Taking a Fresh Look at the Physics of Radiometers.
in [Physics]
|
Prev: Come on creative minds solve this fiasco in the Gulf of Mexico if ?you can
Next: Dark Energy: The problem with Einstein's Cosmological Constant is that there's no physics behind it
From: kado on 15 Jun 2010 20:04 On Jun 15, 5:17 am, "J. Clarke" <jclarke.use...(a)cox.net> wrote: > snip > > Flunked remedial reading did you? It is you that needs to understand the meanings of the words. Your stupid response also demonstrates that you have absolutely no clue about physics or Einstein's Special Theory of Relativity. D.Y.K.
From: NoEinstein on 15 Jun 2010 23:35 On Jun 15, 8:04 pm, "k...(a)nventure.com" <k...(a)nventure.com> wrote: > Dear Kato: Starting with me, no thinking human being will ever again have any reason to understand SR, nor Einstein. I disproved both as easily as snapping my fingers! The Law of the Conservation of Energy, and yours truly, have smashed over a century of the Dark Ages of Einstein. Read any of the links, below, to understand why. NoEinstein Where Angels Fear to Fall http://groups.google.com/group/sci.physics/browse_frm/thread/8152ef3e... Last Nails in Einstein's Coffin http://groups.google.com/group/sci.physics.relativity/browse_frm/thre... Pop Quiz for Science Buffs! http://groups.google.com/group/sci.physics/browse_frm/thread/43f6f316... An Einstein Disproof for Dummies http://groups.google.com/group/sci.physics/browse_thread/thread/f7a63... Another look at Einstein http://groups.google.com/group/sci.physics/browse_frm/thread/41670721... Three Problems for Math and Science http://groups.google.com/group/sci.physics/browse_thread/thread/bb07f30aab43c49c?hl=en Matter from Thin Air http://groups.google.com/group/sci.physics/browse_thread/thread/ee4fe3946dfc0c31/1f1872476bc6ca90?hl=en#1f1872476bc6ca90 Curing Einsteins Disease http://groups.google.com/group/sci.physics/browse_thread/thread/4ff9e866e0d87562/f5f848ad8aba67da?hl=en#f5f848ad8aba67da Replicating NoEinsteins Invalidation of M-M (at sci.math) http://groups.google.com/group/sci.math/browse_thread/thread/d9f9852639d5d9e1/dcb2a1511b7b2603?hl=en&lnk=st&q=#dcb2a1511b7b2603 Cleaning Away Einsteins Mishmash http://groups.google.com/group/sci.physics/browse_thread/thread/5d847a9cb50de7f0/739aef0aee462d26?hl=en&lnk=st&q=#739aef0aee462d26 Dropping Einstein Like a Stone http://groups.google.com/group/sci.physics/browse_thread/thread/989e16c59967db2b?hl=en# Plotting the Curves of Coriolis, Einstein, and NoEinstein (is Copyrighted.) http://groups.google.com/group/sci.physics/browse_thread/thread/713f8a62f17f8274?hl=en# Are Jews Destroying Objectivity in Science? http://groups.google.com/group/sci.physics/browse_thread/thread/d4cbe8182fae7008/b93ba4268d0f33e0?hl=en&lnk=st&q=#b93ba4268d0f33e0 The Gravity of Masses Doesnt Bend Light. http://groups.google.com/group/sci.physics/browse_thread/thread/efb99ab95e498420/cd29d832240f404d?hl=en#cd29d832240f404d KE = 1/2mv^2 is disproved in new falling object impact test. http://groups.google.com/group/sci.physics/browse_thread/thread/51a85ff75de414c2?hl=en&q= Light rays dont travel on ballistic curves. http://groups.google.com/group/sci.physics/browse_thread/thread/c3d7a4e9937ab73e/c7d941d2b2e80002?hl=en#c7d941d2b2e80002 A BLACK HOLE MYTH GETS BUSTED: http://groups.google.com/group/sci.physics/browse_thread/thread/a170212ca4c36218?hl=en# SR Ignored the Significance of the = Sign http://groups.google.com/group/sci.physics/browse_thread/thread/562477d4848ea45a/92bccf5550412817?hl=en#92bccf5550412817 Eleaticus confirms that SR has been destroyed! http://groups.google.com/group/sci.math/browse_thread/thread/c3cdedf38e749bfd/0451e93207ee475a?hl=en#0451e93207ee475a NoEinstein Finds Yet Another Reason Why SR Bites-the-Dust! http://groups.google.com/group/sci.physics/browse_thread/thread/a3a12d4d732435f2/737ef57bf0ed3849?hl=en#737ef57bf0ed3849 NoEinstein Gives the History & Rationale for Disproving Einstein http://groups.google.com/group/sci.physics/browse_thread/thread/81046d3d070cffe4/f1d7fbe994f569f7?hl=en#f1d7fbe994f569f7 There is no "pull" of gravity, only the PUSH of flowing ether! http://groups.google.com/group/sci.physics/browse_thread/thread/a8c26d2eb535ab8/efdbea7b0272072f?hl=en& PD has questions about science. Can any of you help? http://groups.google.com/group/sci.physics/browse_thread/thread/4a2edad1c5c0a4c1/2d0e50d773ced1ad?hl=en& 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 > > On Jun 15, 5:17 am, "J. Clarke" <jclarke.use...(a)cox.net> wrote: > > > > snip > > > Flunked remedial reading did you? > > It is you that needs to understand the meanings of the words. > > Your stupid response also demonstrates that you have absolutely no > clue > about physics or Einstein's Special Theory of Relativity. > > D.Y.K.
