Taking a Fresh Look at the Physics of Radiometers.
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From: Tim BandTech.com on 9 Jun 2010 14:20 On Jun 7, 8:58 pm, "Sue..." <suzysewns...(a)yahoo.com.au> wrote: > 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_radiometerhttp://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. Well, if the photon momentum is taken to be coming from the entire photon energy, then there is no room for the rotational quality to contain more energy. This is the logical trouble with the existing theory. As you say, we should take freedoms in describing these things. Ella Fitzgerald may be going a bit far, but as you take interest in gravitation there is room for a photon relationship with gravitation to provide gravitational shadowing, which could then provide the dark matter resolution. - Tim
From: Timo Nieminen on 9 Jun 2010 15:40 On Jun 10, 4:10 am, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > On Jun 8, 3:00 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > > On Jun 8, 9:34 pm, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > > > > On Jun 8, 2:30 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > > 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.) > > > > Is this your new careful use of language Timo? This 'no' is to mean > > > that you wish to discuss the reflected case. Well, I am discussing the > > > absorbed case. So please do not change the argument as a means of > > > falsifying my argument. > > > So, be clear! We have been discussing the case of reflection as well, > > not just absorption. If you're restricting yourself here to absorption > > only, say so. > > > In the reflecting case, there can be a transfer of momentum with no > > transfer of energy. You can have transfer of momentum without > > absorbing the photon. If you absorb the photon, yes, you transfer all > > of the momentum and energy. Most of the energy (all of the energy, for > > a stationary absorber) _doesn't_ become work done on the absorber. > > > This is the next step (2a,b) in the other line of this thread, let us > > deal with it there. > > > > This 'No' above can be miscontrued. I will > > > take this wording as support of my argument, for there is no actual > > > falsification content provided here. > > > That's very sloppy. A lack of falsification is not support (for > > example, a complete lack of comment is also a complete lack of > > support). > > > > > Bounce a ball off a wall. KE_in = KE_out. No loss of KE. Change in > > > > momentum = 2 * momentum_in. > > I'm almost willing to accept this criticism, but I see that you are > obfuscating the situation more than I am. I asked a simple question > within a position on photon momentum and photon energy, and you > invalidate the question, and then go on to say that I haven't been > clear. > > Here at least we were discussing one simple point, rather than > branching off into three or more methods. It is a point worth > considering, since the computation that granted the photon momentum > took its entire energy e=hf, or at least this is my best > understanding. It then follows that when that photon is absorbed, that > its entire energy is absorbed. This is merely the reversal of the > equations which got us the momentum in the first place, and if we did > not recover all of the energy, then we have a break with energy > conservation. But using the standard theory for photon momentum or radiation momentum, if you know the photon energy, you know the photon momentum. If you know the wave energy flux and speed, you know the momentum flux. If you know the photon momentum, you know the photon energy; if you know the wave momentum flux and speed, you know the wave energy flux. But the energy isn't momentum, and you can change the momentum - that is, exert a force - without changing the energy. They are _different_. More than that, if we use the energy to calculate the momentum, this is just a _calculation_ that _we're_ doing. This doesn't affect the energy, doesn't affect the momentum. The momentum is there even if we don't calculate it, and the energy is there after we calculate it. We aren't turning energy into momentum, or momentum into energy - to be worried that we wouldn't "recover" the energy and violate conservation of energy is meaningless in this context; we're not doing _anything+ to the energy in the _calculation_. If the photon/wave is reflected or absorbed, then the reflection or absorption can do something to the energy. And as you can see from the posted calculation, we use conservation of energy to calculate what happens to the energy, if anything. > That this balance of energy is heat suggests within a cannonball model > (which you are so fond of) that an incredible amount of that directed > energy, all of which was used to compute a momentum, has turned to > heat, with such a slight amount becoming kinetic as to be extremely > difficult to notice. If we had, say, a large lead ball hanging on a > string in stasis, and sent a tiny steel ball into the lead ball at > high velocity then we would observe some heating, but too achieve the > level of heating that light achieves will be quite some trick to mimic > in the terms of massive collisions. Not at all. Wikipedia tells me that the energy of a typical 5.56mm NATO bullet when fired is 1.7kJ. Shoot them into a massive target at a little under 1 round per second, and you achieve approximately the same heating. (Not the same force as with light! Just the same heating.) > The fact that much of this energy > could have been turned into electricity as well poses an even more > intricate model. Likewise, that we might have chosen to reflect the > energy with so little interaction goes in complete opposition to this > idea of the absorber's collision, and the fact that matter varying > from a sooted up plate to a polished plate could affect such a change > in behavior requires much more dynamics. Which means what? Electromagnetically, the polished plate is a good conductor, and presents a very high impedance mis-match to the wave/ photon - high reflectivity results. The sooted plate is very different electromagnetically. High enough conductivity to give high losses, but much smaller impedance change, so not so reflective. We can model reflection from metal or soot reasonably well (at least for isolated bits of soot - there are some practical difficulties for a real sooty surface, but an ideal flat smooth sheet of soot would be simple). If you care about the dynamics, it can be described, electromagnetically. In either case, you can find the Lorentz force acting on the induced currents in the material, and the total force is equal to the radiation pressure. So, we have a choice, we can either use the details of the dynamics, or use conservation of energy and momentum, and we get the same answer. And to comment on something in your reply to Sue: "Well, if the photon momentum is taken to be coming from the entire photon energy, then there is no room for the rotational quality to contain more energy." We're not turning the energy into the momentum, calculating the momentum doesn't get rid of the energy, the energy is still there. It hasn't been used up and become unavailable for turning into angular momentum - especially since we don't turn the energy into angular momentum any more than we turn it into momentum. Take a pulse of light with energy A, momentum B, angular momentum C, and combine it with a pulse of energy A, momentum B, angular momentum - C, the final pulse has a total energy of 2A, total momentum 2B, and total angular momentum 0. The final absence of angular momentum doesn't change the energy at all.
