Taking a Fresh Look at the Physics of Radiometers.
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From: Sue... on 7 Jun 2010 20:58 On Jun 7, 6:42 pm, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > On Jun 7, 5:55 pm, "Sue..." <suzysewns...(a)yahoo.com.au> wrote: > > > On Jun 7, 7:41 am, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > > > Beyond this thermodynamics remains open in my book. > > > This is a bit disturbing at this point in the thread. > > Reynolds explained Crooks radiometer with > > ~thermodynamics~ . > > > Your discussion with Timo seems to be about > > Nichols radiometer. The distinction was > > made earlier in the thread but you will be > > talking past one another if you are not > > about the same device and effect. > > Thanks for the attempt at resolution, but we are on to other aspects > now. > He seems to think that when you attribute all of a photons energy to > momentum that somehow the energy is independent of that momentum. I'm > not going to buy that, but if you can falsify either of us then I > think that input is very welcome. I'll take your side because radiation~suction fits an induction gravity mechanism better. (Timo can conscript a few students if he thinks we are ganging up on him.) > We're really not beyond anything > more than e = h f , e = m c c , > and such simple product relationships. I don't see how e = hf applies where there may be no atomic absorption. <<The requirements of energy and momentum conservation generally forbid the absorption of photons by free carriers, and the process can only take place by interband transitions or with the assistance of phonon absorption or emission. >> http://www.colin-baxter.com/academic/research/downloads/prl063802.pdf > http://en.wikipedia.org/wiki/Nichols_radiometer http://en.wikipedia.org/wiki/Crookes_radiometer > > > Apologies if I am covering old ground but > > it is a long thread and I got here late. > > Hey Sue, no problem. I guess one of the key points is that the > radiometer itself is not quite what most of the discussion is about. > Isolation of radiation pressure from the radiation is more like it. > What I now understand and had overlooked for much of the thread is > that the radiation pressure is merely the photon momentum, as is > overlooked at > http://en.wikipedia.org/wiki/Radiation_pressure > and likely elsewhere. Nichols work is here: > http://books.google.com/books?id=8n8OAAAAIAAJ&pg=RA5-PA329#v=onepage&... <<Theory It may be shown by electromagnetic theory, by quantum theory, or by thermodynamics, making no assumptions as to the nature of the radiation, that the pressure against a surface exposed in a space traversed by radiation uniformly in all directions is equal to one third of the total radiant energy per unit volume within that space.>> http://en.wikipedia.org/wiki/Radiation_pressure Hmmm... 1/3 is a pretty nice number and the statement is very inverse square-ish. Is it too late to switch to Timo's team? I am a sore looser. I see the traversed volume here: http://hyperphysics.phy-astr.gsu.edu/hbase/forces/isq.html But all of it, half of it or 1/3 of it is not doing anything for me unless we put some gas in it and give it a temperature. Yeah... That's the steam. Eggs explode in my microwave oven because of radiation pressure. > which is actually linked to in that wiki you just gave. > Some if his argumentation is quite poor imo. There is a 1933 paper by > a woman Bell that I do not have access to which claims to resolve the > study down to 10E-6 torr. I posted that link a few days ago here. In fairness, I should read about Timo's light-bullets a bit closer before we declare victory. Photons (Phonons?) can be a pretty good model translating angular momentum in a dielectric. I am becoming sceptical however because acoustic radiation pressure is lumped in, apparently as the same effect. If it is just molecules in the traversed volume jiggling more, induction gravity should be unscathed and it really shouldn't matter how you describe the heating process. light bullets, flaming arrows, or Ella Fitzgerald on Memorex. Sue... > > - Tim
