Taking a Fresh Look at the Physics of Radiometers.
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From: Timo Nieminen on 6 Jun 2010 21:02 On Sun, 6 Jun 2010, Tim BandTech.com wrote: > On Jun 6, 4:44 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > On Jun 6, 11:05 pm, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > > > > > > > > > Well, I have gone back to check what I remember pretty well that you > > > wrote, and you state that it will not be observed in perfect vacuum: > > > "Perfect vacuum, no. Very good vacuums, yes, especially with atom > > > trapping. I've seen classical experiments done in vacuum (can't recall > > > how good), where absorbing particles were blasted by short pulses of > > > light, to measure their radiation pressure cross-sections." > > > > > in response to my statement: > > > "I'm sorry but the effect in a perfect vacuum has not been clearly > > > demonstrated. [...] > > > > No, I said the experiment hasn't been *done* in a perfect vacuum. > > Because we can't achieve a perfect vacuum in the lab. Technically, you > > were correct. > > > > We also haven't done an experimental test of Newton's 1st law, because > > we don't have a friction free, viscous drag free, force free > > laboratory. This doesn't mean that I think that Newton's 1st law > > doesn't work in perfect vacuum. > > > > It's an important point. How do we know when we can extrapolate a good > > vacuum result (or poor vacuum, or atomospheric result) to perfect > > vacuum? It's important to know (by measuring) the effect of what is > > there, e.g., damping due to viscous drag, drag due to convective flow, > > the radiometer force, the effect of the gas on the temperature > > difference between the two sides of the vane, etc. > > > > > I have been avoiding your references to optical tweezers and light > > > traps, partially because I have only limited familiarity with those. > > > > They're the main modern application of radiation pressure. *Thousands* > > of experiments have been done. Early experiments to show that the > > method works, and to explore the theoretical principles, experiments > > to use the force e.g. for restraining or moving live cells, > > experiments to quantitatively measure forces between biomolecules. > > > > > Anyway, we are in agreement that radiation pressure should be > > > measurable at 10E-11 torrs using a solid plate device. > > > > In principle. Whether the force is too small to measure in practice is > > the question, and this depends on the particular setup. > > > > > ... and > > > possibly freed electrons as one of the links in support of my position > > > mentions. When we start thinking of those electrons then the tweezers > > > and traps start to make more sense I think, but I cannot argue on > > > these devices without doing some research. > > > > Freed electrons are not responsible for the force in general. The atom > > trapping experiments are quite conclusive on this (since freeing an > > electron would result in ionisation, which would result in the ion not > > being trapped). (Perhaps in some specific experiment? In the modern > > laser experiments, there isn't enough photon energy for (significant) > > photoejection of electrons. Given 1/2 of the photon energy used to > > eject the electron, the rest becoming its KE, electron ejection would > > give about 400 times as much force per photon. If this was significant > > in the various tweezers experiments, it would be observed, as it would > > thoroughly overcome the trapping force. It would also give an > > interesting wavelength dependence.) > > > > > If this experiment has been done why can't we find it? If it has not > > > been done then why not? > > > If Nichols work were falsifiable then we should see a next gen, no > > > different than Nichols followed Crooks. > > > > > Ahah! I just found one! > > > http://iopscience.iop.org/0959-5309/45/2/315 > > > Reads pretty much like some of this discussion, but in 1933. > > > He calls Nichols and Hull's work a 'paper dagger'. > > > He gets down to 10E-6 torr. > > > I can't access the whole paper, but the intro reads easy. > > > > So, a later and better repeat of Nichols and Hull. N&H (or the > > widespread quoting of their result as "definitive") are criticised > > since the discrepancy between theory and experiment was 10% (due to > > the effect of the gas). Hull had a paper a few years after N&%, 1905 > > iirc, on the elimination of gas action, or similar title. This might > > give some idea of what N&H thought at the time. Bell (and Green) had > > another paper the next year, with Hull as 1st author. (I briefly read > > the paper you link above some years ago, but not the others). > > > > > The > > > doubling claim still troubles me. > > > > > I guess if you want to argue the doubling effect some more I can try, > > > > The doubling is no more, no less, than you have with similar doubling > > in the reflection of ping pong balls, billiard balls, etc. The impulse > > needed to reverse the momentum of an object is double that needed to > > stop the momentum of an object (straight from Newton 2). Acting in the > > same time, the reflection force must be double the stopping force. > > No. This is not a clean analogy. In order for a receiver to return a > billiard ball to a sender at the same energy the receiver must not > accelerate. No work can be done on the receiver, Not "not accelerate", but "not move". Work = F*d, power = F*v. So consider a stationary reflector, of billiard balls or photons. Reflected balls/photons have the same energy as the incident ones, so from conservation of energy, no work is being done on the reflector. The momentum is still reversed, so from the conservation of momentum, there must be force. Power = F*v means that these aren't incompatible. Then look at this in a coordinate system where the reflector is moving (but not accelerating). > and this is why I > claim this portion of this subject farcical. Is Newtonian mechanics farcical? Is the conservation of momentum farcical? > If momentum is imparted > upon the receiver, then in order to return the ball with the same > energy to the sender the receiver will have to expend energy. There is > no free lunch, though I'm getting awfully close to it. If you want, do the analysis for perfectly elastic reflection of balls, say of mass m1, from a plate of mass m2. Use conservation of momentum and energy. Compare with "absorption" of completely sticky balls. If you don't believe the prediction of what happens from Newtonian mechanics and conservation of energy and momentum, then I don't think there's much point in continuing this. The reflection/"absorption" of balls isn't exactly the same as reflection/absorption of light, but the basic principles are the same. For the balls case, don't start with the plate at rest, but with the centre of mass of ball+plate at rest. Then look at what these results become when the plate is initially at rest (and moving at the end). > > > but I'd rather go over to the black body stuff and try to understand > > > just how literally the photon momentum is taken. If it's all in e=hf > > > then it is a lie, because the 1300 watts per square meter of sun would > > > be observed as a mechanical force. > > > > Why would it be observed? 