From: Timo Nieminen on 16 Jun 2010 16:59 On Jun 17, 12:53 am, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > On Jun 15, 5:40 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > So, what fraction of the KE of the bullet going into heating would be > > too much? > > > Why is it OK for most of the KE of a bullet to end up heating the > > target, but not the energy of a photon? No answer? > > > > Again, back to this because you keep refusing to comment on the > > > > redshift/blueshift argument. Note well that this gives you work > > > > without absorption (which invalidates your objection). > > > > No, Timo. > > > "No" means what? Are you saying that the redshift/blueshift argument > > _doesn't_ give you work without absorption? If, so explain. Just "no" > > is an insufficient argument. > > Hmmm. So by splitting my text in two here you think that you can > criticize the first line? > The following lines clearly state why I say no: What are you saying "no" to? My directly preceding statement to your "no" was that redshift/blueshift on reflection gives you work without absorption. > > > There is no redshift or blue shift within the Nichols > > > radiometer experiment. The radiometer is initially static, and upon > > > shining light it will arrive in a new static position. No velocity > > > analysis is needed, other than the propagation of the light itself. > > > Yes, and NO WORK is done on the static vane. > > I can partially agree with this statement, You "partially" agree that no work is done on the static vane? So, is _some_ work done on the static vane? How much? If you think no work is done, why "partially"? > but if we're discussing a > black vane then we should also admit that it is heating up, and while > work is not being done, work could be being done. For instance if we > replace the black vane with a solar cell. Irrelevant to radiation pressure. Also irrelevant to a perfect reflector. > > > If > > > you would like to pick up your argument on a blue shifted reference > > > frame then I am willing to go back to that analysis. Neither of us has > > > budged much on these other issues. We're pretty much where we started > > > out, except for your quick reversal on the momentum doubling. > > > What reversal on "momentum doubling"? > > > > Remember > > > 1300 watts in, 1300 watts out? Let's argue this on a lab table, where > > > the effect is understood to take place. Where is the redshift or > > > blueshift? > > > See the details in the 2 posts with the details. First, look at > > reflection from a static reflector, no redshift/blueshift, no change > > in energy of the light from incident to reflected. The momentum of the > > light is reversed, since the direction is reversed. (So far, no > > information is obtained as to what this momentum is; in principle, > > from this part alone, you could have zero momentum and no force.) Look > > at _exactly the same thing_ in a coordinate system where the beam and > > reflector are moving, Then redshift, blueshift, change in energy of > > the light. Where does the energy go? The heating is the same as in the > > stationary case; the only candidate is work (which _is_ dependent on > > choice of coordinate system). If work is done, a force is being > > exerted. From the force, find the momentum flux, = power/speed of > > wave. > > > The redshift and blueshift are due to looking at the reflection in a > > moving coordinate system. > > I just don't see any argument here. We can remove the reflector > apparatus and just consider a light source and get an energy > difference in a relative frame at some velocity. This doesn't tell you anything about force, because there is _no_ force. The momentum of the beam isn't changing, since nothing is changing it, and thus there is no force. With no force, what can we learn about work or momentum flux of the beam? > This is the Doppler > shift of light. It is a puzzle in and of itself, but I fail to see how > it matters to the problem at hand. Since the velocity of the > introduced reference frame is arbitrary you are free to misinterpret > the energy difference either way aren't you? "Misinterpret" - why? Note that we also see a difference in the KE of a classical particle. > > > > Would a bullet exert a force on a perfect reflector? > > > > Yes, but you should define what you mean here by perfect reflector. > > > No loss of mechanical energy during reflection, a "perfectly elastic" > > collision as it is usually described in mechanics. That is, no > > heating. That's for a bullet; for a photon, "no heating" works well, > > too. > > > > > Would a bullet exert a force on a stationary perfect reflector? > > > > Yes. > > > There's no change in energy in this case. If we aren't "turning energy > > into force", where does the force come from? > > This perfect reflector will need to be infinitely massive in order to > remain stationary. Or have another restraining force acting on it. But, yes, you can consider the limit as mass->infinity. > Even if it is such that its acceleration nears zero we still can have > a force > F = m a > where the mass is extremely large (infinite) and the acceleration is > extremely small (zero). Yes. This isn't anything new. We see this with collisions in classical mechanics. This is why most of the KE of the bullet goes into heating. > The electrical qualities of all material are completely > overlooked within the classical analysis, yet here we have the > opportunity to treat the electrical as fundamental and we have not. On a classical level, you have conductivity and polarisability. As stated many times already in this thread, you can get the same results for radiation pressure either from conservation of energy and momentum, or from direct calculation