Taking a Fresh Look at the Physics of Radiometers.
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From: Tim BandTech.com on 7 Jun 2010 09:43 On Jun 7, 1:41 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > 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. I had a typo on the constant c. Now I am in agreement. Nice to see you are serious. > > > 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? Let's see now, I've got a momentum of 4.3E-6 kg m / s but you've got a figure of 4E-6 kg m / s / s . We have a units mismatch on momentum versus newtons, but most importantly each of these figures includes the dimension of kilograms, and these kilograms are travelling at the speed of light, and furthermore since we used the relativistic e = m c c to get here there is no reason not to use it to get back a usual amount of work, otherwise we are discussing an impossibility, right? The math simply inverts, and we wind back up at the 1300 watt figure, which is expressible in units through 1 J = 1 N m = 1 W s and so I must ask you what happened to all of the energy that we started the calculation with? Where is it? We had 1.3 kJ at the beginning of the calculation and now over one second you are claiming to have just something tiny. Well, multiply by c and you'll discover that the energy is all there in that momentum figure. When does the usage of relativity end? only after the thing travelling at the speed of light is absorbed; not before. Isn't this coherent? This really suggests that what is regarded as radiation pressure might better be interpreted as photon momentum, whereas this figure that we are computing above is a fraud. > > > 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. I understand that the K = m v v / 2 does not hold within relativity, and that there is another term, but on top of this we will not get any sense at c. Why move the chase over here? We already are using E = m c c to get momentum, and that is bad enough. To mix the two even more does not seem appropriate. Already we are considering a Newtonian momentum figure on something that is claimed to break Newtonian mechanics, but then you don't care to perform the reversal do you? Again Timo, there should be a more direct falsification. You ask for yet more construction here and fail to falsify my own argument. Your argument is claiming somewhat that the energy has disappeared. It has not, and the math which has generated the argument holds, and its reversal should be considered, since otherwise the conservation of energy will be destroyed. > > 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? Look Timo, this is hardly the right place to be asking this question. You are the one who set up the computations above, and we've considered a one second pulse of laser light. This is 1.3kJ that we've converted to momentum through e=hv and e=mcc. That you criticize this figuring is to say that you are critical of existing theory. > > 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? Yes, at the speed c. You seem to think that the math is a one-way framework. If this is the case then this is perhaps how we come to disagreement. Why don't you answer this question yourself. Clearly we fed the equations all of this energy. If we go back through them we will get 1300 watts on the other side. > > 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'm feeling more secure that the term 'radiation pressure' is a misnomer. We've just derived it as photon momentum, so what is the difference? One is distributed over a specific area whereas the other is a conglomeration. This is nearly a quip on the photon itself. Mostly I'm very interested in this topic, and not interested in studying it via a mimic's approach. You are great to spend so much time on this. I'm working on a new method, but I don't have anything substantial to share yet. There are a number of places that I see frailties in existing theory, and herabouts they are tied together. - Tim > > > 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 10:00 On Jun 7, 7:41 am, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > On Jun 7, 3:37 am, "Sue..." <suzysewns...(a)yahoo.com.au> wrote: > > Thanks snip > > > 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. > > Heckenberg http://arxiv.org/abs/physics/0310022 > > > See also:http://en.wikipedia.org/wiki/Van_der_Waals_forcehttp://en.wikipedia.o... > > > 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. They are always at work but for the light-mill I assumed convection currents explained it simply. If not, long range induction forces should not not be overlooked. Particularly for better vacuums, solar sails, or even the fundamental mechanism of inertia. > 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. No doubt, Maxwell and Einstein overlooked something that W.Weber and F.London picked up on. (VdW) > 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). That is funny. So I am clapping with one hand. > 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. Persistent? Thank you. :-) It is a quality learned from my mother who did not realise that she was not Jewish, Cass Elliot was not attractive and cream cheese is not fat-free even on a bagel. Sue... > > - 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 > >