4 (or 9) micronewtons per square meter? > > Compare this with the gravitational force on a practical plate of the > > same area. (Perhaps also compare this with forces due to Brownian > > motion?) > > > > > This is likely a matter of > > > interpretation. As I approach this as an open problem, then it seems > > > blatantly apparent that what we just claimed as an isolational > > > experiment on radiation pressure can actually be taken as a claim to > > > have isolated photon momentum as much more miniscule than the e=hf > > > derivation. This then could be twisted into photon mass, and as it is > > > such a slight figure, then all the better. > > > > You keep saying that this is problem, but I don't see why. Why is it a > > problem? E=hf gives a very, very, small force. > > No. 1300 watts on a square meter is nearly two horsepower. > This is not a trivial amount of power. This is the energy in e=hf that > I am discussing. In one second of time we have 1.3 kiloJoules of > energy on a square meter, if all that energy is absorbed. Obviously > this is not momentum, so any blackbody interpretation which treats it > as such is a fraud. Yes, it isn't momentum. And it isn't force. It's power (and, over 1 second, an amount of energy). > > E = hf, energy of photon, momentum = E/c = E/(f*lambda) = h/lambda, > > the standard QM result. h is small; this isn't much momentum. > > It's 1300 watts of power per square meter from the sun at the earth's > surface. > h doesn't matter. It's a large quantity of photons. > "A person having a mass of 100 kilograms who climbs > a 3 meter high ladder in 5 seconds is > doing work at a rate of about 600 watts." > - http://en.wikipedia.org/wiki/Watt > > And so we should be getting quite a push when we hold a mirror up to > the sun under the standard photon momentum argument. It is clearly a > misinterpretation and disambiguating this failure is not happening on > this thread. If h doesn't matter, why did you introduce h? Do it without h then. P = 1300W, so F=P/c = 9 micronewtons. I wouldn't call this "quite a push". 1300W incident on a reflecting object _doesn't_ mean that 1300W of work will be done. Or even the same on a perfect absorber. Again, do the calculation for reflection/"absorption" of balls. To repeat yet again, the predicted force is 9 micronewtons per square metre. Why do you think this should be easy to observe? > > This is small enough so it isn't casually observable. Observed by > > Nichols and Hull (to about 10%, according to Bell and Green), and > > others since. Observed in special circumstances before and since > > (e.g., as contributor to comet tails). > > > > > The radiation pressure is not unlike electron spin, in that it is a > > > small effect that is difficult to notice and for me, difficult to > > > understand. Incidentally, the p=u/3 that I see in other sources is not > > > the equation you rely upon. > > > > The 1/3 is for omni-directional radiation, such as blackbody > > radiation. What is done experimentally usually uses a beam of some > > kind. For a plane wave, or a parallel (i.e., collimated) beam, p=P/c > > (or p=nP/c=P/v in a medium). For a focussed beam, it is a little less > > (and this difference is seen, and is essential to 3D trapping in > > optical tweezers). > > > > And of course, u=P/c. > > > > > and for me, difficult to > > > understand. > > > > You're not the first. Until Maxwell's calculation of electromagnetic > > radiation pressure, it was generally assuming that there was no > > radiation pressure, that no wave would exert a non-zero time-averaged > > force (since it has no mass). In the 1800s, the lack of observabed > > wave pressure was used as an argument for the wave theory of light and > > against corpuscular theories. One of the last works of the late Thomas > > Gold was a short note on solar sails and why they don't work. You are > > not alone. > > > > Maxwell's result (and Poynting's and Heaviside's) was specifically > > electromagnetic, so N. A. Umov's general result (not given the > > attention it deserves in the West) in 1874 was also a significant > > advance. Then Einstein in 1905 with E=mc^2. > > While you bring it up I had better just run my own simple thinking on > photon momentum against this e=mcc. A photon has energy > e = h f > where f is the frequency, not worrying about any 2pi factor. > The velocity of the photon is > v = c > and so using > e = m c c > we have > h f = m c c > and so the photon's equivalent mechanical momentum > p = m v > can be expressed as > p = m c = e / c = h f / c. .... which is the same as p = f/lambda as already stated earlier. > This also leads to a mass expression for the photon, or at least an > equivalent mass for the photon. For some definitions of "mass", this is mass, what is sometimes called "relativistic mass". For the usual modern technical definition of mass (which is rest mass), it isn't mass. It does tell you the > This expression takes what I believe > to be the entire energy of the photon and expresses a mechanical > momentum. Thus the 1300 watts of solar power should provide quite some > motive force if the translation is accurate. Thus far we do not see > any such mechanical force, Do the actual calculation, and you will see that the force is small. Note the 1/c in your expression; c is a large number. > and instead see the ability to absorb a > large amount of this as heat or with less efficiency to turn it into > eletricity with special diodes, called solar cells. Just as we would see with the collision of lightweight balls sticking to a massive object. Do the calculation, using Newtonian mechanics, for a stream of Newtonian particles moving at c, with kinetic energy (for each particle) equal to hf for, say, 500nm light, and total KE per second in the stream of 1300W. Do this for reflection or "absorption" of the particles. Don't just say that 1300W must give a huge force, do the calculation instead. > The last two > phenomena are acceptable from AC principles, whereas the mechanical > momentum claim is not. Compare with the result of the suggested Newtonian calculation above. Is it still not acceptable? -- Timo
