Taking a Fresh Look at the Physics of Radiometers.
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From: NoEinstein on 11 Jun 2010 23:26 On Jun 11, 2:24 pm, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > Dear Tim: Light isn't different things determined by "wave length" (a misnomer). The determining factors of the energy of the light is how closely the standard size photons are spaced apart, and how many trains of photons are arriving. The materials hit by light respond to the frequency of the impacts, at valance levels within the atoms which correspond to each particular frequency. The photons which get "reflected" (re emitted) are never the same photons. There are no sine curves relative to light, because light isn't waves. It's just high speed tangles of ether too small to be able to give off photons. Gamma rays started out like photons. But because of the very high energies of emission, tended to gain additional ether in transit, until the mass got great enough to emit occasional photons. Gamma rays straddle the fence between energy and mass. I laugh hearing that some suppose photons (mass-less) carry additional energy in their spin. If anyone thinks so, they must have better measuring devices than I know about. In addition to the hundreds of descriptions of my New Science, here, I have several dozen, as yet unpublished, essays on the physics of light. There are no equations. I'll leave those to others to write. NoEinstein > > On Jun 11, 1:53 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > On Jun 11, 7:38 am, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > > > Dear Tim: The 'spin' of a NATO (or other) bullet isn't some huge > > additional momentum. If the KICK of the gun is 30 pounds, and the > > inertia of the gun is 10 pounds (the static weight), then the total KE > > is just 40 pounds, including the angular "KE" of the spinning bullet. > > That spin is produced by sacrificing some of the forward momentum. > > The effect is to rotate the gun in the direction opposite to the spin > > of the bullet. It isn't the KE nor the momentum of the bullet causing > > the penetration as much as the REDUCTION in the strengths of the > > materials that are hitwhich don't "like" rapidly applied loads, nor > > loads of more than one stress type at the same time. NoEinstein > > > > On Jun 10, 10:31 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > > On Thu, 10 Jun 2010, Tim BandTech.com wrote: > > > > > On Jun 10, 3:19 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > > > On Jun 10, 9:43 pm, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > > > > > > > On Jun 9, 3:40 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > > > > > On Jun 10, 4:10 am, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > > > > > > > > > > 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.) > > > > > > > > Here again I see your obfuscatory tactic. Firstly you falsify and in > > > > > > > your conclusion you agree. > > > > > > > You were quite specific: "level of heating that light achieves will be > > > > > > quite some trick to mimic". This level of heating is easy to mimic. > > > > > > > > The amount of heating that light is capable > > > > > > > of when absorbed versus the work that can be done mechanically due to > > > > > > > that absorption are remarkable in comparison to your NATO bullet. > > > > > > > So, you want to change "level of heating" to "amount of heating versus > > > > > > work"? If you meant this in the first place, you weren't clear enough. > > > > > > To criticise my reply to your actual original words on the basis of > > > > > > your _changed_ version lies somewhere on the scale from weaseling to > > > > > > complete bullshit. > > > > > > I'll have to own here that I should have used the word 'relative' > > > > > within the context, but I see it is fairly easy to interperet since > > > > > the context of the whole argument is still present. Hell, I can match > > > > > the heat of sunlight rubbing some steel on a rock. The lead of the > > > > > bullet will melt on impact. Your interpretation of my writing is > > > > > clearly not coherent at many levels. > > > > > Your original argument wasn't coherent, then. As far as I could tell, you > > > > thought it improper that almost all of the energy should go into heating, > > > > not work. > > > > Jeeze. I pity any onlooker who attempts to decode your own position. > > > Now, via double crossing, I must admit that I do agree with your > > > statement above and that this is exactly what I've been discussing all > > > along. > > > > > We see the same thing in Newtonian mechanics for a light object in > > > > inelastic