of the forces, using the Lorentz force. In the latter, you use the conductivity and polarisability to find the induced currents and polarisation. This isn't "completely overlooked". > Then too, within descriptions of spectral analysis that I have read, > there is no attempt as NoEinstein makes to define the detail of the > mechanism by which photon absorption raises an electron's energy. We > are just supposed to believe that it pops in or pops out. Shouldn't we > be able to get a bit further than this? Here we are dancing very close > to it. Perhaps some of your work should be to bring that old theory > into more detail. Quantum electrodynamics, QED, deals with this. About the simplest full calculation for a common real material is for the hydrogen atom. For this, it works very well (i.e., very well quantitatively). If you were to look at an example of this calculation, you will see why "descriptions of spectral analysis" present a simplified picture. > > > Can't you see that the standard theory of photon interaction has to do > > > with electron interaction? The heat must come from somewhere. How 1300 > > > watts can get blown in a 1mx1m black body into pure heat while doing > > > miniscule work is actually a pretty huge problem, > > > It's understood. In the context of classical EM theory, the material > > is described by conductivity sigma, current will be J=sigma*E, work > > done on the moving charge will be W=J.E=sigma*E^2. There will also be > > the magnetic Lorentz force, F=JxB. > > > Turn on an electic heater, and you can have an example of 1300W (in > > principle, although 1kW, 2kW, 2.4kW are the usual figures for heaters > > here) into "pure heat", exactly the same type of Ohmic heating. > > > Why would this be a "pretty huge problem"? It's been understood at > > this level for over a century. There's a reasonable understanding of > > what conductivity means on an atomic scale - the picture of thermal > > free electrons colliding with each other and the positive ions is OK, > > too. > > Well, I think we'd be quite pleased to have even half of the sun's > heat turned into electricity. > We are not stuck with that heat necessarily. Most applications attempt > to avoid heating. > We already have fusion power at a safe distance; we just haven't > harnessed it fully yet. Nice thoughts, but irrelevant to radiation pressure. > As you've stated some emag above then I think it is true that you > understand that the mechanical arguments are merely an analogy to > light's behavior, and that the actual photon momentum problem is not > as mechanical as is claimed. The question of how the light is heating > is quite relevant, and rather than looking the other way we should try > to get to the bottom of it. No, it isn't relevant. That you can get radiation forces without heating shows it's irrelevant. How we get the heating is known - it's just F=qE, acceleration of the electron, and then loss of this gained KE in collisions with other electrons and the rest of the material. What is there to get to the bottom of? Well, of course, you might like to do something useful with the absorbed energy, other than heating. So might others. There is a large amount of research being done on solar cells and photosynthesis. But this is doubly irrelevant to the original point of radiation pressure. > > > especially if you > > > attribute all of that energy to mechanical momentum. Now, I understand > > > that these words are a bit loose, but the fact remains that photon > > > momentum is derived directly from the energy in e=hf, which I think we > > > both presume is the entire photon energy. > > > If the heating bothers you that much, consider reflection, perfect > > reflection with no heating. Look at the force there as compared with > > the absorbing case (twice as much), and the heating there (zero). > > Again, I'll invoke the no work argument on the perfect reflector. > Also, as far as we know this perfect reflector does not exist. > So, with a 99% reflective mirror what right do we have to claim that > the pittance that is the photon momentum is not driven by the absorbed > light, with less heating going on? No, this is just wrong. That is, "no work on a perfect reflector" is wrong. "No heating on a perfect reflector" is correct. Back to redshift/blueshift. Consider the stationary perfect reflector. Energy in = energy out, no work. Perfect reflection, no heating, no absorption. Same thing with either classical particles, photons, or classical light. Looking at it from a moving coordinate system, and you no longer have energy in = energy out. Still, you have no heating - changing the coordinate system doesn't change the temperature. Work is done, with no absorption, no heating. Same thing with classical particles: completely elastic collisions, no heating, no loss of mechanical energy, and work is done. > > You get the same with collisions by classical particles. Reflection > > (elastic collision), no heating, twice the impulse; absorption > > (completely inelastic "sticky" collision), most of the KE goes into > > heating for for a low mass fast particle. > > In either case we have failed to investigate the mechanism of the > heating. Insofar as matter is composed of electrical material then I > think it is quite impressive to look at this heating as electrical > energy. You are happy to put a bar here where you will not go beyond, > yet this is deep. It's irrelevant to the original point. Note that non-electromagnetic forces are needed when looking at inter- atomic forces in real materials. It isn't enough to say "matter is made of charged particles, therefore it's all electrical". It's a little more complicated (but still irrelevant). > > You don't convert the energy into momentum