From: Tim BandTech.com on 6 Jun 2010 23:44 On Jun 6, 9:02 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > On Sun, 6 Jun 2010, Tim BandTech.com wrote: > > On Jun 6, 4:44 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > No. This is not a clean analogy. In order for a receiver to return a > > billiard ball to a sender at the same energy the receiver must not > > accelerate. No work can be done on the receiver, > > Not "not accelerate", but "not move". Work = F*d, power = F*v. > > So consider a stationary reflector, of billiard balls or photons. > Reflected balls/photons have the same energy as the incident ones, so from > conservation of energy, no work is being done on the reflector. The > momentum is still reversed, so from the conservation of momentum, there > must be force. Power = F*v means that these aren't incompatible. > > Then look at this in a coordinate system where the reflector is moving > (but not accelerating). > > > and this is why I > > claim this portion of this subject farcical. > > Is Newtonian mechanics farcical? Is the conservation of momentum farcical? > > > If momentum is imparted > > upon the receiver, then in order to return the ball with the same > > energy to the sender the receiver will have to expend energy. There is > > no free lunch, though I'm getting awfully close to it. > > If you want, do the analysis for perfectly elastic reflection of balls, > say of mass m1, from a plate of mass m2. Use conservation of momentum and > energy. Compare with "absorption" of completely sticky balls. > > If you don't believe the prediction of what happens from Newtonian > mechanics and conservation of energy and momentum, then I don't think > there's much point in continuing this. Light is not the same as massive objects colliding. I've merely pointed out where the analogy breaks down. In effect again, there will be no perfect reflector, except in the mechanical case of a grounded plate which is not free to accelerate. I'm not breaking any physical laws here, and if I were you should be able to find the flaw in my wording and falsify directly rather than building more argument to sift through. I've been through this one already with another guy about passing a ball back and forth in space. It is not at all the same problem, since light will go at speed c, and only speed c, staying within that existing theory. Yes, now we can gyrate onto the redshift concept, but it just isn't there for the practical problem. Taking photon momentum literally, then we should consider what 1300 watts is in terms of mechanical systems. I did this in the quote of my last post from the wiki. > > The reflection/"absorption" of balls isn't exactly the same as > reflection/absorption of light, but the basic principles are the same. > For the balls case, don't start with the plate at rest, but with the > centre of mass of ball+plate at rest. Then look at what these results > become when the plate is initially at rest (and moving at the end). > > > > > but I'd rather go over to the black body stuff and try to understand > > > > just how literally the photon momentum is taken. If it's all in e=hf > > > > then it is a lie, because the 1300 watts per square meter of sun would > > > > be observed as a mechanical force. > > > > Why would it be observed? 4 (or 9) micronewtons per square meter? > > > Compare this with the gravitational force on a practical plate of the > > > same area. (Perhaps also compare this with forces due to Brownian > > > motion?) > > > > > This is likely a matter of > > > > interpretation. As I approach this as an open problem, then it seems > > > > blatantly apparent that what we just claimed as an isolational > > > > experiment on radiation pressure can actually be taken as a claim to > > > > have isolated photon momentum as much more miniscule than the e=hf > > > > derivation. This then could be twisted into photon mass, and as it is > > > > such a slight figure, then all the better. > > > > You keep saying that this is problem, but I don't see why. Why is it a > > > problem? E=hf gives a very, very, small force. > > > No. 1300 watts on a square meter is nearly two horsepower. > > This is not a trivial amount of power. This is the energy in e=hf that > > I am discussing. In one second of time we have 1.3 kiloJoules of > > energy on a square meter, if all that energy is absorbed. Obviously > > this is not momentum, so any blackbody interpretation which treats it > > as such is a fraud. > > Yes, it isn't momentum. And it isn't force. It's power (and, over 1 > second, an amount of energy). > > > > E = hf, energy of photon, momentum = E/c = E/(f*lambda) = h/lambda, > > > the standard QM result. h is small; this isn't much momentum. > > > It's 1300 watts of power per square meter from the sun at the earth's > > surface. > > h doesn't matter. It's a large quantity of photons. > > "A person having a mass of 100 kilograms who climbs > > a 3 meter high ladder in 5 seconds is > > doing work at a rate of about 600 watts." > > -http://en.wikipedia.org/wiki/Watt > > > And so we should be getting quite a push when we hold a mirror up to > > the sun under the standard photon momentum argument. It is clearly a > > misinterpretation and disambiguating this failure is not happening on > > this thread. > > If h doesn't matter, why did you introduce h? > > Do it without h then. P = 1300W, so F=P/c = 9 micronewtons. I wouldn't > call this "quite a push". > > 1300W incident on a reflecting object _doesn't_ mean that 1300W of work > will be done. Or even the same on a perfect absorber. Again, do the > calculation for reflection/"absorption" of balls. > > To repeat yet again, the predicted force is 9 micronewtons per square > metre. Why do you think this should be easy to observe? We've moved to photon momentum here, and not radiation