collision with a heavy object. Qualitatively, the same type of > > > > thing, most of the energy going into heating. > > > > It is a peculiar thing to discuss an inelastic collision this way. > > > There are more dynamics here and the options as to what happens with > > > the energy are numerous. For instance, if we could design a spring > > > with a ratchet to capture the projectile then we would only heat by > > > the inefficiency of the spring device, rather than by the objects > > > entire kinetic energy. Likewise as I stated before, this capture > > > mechanism could as well generate electricity, and of course this is > > > the quest of many right now; to capture electricity from the sun, and > > > to do it at a better margin than 20%. We cannot grant the heat clause > > > unconditionally, and instead want the mechanism of that heating. It is > > > not so difficult to see electron disturbance; literally microcurrents > > > flowing in short circuit fashion which provide the heating effect. > > > That these charge concerns are so nearby to the Nichols device is a > > > caveat on the operation of that device. That your light traps are so > > > behaved is likewise a caveat. Some would like to free these electrons > > > to take a larger path. Others would like to ignore them for their > > > inconvenient truth. > > > > > > The context of the discussion for me revolves around the photon energy > > > > > and how we can come to attribute the photon momentum to the photon > > > > > energy without concern for such things as angular momentum. > > > > > We're not attributing the photon momentum to the photon energy. > > > > > For a moving bullet, does the kinetic energy cause the momentum? Does the > > > > momentum cause the kinetic energy? Do we attribute one to the other? > > > > > For the photon, consider a spin +1 photon. How much of its energy is "in" > > > > the angular momentum? > > > > > Be specific: consider a 500nm photon, with hbar angular momentum. We can > > > > write down 3 numbers: its energy, its momentum, and its angular momentum. > > > > How are these 3 related to each other, showing appropriate concern for > > > > angular momentum? > > > > > How about for a 5Hz photon? Do the same. Does this mean that a 5Hz photon > > > > shouldn't have any momentum? > > > > > The ratios or values of energy, momentum, and angular momentum, as > > > > commonly stated for photons, come straight out of classical > > > > electromagnetic theory. Are these ratios or values wrong? Yes or no, no > > > > handwaving, no waffle, just a straight answer. > > > > From what I've read of the quantum stuff the angular momentum is not > > > so easy to observe. They even go so far as to grant a photon an > > > orbital angular momentum, though the orbital context is not present in > > > freely propagating light. And yet the light as an oscillation is still > > > a coherent concept, and so the rotation momentum that I am considering > > > is more like your NATO bullet spinning. > > > > As I recall the 5Hz photon will have the same quantum angular momentum > > > than will the 500nm photon, but will have less total energy via e=hf. > > > > As you seek a straight answer I must ask you how crooked is the path > > > of modern theory? Yes, we'd all like a straight answer, yet we do not > > > actually have one yet. This is not to say that we should give up. We > > > seek a straight answer, but I must admit that I do not yet have one. > > > Simply eating particle/wave duality will deny a straight answer, but > > > beyond this it seems that in the accumulation the overlap of radiation > > > pressure with photon momentum has gotten lost in the shuffle. This is > > > a great topic, and so difficult to discuss that I believe you and I > > > both deserve credit, even while this discussion erodes. > > > > - Tim- Hide quoted text - > > > > - Show quoted text -- Hide quoted text - > > > > - Show quoted text - > > That seems alright to me. When we get into photons we're talking about > frequencies of > 6E14 Hertz > for Timo's 500 nm light source. If this is a rotational figure then > clearly we've got quite some difference to the NATO bullet. Now, I'll > admit that this crux of claiming a rotational component inherent to > the energy of the light is incomplete, but this is still a fairly > coherent way of thinking. Thinking on these terms nobody ever seems to > puzzle over the wavelength character of the photon too terribly much > beyond the spectroscopic aspects. It is a shame that the light > wavelength has not taken more of a literal interpretation than a > sinewave. Still, this is how I come to express that frequency and it > is pretty standard theory. I'm for breaking into this, but my position > is incomplete. I'm just trying to crack the thing open enough to fit a > wedge in. A shim is a wonderful thing, and the slight photon momentum > must be something like a shim itself. At least noone will quibble over > lights ability to shimmer. I don't reccommend shimmying up all the way > to the top of modern theory, for if the thing fails it will be a long > fall down. I'm sure you remember Humpty Dumpty, but that is another > allusion altogether. > > - Tim- Hide quoted text - > > - Show quoted text -