From: NoEinstein on 7 Jun 2010 15:47 On Jun 6, 4:58 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > I had intended to include the following links that relate to the time of travel of photons around light courses defined by mirrors. NE Where Angels Fear to Fall http://groups.google.com/group/sci.physics/browse_frm/thread/8152ef3e... 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 > > On Jun 5, 6:29 pm, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > > Dear Tim: In my invalidation of the M-M experiment, I chose to look > at light in S L O W motion. I realized that the easiest way to > understand what was happening with the light was to consider just a > SINGLE photon racing along each of the two light courses. That is so > much simpler than trying to say, as Timo does, that the light red > shifts or blue shifts. Einstein declared that light never blue > shifts, because that would require the photons to exceed 'c'. Using > simple 9th grade algebra, I have proved, conclusively, that light > velocity is: V = 'c' plus or minus v (or the velocity of the light > source). Each time Timo talks about... a reflector, he could just as > well be talking about the mirrors in the M-M experiment. Below are > the links which explain these things. Also, I have determined that > there is always a slight "Friction of Reflection" which red-shifts all > reflected light. Some of the energy imparted to the reflector > converts to heat that isn't instantaneously re emitted with the > reflected light. That necessitates that reflected light will be red > shifted, otherwise the Law of the Conservation of Energy is violated. > NoEinstein > > > > > > > On Jun 4, 5:40 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > On Jun 4, 9:59 pm, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > > > > On Jun 3, 5:59 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > > On Jun 4, 7:09 am, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > > > > > > Here is another link in support of the challenge to radiation > > > > > > pressure's validity: > > > > > > "It is therefore much more likely that in a given case the apparent > > > > > > 'radiation pressure' is caused either by thermal surface effects or > > > > > > electrons which are released from the surface by the radiation." > > > > > > -http://www.physicsmyths.org.uk/#radpress > > > > > > That is a _really_ bad webpage! Just consider its claim: "Even if one > > > > > assumes a momentum, a radiation pressure force could only be caused by > > > > > a momentum change dp/dt, but this is not possible because the speed of > > > > > light c has to be constant" (1) Direction of motion matters when it > > > > > comes to momentum, (2) refractive index. > > > > > Yes, and this is very similar to the arguments that I have provided.. > > > > ! I haven't seen you directly arguing such complete nonsense. That web > > > page was saying that the direction of motion doesn't affect momentum, > > > that if an object changes direction and goes, e.g., in the exact > > > opposite direction, there is no change in momentum. Before you claim > > > that the argument given there is very similar to yours, you should > > > understand just how completely defective that argument is. > > > > > The conservation of momentum is a strict principle, one that you have > > > > already cast aside in your argument about changing momenta as media > > > > change in dielectric quality > > > > ?? Conservation of momentum is why, if the momentum flux of the beam > > > changes, there must be a matching force on the surface, or, if there > > > is a force on the surface of the liquid (as we observe > > > experimentally), then the momentum flux of the beam must change? > > > > Why do you say this "casts aside" conservation of momentum? It's > > > directly based on the conservation of momentum. > > > > > You already accepted once the farce of > > > > the reflector as a doubling agent of radition pressure effects, and > > > > then go back to supporting it. What do you have to say about > > > > conservation of momentum? > > > To support my claim above I quote from your own words > > > "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 > > > I don't mean to be crass but I do mean to use this medium with > > accountability, which is one of its strong points. I would like you to > > find such falsifications in my own writing. > > > The radiation pressure effect is claimed to be operant even with no > > initial relative velocity between the source and reflector. Further, > > as far as I can tell none of the existing work on radiation pressure > > relies upon any red shift argument as the balance of energy. Simply > > put, if the perfect reflector reflects all of the