pressure. I guess we're near to a stalemate on this, because of the quantity of repetitions that we are going onto. I encourage you to falsify me as I have falsified you, and remind you of your own admission: "If the reflector is moving, there will be a Doppler shift, and energy in will be different from energy out. If the reflector is stationary, then you could have, e.g., 1300W in and 1300W out. But no work would be done." - http://groups.google.com/group/sci.physics/msg/f064666482b7c3b5 This is essentially the same thing that I am communicating here. > > > > This is small enough so it isn't casually observable. Observed by > > > Nichols and Hull (to about 10%, according to Bell and Green), and > > > others since. Observed in special circumstances before and since > > > (e.g., as contributor to comet tails). > > > > > The radiation pressure is not unlike electron spin, in that it is a > > > > small effect that is difficult to notice and for me, difficult to > > > > understand. Incidentally, the p=u/3 that I see in other sources is not > > > > the equation you rely upon. > > > > The 1/3 is for omni-directional radiation, such as blackbody > > > radiation. What is done experimentally usually uses a beam of some > > > kind. For a plane wave, or a parallel (i.e., collimated) beam, p=P/c > > > (or p=nP/c=P/v in a medium). For a focussed beam, it is a little less > > > (and this difference is seen, and is essential to 3D trapping in > > > optical tweezers). > > > > And of course, u=P/c. > > > > > and for me, difficult to > > > > understand. > > > > You're not the first. Until Maxwell's calculation of electromagnetic > > > radiation pressure, it was generally assuming that there was no > > > radiation pressure, that no wave would exert a non-zero time-averaged > > > force (since it has no mass). In the 1800s, the lack of observabed > > > wave pressure was used as an argument for the wave theory of light and > > > against corpuscular theories. One of the last works of the late Thomas > > > Gold was a short note on solar sails and why they don't work. You are > > > not alone. > > > > Maxwell's result (and Poynting's and Heaviside's) was specifically > > > electromagnetic, so N. A. Umov's general result (not given the > > > attention it deserves in the West) in 1874 was also a significant > > > advance. Then Einstein in 1905 with E=mc^2. > > > While you bring it up I had better just run my own simple thinking on > > photon momentum against this e=mcc. A photon has energy > > e = h f > > where f is the frequency, not worrying about any 2pi factor. > > The velocity of the photon is > > v = c > > and so using > > e = m c c > > we have > > h f = m c c > > and so the photon's equivalent mechanical momentum > > p = m v > > can be expressed as > > p = m c = e / c = h f / c. > > ... which is the same as p = f/lambda as already stated earlier. > > > This also leads to a mass expression for the photon, or at least an > > equivalent mass for the photon. > > For some definitions of "mass", this is mass, what is sometimes called > "relativistic mass". For the usual modern technical definition of mass > (which is rest mass), it isn't mass. > > It does tell you the > > > This expression takes what I believe > > to be the entire energy of the photon and expresses a mechanical > > momentum. Thus the 1300 watts of solar power should provide quite some > > motive force if the translation is accurate. Thus far we do not see > > any such mechanical force, > > Do the actual calculation, and you will see that the force is small. Note > the 1/c in your expression; c is a large number. > > > and instead see the ability to absorb a > > large amount of this as heat or with less efficiency to turn it into > > eletricity with special diodes, called solar cells. > > Just as we would see with the collision of lightweight balls sticking to a > massive object. > > Do the calculation, using Newtonian mechanics, for a stream of Newtonian > particles moving at c, with kinetic energy (for each particle) equal to hf > for, say, 500nm light, and total KE per second in the stream of 1300W. OK. I just did set it up in a spreadsheet. Each photon at 500nm has energy of 5.03E-019 Joules The total number of photons in one second that will stream is 2.58E+021 photons The momentum per photon is 1.32E-027 kg m / s The momentum for one seconds worth is 3.42E-006 kg m / s I agree that the momentum is small, but this does nothing to diminish the power argument and the net energy of this momentum when absorbed. The momentum figure is operant at the speed of light and damn well better be small or we'll all be vaporized. It's tough enough already being out in the sun all day. Rather than getting pushed over by the sunlight we're getting heated by it instead. There is no photon momentum in a literal sense. So long as the momentum is derived from the power then there is nothing wrong with my own logic. The power is still there, even in this slight figure, because momentum is not the same as energy or power. It's still 1300 watts, and a one second pulse of this is quite some energy, though I was just looking at lasers and most can only pulse this kind of power for shorter durations. Energy takes many forms, and we see that they can be converted, but this does not mean that they actually are converted, though we can still write the equation. Somehow the physics culture has come to take the photon momentum as fundamental but instead it is a trick, and we may as well give them mass from another trick by the same trickster. Is my logic a trick? No, I do not think so. If it is then I do wish you would expose it as such via falsification rather than via more construction. We've already constructed the things that we need. There is no need for any more. > > Do this for reflection or "absorption" of the particles. Don't just say > that 1300W must give a huge force, do the calculation instead. Look, 1300 watts is a very sound figure. It means nothing more or nothing less. If anything it is just a matter of how long you have the 1300 watts for, and if we presume to consider one second, well, then we have an energy figure. The fact is that the interpretation of photon momentum places all of the energy of the photon into momentum, and there is a lack of truth in this, yet it has come to be a foundational concept for much thinking. It is easily disproven, and I have done so here. I am still open to direct falsification of my statements. The oddities of conjoining this with radiation pressure, and the numerous interpretations possible are baffling. - Tim > > > The last two > > phenomena are acceptable from AC principles, whereas the mechanical > > momentum claim is not. > > Compare with the result of the suggested Newtonian calculation above. Is > it still not acceptable? > > -- > Timo