From: Tim BandTech.com on 7 Jun 2010 23:38 On Jun 7, 3:55 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > 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. Newton probably had a deep appreciation for light. He could tell there was energy there as the sun warmed things. Still, how much energy was probably quite an open puzzle. It would be possible to put a lower bound on the energy by observing the heating, but was this all of the energy? No, says modern theory, there is in addition to energy of the photon itself a radiation pressure capable of providing a minor propulsion to matter in addition to its radiant energy. Is this an accurate portrayal of the modern theory? This I will argue is a twisted interpretation, especially based on the mathematics that you requested, and which until I did it had no idea that radiation pressure actually is photon momentum, which is the entire theoretical photon energy, and hence my claim of twistedness. > > > > > 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. No, not at all. As I just stated in the 'twisted' paragraph it is that all of these figures are the same energy. To represent them as distinct things is breaking down under our careful analysis, which from my side was to attempt to isolate them. > > 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? Alright, I guess I'll have to entertain you here. K = m v v / 2 : kinetic energy e(photon) = K : all of photon's energy kinetic v(photon) = c : velocity of photon p(photon) = m v = 2 K / c = 2 h f / c . I guess this is what you are talking about. I have no idea why you are pressing for this. > > 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? The e = h f equation is not fundamental is it? It is derived. If we were to simply suppose that each photon carried half the energy of this claim then we would merely need to double the quantity of photons to have an equivalent theory... up to a point. I'm not feeling solid about this argument but you are asking me to think about it, so there you have it. Newton did plenty with light, but I don't really know about any Newtonian photon, or energy computation on light. Here is some solid Newtonian thinking: "RULE 1 We are to admit no more causes of natural things than such as are both true and sufficient to explain their appearances. To this purpose the philosophers say that Nature does nothing in vain, and more is in vain when less will serve; for Nature is pleased with simplicity, and affects not the pomp of superfluous causes." - Isaac Newton, Principia Mathematica; System Of The World; Rules Of Reasoning In Philosophy; translation by Andrew Motte > > 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. I'm not following any larger argument on this information. I see that you claim an equivalence between light and massive objects, but then that there are discrepancies too. > > > 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. Actually we've just recently exposed that radiation pressure is photon momentum in the e=hf, e=mcc sense. > > (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). Yes, I'll go along with this, except to expose that by not reversing the equations we are getting a figure that is supposedly experimentally verifiable (the 4.6E-6 figure on 1370 Watts), and so in some regards my earlier question about when the relativistic conversions end might be important. > > (3) Conservation of energy and momentum gives a prediction of > radiation pressure. Nah. Photon momentum is radiation pressure. Radiation pressure is a misnomer. > > (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. I'd like to understand this but don't yet. Back to Maxwell's Treatise another day... > > 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! When the momentum is absorbed is not the energy likewise absorbed? > > Seriously, it's time to do the Newtonian calculation (at v=c, or v=c/2 > if you prefer). I think I did this above, but I'm not quite sure what you are getting at. So lead me on some more up there and I'll try to be a good dog. > > > > 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. Well, to me this is becoming the most interesting part of the analysis. On the one hand the math we used to generate the photon momentum takes into account all of the photon's energy, yet upon absorbing the photon we do not see this come out as mechanical energy. This math actually does match experiment as you substantiate elsewhere, and so we should accept this one way paradigm, but this is not a completed theory then, for no discussion like this has been undertaken. To reverse the math we simply admit that when a photon hits a cold black surface that it's energy is transferred into that surface. Knowing the momentum of the photon we simple backtrack the math and get a small energy figure, and based on our earlier figures we will find 1300 watts of energy was absorbed. Because we put all of this energy in as momentum it follows that within the photon this became kinetic energy, and that the surface therefor receives 1300 watts of mechanical work. This is where you should falsify I believe. I was just reading a wiki page on momentum and they claim that static electric fields have momentum. Do you balk at this? At this point I have no idea what to really believe. This is all a wake up call for me. I am so confused