beams light back > > toward the source then it cannot have absorbed any energy and hence it > > will not accelerate. > > > Next, if we do take the accepted photon momentum derived from > > e = h f > > then we would witness the entire 1300 watts of power on a square meter > > of mirror, which should do quite some work according to this accepted > > momentum equation, particularly for those fond of a doubling effect. > > Clearly this is not the source of radiation pressure. Even if we go to > > a black body which absorbs the energy, we must admit that there is no > > effect; The energy is converted into heat, to be reradiated, and if we > > assume that this reradiation is isotropic then we need not bother with > > any directed effect from it within the already corrupted concept of > > photon momentum. All the while you stand by the photon momentum > > equation, even while it fails to provide any momentum. This argument > > on momentum does not necessarily have anything to do with the claimed > > existence of radiation pressure. > > > The heating effect is nearby to the redshift concept, but again, we > > are not necessarily talking about radiation pressure here. This is an > > altogether different effect as far as I can tell. Wouldn't we have to > > get into thermodyamics more and Planck's energy distribution to > > discuss this effect? You see, it is a bit crazy how many sidelines > > there are, and so long as we wish to discuss the radiometer as a means > > of measuring radiation pressure, well, we are not really discussing a > > radiometer any more. These fine points you seem not to have > > acknowledged though I've stated them several times. You have pretty > > much accepted (I believe) that there will be no effect in the best > > vacuum labs available. Isn't this enough to admit that you do not > > believe that radiation pressure exists? > > > As I see it the problem of human as scientist is embedded within this > > discussion. Upon an 'accepted' theory being posited, for one to break > > away from that norm places one as a potential quack. And yet, without > > this freedom of thought, the quality of science will suffer. Such a > > person has relinquished their freedom to falsify, and in doing so has > > guaranteed their success within a system where mimicry is a merit. I > > respect your rights, particularly here on this free medium, but I > > would point out to you that as a partially accurate sounding board to > > existing theory you are in some regards helping to disprove that > > theory. > > > As you argue for the radiation pressure on a perfect reflector in > > outer space why would you deny that it can be provided in a lab at > > 10E-11 torrs with a 10kW laser? Should this be enough to turn the > > spindle of a Crooks style radiometer, or is the friction too great? > > Really, I must admit that I am growing cynical here, if for no other > > reason than the fact that your response to this question will be > > miserable to read. > > > You are a strong poster Timo and I don't mean you any harm. I can't > > have this discussion without you. I find it poor that people who claim > > to be scientific will not apply themselves at this level. I ask that I > > be falsified, and you are the only one who steps forward. I am still > > open to such falsification, and it could turn on some minor point or > > discrepancy that we are overlooking. Still, could it be the other way > > around too? One must give ones self enough credit to declare something > > false if one is to give one's self enough credit to declare something > > to be true. Otherwise mimicry ensues, and this is a transparent human > > factor inculcated within the schooling. I give you as much credit as I > > can but attempt to hold you accountable. Thanks, Timo, for hanging in > > there. I half take back my prediction on your response. > > > - Tim > > > > You don't think the momentum of an object that reverses direction > > > changes? You don't think that a force (i.e., a change in momentum as > > > per Newton) is needed to change the direction of an object? > > > > If one instead, at least provisionally, accepts Newton's laws of > > > motion as correct, if an object with momentum p reverses direction, > > > while maintaining the same speed, the final momentum is -p, for a > > > change in momentum of -2p, with an average force of -2p/t where t is > > > the time over which the force that causes the change in direction is > > > applied. If instead, the object is "absorbed", has a sticky collision, > > > then the change in momentum is -p, only half of the "reflection" > > > change in momentum. What's so mysterious about radiation pressure on a > > > reflector being double?- Hide quoted text - > > > - Show quoted text -- Hide quoted text - > > > - Show quoted text -- Hide quoted text - > > - Show quoted text -