From: Timo Nieminen on 7 Jun 2010 01:41 On Sun, 6 Jun 2010, Tim BandTech.com wrote: > On Jun 6, 9:02 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: [big cut for brevity, the core seems to be:] [E=hf] > > > This expression takes what I believe > > > to be the entire energy of the photon and expresses a mechanical > > > momentum. Thus the 1300 watts of solar power should provide quite some > > > motive force if the translation is accurate. Thus far we do not see > > > any such mechanical force, > > > > Do the actual calculation, and you will see that the force is small. Note > > the 1/c in your expression; c is a large number. > > > > > and instead see the ability to absorb a > > > large amount of this as heat or with less efficiency to turn it into > > > eletricity with special diodes, called solar cells. > > > > Just as we would see with the collision of lightweight balls sticking to a > > massive object. > > > > Do the calculation, using Newtonian mechanics, for a stream of Newtonian > > particles moving at c, with kinetic energy (for each particle) equal to hf > > for, say, 500nm light, and total KE per second in the stream of 1300W. > > OK. I just did set it up in a spreadsheet. Each photon at 500nm has > energy of > 5.03E-019 Joules > The total number of photons in one second that will stream is > 2.58E+021 photons > The momentum per photon is > 1.32E-027 kg m / s > The momentum for one seconds worth is > 3.42E-006 kg m / s > I agree that the momentum is small, but this does nothing to diminish > the power argument and the net energy of this momentum when absorbed. OK, this will do for a start. This _isn't_ the suggested calculation, but it _is_ a start. Note that E = hf = 3.98e-19J, so you underestimate photon flux by 20%, and this the momentum by 20%. With the right E, this brings you back to to the already-mentioned 4 or 9 micronewtons. > The momentum figure is operant at the speed of light and damn well > better be small or we'll all be vaporized. It's tough enough already > being out in the sun all day. Rather than getting pushed over by the > sunlight we're getting heated by it instead. Why would 4 or 9 (or 3 or 7) micronewtons push you over? Do you expect to feel this force? > being out in the sun all day. Rather than getting pushed over by the > sunlight we're getting heated by it instead. There is no photon > momentum in a literal sense. There's a reason why I suggested the calculation: > > Do the calculation, using Newtonian mechanics, for a stream of Newtonian > > particles moving at c, with kinetic energy (for each particle) equal to hf > > for, say, 500nm light, and total KE per second in the stream of 1300W. Try this! Don't do this for real photons, do this for Newtonian particles with the same kinetic energy as the photons. > So long as the momentum is derived from the power then there is > nothing wrong with my own logic. The power is still there, even in > this slight figure, because momentum is not the same as energy or > power. It's still 1300 watts, and a one second pulse of this is quite > some energy, though I was just looking at lasers and most can only > pulse this kind of power for shorter durations. Energy takes many > forms, and we see that they can be converted, but this does not mean > that they actually are converted, though we can still write the > equation. Somehow the physics culture has come to take the photon > momentum as fundamental but instead it is a trick, and we may as well > give them mass from another trick by the same trickster. > > Is my logic a trick? > No, I do not think so. If it is then I do wish you would expose it as > such via falsification rather than via more construction. We've > already constructed the things that we need. There is no need for any > more. Power isn't force, energy isn't force, work isn't force. As far as I can tell, your argument here is solely that 1300W is a lot of power, and if light has momentum, 1300W of light must have lots of momentum, and since we're not pushed around (as far as we can notice) by sunlight, therefore light doesn't have momentum. You did the calculation above, and you can see that the force would only be micronewtons. Isn't this a direct falsification of any "1300W must correspond to a large force" argument? If you don't like to do a calculation for photons, or light, try it for classical Newtonian particles of the same energy and speed. (You should get a force double that of light, 9 micronewtons for "absorption", and 17 micronewtons for reflection. That's for KE coming in at 1300W.) > > Do this for reflection or "absorption" of the particles. Don't just say > > that 1300W must give a huge force, do the calculation instead. > > Look, 1300 watts is a very sound figure. It means nothing more or > nothing less. If anything it is just a matter of how long you have the > 1300 watts for, and if we presume to consider one second, well, then > we have an energy figure. The fact is that the interpretation of > photon momentum places all of the energy of the photon into momentum, > and there is a lack of truth in this, yet it has come to be a > foundational concept for much thinking. It is easily disproven, and I > have done so here. First, what does "the interpretation of photon momentum places all of the energy of the photon into momentum" mean? The standard accepted physics theories of radiation pressure, whether by classical light or photons, don't say this. In particular, who says that energy is momentum, or that momentum is energy? Second, you calculated yourself that the momentum of 1300J worth of photons is 3.4e-6 kg.m/s (OK, 20% too low, but let us not worry about a little error.) Is this 3.4e-6 kg.m/s "all of" the energy of that 1300J of photons? Which did you disprove? If the former, it's already accepted that energy and momentum are different things; this doesn't say anything useful about radiation pressure. If the latter, how did you disprove it? > I am still open to direct falsification of my > statements. The oddities of conjoining this with radiation pressure, > and the numerous interpretations possible are baffling. -- Timo