From: NoEinstein on 11 Jun 2010 23:34 On Jun 11, 4:17 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > Dear Timo: You know enough about physics to bluff with the best of them. But you lack sufficient focus to understand very many of the particulars. It should suffice to say that the energy IN must = the energy OUT. Figuring out what portion of the energy when where is madness. All high energy particle physics, like that fiasco in France, is concerned with "matter" which doesn't really matter! NoEinstein > > On Jun 11, 9:38 pm, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > > > On Jun 10, 10:31 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > On Thu, 10 Jun 2010, Tim BandTech.com wrote: > > > > On Jun 10, 3:19 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > > On Jun 10, 9:43 pm, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > > > > > > On Jun 9, 3:40 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > > > > On Jun 10, 4:10 am, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > > > > > > > > > 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.) > > > > > > > Here again I see your obfuscatory tactic. Firstly you falsify and in > > > > > > your conclusion you agree. > > > > > > You were quite specific: "level of heating that light achieves will be > > > > > quite some trick to mimic". This level of heating is easy to mimic. > > > > > > > The amount of heating that light is capable > > > > > > of when absorbed versus the work that can be done mechanically due to > > > > > > that absorption are remarkable in comparison to your NATO bullet. > > > > > > So, you want to change "level of heating" to "amount of heating versus > > > > > work"? If you meant this in the first place, you weren't clear enough. > > > > > To criticise my reply to your actual original words on the basis of > > > > > your _changed_ version lies somewhere on the scale from weaseling to > > > > > complete bullshit. > > > > > I'll have to own here that I should have used the word 'relative' > > > > within the context, but I see it is fairly easy to interperet since > > > > the context of the whole argument is still present. Hell, I can match > > > > the heat of sunlight rubbing some steel on a rock. The lead of the > > > > bullet will melt on impact. Your interpretation of my writing is > > > > clearly not coherent at many levels. > > > > Your original argument wasn't coherent, then. As far as I could tell, you > > > thought it improper that almost all of the energy should go into heating, > > > not work. > > > Jeeze. I pity any onlooker who attempts to decode your own position. > > Now, via double crossing, I must admit that I do agree with your > > statement above and that this is exactly what I've been discussing all > > along. > > So, you agree that "improper that almost all of the energy should go > into heating, > not work" is a fair summary of your original argument? > > Then what complaint do you have against the bullet example? In the > bullet example, almost all of the energy goes into heating. Surely > this is just as "improper". If it's OK for most of the KE of a bullet > to end up heating the target, rather than doing work, why is somehow > wrong for a photon? > > Note well that in both the bullet and photon cases, you get a force if > the bullet/photon is reflected with no loss of energy. It's simpler to > consider the reflection case, since you avoid the complications you > get into below. > > > > > > > > We see the same thing in Newtonian mechanics for a light object in > > > inelastic collision with a heavy object. Qualitatively, the same type of > > > thing, most of the energy going into heating. > > > It is a peculiar thing to discuss an inelastic collision this way. > > There are more dynamics here and the options as to what happens with > > the energy are numerous. For instance, if we could design a spring > > with a ratchet to capture the projectile then we would only heat by > > the inefficiency of the spring device, rather than by the objects > > entire kinetic energy. Likewise as I stated