From: NoEinstein on 7 Jun 2010 15:53 On Jun 6, 7:55 pm, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > Dear Tim: Once I can get further... off of square two with my New Science, there are "infinite" avenues for experiments, research, and just plain fun! I know that there are other rational thinkers, like you, amid the thousands reading my posts. Come on, readers (the rational ones), let your presence be known! NoEinstein > > On Jun 6, 4:39 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > On Jun 5, 3:25 pm, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > > > Dear Tim: Since I replied to this same reply of yours, earlier, I've > > realized that the Moon is being held in orbit about the Sun; and being > > held in orbit about the Earth, by the radiations exchanged among those > > three. Lunar eclipses don't send the moon flying out on its tangent, > > probably because the ether pressure "reserve" is greater than the > > ether pressure lost due to being in the shadow. The shadow must be > > maintained long enough so the ether pressure reserve gets used up. > > > The following might be considered science 'fiction', now, but one day > > mankind may be able to maneuver, say, the planet Mars completely out > > of the solar system. This will require having an artificial > > satellite, with a huge reserve of FUSION power, to provide the "solar" > > energy to maintain life on the revolving planet. The destination > > would be a star young enough to sustain human life for a few billion > > years. And... it might be possible, by controlling solar flares and > > sunspots, to turn the Sun into a propulsion system to carry the entire > > solar system close enough to another 'younger' star, to facilitate the > > relocation. Forgive my "imagined" science. Real physics would have > > to apply there, too! NoEinstein > > Yowser ! > > The sun propulsion thing sounds fascinating. > > - Tim- Hide quoted text - > > - Show quoted text -
From: Timo Nieminen on 7 Jun 2010 15:55
On Jun 7, 11:43 pm, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > On Jun 7, 1:41 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > 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. > > I had a typo on the constant c. Now I am in agreement. Nice to see you > are serious. > > > > 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? > > Let's see now, I've got a momentum of 4.3E-6 kg m / s > but you've got a figure of 4E-6 kg m / s / s . > We have a units mismatch on momentum versus newtons, Momentum per second has the units of force. Force times time is impulse, same units as momentum. Per Newton 2, force is the rate of transfer of momentum. > but most > importantly each of these figures includes the dimension of kilograms, > and these kilograms are travelling at the speed of light, and > furthermore since we used the relativistic > e = m c c > to get here there is no reason not to use it to get back a usual > amount of work, otherwise we are discussing an impossibility, right? > The math simply inverts, and we wind back up at the 1300 watt figure, > which is expressible in units through > 1 J = 1 N m = 1 W s > and so I must ask you what happened to all of the energy that we > started the calculation with? Where is it? We didn't do anything with the energy; it's still there. > We had 1.3 kJ at the > beginning of the calculation and now over one second you are claiming > to have just something tiny. Well, multiply by c and you'll discover > that the energy is all there in that momentum figure. Momentum _isn't_ energy, energy _isn't_ momentum. > When does the > usage of relativity end? only after the thing travelling at the speed > of light is absorbed; not before. Isn't this coherent? When does the usage of relativity begin? You get the same kind of result using Newtonian mechanics and tiny particles. > This really suggests that what is regarded as radiation pressure might > better be interpreted as photon momentum, whereas this figure that we > are computing above is a fraud. > > > > > > 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. > > I understand that the > K = m v v / 2 > does not hold within relativity, and that there is another term, but > on top of this we will not get any sense at c. Why move the chase over > here? Your main complaint with the result above is that you have difficulty reconciling the large energy with the small momentum, the large power with the small force. Do the calculation for Newtonian particles, and you get the same type of qualitative result (even similar numbers, just different by a factor of 2). Do you disbelieve this result? Note well, the _same kind_ of result. If you photon momentum calculation makes you think it's all rubbish, will getting the same kind of result with a Newtonian calculation make you think that Newtonian mechanics is all rubbish? Yes, of course (1/2)mv^2 is known to be incorrect at v=c. If this bothers you, do the calculation for v=c/2, where Newtonian mechanics is a reasonable (even if no longer very accurate) approximation. > We already are using > E = m c c > to get momentum, and that is bad enough. To mix the two even more does > not seem appropriate. Already