From: Sue... on 7 Jun 2010 03:37 On Jun 6, 7:49 pm, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > On Jun 6, 4:44 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > On Jun 6, 11:05 pm, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > > > > Well, I have gone back to check what I remember pretty well that you > > > wrote, and you state that it will not be observed in perfect vacuum: > > > "Perfect vacuum, no. Very good vacuums, yes, especially with atom > > > trapping. I've seen classical experiments done in vacuum (can't recall > > > how good), where absorbing particles were blasted by short pulses of > > > light, to measure their radiation pressure cross-sections." > > > > in response to my statement: > > > "I'm sorry but the effect in a perfect vacuum has not been clearly > > > demonstrated. [...] > > > No, I said the experiment hasn't been *done* in a perfect vacuum. > > Because we can't achieve a perfect vacuum in the lab. Technically, you > > were correct. > > > We also haven't done an experimental test of Newton's 1st law, because > > we don't have a friction free, viscous drag free, force free > > laboratory. This doesn't mean that I think that Newton's 1st law > > doesn't work in perfect vacuum. > > > It's an important point. How do we know when we can extrapolate a good > > vacuum result (or poor vacuum, or atomospheric result) to perfect > > vacuum? It's important to know (by measuring) the effect of what is > > there, e.g., damping due to viscous drag, drag due to convective flow, > > the radiometer force, the effect of the gas on the temperature > > difference between the two sides of the vane, etc. > > > > I have been avoiding your references to optical tweezers and light > > > traps, partially because I have only limited familiarity with those. > > > They're the main modern application of radiation pressure. *Thousands* > > of experiments have been done. Early experiments to show that the > > method works, and to explore the theoretical principles, experiments > > to use the force e.g. for restraining or moving live cells, > > experiments to quantitatively measure forces between biomolecules. > > > > Anyway, we are in agreement that radiation pressure should be > > > measurable at 10E-11 torrs using a solid plate device. > > > In principle. Whether the force is too small to measure in practice is > > the question, and this depends on the particular setup. > > > > ... and > > > possibly freed electrons as one of the links in support of my position > > > mentions. When we start thinking of those electrons then the tweezers > > > and traps start to make more sense I think, but I cannot argue on > > > these devices without doing some research. > > > Freed electrons are not responsible for the force in general. The atom > > trapping experiments are quite conclusive on this (since freeing an > > electron would result in ionisation, which would result in the ion not > > being trapped). (Perhaps in some specific experiment? In the modern > > laser experiments, there isn't enough photon energy for (significant) > > photoejection of electrons. Given 1/2 of the photon energy used to > > eject the electron, the rest becoming its KE, electron ejection would > > give about 400 times as much force per photon. If this was significant > > in the various tweezers experiments, it would be observed, as it would > > thoroughly overcome the trapping force. It would also give an > > interesting wavelength dependence.) > > > > If this experiment has been done why can't we find it? If it has not > > > been done then why not? > > > If Nichols work were falsifiable then we should see a next gen, no > > > different than Nichols followed Crooks. > > > > Ahah! I just found one! > > > http://iopscience.iop.org/0959-5309/45/2/315 > > > Reads pretty much like some of this discussion, but in 1933. > > > He calls Nichols and Hull's work a 'paper dagger'. > > > He gets down to 10E-6 torr. > > > I can't access the whole paper, but the intro reads easy. > > > So, a later and better repeat of Nichols and Hull. N&H (or the > > widespread quoting of their result as "definitive") are criticised > > since the discrepancy between theory and experiment was 10% (due to > > the effect of the gas). Hull had a paper a few years after N&%, 1905 > > iirc, on the elimination of gas action, or similar title. This might > > give some idea of what N&H thought at the time. Bell (and Green) had > > another paper the next year, with Hull as 1st author. (I briefly read > > the paper you link above some years ago, but not the others). > > > > The > > > doubling claim still troubles me. > > > > I guess if you want to argue the doubling effect some more I can try, > > > The doubling is no more, no less, than you have with similar doubling > > in the reflection of ping pong balls, billiard balls, etc. The impulse > > needed to reverse the momentum of an object is double that needed to > > stop the momentum of an object (straight from Newton 2). Acting in the > > same time, the reflection force must be double the stopping force. > > No. This is not a clean analogy. In order for a receiver to return a > billiard ball to a sender at the same energy the receiver must not > accelerate. No work can be done on the receiver, and this is why I > claim this portion of this subject farcical. If momentum is imparted > upon the receiver, then in order to return the ball with the same > energy to the sender the receiver will have to expend energy. There is > no free lunch, though I'm getting awfully close to it. > > > > > > but I'd rather go over to the black body stuff and try to understand > > > just how literally the photon momentum is taken. If it's all in e=hf > > > then it is a lie, because the 1300 watts per square meter of sun would > > > be observed as a mechanical force. > > > Why would it be observed? 