before, this capture > > mechanism could as well generate electricity, and of course this is > > the quest of many right now; to capture electricity from the sun, and > > to do it at a better margin than 20%. We cannot grant the heat clause > > unconditionally, and instead want the mechanism of that heating. It is > > not so difficult to see electron disturbance; literally microcurrents > > flowing in short circuit fashion which provide the heating effect. > > That these charge concerns are so nearby to the Nichols device is a > > caveat on the operation of that device. That your light traps are so > > behaved is likewise a caveat. Some would like to free these electrons > > to take a larger path. Others would like to ignore them for their > > inconvenient truth. > > Yes, we can explain the heating when an EM wave is absorbed in terms > of electrodynamics. Are you seeking to attribute the observed forces > to electron ejection? The atom trapping experiments show it isn't so, > very clearly. Given that the observed forces are predicted by > conventional theory to within the experimental error (i.e., better > than 10% in the less accurate experiments, 0.1% in good experiments), > it doesn't appear to be significant in the macroscopic experiments > either. Does the magnetic force on a current-carrying wire depend on > electron ejection? If not, why should the forces due to an EM wave? > > > > > > > > > The context of the discussion for me revolves around the photon energy > > > > and how we can come to attribute the photon momentum to the photon > > > > energy without concern for such things as angular momentum. > > > > We're not attributing the photon momentum to the photon energy. > > > > For a moving bullet, does the kinetic energy cause the momentum? Does the > > > momentum cause the kinetic energy? Do we attribute one to the other? > > > > For the photon, consider a spin +1 photon. How much of its energy is "in" > > > the angular momentum? > > > > Be specific: consider a 500nm photon, with hbar angular momentum. We can > > > write down 3 numbers: its energy, its momentum, and its angular momentum. > > > How are these 3 related to each other, showing appropriate concern for > > > angular momentum? > > > > How about for a 5Hz photon? Do the same. Does this mean that a 5Hz photon > > > shouldn't have any momentum? > > > From what I've read of the quantum stuff the angular momentum is not > > so easy to observe. They even go so far as to grant a photon an > > orbital angular momentum, though the orbital context is not present in > > freely propagating light. And yet the light as an oscillation is still > > a coherent concept, and so the rotation momentum that I am considering > > is more like your NATO bullet spinning. > > > As I recall the 5Hz photon will have the same quantum angular momentum > > than will the 500nm photon, but will have less total energy via e=hf. > > Keep it simple,. don't worry about orbital angular momentum. If you > can't answer the simple question, why make it more complicated. > > Don't get sidetracked into pretty pictures of photons, either, or long > philosophical tracts. Answer the simple question first. > > If the momentum and/or angular momentum are attributed to the energy, > how much of each, in each case, for the 500nm photon and the 5Hz > photon? Angular momentum hbar, E=hf, p=h/lambda. Write down the > numbers, and say how much of the energy is "in" the angular momentum > (or whatever you mean, when you say that having angular momentum means > that less energy must be "available" for the momentum). > > (There are 2 potential nuggets to be had here, if you bother to do > it.) > > [moved] > > > > The ratios or values of energy, momentum, and angular momentum, as > > > commonly stated for photons, come straight out of classical > > > electromagnetic theory. Are these ratios or values wrong? Yes or no, no > > > handwaving, no waffle, just a straight answer. > > > As you seek a straight answer I must ask you how crooked is the path > > of modern theory? Yes, we'd all like a straight answer, yet we do not > > actually have one yet. This is not to say that we should give up. We > > seek a straight answer, but I must admit that I do not yet have one. > > If you can't say straight-out that it's right or wrong, from what > information are you arguing? > > It's simple in a way. We can, directly from Maxwell's equations, by > finding the induced currents and dielectric polarisation, find the > force acting on the current and polarisation, via the Lorentz force, > and find the work done on the current by the field (which is the > heating). > > > beyond this it seems