as to why you would deny any connection between momentum and energy, and it is clear that to uphold existing theory we cannot admit the 1300 watts of mechanical work, otherwise we'll really have high power solar mechanics going on, yet this is what the math would do. There must be a glitch imo. > > 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? No. You're speaking with a lack of concession for the fact that when the momentum rises so does the kinetic energy. Instead of admitting this you say: "You get a different number, with different units. One can be small, and the other large." Now if you added the concession that I recommend then you refutation here withers. > > > 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 Well Timo, one thing at a time, perhaps two. I am sorry I have been so rude, but I will be happy to take up the blueshift argument sometime soon. After enough cycles of repetitious banter we should not engage in more of the same, unless there is some avenue to change it up a little. Well, I have suggested such an avenue, but I don't honestly have a mechanism yet. To fully go here I think we'll need to break e=hf into two parts; one rotational and one directed. Still, I am not up to this yet, and obviously it won't go far with you. Maybe we could drop this for now and go to the blue shift argument, which I can pick up right where you left off. Still, if you have something new to add on to this excessively repetitious and long post then please do, but really if we're both just repeating ourselves we should call a stalemate; temporarily at least. I do understand that momentum and kinetic energy are not identical, but that they are related. Yes, they are different, but not so terribly different. And yet experiment agrees with you, but the theory does not; not when we attempt to extract the 1300 watts as mechanical energy. In a few decades we may have a nanotechnology that will do this. First for coherent light, and then if we are lucky for the noisy sunlight. Even a 50% efficient mechanical or electrical figure would be fine. - Tim > > 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.
From: Timo Nieminen on 8 Jun 2010 02:30 On Mon, 7 Jun 2010, Tim BandTech.com wrote: > On Jun 7, 3:55 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: [a very quick point, will return later] > > Of course the energy hasn't disappeared! Momentum isn't energy, energy > > isn't momentum! They're not the same thing! > > When the momentum is absorbed is not the energy likewise absorbed? No! (I think "absorbed" is the wrong word here.) Bounce a ball off a wall. KE_in = KE_out. No loss of KE. Change in momentum = 2 * momentum_in.
From: kado on 8 Jun 2010 04:50 On Jun 7, 1:59 pm, Sam Wormley <sworml...(a)gmail.com> wrote: > > > Relativity theory is self consistent and has no contradictions. > In fact, there has yet to be an observation that contradicts a > prediction of relativity. > > While Einstein's Special Relativity is consistent in itself, it is not without its problems. And your second sentence is wrong. The Eotvos experiment by R. H. Dicke at Princeton demonstrated that mass is not relativistic. Furthermore, there is the twin paradox. Now don't give me all the BS that the relativistists use to try to resolve this paradox. Just tell me what twin is younger than the other when they meet after the traveling twin has completed his/her trip and both are at rest in what Einstein called the 'stationary system' on page 40 of "The Principle of Relativity" copyrighted by Lorentz, Einstein, Minkowski, and Weyl. Moreover, there is the paradox of rapidly rotating discs (like a Frisbee) moving near the speed of light, wherein the physical form must constantly change due to the tensor mechanics. And these physcal changes must occur without generating heat, or violate the Laws of Thermodynamics. D.Y.Kadoshima
From: kado on 8 Jun 2010 04:58
On Jun 7, 1:04 pm, Sam Wormley <sworml...(a)gmail.com> wrote: > > From the quantum mechanical perspective, > > 1. photons are emitted (by charged particles) > 2. photons propagate at c > 3. photons are absorbed (by charged particles) > > Photon momentum > p = hν/c = h/λ > > Photon Energy > E = hν I would like to point out an inconsistency in your post about the photon momentum Momentum is a mathematically calculated dynamic property of an uniformly moving mass, i.e., momentum is mass times velocity by definition. The photon is commonly given as a 'massless' quality, i.e. a photon has no mass. Thus any idea of photon momentum is an oxymoron. Please do not think that I'm picking on you. I just choose to respond to errors or inconsistencies your posts, and choose decline to respond to the pure BS of others D.Y. Kadoshima |