4 (or 9) micronewtons per square meter? > > Compare this with the gravitational force on a practical plate of the > > same area. (Perhaps also compare this with forces due to Brownian > > motion?) > > > > This is likely a matter of > > > interpretation. As I approach this as an open problem, then it seems > > > blatantly apparent that what we just claimed as an isolational > > > experiment on radiation pressure can actually be taken as a claim to > > > have isolated photon momentum as much more miniscule than the e=hf > > > derivation. This then could be twisted into photon mass, and as it is > > > such a slight figure, then all the better. > > > You keep saying that this is problem, but I don't see why. Why is it a > > problem? E=hf gives a very, very, small force. > > No. 1300 watts on a square meter is nearly two horsepower. > This is not a trivial amount of power. This is the energy in e=hf that > I am discussing. In one second of time we have 1.3 kiloJoules of > energy on a square meter, if all that energy is absorbed. Obviously > this is not momentum, so any blackbody interpretation which treats it > as such is a fraud. > > > > > E = hf, energy of photon, momentum = E/c = E/(f*lambda) = h/lambda, > > the standard QM result. h is small; this isn't much momentum. > > It's 1300 watts of power per square meter from the sun at the earth's > surface. > h doesn't matter. It's a large quantity of photons. > "A person having a mass of 100 kilograms who climbs > a 3 meter high ladder in 5 seconds is > doing work at a rate of about 600 watts." > -http://en.wikipedia.org/wiki/Watt > > And so we should be getting quite a push when we hold a mirror up to > the sun under the standard photon momentum argument. It is clearly a > misinterpretation and disambiguating this failure is not happening on > this thread. > > > > > This is small enough so it isn't casually observable. Observed by > > Nichols and Hull (to about 10%, according to Bell and Green), and > > others since. Observed in special circumstances before and since > > (e.g., as contributor to comet tails). > > > > The radiation pressure is not unlike electron spin, in that it is a > > > small effect that is difficult to notice and for me, difficult to > > > understand. Incidentally, the p=u/3 that I see in other sources is not > > > the equation you rely upon. > > > The 1/3 is for omni-directional radiation, such as blackbody > > radiation. What is done experimentally usually uses a beam of some > > kind. For a plane wave, or a parallel (i.e., collimated) beam, p=P/c > > (or p=nP/c=P/v in a medium). For a focussed beam, it is a little less > > (and this difference is seen, and is essential to 3D trapping in > > optical tweezers). > > > And of course, u=P/c. > > > > and for me, difficult to > > > understand. > > > You're not the first. Until Maxwell's calculation of electromagnetic > > radiation pressure, it was generally assuming that there was no > > radiation pressure, that no wave would exert a non-zero time-averaged > > force (since it has no mass). In the 1800s, the lack of observabed > > wave pressure was used as an argument for the wave theory of light and > > against corpuscular theories. One of the last works of the late Thomas > > Gold was a short note on solar sails and why they don't work. You are > > not alone. > > > Maxwell's result (and Poynting's and Heaviside's) was specifically > > electromagnetic, so N. A. Umov's general result (not given the > > attention it deserves in the West) in 1874 was also a significant > > advance. Then Einstein in 1905 with E=mc^2. > > While you bring it up I had better just run my own simple thinking on > photon momentum against this e=mcc. A photon has energy > e = h f > where f is the frequency, not worrying about any 2pi factor. > The velocity of the photon is > v = c > and so using > e = m c c > we have > h f = m c c > and so the photon's equivalent mechanical momentum > p = m v > can be expressed as > p = m c = e / c = h f / c. > This also leads to a mass expression for the photon, or at least an > equivalent mass for the photon. This expression takes what I believe > to be the entire energy of the photon and expresses a mechanical > momentum. Thus the 1300 watts of solar power should provide quite some > motive force if the translation is accurate. Thus far we do not see > any such mechanical force, and instead see the ability to absorb a > large amount of this as heat or with less efficiency to turn it into > eletricity with special diodes, called solar cells. The last two > phenomena are acceptable from AC principles, whereas the mechanical > momentum claim is not. All that we can hope for in an AC situation is > to shake the matter about quite alot, which it traditionally thought > of as heat. Well, this is a loose, but consistent logic with observed > systems. Photon momentum is wrongly addressed as a mechanical vector > of a static variety, and would be better expressed in an AC form. Yes, > it propagates in a direction (though often from omnidirectional > sources), but this does not inherently imply a force vector ala > Newton. This vector marks a passage of energy that is obviously not at > all to do with physical momentum. Therefor the equations in use are > too loosely presented, as a high school student will easily get a > photon mass out of the product relationships. We are doing little > better than this sort of derivation; we merely abide by 'photons don't > have mass' ruling, and that ruling is all that keeps us from going > over the edge on that side. You are trying to work with half a tweezers. My spell chequer won't let me type the singular form. Probably because the notion of half a tweezers is absurd. :-)) <<Optical tweezers can be used for three-dimensional manipulation of transparent particles around 1100 ñm in diameter, or other small particles which behave as reactive dipoles. The single-beam gradient optical trap consists of a single laser beam, tightly focused to create a very strong field gradient both radially and axially, which acts on polarisable particles to cause a dipole force. Polarisable particles are attracted to the strongest part of the field, at the beam focus, due to the gradient force. A scattering force results from momentum transfer to the particle when light is scattered by it. Under the right conditions the gradient force can balance the scattering and gravity forces, to trap particles three-dimensionally in the laser beam. If the laser beam is not tightly focused, the axial component of the gradient force will be weak, and only radial trapping will be possible [1, 4].>> "Optical trapping of absorbing particles" Authors: H. Rubinsztein-Dunlop, T. A. Nieminen, M. E. J. Friese, N. R. Heckenberg http://arxiv.org/abs/physics/0310022 See also: http://en.wikipedia.org/wiki/Van_der_Waals_force http://en.wikipedia.org/wiki/London_dispersion_force http://en.wikipedia.org/wiki/Induced_gravity Sue... > We have instead barreled over the edge on > the other side. This is the current state. The creature called a > lemming comes to mind, and we are perhaps little better than them at > some base level. In fact, we provably are designing our own cliff and > it's collapse on a very grand scale. To presume that physicists are > somehow better than this overlooks a fundamental problem of human > existence. Likewise for mathematicians. We come from a blind state of > a blank slate and without the mimicry we would have nothing, but with > it we are likewise stuck. > > - Tim