From: Timo Nieminen on 9 Jun 2010 16:17 On Jun 9, 9:08 am, "Sue..." <suzysewns...(a)yahoo.com.au> wrote: > > > Anyway, this is the simple theoretical framework. The key result, > > 1b(iii), is sufficient to show that a wave has momentum, which is > > sufficient to show that it will exert a force if reflected, refracted, > > or absorbed. 2a,b show that this is all consistent with absorption of > > energy and heating. > > I am all for exploiting an atomic oscillator's > ability to accurately meter out chunks of energy > wherever it is helpful. And your method may indeed > be helpful to those who live and breathe quantum > mechanics and statistics. Doesn't exploit it, doesn't depend on it. If you could provide a purely classical wave, 1b(iii) would still work perfectly well. Light, radio, sound, elastic waves, if available as a purely classical continuous wave, would in principle work fine. That it also works for quantised waves is nothing more then the correspondence principle, that quantum mechanics gives us what we're used to getting, in the appropriate limit. > Qualitatively, I can't agree with your statement: > > "1b(iii), is sufficient to show > that a wave has momentum, which is sufficient > to show that it will exert a force if reflected, > refracted, or absorbed." > > I would find some words you only alluded to > and conclude differently. > > The wave has angular momentum. Can be true (isn't necessarily true), but irrelevant to radiation forces. (Not, of course, to radiation torques.) > The gas in the traversed volume exerts the force. See both radiation forces and gas forces separately, with sufficient time resolution, 'nuff said! http://www.sciencemag.org/cgi/content/abstract/science.1189403
From: NoEinstein on 9 Jun 2010 16:18 On Jun 8, 7:32 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > Dear Timo: Photons do not exert pressure, period. Mass-less energy never has an associated force. NE > > On Jun 8, 9:09 pm, "Sue..." <suzysewns...(a)yahoo.com.au> wrote: > > > On Jun 8, 6:31 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > (Not 1/3 since not omnidirectional.) > > > I am interpreting the 1/3 to be an allowance for > > the fact that the pyramidal volumes are > > only confined at the ends. Areas > > of 1/9 are easly seen in this figure. > > You'll have to imagine the lines > > for 1/3. > > >http://hyperphysics.phy-astr.gsu.edu/hbase/forces/isq.html > > > Risking accusation of practising numerology > > without a licence, the factor 1/3 appears > > prominently here: > > >http://en.wikipedia.org/wiki/Boltzmann_constant > > Sure, the atoms in a gas are moving omni-directionally. Compare with > pressure due to a directed stream of particles. Geometry doesn't care > about the fine details, such as whether we're talking about atoms or > photons, 1/3 will do in either case. (Exercise: what is the factor > equivalent ot the 1/3 for an N-dimensional space?)- Hide quoted text - > > - Show quoted text -
From: NoEinstein on 9 Jun 2010 16:23
On Jun 8, 1:19 pm, Sam Wormley <sworml...(a)gmail.com> wrote: > Dear Sam: I've disproved SR. There is no such thing as space-time variance due to velocity or nearness to mass. You should make your own post on... relativity-stupidity; you should get some good conversations, there. NE > > On 6/8/10 3:50 AM, k...(a)nventure.com wrote: > > > 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. > > And just how do you see the "twin paradox" as a problem for > special relativity? Be specific. |