that in the accumulation the overlap of radiation > > pressure with photon momentum has gotten lost in the shuffle. > > No. Or if it did, stop shuffling it. It isn't an overlap, one is the > rate of change of the other. > > For a beam of light, the radiation force is the change in the momentum > flux when the beam is reflected, refracted, or absorbed. For a photon, > the impulse imparted to a reflector, refractor, or absorber, is the > change in the photon momentum when it is reflected, refracted, or > absorbed. > > If light has momentum, there must be a force when this momentum > changes. If light can exert a force, it must have momentum. The simple > versions of the derivation of radiation pressure make use of this > explictly, which isn't what I would call "getting lost". > > This is just Newton's laws of motion, just a statement of the > conservation of momentum. (It isn't a full application of Newtonian > mechanics, with Newton's momentum = density * volume * velocity, but > Newton's laws of motion are fine.) > > Newton 2 works for radiation, for waves, for photons. Force is the > rate of change of momentum, imoulse is the change in momentum. This > isn't lost. If you prefer to think about photon momentum or wave > momentum instead of force or radiation pressure, go ahead. But don't > forget Newton 2!- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -
From: NoEinstein on 11 Jun 2010 23:45 On Jun 11, 5:35 pm, "Sue..." <suzysewns...(a)yahoo.com.au> wrote: > Dear Sue: Are... your solar sails white or black? If they are white there should be no thrust, just like the Crooke Radiometer doesn't rotate with the white side trailing. If they are black, the heating "may" find enough atoms to kick back to cause a thrust. The best argument against photon thrust is the fact that all that solar energy doesn't send the planet Mercury flying away. Photon exchange allows the ether envelope to have more pressure on the opposed sides than on the facing sides. That's why objects in space will attract, gravitationally. But note: There can be no gravity between objects if the ether is discontinuous, as it generally is between galaxies. NoEinstein > > On Jun 11, 7:07 am, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > > > On Jun 9, 5:55 pm, "Sue..." <suzysewns...(a)yahoo.com.au> wrote: > > > > 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.org/wiki/Boltzmann_constant > > > > > > > > > > > > > > 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. > > ============== > > > > > Hmmm... I thought by going to 10E-6 torrs that the temperature effects > > were being diminished. Likewise sound does not travel in a vacuum- > > although this very word 'vacuum' is pretty dubious. As soon as you put > > anything in a vacuum it cannot be a vacuum anymore. So as soon as we > > plop down a Nichols device within the vacuum we ought perhaps to state > > that the Nichols device has been isolated, rather than use the vacuum > > terminology. Still, to what degree is the isolation perfect? Just tap > > the lab table and you've got your sound transmission again. > > Metallic vanes and glass surfaces can spoil a vacuum too. > > I don't know who went there and measured it but I have > some information that if you want vacuum better than > 1 atom per cc you have to visit another universe. > > But let's say for the sake of argument that you could > sweep ALL the gas out of a Nichols device. The electrons > in the vanes and the electons in the glass will still > jiggle from illumination and a few will still try > to fill the space with even greater repulsion from > the protons in their host matter. > > Electron valves (vacuum tubes) don't malfunction > if we take too much gas out. Eh? > > http://en.wikipedia.org/wiki/Thermionic_emissionhttp://en.wikipedia.org/wiki/Paschen%27s_law > > > > > > > > > > >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 as 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. > > > Well, I looked at some Hulse-Taylor links. I suppose if it is true > > that gravitational shadowing is an effect then the planetary orbits > > will take on more dynamics and might help explain all of the strange > > angles of rotational axes, magnetic axes, maybe even how they can be > > in such a clean plane. That would be quite some fine tuning and would > > put the inner planets as the leaders. Anyway Sue it's nice to see you > > consider this idea, though cryptically. > > I don't buy into the notions about quadrupole moments from > a monopolar ~antenna~. So induced gravity explains Hulse-Taylor > and a quite LIGO very nicely, (tho only qualitativly) :-( > > > > > Here is a simple question for you: