somewhat what I was arguing for. It's known, and published many times (also mentioned many times earlier in this thread). Here is a recent example from the education literature: http://arxiv.org/abs/0807.1310 This also discusses some of the problems with some of the common treatments of the problem. -- Timo
From: Tim BandTech.com on 19 Jul 2010 13:22
On Jul 18, 6:56 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > On Sun, 18 Jul 2010, Tim BandTech.com wrote: > > On Jul 17, 8:27 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > On Jul 18, 12:31 am, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > > > > On Jul 16, 10:11 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > > On Jul 17, 11:09 am, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > > > > > > On Jul 16, 8:46 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > > > > > We have: > > > > > > > > (a) The power of an EM beam (i.e., the energy flux) is affected by > > > > > > > redshift or blueshift. > > > > > > > > (b) The cycles don't disappear. (I don't see why you think this is > > > > > > > relevant, but let me leave it here on the list.) > > > > > > > > (c) P = F.v > > > > > > > > Consider a stationary perfect reflector reflecting a monochromatic > > > > > > > collimated beam of power P0 and frequency f0. By "perfect reflector", > > > > > > > I mean that no energy is absorbed by the reflector. By "absorbed", I > > > > > > > mean that none of the incident energy is transferred to the internal > > > > > > > energy of the reflector. "Internal energy" most importantly includes > > > > > > > thermal energy, but could include other forms of energy, such as > > > > > > > electronic excitations leading to fluorescence, etc. "Internal energy" > > > > > > > does not include kinetic energy due to motion or rotation of the > > > > > > > object (i.e., the reflector, in this case). > > > > > > > > We have v = 0, and, from P = F.v, no work is done, regardless of what > > > > > > > the force F is. > > > > > > > > Since no energy is absorbed (since we are considering a perfect > > > > > > > reflector), and no work is being done, the reflected beam must be of > > > > > > > power P0. Since the reflector is stationary, there is no Doppler > > > > > > > shift, and the reflected beam is of power f0. > > > > > > > > From conservation of energy, > > > > > > > > P_out = P_in - rate of increase in internal energy - rate of increase > > > > > > > of KE > > > > > > > > P_out = P0 - 0 - F.v > > > > > > > > P_out = P0 - 0 - F.0 = P0 > > > > > > > This is an argument for no acceleration of the reflector. This is what > > > > > > I've been arguing for nearly from the start. Aren't you now > > > > > > contradicting the Nichols results? > > > > > > No. > > > > > > OK, we need to stop here; no point in going past this at the moment. > > > > > > Let DP = the change in power of the beam on reflection (D as a > > > > > substitute for delta). > > > > > > For a stationary perfect reflector, DP = 0. > > > > > > We have, above, -DP = absorbed power + rate of doing work. Since the > > > > > reflector is a perfect reflector, absorbed power = 0. This reduces to > > > > > > DP = - rate of doing work = -F.v > > > > > > Since DP = 0, and v = 0, F must satisfy > > > > > > 0*F = 0. > > > > > > You have been claiming that this means that F must be zero, I have > > > > > been claiming that F can be non-zero and still satisfy this. > > > > > > Let me give an example. Take F = 1kN. (DP = 0W, v = 0 m/s.) > > > > > > 0*1000 = 0, as required. > > > > > > Try this with any number for F (that is, any finite force, zero, > > > > > positive, or negative), and you will see that 0*F=0 is satisfied for > > > > > any value of F. > > > > > > It is not valid to argue that 0*F = 0 means that F=0. > > > > > > OK so far? If not, why isn't the numerical example showing that F=1kN > > > > > is compatible with DP = 0 sufficient? > > > > > But what are you arguing for? You've essentially led us down this path > > > > of incomplete construction and landed at an incomplete result. > > > > You have repeated claimed that DP = 0 means that F = 0. You have made > > > these claims for a stationary reflector, that the lack of absorption > > > and the lack of Doppler shift means that there can't be any force. > > > > It comes down to DP = -F.v, in the absence of absorption. Your claim > > > has, repeatedly, been that DP=0 requires F=0. This is an incorrect > > > claim, and the above is an attempt to show that it is incorrect. Since > > > we have returned to this claim for the 3rd or 4th time (Why? Isn't it > > > obvious that 0=0*F allows non-zero F?), it needs to be dealt with > > > before proceeding. > > > > > According to your analysis F can be anything, and this is just like > > > > the static case of a person trying to push over a boulder. As hard as > > > > we may push, if the boulder does not shift, then no work was done. > > > > No, you are overgeneralising. P = F.v works even if the boulder is > > > accelerating. It does not need to be fixed in place. It is sufficient > > > that v(t=t1)=0 for P(t1)=0. If we also have DP(t1)=0, then DP = -F.v > > > is still satisfied, for any F. > > > > Yes, that is the point, that F _can_ be anything. You have been > > > arguing that, under the same conditions, you _must_ have F=0. > > > Ahh. If the boulder is in outer space, then yes, the force must be > > zero. If the force is even slight then the boulder accelerates. > > Likewise for a free reflector at steady zero velocity the force must > > be zero. > > Since the rate of doing work by a force is P=F.v, your claim is equivalent > to claiming that 0*F=0 requires F=0. > > Counterexample: 0*1000=0. > > > I don't understand how the discussion has degenerated this > > badly. > > I don't know either. P=F.v is simple mathematics, 0*F=0 for any finite F > is elementary mathematics. You even agree(d) with P=F.v. > > If you are still claiming