Doppler energy extraction: new solar cell technology
in [Physics]
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From: Timo Nieminen on 15 Jul 2010 18:17 On Thu, 15 Jul 2010, Tim BandTech.com wrote: > On Jul 14, 7:24 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > On Wed, 14 Jul 2010, Tim BandTech.com wrote: > > > If you have a clean argument that denies this conclusion then why > > > don't you build it completely here? > > > > The whole argument has been presented; what has followed has been trying > > to find out where you disagree with it. > > > > It doesn't help that you will - for post after post - refuse to answer > > direct questions intended to find out. It doesn't help that you will post > > multiple paragraphs of waffle instead of a simple "yes" or "no". > > > > For example, do you agree that P=Fv is correct? (Here, P is the rate of > > doing work on an object moving at v by a force F. Since this rate of > > doing work is a power, I've used the traditional "P".) > > I agree that under some conditions that the equation you use is > correct. A more careful expression is: > P(t) = F(t) dot v(t) > where F and v are vectors, generally in (x,y,z) space. Indeed, it is a scalar product of 2 vectors, since both F and v are vector quantities, and P is a scalar quantity. If you wish, we can write it as F.v or F dot v or dot(F,v) or whatever. But "under some conditions" is not enough for us to proceed. Given our "progress" so far, I think it is important to have a more concrete agreement (or disagreement) on this, or else we will simply return here later. Is P=F.v the correct expression for the rate of doing work by force F on an object moving at velocity v? Yes or no? > However, if we > take a beam of light whose intensity over some small area is 1000 > watts, we should not expect to receive 1000 watts of mechanical work > from that light source. Indeed, we can show that a 1kW beam will not do 1kW of work. In P=F.v, P is the rate of doing work on an object, not the power of the beam, and not equal to the power of the beam except perhaps in extraordinary circumstances, if ever. > > If yes, we can move on to the next step. > > > > If no, can you provide a falsification, an opposing derivation, or such? > > > > We have, so far, agreed: > > > > (a) The power of an EM beam (i.e., the energy flux) is affected by > > redshift or blueshift. > > > > (b) The cycles don't disappear. (I don't see why you think this is > > relevant, but let me leave it here on the list.) > > > > Can we add (c) P=Fv to the list? -- Timo
From: Tim BandTech.com on 16 Jul 2010 07:46 On Jul 15, 6:17 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > On Thu, 15 Jul 2010, Tim BandTech.com wrote: > > On Jul 14, 7:24 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > On Wed, 14 Jul 2010, Tim BandTech.com wrote: > > > > If you have a clean argument that denies this conclusion then why > > > > don't you build it completely here? > > > > The whole argument has been presented; what has followed has been trying > > > to find out where you disagree with it. > > > > It doesn't help that you will - for post after post - refuse to answer > > > direct questions intended to find out. It doesn't help that you will post > > > multiple paragraphs of waffle instead of a simple "yes" or "no". > > > > For example, do you agree that P=Fv is correct? (Here, P is the rate of > > > doing work on an object moving at v by a force F. Since this rate of > > > doing work is a power, I've used the traditional "P".) > > > I agree that under some conditions that the equation you use is > > correct. A more careful expression is: > > P(t) = F(t) dot v(t) > > where F and v are vectors, generally in (x,y,z) space. > > Indeed, it is a scalar product of 2 vectors, since both F and v are vector > quantities, and P is a scalar quantity. If you wish, we can write it as > F.v or F dot v or dot(F,v) or whatever. > > But "under some conditions" is not enough for us to proceed. Given our > "progress" so far, I think it is important to have a more concrete > agreement (or disagreement) on this, or else we will simply return here > later. > > Is P=F.v the correct expression for the rate of doing work by force F on > an object moving at velocity v? > > Yes or no? Yes. > > > However, if we > > take a beam of light whose intensity over some small area is 1000 > > watts, we should not expect to receive 1000 watts of mechanical work > > from that light source. > > Indeed, we can show that a 1kW beam will not do 1kW of work. In P=F.v, P > is the rate of doing work on an object, not the power of the beam, and not > equal to the power of the beam except perhaps in extraordinary > circumstances, if ever. > > > > If yes, we can move on to the next step. > > > > If no, can you provide a falsification, an opposing derivation, or such? > > > > We have, so far, agreed: > > > > (a) The power of an EM beam (i.e., the energy flux) is affected by > > > redshift or blueshift. > > > > (b) The cycles don't disappear. (I don't see why you think this is > > > relevant, but let me leave it here on the list.) > > > > Can we add (c) P=Fv to the list? > > -- > Timo