Is radiation pressure a unique > > effect from photon momentum? > > Qualitatively no. Because photons are pseudoparticles. > But they are powerful modeling tools. If the maths > is rigourous and makes better predictions there is > every reason to favour their use in explaining an > phenomena. > > If atom "D" absorbs a unit of energy with some > causal link to atom "S" loosing a unit of energy, > call it a photon. > > But don't try to say too much about the path. > That requires classical electromagnetism. > > > My answer is no, but I am still not > > getting much clear feedback on this, and as I read there is little > > direct connection made, though the math appears to work out directly. > > This is based on a wiki link number of the sun's radiation pressure, > > so I'm not feeling so solid. As I recall it is 4.6E-6 Pascals of solar > > radiation pressure here on the earth's surface from 1375 watts of > > solar radiation energy per square meter. It's rather alot of heat for > > a little kick. I'm trying to give this theory a kick in the pants. > > Speaking of panting, I'd better see what Timo has been up to on this > > thread. > > Well... Timo threw a nearly atomic scale experiment at > us with the tweezers, so we have to keep him honest and make > sure it scales to Nichols radiometers and solar-sails. I don't > see any thing about Boltzmann constant or traversed > volume in his most recent post. > > I won't say his technique can't leapfrog all that, it may. > But I am not seeing it yet. > > Sue... > > > > > > > - Tim- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -
From: NoEinstein on 11 Jun 2010 23:50 On Jun 11, 9:24 pm, "Sue..." <suzysewns...(a)yahoo.com.au> wrote: > Dear Sue: Being authoritative isn't stating science fiction, but science facts. Mass-less photons impart no kick upon hitting an object and being re emitted by an object. You are talking out your ear. NoEinstein > > On Jun 11, 4:17 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > On Jun 11, 9:38 pm, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > > > > On Jun 10, 10:31 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > On Thu, 10 Jun 2010, Tim BandTech.com wrote: > > > > > On Jun 10, 3:19 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > > > On Jun 10, 9:43 pm, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > > > > > > > On Jun 9, 3:40 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > > > > > On Jun 10, 4:10 am, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > > > > > > > > > > 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.) > > > > > > > > Here again I see your obfuscatory tactic. Firstly you falsify and in > > > > > > > your conclusion you agree. > > > > > > > You were quite specific: "level of heating that light achieves will be > > > > > > quite some trick to mimic". This level of heating is easy to mimic. > > > > > > > > The amount of heating that light is capable > > > > > > > of when absorbed versus the work that can be done mechanically due to > > > > > > > that absorption are remarkable in comparison to your NATO bullet. > > > > > > > So, you want to change "level of heating" to "amount of heating versus > > > > > > work"? If you meant this in the first place, you weren't clear enough. > > > > > > To criticise my reply to your actual original words on the basis of > > > > > > your _changed_ version lies somewhere on the scale from weaseling to > > > > > > complete bullshit. > > > > > > I'll have to own here that I should have used the word 'relative' > > > > > within the context, but I see it is fairly easy to interperet since > > > > > the context of the whole argument is still present. Hell, I can match > > > > > the heat of sunlight rubbing some steel on a rock. The lead of the > > > > > bullet will melt on impact. Your interpretation of my writing is > > > > > clearly not coherent at many levels. > > > > > Your original argument wasn't coherent, then. As far as I could tell, you > > > > thought it improper that almost all of the energy should go into heating, > > > > not work. > > > > Jeeze. I pity any onlooker who attempts to decode your own position. > > > Now, via double crossing, I must admit that I do agree with your > > > statement above and that this is exactly what I've been discussing all > > > along. > > > So, you agree that "improper that almost all of the energy should go > > into heating, > > not work" is a fair summary of your original argument? > > > Then what complaint do you have against the bullet example? In the > > bullet example, almost all of the energy goes into heating. Surely > > this is just as "improper". If it's OK for most of the KE