or force, you can change > > the momentum without changing the energy. The features that you bring > > up as "problems" are the same in classical mechanics. > > OK Timo. I'll have to accept that this is true, and that my only > quibble can be over how the radiation turns into heat in the case of > the absorber. The details don't matter for the main conclusion. In any case, since you can have forces without any heating (in an ideal case), heating itself is largely irrelevant to the main point of radiation forces. > I do now see that when we capture the kinetic energy as > in a perfect spring with a ratchet that we will only see a mechanical > push of the momentum at the target, but that the spring is now wound > for more mechanical work to take place. For the real problem, welcome to the wonderful world of quantum mechanics and QED. Solar cells is an active research field; if you want to read about mechanisms for capturing energy from the sun, there;s plenty out there. > > > Why you would hinge your argument on electron ejection is not clear. > > > All that we need is a dipole; particularly a dipole fueled by light > > > energy. > > > You're the one who brought up electron ejection. "Fueled by" light > > energy is a bad choice of language. "Fuel" is used up in most examples > > of its use. For light in a lossless dielectric, the energy isn't used > > up by the induced polarisation of the medium. > > Yes, I did bring it up, and you have stuck with it. I've continued to say that there are experiments completely ruling out electron ejection as the mechanism of the force. How is this "hinging my argument on electron ejection"? > Light does get > used up, except in the clearest media. So consider the clearest media. Consider two media of the same refractive index, but with different absorption coefficients. Same amount of induced polarisation, different loss of energy. The energy isn't being "used up" by the induced polarisation. Consider a spring. Compress the spring, and release it. You can get all of the energy that went into compressing it back out, for an ideal spring. Try this with a real spring, and you will see that there is some loss. Does the function of the spring _depend_ on the loss? > The optical cable relies on TIR > and gets a high figure at a single wavelength. TIR is irrelevant to the loss in the material. Scattering losses due to defects in the fibre wall and the inhomogeneities are irrelevant. Yes, there is some - very little - heating. So what? Why does it matter? > The plate glass window > does not do so well in its application. And a glass of Coke does even worse, optically. And? > The dipole that I am > suggesting is not the AC dipole of the dielectric material. I am > suggesting that if absorbed photons create a DC dipole by steadily > kicking electrons in the direction of the original light propagation, > that then a DC dipole is formed, without electron ejection. This then > could be the mechanism of the tweezer experiments that you do, and of > the old time radiometer in vacuum. As already said, you can calculate radiation pressure on an absorber by finding the induced currents, and then finding the force acting on them via the Lorentz force. This gives you a time-averaged force in the direction of propagation. Of course absorption forces act by pushing the current in the direction of propagation. Yes, no electron ejection needed. It doesn't matter whether or not a DC dipole is formed. The internal forces due to the dipole add up to zero. What you need is the force acting on the current. It isn't the atomic-scale mechanism of the force in optical tweezers - this is the Lorentz force acting on the polarisation current. Similar, but the charges remain bound. This isn't a mystery, this is known, this has been published. > Well, the amount that we are using up is so small that we cannot even > find a reflector that will return the difference. I suppose we should > go over to the Crookes device to make this claim, where steady work is > done. So far we have no figure on the friction of the bearing. Anyway, > the work done will be terribly small relative to the power received if > we could get the Crookes working under strong vacuum. You've dodged > the fact that we've atributed the photon's entire energy to the > mechanical realm. I just don't buy it. It may merely be an analogy > rather than a fact. We haven't attributed "the photon's entire energy to the mechanical realm". (Or have we? Define "mechanical realm" clearly and meaningfully, and likewise "attributed", and we can see.) > > Given a high enough temperature, and the right material for the > > radiometer vane, it might be an observable effect. It isn't an > > explanation for the general body of experiments, because in at least > > some of them (many of them), it won't contribute. > > Well, you've given a little bit of room here. At the singular position > where one photon is absorbed we may well have high heating, > particularly due to the very slow rate of thermal conduction. This > would not be a steady state effect. If this is correct then there > would be a slight depletion of electrons at the lighted side, and not > necessarily any charge difference at the back side of a metal plate. Compare the photon energy with the energy needed for electron ejection. Calculate the expected force. Compare with observed radiometer forces. Then you might be able to say something about your "If this is correct". Not relevant to radiation forces, though. > > > > Do the calculation. Quantitatively. With numbers. Just answer the > > > > question, don't get sidetracked into pretty pictures of photons or > > > > long philosophical tracts. Answer the simple question first! > > > > OK. I've gone back to look at