From: Timo Nieminen on 16 Jul 2010 08:46 We have: (a) The power of an EM beam (i.e., the energy flux) is affected by redshift or blueshift. (b) The cycles don't disappear. (I don't see why you think this is relevant, but let me leave it here on the list.) (c) P = F.v Consider a stationary perfect reflector reflecting a monochromatic collimated beam of power P0 and frequency f0. By "perfect reflector", I mean that no energy is absorbed by the reflector. By "absorbed", I mean that none of the incident energy is transferred to the internal energy of the reflector. "Internal energy" most importantly includes thermal energy, but could include other forms of energy, such as electronic excitations leading to fluorescence, etc. "Internal energy" does not include kinetic energy due to motion or rotation of the object (i.e., the reflector, in this case). We have v = 0, and, from P = F.v, no work is done, regardless of what the force F is. Since no energy is absorbed (since we are considering a perfect reflector), and no work is being done, the reflected beam must be of power P0. Since the reflector is stationary, there is no Doppler shift, and the reflected beam is of power f0. From conservation of energy, P_out = P_in - rate of increase in internal energy - rate of increase of KE P_out = P0 - 0 - F.v P_out = P0 - 0 - F.0 = P0 (We could write the rate of increase of internal energy in terms of an absorption coefficient a as a*P0. For a perfect reflector, a=0.) Note well that the reflected beam is of power P0 and frequency f0 for _any_ force F, zero or non-zero, that the beam might exert on the reflector. The perfect reflector, with no absorption, is an idealisation. It doesn't matter that it can't be achieved with a real reflector; we are considering an idealised thought experiment here, just as we consider frictionless surfaces, rigid bodies, inextensible strings, light strings, monochromatic collimated beams, and so on. We can get usefully close to the idealisation with a real reflector, so a reasonable approximation of the thought experiment can be realised. If you feel that we come to some wrong result due to these idealisations, do say so, but preferably with some concrete details of why. The usage of "perfect reflector" and "absorbed" above are consistent with standard usage in physics. These are technical terms, and do not necessarily mean the same thing as the same terms in everyday English. It will be important to use these terms consistently, and not change their meanings in the middle of the discussion. Do these terms need further clarification? (Yes/no? If yes, what do you suggest?) For the next step, it will be necessary to consider relatively moving reference frames. It will be sufficient to use Galileian relativity. In principle, we could also use Lorentz/Einstein relativity (i.e., SR), taking limits to first order in v/c if we wish. Under Galileian relativity, mass, force, acceleration, and internal energy are invariant under change of inertial reference frame (i.e., they are the same in all inertial reference frames). Velocity of an object, and its KE, are not invariant under such transformations. (Also true to first order in v/c in SR.) Agreed?
From: Tim BandTech.com on 16 Jul 2010 21:09 On Jul 16, 8:46 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > We have: > > (a) The power of an EM beam (i.e., the energy flux) is affected by > redshift or blueshift. > > (b) The cycles don't disappear. (I don't see why you think this is > relevant, but let me leave it here on the list.) > > (c) P = F.v > > Consider a stationary perfect reflector reflecting a monochromatic > collimated beam of power P0 and frequency f0. By "perfect reflector", > I mean that no energy is absorbed by the reflector. By "absorbed", I > mean that none of the incident energy is transferred to the internal > energy of the reflector. "Internal energy" most importantly includes > thermal energy, but could include other forms of energy, such as > electronic excitations leading to fluorescence, etc. "Internal energy" > does not include kinetic energy due to motion or rotation of the > object (i.e., the reflector, in this case). > > We have v = 0, and, from P = F.v, no work is done, regardless of what > the force F is. > > Since no energy is absorbed (since we are considering a perfect > reflector), and no work is being done, the reflected beam must be of > power P0. Since the reflector is stationary, there is no Doppler > shift, and the reflected beam is of power f0. > > From conservation of energy, > > P_out = P_in - rate of increase in internal energy - rate of increase > of KE > > P_out = P0 - 0 - F.v > > P_out = P0 - 0 - F.0 = P0 > This is an argument for no acceleration of the reflector. This is what I've been arguing for nearly from the start. Aren't you now contradicting the Nichols results? > (We could write the rate of increase of internal energy in terms of an > absorption coefficient a as a*P0. For a perfect reflector, a=0.) > > Note well that the reflected beam is of power P0 and frequency f0 for > _any_ force F, zero or non-zero, that the beam might exert on the > reflector. > > The perfect reflector, with no absorption, is an idealisation. It > doesn't matter that it can't be achieved with a real reflector; we are > considering an idealised thought experiment here, just as we consider > frictionless surfaces, rigid bodies, inextensible strings, light > strings, monochromatic collimated beams, and so on. We can get > usefully close to the idealisation with a real reflector, so a > reasonable approximation of the thought experiment can be realised. If > you feel that we come to some wrong result due to these idealisations, > do say so, but preferably with some concrete details of why. > > The usage of "perfect reflector" and "absorbed" above are consistent > with standard usage in physics. These are technical terms, and do not > necessarily mean the same thing as the same terms in everyday English. > It will be important to use these terms consistently, and not change > their meanings in the middle of the discussion. > > Do these terms need further clarification? (Yes/no? If yes, what do > you suggest?) > > For the next step, it will be necessary to consider relatively moving > reference frames. It will be sufficient to use Galileian relativity. > In principle, we could also use Lorentz/Einstein relativity (i.e., > SR), taking limits to first order in v/c if we wish. > > Under Galileian relativity, mass, force, acceleration, and internal > energy are invariant under change of inertial reference frame (i.e., > they are the same in all inertial reference frames). Velocity of an > object, and its KE, are not invariant under such transformations. > (Also true to first order in v/c in SR.) Agreed? I guess so. You are not yielding a complete argument, but I am willing to go along. So far you seem to be supporting the idea that there is no theoretical support for the radiation pressure acceleration effect, from energy conservation of a perfect reflector. - Tim