of a bullet > > to end up heating the target, rather than doing work, why is somehow > > wrong for a photon? > > > Note well that in both the bullet and photon cases, you get a force if > > the bullet/photon is reflected with no loss of energy. It's simpler to > > consider the reflection case, since you avoid the complications you > > get into below. > > > > > We see the same thing in Newtonian mechanics for a light object in > > > > inelastic collision with a heavy object. Qualitatively, the same type of > > > > thing, most of the energy going into heating. > > > > It is a peculiar thing to discuss an inelastic collision this way. > > > There are more dynamics here and the options as to what happens with > > > the energy are numerous. For instance, if we could design a spring > > > with a ratchet to capture the projectile then we would only heat by > > > the inefficiency of the spring device, rather than by the objects > > > entire kinetic energy. Likewise as I stated before, this capture > > > mechanism could as well generate electricity, and of course this is > > > the quest of many right now; to capture electricity from the sun, and > > > to do it at a better margin than 20%. We cannot grant the heat clause > > > unconditionally, and instead want the mechanism of that heating. It is > > > not so difficult to see electron disturbance; literally microcurrents > > > flowing in short circuit fashion which provide the heating effect. > > > That these charge concerns are so nearby to the Nichols device is a > > > caveat on the operation of that device. That your light traps are so > > > behaved is likewise a caveat. Some would like to free these electrons > > > to take a larger path. Others would like to ignore them for their > > > inconvenient truth. > > > Yes, we can explain the heating when an EM wave is absorbed in terms > > of electrodynamics. Are you seeking to attribute the observed forces > > to electron ejection? The atom trapping experiments show it isn't so, > > very clearly. Given that the observed forces are predicted by > > conventional theory to within the experimental error (i.e., better > > than 10% in the less accurate experiments, 0.1% in good experiments), > > it doesn't appear to be significant in the macroscopic experiments > > either. Does the magnetic force on a current-carrying wire depend on > > electron ejection? If not, why should the forces due to an EM wave? > > > > > > The context of the discussion for me revolves around the photon energy > > > > > and how we can come to attribute the photon momentum to the photon > > > > > energy without concern for such things as angular momentum. > > > > > We're not attributing the photon momentum to the photon energy. > > > > > For a moving bullet, does the kinetic energy cause the momentum? Does the > > > > momentum cause the kinetic energy? Do we attribute one to the other? > > > > > For the photon, consider a spin +1 photon. How much of its energy is "in" > > > > the angular momentum? > > > > > Be specific: consider a 500nm photon, with hbar angular momentum. We can > > > > write down 3 numbers: its energy, its momentum, and its angular momentum. > > > > How are these 3 related to each other, showing appropriate concern for > > > > angular momentum? > > > > > How about for a 5Hz photon? Do the same. Does this mean that a 5Hz photon > > > > shouldn't have any momentum? > > > > From what I've read of the quantum stuff the angular momentum is not > > > so easy to observe. They even go so far as to grant a photon an > > > orbital angular momentum, though the orbital context is not present in > > > freely propagating light. And yet the light as an oscillation is still > > > a coherent concept, and so the rotation momentum that I am considering > > > is more like your NATO bullet spinning. > > > > As I recall the 5Hz photon will have the same quantum angular momentum > > > than will the 500nm photon, but will have less total energy via e=hf. > > > Keep it simple,. don't worry about orbital angular momentum. If you > > can't answer the simple question, why make it more complicated. > > > Don't get sidetracked into pretty pictures of photons, either, or long > > philosophical tracts. Answer the simple question first. > > > If the momentum and/or angular momentum are attributed to the energy, > > how much of each, in each case, for the 500nm photon and the 5Hz > > photon? Angular momentum hbar, E=hf, p=h/lambda. Write down the > > numbers, and say how much of the energy is "in" the angular momentum > > (or whatever you mean, when you say that having angular momentum means > > that less energy must be "available" for the