that F must be zero when P=0, even if v=0, > please give some support for the claim, rather than simply repeating it. > > > > For the instantaneous power to be returned entirely, it is sufficient > > > to have v=0. We have 3 cases: > > > Sufficient, but more importantly the acceleration due to that power > > must be zero, and v can actually be anything. Why you would care to > > construct the special case rather than the general case is puzzling, > > but acceptable. Why do we have three cases? Is it because you wish to > > confuse the subject? > > > > (a) We have v=0 instantaneously only. The object is accelerating, but > > > v(t) = 0 for some t=t1. At t=t1, we have DP(t1)=0. > > > No. There is no acceleration. > > Prove it. It is possible to have v=0 and a=nonzero at some time t1. Do you > deny this? > > > > > > (b) The object is not accelerating. This will occur if (i) > > > F_reflection = 0, and F_total = F_reflection, (ii) mass is infinite, > > > or (iii) F_total = 0 and F_reflection is non-zero, which requires > > > F_other = -F_reflection, some restraining force preventing the > > > acceleration. > > > > Note well that infinite mass only means a=0, not v=0. In particular, > > > infinite mass can be accompanied by non-zero v, and non-zero DP. That > > > is, infinite mass _does not_ mean that all incident power will be > > > returned. > > > Well, now you are trying to unhinge your own v=0, and that is not > > healthy. > > The a=0 that I mention should have been the crux up front. I > > have only said this perhaps a hundred times here, on the perfect > > reflector. What is this lameness that comes over you? Why do you dodge > > all trails leading back to Nichols? It is as if you will come out with > > some small fact in a decapost or two that will provide an answer. I'd > > like some straight talk, and I'd like it in your next post, please. > > Why do you dodge elementary mathematics? > > You are continuing to claim that 0*F=0 requires F=0! > > You agreed that the rate of doing work on an object is P=F.v. > > Do you disagree with that now? > > You didn't disagree with > > P_out = P_in - rate of increase in internal energy - rate of increase of > KE > > which gives us > > DP = -F.v > > before. Do you disagree with this now? > > DP = 0, no change in power due to reflection, gives us > > F.v=0. > > For a stationary reflector, v=0. So, we have 0*F=0. > > You disagree, and claim that F must be zero. > > Can you show that non-zero F leads to a contradiction? That non-zero F > doesn't satisfy 0*F=0? > > ********** THIS IS THE KEY POINT HERE! **************** > To summarise: > > At some time t, we have no change in power on reflection (DP(t)=0), no > absorption (rate of increase in internal energy = 0), v(t)=0. > > F(t) can be any finite force, and conservation of energy is still > satisfied at time t. > > Yes or no? If no, support your position. No. For a reflector in free space the force upon it must be zero in order for it not to accelerate. Any arbitrary force will certainly cause the reflector to accelerate, and this is work done on the reflector. This work requires a source, and I have established that the perfect reflector cannot use the light as this source. You've chopped up your argument into such little pieces that there is no coherence. The argument that I make here is simply F = m a As much as I try to cleanse the argument you hold your ground, and your position has eroded to a pitiful stance. You are not a scientist in this moment. We are all humans, and you seem caught up in winning a struggle against me rather than addressing the problem at hand. Why else would you focus on such a degenerate statement as 0 * F = 0 and insist that it provides arbitrary force? Simply substitute one Newton of force on the reflector, and then we will have to ask what the source of this force is? We can freely substitute any value into your wonderful equation in any units and have perfect nonsense. Is this your 12 year old son using your account that I am speaking with? - Tim > ************************** > > > > Deal with the basics first. The basics - conservation of energy and > > > momentum - still work with all the complicated details added. > > > I find this laughable. I don't care to smear you, yet I am nearly > > forced to by the weakness of your argument. You've derived a force > > that can be anything. > > No, I haven't derived a force. This isn't a derivation. It's a refutation > of your repeated claim that DP=0 requires F=0. > > If we can't get past this point - just relying on conservation of energy, > basic classical mechanics, and elementary mathematics - what chance do we > have with something more complicated? From past experience, you would just > repeat your "but the force must be zero", and we'd return to here anyway. > > Deal with it now, not in 26 more posts. > > > Shouldn't we admit that there is need of a finite force? > > That F must be finite has already been stated explicitly. After all, we > have 0*F=0, which is satisfied for any number F. What is 0*infinity? > > > Let's consider a small reflective plate in free space, as in outer > > space. Beyond your > > P = f . v > > we also have the familiar > > f = m a > > and so we may as well write > > P = m a . v > > and so to overlook the acceleration will be a mistake. > > So, if P=0, and v=0, we have > > 0*a = 0. > > Are you claiming that this means that a must be zero? > > Btw, you have restricted this to a special case here, where the force F > being considered is the total force acting on the object. It isn't > necessary to make this assumption, and it isn't possible to make this > assumption when modelling some of the relevant experiments. > > > > (Cut the rest for now. There isn't any big secret in the classical EM > > > picture of reflection, either by a dielectric or a conductor - there > > > is a change in wave impedance, there is > > ... > > read more » |