From: Tim BandTech.com on 7 Jun 2010 07:41
On Jun 7, 3:37 am, "Sue..." <suzysewns...(a)yahoo.com.au> wrote: > On Jun 6, 7:49 pm, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > > > On Jun 6, 4:44 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > On Jun 6, 11:05 pm, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: <snip> > > > Maxwell's result (and Poynting's and Heaviside's) was specifically > > > electromagnetic, so N. A. Umov's general result (not given the > > > attention it deserves in the West) in 1874 was also a significant > > > advance. Then Einstein in 1905 with E=mc^2. > > > While you bring it up I had better just run my own simple thinking on > > photon momentum against this e=mcc. A photon has energy > > e = h f > > where f is the frequency, not worrying about any 2pi factor. > > The velocity of the photon is > > v = c > > and so using > > e = m c c > > we have > > h f = m c c > > and so the photon's equivalent mechanical momentum > > p = m v > > can be expressed as > > p = m c = e / c = h f / c. > > This also leads to a mass expression for the photon, or at least an > > equivalent mass for the photon. This expression takes what I believe > > to be the entire energy of the photon and expresses a mechanical > > momentum. Thus the 1300 watts of solar power should provide quite some > > motive force if the translation is accurate. Thus far we do not see > > any such mechanical force, and instead see the ability to absorb a > > large amount of this as heat or with less efficiency to turn it into > > eletricity with special diodes, called solar cells. The last two > > phenomena are acceptable from AC principles, whereas the mechanical > > momentum claim is not. All that we can hope for in an AC situation is > > to shake the matter about quite alot, which it traditionally thought > > of as heat. Well, this is a loose, but consistent logic with observed > > systems. Photon momentum is wrongly addressed as a mechanical vector > > of a static variety, and would be better expressed in an AC form. Yes, > > it propagates in a direction (though often from omnidirectional > > sources), but this does not inherently imply a force vector ala > > Newton. This vector marks a passage of energy that is obviously not at > > all to do with physical momentum. Therefor the equations in use are > > too loosely presented, as a high school student will easily get a > > photon mass out of the product relationships. We are doing little > > better than this sort of derivation; we merely abide by 'photons don't > > have mass' ruling, and that ruling is all that keeps us from going > > over the edge on that side. > > You are trying to work with half a tweezers. My spell chequer > won't let me type the singular form. Probably because the > notion of half a tweezers is absurd. :-)) > > <<Optical tweezers can be used for three-dimensional manipulation > of transparent particles around 1100 ñm in diameter, or other > small particles which behave as reactive dipoles. > The single-beam gradient optical trap consists of a single > laser beam, tightly focused to create a very strong field > gradient both radially and axially, which acts on polarisable > particles to cause a dipole force. Polarisable particles are > attracted to the strongest part of the field, at the beam focus, > due to the gradient force. A scattering force results from > momentum transfer to the particle when light is scattered by it. > Under the right conditions the gradient force can balance the > scattering and gravity forces, to trap particles three-dimensionally > in the laser beam. If the laser beam is not tightly focused, the > axial component of the gradient force will be weak, and only > radial trapping will be possible [1, 4].>> > "Optical trapping of absorbing particles" > Authors: H. Rubinsztein-Dunlop, T. A. Nieminen, M. E. J. Friese, N. R. > Heckenberghttp://arxiv.org/abs/physics/0310022 > > See also:http://en.wikipedia.org/wiki/Van_der_Waals_forcehttp://en.wikipedia.org/wiki/London_dispersion_forcehttp://en.wikipedia.org/wiki/Induced_gravity > > Sue... Thanks Sue. I suggested the electric forces were at work in the trap a few posts ago, and here you have Timo stating it in prior work. As we discuss AC and DC principles, we are probably going to agree that the electron is a constant charge, and so in effect is a DC concept. On top of this the idea of stopping an electon does not exist as far as I can tell. On top of this when we allow for electron spin then we have gone beyond Maxwell, and even Einstein, and the raw charge which Maxwell managed his equations on does not exist. In effect this is the ultimate Maxwellian behavior, but denied charge as fundamental to magnetism as in a displacement current model. Beyond this thermodynamics remains open in my book. I'm not sure if you are insinuating something more specific with the tweezer (no spell checker here other than myself). I do appreciate the electrical nature of the balance, though I can't say that I understand the existing theory. I need to understand the atomic level better, so I appreciate your persistent linkage into VdW and so forth. - Tim > We have instead barreled over the edge on > > the other side. This is the current state. The creature called a > > lemming comes to mind, and we are perhaps little better than them at > > some base level. In fact, we provably are designing our own cliff and > > it's collapse on a very grand scale. To presume that physicists are > > somehow better than this overlooks a fundamental problem of human > > existence. Likewise for mathematicians. We come from a blind state of > > a blank slate and without the mimicry we would have nothing, but with > > it we are likewise stuck. > > > - Tim |