Taking a Fresh Look at the Physics of Radiometers.
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From: kado on 8 Jun 2010 17:05 On Jun 8, 10:19 am, Sam Wormley <sworml...(a)gmail.com> wrote: > 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. 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'?" D.Y.K.
From: Timo Nieminen on 8 Jun 2010 17:36 On Jun 8, 8:26 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > Next will be: 2a. Absorption by a stationary absorber, and 2b. > Absorption by a moving absorber. To continue: 2a. Stationary absorber. (i) Classical particle. E_in = E = (1/2)mv^2, E_out = 0, p_in = mv = 2E/v, p_out = 0. I = mv = 2E/v, F = mvN = 2EN/v. The force is 1/2 of the force on a reflecting plate. Since there is no motion of the reflector, no work is done. So, E_in becomes energy other than mechanical energy. part from losses due to acoustic waves (in the air and the plate) which are potentially lost from the plate/particle system, this energy will result in heating of the reflecting plate (+ the particle that is now stuck to it). So, rate at which the plate is heated is EN. (ii) Photon. E_in = E = hf, E_out = 0, p_in = h/lambda = E/c, p_out = 0. I = h/lambda = E/c, F = hN/lambda = EN/c. The force is 1/2 of the force on a reflecting plate. Again, no work, so all of this energy goes to heat the plate, at a power of EN. (iii) Classical wave. P_in = P0 P_out = 0 Without assuming a particular momentum, all we have is that no work is done, so the heating power is P0. Since no work is done, we don't have any information about the force yet. 2b. Moving absorber, moving at v_cs (as before, just a change to a moving coordinate system). (i) classical particle. E_in = (1/2)m(v+v_cs)^2, E_out = (1/2)m v_cs^2, p_in = m(v +v_cs), p_out = m v_cs. E_in - E_out = (1/2)mv^2 + m v v_cs = E + m v v_cs I = p_in - p_out = mv = 2E/v, as for stationary absorber. F = mvN = 2EN/v, as for stationary absorber. The rate of energy transfer to the absorber is: (E_in - E_out)N = EN + m v v_cs N = heating power + m v v_cs N (since the heating power must be unaffected by our change in coordinate system). The remainder of the power transfer must be the rate of doing work, which gives us the force by F = P/v = m v v_cs N / v_cs = mvN = 2EN/v, as for the stationary absorber. For v_cs << v, almost all of the incident kinetic energy goes towards heating the absorber, not doing work on the absorber. (ii) Photon. Blueshifted by (1+v_cs/c). E_in = E(1+v_cs/c) = hf(1+v_cs/c), E_out = ?, p_in = h(1+v_cs/c)/lambda = E(1+v_cs/c)/c, p_out = ?. We don't end up with a photon anymore. With the particle sticking to the plate, the particle still had KE and momentum when it was stuck to a moving absorber. If we assume that E_out = 0, p_out = 0, we have Rate of transfer of energy = E_in N = heating power + hfN v_cs/c = heating power + EN v_cs/c The leftover power, beyond the heating, must be the rate of doing work on the absorber. So, F = power/v = EN v_cs/(c v_cs) = EN/c, the same as for the stationary absorber. (p_in - p_out)N = F + EN v_cs/c^2 So, the momentum of the absorber is increasing, by an extra amount. We could have assumed that P_out was zero for the classical particle case, and we would have gotten the same result. In the particle case, the increase in momentum is due to the increase in mass (still moving at v_cs) due to the particles sticking to the absorber. Here, we are increasing the inertia of the absorber (I will just call it "inertia", to not need to bring definitions of mass, relativistic mass, rest mass, etc into it, especially since we still being classical.) The rate of increase of inertia is (rate of extra increase in momentum)/v_cs = EN/c^2, = heating power/c^2. So, the thermal energy we are adding to the absorber has inertia E/c^2. (iii) Classical wave. Blueshifted by (1+v_cs/v_w). P_in = P0(1+v_cs/v_w) P_out = 0? Rate of heating ust be the same as in the coordinate system where the absorber was stationary. Rate of transfer of energy = rate of heating + rate of doing work P_in - P_out = P0(1+v_cs/v_w) = rate of heating + P0 v_cs/v_w F = P/v = P0 v_cs/(v_w v_cs) = P0/v_w. This is the radiation pressure force, half of what we had for the reflector. From the reflecting case, we had momentum flux = p_flux = P0/v_w. Here, we have p_flux = P0(1+v_cs/v_w)/v_w = force + extra increase in momentum. Rate of increase of inertia = extra increase in momentum / v_cs = P0/ v_w^2. 3. In summary, The momentum flux of a classical wave of power P and speed v_w is p_flux = P/v_w. The force exerted on a reflector is 2 p_flux = 2P/v_w. The force exerted on an absorber is p_flux = P/v_w. For a stationary reflector, no energy is absorbed; there is no heating. For a moving reflector, there is a change in energy of the wave due to redshift/blueshift; this is the work done on the reflector and there is no heating. For an absorber, moving or stationary, there is heating. For a moving absorber, some of the energy goes into work done on the absorber, the rest is heating. For v << v_w, most of the energy does into heating. For the wave or photon, it's almost the same as the case for the classical particle. There are two main differences: (i) The energy-momentum relationship is different for the classical particle; the classical particle stream exerts twice as much force. (ii) Energy has inertia. Arguments against the existence of radiation pressure/force/momentum based on the ratio of momentum to energy, or work done to heating, apply (almost) equally to the classical particle case - the numbers are the same except for a factor of 2. 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.
From: Sam Wormley on 8 Jun 2010 19:00 On 6/8/10 4:05 PM, kado(a)nventure.com wrote: > On Jun 8, 10:19 am, Sam Wormley<sworml...(a)gmail.com> wrote: >> 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. > > > 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'?" > > > D.Y.K. 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_paradox.html
From: Sue... on 8 Jun 2010 19:08 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: Tim BandTech.com on 9 Jun 2010 14:10
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. 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. 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. - Tim |