Taking a Fresh Look at the Physics of Radiometers.
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From: NoEinstein on 9 Jun 2010 17:18 On Jun 8, 7:08 pm, "Sue..." <suzysewns...(a)yahoo.com.au> wrote: > Dear Sue: There is no... wrestling with this fact: Light is composed of trains of photons (energy) which need no medium for their propagation. The ether nurtures the light on its way, never drags it! NE > > On Jun 8, 5:36 pm, Timo Nieminen <t...(a)physics.uq.edu.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. > > 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. > The gas in the traversed volume exerts the force. > > Its a minor point for those that already > understand how the media behaves for > sound and light and want to unleash > the power of statistics effectively. > I am sure it is clever to the nth degree. > > For those still wrestling with Newton's > corpuscle, as a viable propagation model, > your version might be a little misleading. > > Sue... > > http://en.wikipedia.org/wiki/Radiation_pressure
From: NoEinstein on 9 Jun 2010 17:27 On Jun 9, 3:40 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > Dear Timo: You are arguing through your hat. Photons are MASS-less, and thus have no momentum. Gamma rays are so energetic that they can give off at least one photon each. Gamma rays are similar to photons, but they have both mass and momentum. NoEinstein > > 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.- Hide quoted text - > > - Show quoted text -
From: Sue... on 9 Jun 2010 17:55 On Jun 9, 2:20 pm, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > 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.or... > > > > > 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. The light is *all* angular momentum until charges in the gas convert it through interaction. Antenna are necessary for any directed energy. (With a minor exception) <<for a highly relativistic charge the radiation is emitted in a narrow cone whose axis is aligned along the direction of motion.>> http://farside.ph.utexas.edu/teaching/jk1/lectures/node33.html > As you say, we should take freedoms in describing these > things. Ella Fitzgerald may be going a bit far, The wiki references indicate the principle is the same for sound so Ella is not "too far" but specifically included. Anything that heats the media is what I understand for the Nichols device. >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. The term shadowing has a negative connotation from some dubious shielding experiments but yes there is mechanism for enhancement or attenuation along a gravitational path so the "shadowing" label seems to stick. It is not and bad as a big red "A" on your blouse and it might even be a source of pride when Mercury and Hulse-Taylor are considered. Sue... > > - Tim
From: Sue... on 9 Jun 2010 19:36 On Jun 9, 4:17 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > 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. I would certainly hope the second case (photon) exploits it because it has a familiar equation. The other two I am not ready to buy into. The classical particle is not at issue. any billiard player can attest to that one. The classical wave has a term "Rate of increase of inertia" that is very troubling, especially to one who harbours notions that inertia and gravity is not fundamental. > > > 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.) My view is that atoms in the path-dielectric respond to illumination with torque and lots of them doing that will try to increase their volume. Even more so if they are illuminated twice due to a reflector in the neighbourhood. > > > 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 I have seen optical tweezers descriptions both as induced dipoles and directed light momentum so that doesn't get you a slam-dunk. (nor does it get the publisher any of my mad-money) I certainly would not scale it up to a solar-sail or even Nichol's radiometer. The volume of the path is a factor in those case. If you have the paper, It might be interesting to see if the traversed volume is considered. If "both radiation forces and gas forces separately" it seems it would be. It might appear as some region relative to the beam-waist rather than traversed-volume but it would be comforting to the numerologists of the world to see the factor 1/3 still applies at small scales to know we are really discussing the same effect. Sue... http://en.wikipedia.org/wiki/Boltzmann_constant http://en.wikipedia.org/wiki/Radiation_pressure
From: NoEinstein on 9 Jun 2010 21:37
On Jun 9, 4:25 pm, BURT <macromi...(a)yahoo.com> wrote: > Dear Burt: If you defined your terms, there is enough in your reply to bluff some science. By why bluff? Gravity is: "Ether flow toward the Earth (or between objects in space) that is replenished by photon or charged particle exchange between the masses." I can say in one sentence what it has taken Einstein over a century just to try to explain. "Theories a true in inverse proportion to the time required to explain them." So, your reply comes up short. NoEinstein > > On Jun 6, 2:04 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > On Jun 5, 8:00 pm, BURT <macromi...(a)yahoo.com> wrote: > > > Dear Burt: Where did you get the notion that circular orbits have no > > gravity? If that were so, then, how are those telecommunications > > satellites held in orbit? I've got gravity nailed as: Flowing ether, > > replenished by photon exchange. Nothing that you've ever said changes > > those facts. NE > > There is a round curve of gravity for energy in a circular orbit. But > there is no strength of gravity to change the motion circular speed. > The strength of gravity does not lie in the curve but in space flow. A > circular orbit has zero gravity strength but a pre speed. You can > quantify the prespeed in space for the circular orbit. Pre-speed is > the motion through space independant of the strength of gravity > pushing it faster or slower. Gravity gives and takes from pre-motion > of falling energy in elliptical orbit > > MItch Raemsch |