From: Timo Nieminen on 16 Jul 2010 22:11
On Jul 17, 11:09 am, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > On Jul 16, 8:46 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > We have: > > > (a) The power of an EM beam (i.e., the energy flux) is affected by > > redshift or blueshift. > > > (b) The cycles don't disappear. (I don't see why you think this is > > relevant, but let me leave it here on the list.) > > > (c) P = F.v > > > Consider a stationary perfect reflector reflecting a monochromatic > > collimated beam of power P0 and frequency f0. By "perfect reflector", > > I mean that no energy is absorbed by the reflector. By "absorbed", I > > mean that none of the incident energy is transferred to the internal > > energy of the reflector. "Internal energy" most importantly includes > > thermal energy, but could include other forms of energy, such as > > electronic excitations leading to fluorescence, etc. "Internal energy" > > does not include kinetic energy due to motion or rotation of the > > object (i.e., the reflector, in this case). > > > We have v = 0, and, from P = F.v, no work is done, regardless of what > > the force F is. > > > Since no energy is absorbed (since we are considering a perfect > > reflector), and no work is being done, the reflected beam must be of > > power P0. Since the reflector is stationary, there is no Doppler > > shift, and the reflected beam is of power f0. > > > From conservation of energy, > > > P_out = P_in - rate of increase in internal energy - rate of increase > > of KE > > > P_out = P0 - 0 - F.v > > > P_out = P0 - 0 - F.0 = P0 > > This is an argument for no acceleration of the reflector. This is what > I've been arguing for nearly from the start. Aren't you now > contradicting the Nichols results? No. OK, we need to stop here; no point in going past this at the moment. Let DP = the change in power of the beam on reflection (D as a substitute for delta). For a stationary perfect reflector, DP = 0. We have, above, -DP = absorbed power + rate of doing work. Since the reflector is a perfect reflector, absorbed power = 0. This reduces to DP = - rate of doing work = -F.v Since DP = 0, and v = 0, F must satisfy 0*F = 0. You have been claiming that this means that F must be zero, I have been claiming that F can be non-zero and still satisfy this. Let me give an example. Take F = 1kN. (DP = 0W, v = 0 m/s.) 0*1000 = 0, as required. Try this with any number for F (that is, any finite force, zero, positive, or negative), and you will see that 0*F=0 is satisfied for any value of F. It is not valid to argue that 0*F = 0 means that F=0. OK so far? If not, why isn't the numerical example showing that F=1kN is compatible with DP = 0 sufficient? > > (We could write the rate of increase of internal energy in terms of an > > absorption coefficient a as a*P0. For a perfect reflector, a=0.) > > > Note well that the reflected beam is of power P0 and frequency f0 for > > _any_ force F, zero or non-zero, that the beam might exert on the > > reflector. > > > The perfect reflector, with no absorption, is an idealisation. It > > doesn't matter that it can't be achieved with a real reflector; we are > > considering an idealised thought experiment here, just as we consider > > frictionless surfaces, rigid bodies, inextensible strings, light > > strings, monochromatic collimated beams, and so on. We can get > > usefully close to the idealisation with a real reflector, so a > > reasonable approximation of the thought experiment can be realised. If > > you feel that we come to some wrong result due to these idealisations, > > do say so, but preferably with some concrete details of why. > > > The usage of "perfect reflector" and "absorbed" above are consistent > > with standard usage in physics. These are technical terms, and do not > > necessarily mean the same thing as the same terms in everyday English. > > It will be important to use these terms consistently, and not change > > their meanings in the middle of the discussion. > > > Do these terms need further clarification? (Yes/no? If yes, what do > > you suggest?) > > > For the next step, it will be necessary to consider relatively moving > > reference frames. It will be sufficient to use Galileian relativity. > > In principle, we could also use Lorentz/Einstein relativity (i.e., > > SR), taking limits to first order in v/c if we wish. > > > Under Galileian relativity, mass, force, acceleration, and internal > > energy are invariant under change of inertial reference frame (i.e., > > they are the same in all inertial reference frames). Velocity of an > > object, and its KE, are not invariant under such transformations. > > (Also true to first order in v/c in SR.) Agreed? > > I guess so. You are not yielding a complete argument, but I am willing > to go along. So far you seem to be supporting the idea that there is > no theoretical support for the radiation pressure acceleration effect, > from energy conservation of a perfect reflector. |