the questions. My gut tells me that to > > > get so little mechanical work out of the light suggests that it has > > > quite alot of torque. This is just turning into noise under standard > > > reception. We actually need a mechanism like this in order to get so > > > little work out, and so much heat. > > > Do the calculation. The numbers matter. > > I suspect the calculation you are thinking of is the mechanical > computation of momentum versus kinetic energy. > p = m v > K = m v v / 2 > for a nonrelativistic case. Already we believe that the m in these > equations is fraudulent for light. I admit that it is impressive how > the light holds up under the mechanical analog, but we will always be > dodging the m in these equations with light. > The argument of the bullet does something for your argument, but > aren't you curious about the mechanism of heating there too? > My own opinion on thermodynamics is that it is an open subject, since > vibrating atoms should conduct at the speed of sound rather than the > sloowww propagation that heat conducts at. I don't mean to open that > can of worms here, but still feel the need to make the statement. Why not just do the calculation? For 500nm and 5Hz photons, what are the energies, momenta, and angular momenta? Energy and momentum, as you did before. Angular momentum, hbar. > > Are these ratios or values from classical EM theory wrong? Yes or no, > > no handwaving, no waffle, just a straight answer. > > I am not strong enough on light polarization to answer this. On the > one hand It seems that if light comes in just the two forms with the h > AM as either right or left then that should be all that we will ever > have. This then means that the longitudinal or linear wave is not > fundamental. Yet we believe that we also have vertically or > horizontally polarized light, and it is not a mixture of the two > circular forms. You can always represent V or H polarisations as a sum of L and R. And the other way around! The maths doesn't say which is fundamental, but you can always say that linearly polarised light is a mixture of the circular polarisations. > I feel very comfortable with the notion that there is > additional geometry in light, especially since the wavelength of the > light plays such a small part within current interpretation. For > instance we do not claim any photon diameter. There is room for > dimension to play a role in light's properties. The wavelength of light plays an essential role in current theories. In the classical theory, it tells you how the wave behaves. In the quantum theories, it tells you how the wavefunction behaves. Why would we claim a photon diameter? More specifically, what effect would it have? How would we measure it? Why would we want to add an un- measured, unknown magic number with unknown effects to current theories? Why expect a photon to be like a little billiard ball? Hanbury Brown's book, "Boffin", has a nice bit on the fun that can result when people lean too much on billiard ball ideas of photons. Good book, read it. ("Hanbury Brown" is his last name.) (This is _way_ irrelevant to the original point.) > > > If I did this is the first time that you've claimed such, and have > > > said nothing here to resolve the misunderstanding. I suspect you will > > > have little to say about this. I see that here you do not deny the > > > instantaneous zero. As I recall your equation will predict four zeros > > > in photon momentum per cycle, so where did the energy go? Well, you'll > > > say, momentum and energy aren't the same thing... And then we can > > > cover the whole loop again. > > > From before: > > > [you] > > > > Within your own analysis of radiation pressure you use cross products > > > of the E field. We should accept that the E field is sinusoidal, and > > > so contains instantaneous zeros. These zeros are points in time at > > > which your effect has gone to zero. > > > [me] > > Yes, and? For a plane polarised beam, you can think of it as a > > succession of blobs of energy, 2 per wavelength, moving at the wave > > speed. It's the blobs of energy that have the momentum (since they > > have the energy). As the direction of each blob is reversed as the > > beam is reflected (or each blob is absorbed, if you prefer to consider > > radiation pressure by absorption), there is a force. Between blobs, > > nothing is happening, so why should there be a force? No energy is > > changing direction then, so the momentum is changing, so a non-zero > > force isn't expected, since that would violate conservation of > > momentum. > > > [new] > > The energy doesn't disappear, it's just moving along with gaps of zero > > energy density in between blobs of energy. To _not_ have instantaneous > > zeros in radiation pressure force due to a linearly polarised beam > > would be a problem, to have them is expected. > > Well, we are half way back to an AC theory here. So you do own that > the classical momentum is just an analogy right? ?? Why say that? More to the point, what is the difference between "classical momentum" and some other kind of momentum? > The photon is rubbing up against the wave in your interpretation. No. Absolutely not. This is still purely classical light, a purely classical EM wave. No photons. > It > is a mistake to claim that the nulls of the wave are the end of a > photon isn't it? Yes, completely wrong. You don't have 2 photons per cycle, in general. > Under this interpretation there would be a lowest > power beam; a train of photons spaced by their wavelength. Yes, and this would be wrong, which is why that interpretation is completely wrong. > I believe > you are breaching the particle/wave duality. They've been melanged > into one, just as radiation pressure and photon