momentum). > > > (There are 2 potential nuggets to be had here, if you bother to do > > it.) > > > [moved] > > > > > The ratios or values of energy, momentum, and angular momentum, as > > > > commonly stated for photons, come straight out of classical > > > > electromagnetic theory. Are these ratios or values wrong? Yes or no, no > > > > handwaving, no waffle, just a straight answer. > > > > As you seek a straight answer I must ask you how crooked is the path > > > of modern theory? Yes, we'd all like a straight answer, yet we do not > > > actually have one yet. This is not to say that we should give up. We > > > seek a straight answer, but I must admit that I do not yet have one. > > > If you can't say straight-out that it's right or wrong, from what > > information are you arguing? > > > It's simple in a way. We can, directly from Maxwell's equations, by > > finding the induced currents and dielectric polarisation, find the > > force acting on the current and polarisation, via the Lorentz force, > > and find the work done on the current by the field (which is the > > heating). > > > > beyond this it seems that in the accumulation the overlap of radiation > > > pressure with photon momentum has gotten lost in the shuffle. > > > No. Or if it did, stop shuffling it. It isn't an overlap, one is the > > rate of change of the other. > > ============= > > > > > For a beam of light, the radiation force is the change in the momentum > > flux when the beam is reflected, refracted, or absorbed. For a photon, > > the impulse imparted to a reflector, refractor, or absorber, is the > > change in the photon momentum when it is reflected, refracted, or > > absorbed. > > The notion certainly didn't come from the sci-fi > shelves. It may be instructive to examine the > differences in two concise statements. > > <<Since the force on a dielectric object is given > by the change in momentum of light induced due to > refraction of the light by the object, the total > force on the object is the difference between the > momentum flux entering the object and that leaving > the object. The total force on an object due to > refraction of light is therefore >> eq3,p2 > "Optical Tweezers: Measuring ... > > read more »- Hide quoted text - > > - Show quoted text -
From: kado on 12 Jun 2010 00:27
On Jun 8, 4:00 pm, Sam Wormley <sworml...(a)gmail.com> wrote: > On 6/8/10 4:05 PM, k...(a)nventure.com wrote: snip > >>> Furthermore, there is the twin paradox. > > >> And just how do you see the "twin paradox" as a problem for > >> special relativity? Be specific. > > > I thought I was. Anyway, a paradox is a self contradictory statement. > > > Special Relativity maintains that: > > > 1. There is no preferred frame of reference (FOR), so any FOR is > > just as valid as any other. So the FOR of the traveling twin is > > just as valid as the FOR of the stay at home twin. > > 2. Time contracts, only contracts, and always contracts, when moving > > near the velocity of light. > > 3. Nature is time symmetrical. > > > So: > > > 1. The traveling twin grows old slower (due to time contraction) > > than does the stay at home twin, and > > 2. The stay at home twin grows old slower than the traveling twin > > (due to time contraction) concurrently (i.e., at the same time) > > because the FOR of one is just as valid as that of the other. > > > Therefore: > > > Each twin is younger than his/her sibling when the traveling twin > > completes his/her trip and they are again together at home, and at > > rest (stationary) in the 'stationary system'. > > > So is this not a contradiction? and > > > So is this not a problem within relativity? > > > That's why I specificly asked: "Which twin is younger that his/her > > sibling when they are both together and at rest in the 'stationary > > system'?" > > > Do you understand that the frames of the twins are not both > inertial frames of reference? > > Physics FAQ: The Twin Paradox > > http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_... You still did not answer my question! Furthermore, I specifically mentioned that I was referring to the point in time and space when they are both together at rest in what Einstein called the 'stationary system'. So both must be in the one and same rest frame of reference, not any inertial frame of reference wherein Einstein believed that time, mass, etc., are relative. Moreover, answering my question about Special Relativity with a Special Relativity response is circular reasoning, and is the same sort of BS that lawyers us to 'prove' that a straight line is curved and a curved line is straight. So please just tell me which twin is younger than the other. D.Y.K. |