we are considering a Newtonian momentum > figure on something that is claimed to break Newtonian mechanics, but > then you don't care to perform the reversal do you? Again Timo, there > should be a more direct falsification. > You ask for yet more > construction here and fail to falsify my own argument. It's been falsified many times already. Once should be enough. Where is the logic in your claim? Your only basis for it above is that a large energy results in a small momentum, so must be wrong. But momentum and energy aren't the same thing! Qualitatively, one gets the same kind of result from Newtonian mechanics; it shouldn't be mysterious (unfamiliar, and offensive to our common sense is OK, since this is outside our common everyday experience of mechanics). The falsifications so far: (1) Radiation pressure is observed experimentally, for RF, light, and acoustic waves at least. (2) The result of high energy/small momentum is exactly as expected from either relativistic or Newtonian mechanics. That energy and momentum are different things is also clear in both mechanics. Both are also observed to give a good description of reality (Newtonian mechanics is it's regime of applicability). (3) Conservation of energy and momentum gives a prediction of radiation pressure. (4) Direct calculation of the forces gives electromagnetic radiation pressure, using classical electromagnetic theory (which is also observed to give a good description of reality). In particular, the same physics that is used to (successfully) describe electric motors etc. results in a prediction of radiation pressure. The results of (3) and (4) are the same, and the same as (2) using relativistic mechanics of zero rest-mass photons. The theoretical models agree with experiment (with 10% for Nichols and Hull, 4% for Bell and Green, 0.1% for R. V. Jones, within 5% for our own absolute measurements). If you choose to outright reject relativistic and Newtonian mechanics, the conservation of energy and momentum, and electric motors, and ignore a diverse range of thousands of experiments, what can be said? (Other than you _not_ being open to falsification!) > Your argument is claiming somewhat that the energy has disappeared. It > has not, and the math which has generated the argument holds, and its > reversal should be considered, since otherwise the conservation of > energy will be destroyed. Of course the energy hasn't disappeared! Momentum isn't energy, energy isn't momentum! They're not the same thing! Seriously, it's time to do the Newtonian calculation (at v=c, or v=c/2 if you prefer). > > > 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? > > Look Timo, this is hardly the right place to be asking this question. > You are the one who set up the computations above, and we've > considered a one second pulse of laser light. This is 1.3kJ that we've > converted to momentum through e=hv and e=mcc. That you criticize this > figuring is to say that you are critical of existing theory. Why claim that I criticise this? It's correct, and I said so (apart from your calculation error from typo). > > 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? > > Yes, at the speed c. You seem to think that the math is a one-way > framework. If this is the case then this is perhaps how we come to > disagreement. Why don't you answer this question yourself. Clearly we > fed the equations all of this energy. If we go back through them we > will get 1300 watts on the other side. Yes, do it the other way. The energy is still all there. We haven't magicced the energy into becoming momentum; momentum and energy are two different things. In Newtonian mechanics, do you get rid of kinetic energy (1/2)mv^2 by calculating the momentum mv? You get a different number, with different units. One can be small, and the other large. Do you think this spells trouble for conservation of energy? > > 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'm feeling more secure that the term 'radiation pressure' is a > misnomer. > We've just derived it as photon momentum, so what is the difference? Force is rate of transfer of momentum. Newton 2 links force and momentum together. If light carries momentum, it must be able to exert force. If light exerts force, it must carry momentum. This would have been the clear and evident result from the redshift/ blueshift thing by now (hopefully), but you've avoided this so far. From conservation of energy, the work-energy theorem, and Galileian invariance, out comes radiation force and momentum. > One is distributed over a specific area whereas the other is a > conglomeration. This is nearly a quip on the photon itself. Mostly I'm > very interested in this topic, and not interested in studying it via a > mimic's approach. You are great to spend so much time on this. I'm > working on a new method, but I don't have anything substantial to > share yet. There are a number of places that I see frailties in > existing theory, and herabouts they are tied together. |