momentum have been > blended. No, no photons yet. Just a classical EM wave. The instantaneous zeros you complained about are in the classical theory; the explanation is in the classical theory. No photons needed, no photons invoked. > > An external field can produce forces on a dipole (and this is the most > > common explanation for the gradient forces on Rayleigh particles in > > optical tweezers). The internal forces won't. (If external radiation > > fields cause such a force, and change the momentum of the particle, > > then they must have carried momentum, and exerted a radiation force. > > If you sat down and looked at the Lorentz force acting on a reflector > > or absorber, this is what you will get.) > > Oh. I guess you've just painted the momentum analogy more clearly to > me here. All mechanical influence is momentum, regardless of the > quality of the source of influence. Well, this is only half of an > answer to me. I do want to understand the heating. Ohm's law, basically, at least for a conducting object, where the field drives free electrons. For atomic transitions, welcome to the quantum problem. (The full one; the various semi-classical simplifications just say that it is so, the energy is absorbed - not much explanation.) > > The "gradient force", basically a dipole in a non-uniform field, acts > > to move the dipole into the highest field (like iron to a magnet, bits > > of paper or thread to an electrically charged object, etc.). > > > The reflection from a small conducting particle (a conducting Rayleigh > > particle) is in the usual pattern for dipole radiation (not isotropic, > > but a doughnut-shaped radiation pattern without a preferred single > > direction), so the reflected light has zero total momentum. The > > incident light which is reflected has non-zero momentum. Result: a > > force in the direction of the incident light. If large enough, it will > > overcome the gradient force. > > > Same thing for highly-reflective non-conducting particles - it's the > > reflectivity that matters, not the conductivity as such. But > > conducting particles tend to be more reflective. > > > > Why the inversion? > > > What inversion? > > Well, the trap doesn't work. That is the inverse of trapping. Don't > you think that charge dynamics could explain this? > I'm having trouble seeing the doughnut shape of radiation that you > describe. Is this scattering that you are talking about? I'm not > seeing how it takes over at small size. Can't we just use a larger > metal particle and consider it? Scattering is beyond me still. I do > appreciate the exercise that you've given me. Thanks. > Under a new geometry it may well be that light can take the simplicity > that you like to give it, yet the interpretation will read quite > differently. A small scatterer acts like a short dipole antenna. The scattered field looks like the field radiated by a dipole antenna. The scaling with size comes about due to the efficiency (or lack thereof) of a short dipole antenna. See doughnut figure at http://en.wikipedia.org/wiki/Dipole_antenna (and the formulae are there too). Yes, this is scattering, Rayleigh scattering. With a larger metal particle, you need to include the higher-order multipoles, and the scattered field is a lot more complicated. Still, reflection forces beat gradient forces, and the particle is pushed along the beam (and not trapped). This doesn't depend on it being conducting, just on it being reflective. The same thing, pushing out of the trap when reflective enough, is seen for non-absorbing, non-conducting particles, too. See http://www.opticsinfobase.org/abstract.cfm?uri=oe-16-19-15039 (don't worry, this is open access). -- Timo
From: NoEinstein on 17 Jun 2010 03:04 On Jun 16, 4:59 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > Dear Timo: If you shoot a bullet through a steel plate, and then feel the metal around the hole, it will be hot. But who, besides a marginal thinker like you supposes that such fact makes a difference in the price of eggs in China? You stay on 'heat' questions, because no one cares, and so there isn't much likelihood of there being data to show you wrong. Your pretending expertise on heat shows you to be a pedant. Einstein was one of those, too. And he absolutely didn't know what he was doing or saying! NoEinstein > > On Jun 17, 12:53 am, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > > > On Jun 15, 5:40 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > So, what fraction of the KE of the bullet going into heating would be > > > too much? > > > > Why is it OK for most of the KE of a bullet to end up heating the > > > target, but not the energy of a photon? > > No answer? > > > > > > Again, back to this because you keep refusing to comment on the > > > > > redshift/blueshift argument. Note well that this gives you work > > > > > without absorption (which invalidates your objection). > > > > > No, Timo. > > > > "No" means what? Are you saying that the redshift/blueshift argument > > > _doesn't_ give you work without absorption? If, so explain. Just "no" > > > is an insufficient argument. > > > Hmmm. So by splitting my text in two here you think that you can > > criticize the first line? > > The following lines clearly state why I say no: > > What are you saying "no" to? My directly preceding statement to your > "no" was that redshift/blueshift on reflection gives you work without > absorption. > > > > > There is no redshift or blue shift within the Nichols > > > > radiometer experiment. The radiometer is initially static, and upon > > > > shining light it will arrive in a new static position. No velocity > > > > analysis is needed, other than the propagation of the light itself. > > > > Yes, and NO WORK is done on the static vane. > > > I can partially agree with this statement, > > You "partially" agree that no work is done on the static vane? So, is > _some_ work done on the static vane? How much? If you think no work is > done, why "partially"? > > > but if we're discussing a > > black vane then we should also admit that it is heating up, and while > > work is not being done, work could be being done. For instance if we > > replace the black vane with a solar cell. > > Irrelevant to radiation pressure. Also irrelevant to a perfect > reflector. > > > > > > > > > If > > > > you would like to pick up your argument on a blue shifted reference > > > > frame then I am willing to go back to that analysis. Neither of us has > > > > budged much on these other issues. We're pretty much where we started > > > > out, except for your quick reversal on the momentum doubling. > > > > What reversal on "momentum doubling"? > > > > > Remember > > > > 1300 watts in, 1300 watts out? Let's argue this on a lab table, where > > > > the effect is understood to take place. Where is the redshift or > > > > blueshift? > > > > See the details in the 2 posts with the details. First, look at > > > reflection from a static reflector, no redshift/blueshift, no change > > > in energy of the light from incident to reflected. The momentum of the > > > light is reversed, since the direction is reversed. (So far, no > > > information is obtained as to what this momentum is; in principle, > > > from this part alone, you could have zero momentum and no force.) Look > > > at _exactly the same thing_ in a coordinate system where the beam and > > > reflector are moving, Then redshift, blueshift, change in energy of > > > the light. Where does the energy go? The heating is the same as in the > > > stationary case; the only candidate is work (which _is_ dependent on > > > choice of coordinate system). If work is done, a force is being > > > exerted. From the force, find the momentum flux, = power/speed of > > > wave. > > > > The redshift and blueshift are due to looking at the reflection in a > > > moving coordinate system. > > > I just don't see any argument here. We can remove the reflector > > apparatus and just consider a light source and get an energy > > difference in a relative frame at some velocity. > > This doesn't tell you anything about force, because there is _no_ > force. The momentum of the beam isn't changing, since nothing is > changing it, and thus there is no force. With no force, what can we > learn about work or momentum flux of the beam? > > > This is the Doppler > > shift of light. It is a puzzle in and of itself, but I fail to see how > > it matters to the problem at hand. Since the velocity of the > > introduced reference frame is arbitrary you are free to misinterpret > > the energy difference either way aren't you? > > "Misinterpret" - why? Note that we also see a difference in the KE of > a classical particle. > > > > > > > > > > Would a bullet exert a force on a perfect reflector? > > > > > Yes, but you should define what you mean here by perfect reflector. > > > > No loss of mechanical energy during reflection, a "perfectly elastic" > > > collision as it is usually described in mechanics. That is, no > > > heating. That's for a bullet; for a photon, "no heating" works well, > > > too. > > > > > > Would a bullet exert a force on a stationary perfect reflector? > > > > > Yes. > > > > There's no change in energy in this case. If we aren't "turning energy > > > into force", where does the force come from? > > > This perfect reflector will need to be infinitely massive in order to > > remain stationary. > > Or have another restraining force acting on it. But, yes, you can > consider the limit as mass->infinity. > > > Even if it is such that its acceleration nears zero we still can have > > a force > > F = m a > > where the mass is extremely large (infinite) and the acceleration is > > extremely small (zero). > > Yes. This isn't anything new. We see this with collisions in classical > mechanics. This is why most of the KE of the bullet goes into heating. > > > The electrical qualities of all material are completely > > overlooked within the classical analysis, yet here we have the > > opportunity to treat the electrical as fundamental and we have not. > > On a classical level, you have conductivity and polarisability. As > stated many times already in this thread, you can get the same results > for radiation pressure either from conservation of energy and > momentum, or from direct calculation of the forces, using the Lorentz > force. In the latter, you use the conductivity and polarisability to > find the induced currents and polarisation. This isn't "completely > overlooked". > > > Then too, within descriptions of spectral analysis that I have read, > > there is no attempt as NoEinstein makes to define the detail of the > > mechanism by which photon absorption raises an electron's energy. We > > are just supposed to believe that it pops in or pops out. Shouldn't we > > be able to get a bit further than this? Here we are dancing very close > > to it. Perhaps some of your work should be to bring that old theory > > into more detail. > > Quantum electrodynamics, QED, deals with this. About the simplest full > calculation for a common real material is for the hydrogen atom. For > this, it works very well (i.e., very well quantitatively). If you were > to look at an example of this calculation, you will see why > "descriptions of spectral analysis" present a simplified picture. > > > > > > > > > Can't you see that the standard theory of photon interaction has to do > > > > with electron interaction? The heat must come from somewhere. How 1300 > > > > watts can get blown in a 1mx1m black body into pure heat while doing > > > > miniscule work is actually a pretty huge problem, > > > > It's understood. In the context of classical EM theory, the material > > > is described by conductivity sigma, current will be J=sigma*E, work > > > done on the moving charge will be W=J.E=sigma*E^2. There will also be > > > the magnetic Lorentz force, F=JxB. > > > > Turn on an electic heater, and you can have an example of 1300W (in > > > principle, although 1kW, 2kW, 2.4kW are the usual figures for heaters > > > here) into "pure heat", exactly the same type of Ohmic heating. > > > > Why would this be a "pretty huge problem"? It's been understood at > > > this level for over a century. There's a reasonable understanding of > > > what conductivity means on an atomic scale - the picture of thermal > > > free electrons colliding with each other and the positive ions is OK, > > > too. > > > Well, I think we'd be quite pleased to have even half of the sun's > > heat turned into electricity. > > We are not stuck with that heat necessarily. Most applications attempt > > to avoid heating. > > We already have fusion power at a safe distance; we just haven't > > harnessed it fully yet. > > Nice thoughts, but irrelevant to radiation pressure. > > > As you've stated some emag above then I think it is true that you > > understand that the mechanical arguments are merely an analogy to > > light's behavior, and that the actual photon momentum problem is not > > as mechanical as is claimed. The question of how the light is heating > > is quite relevant, and rather than looking the other way we should try > > to get to the bottom of it. > > No, it isn't relevant. That you can get radiation forces without > heating shows it's irrelevant. How we get the heating is known - it's > just F=qE, acceleration of the electron, and then loss of this gained > KE in collisions with other electrons and the rest of the material. > What is there to get to the bottom of? > > Well, of course, you might like to do something useful with the > absorbed energy, other than heating. So might others. There is a large > amount of research being done on solar cells and photosynthesis. But > this is doubly irrelevant to the original point of radiation pressure. > > > > > especially if you > > > > attribute all of that energy to mechanical momentum. Now, I understand > > > > that these words are a bit loose, but the fact remains that photon > > > > momentum is derived directly from the energy in e=hf, which I think we > > > > both presume is the entire photon energy. > > > > If the heating bothers you that much, consider reflection, perfect > > > reflection with no heating. Look at the force there as compared with > > > the absorbing case (twice as much), and the heating there (zero). > > > Again, I'll invoke the no work argument on the perfect reflector. > > Also, as far as we know this perfect reflector does not exist. > > So, with a 99% reflective mirror what right do we have to claim that > > the pittance that is the photon momentum is not driven by the absorbed > > light, with less heating going on? > > No, this is just wrong. That is, "no work on a perfect reflector" ... > > read more »- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -
From: Timo Nieminen on 17 Jun 2010 04:44
On Jun 17, 12:53 am, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > > The question of how the light is heating > is quite relevant, and rather than looking the other way we should try > to get to the bottom of it. Already mentioned the classical picture of the heating by light as essentially Ohm's law; augment with a microscopic picture of the conduction electrons in thermal motion if you wish (their mean thermal speed, btw, vastly exceeds their mean drift speed due to an applied field). As for absorption of light by atoms, one of the simplest, clearest, and available treatments is Henderson et al., "How a Photon Is Created or Absorbed", J. Chem. Ed.: http://jchemed.chem.wisc.edu/JCEWWW/Articles/DynaPub/DynaPub.html This looks at the dynamics that are usually ignored. While this is one of the simplest and clearest, I don't think it's very simple or clear, just more so than alternatives. This is only half of the story, as it still doesn't explain how the energy ends up as heat. The answer to that is collisional de- excitation. That is, the excited atom (with extra energy after absorbing a photon) loses that energy, or at least some of it, in a collision with another atom. After the collision, a good part of this energy ends up as the overall KE of the atoms involved, rather than as the energy of the original excited electron, so isn't available for emission of a photon (of the original energy). I think that focussing on the heating by light, especially in detail, just distracts from the original issue of radiation momentum and pressure. In particular, the fundamental point is that moving energy means that you have momentum, even if the energy is being moved by a wave with no transfer of mass. E.g., by an EM wave, or in quantum terms, photons of zero rest mass, but also by water waves, acoustic waves etc, where the material in which the wave propagates doesn't move from point A to point B. This is a key advantage of looking at it in terms of conservation of energy and momentum, rather than the details of the interaction between a specific type of wave and a specific type of matter. The conservation principles are general, and useful whether the moving energy is wave energy as above, KE of moving matter, thermal energy, or whatever, and also whether we're talking Newtonian physics or relativistic physics, or classical or quantum. The downside is that it doesn't tell you about those details, even if you're interested in them. -- Timo |