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Where Does the Power Go?
On Mon, 9 Oct 2006 00:06:05 -0800, "Dana" wrote:
I see that guy has not answered yet about the temp Hi Dana, It was a sucker punch anyway. Cecil doesn't have the experience in optics to answer it with any particular immediacy or accuracy, nor do 99% of those in this group (lurkers excluded). I've got an extensive career in the field that includes 5 patents in photonic applications. I've also designed a number of proprietary photonic devices that measure blood-gas chemistry, O2 saturation, pH, and the Stroud Moment (as it relates to mental acuity through visual testing). 73's Richard Clark, KB7QHC |
Where Does the Power Go?
Richard Clark wrote:
On Mon, 9 Oct 2006 00:06:05 -0800, "Dana" wrote: I see that guy has not answered yet about the temp Hi Dana, It was a sucker punch anyway. Cecil doesn't have the experience in optics to answer it with any particular immediacy or accuracy, nor do 99% of those in this group (lurkers excluded). I've got an extensive career in the field that includes 5 patents in photonic applications. I've also designed a number of proprietary photonic devices that measure blood-gas chemistry, O2 saturation, pH, and the Stroud Moment (as it relates to mental acuity through visual testing). 73's Richard Clark, KB7QHC Richard, In the interest of fairness, Cecil said he would be taking a motorcycle trip over the long weekend. At 3 MHz I get 144 micro Kelvin by multiplying frequency by Planck's constant and dividing by Boltzman's. My daughter had a similar problem in her High School freshman Earth Science class a few weeks ago. It might not be entirely fair to translate a lack of response to a lack of ability. There are other factors as well, among them being lack of even the slightest amount of interest, and difficulty deciphering what the heck you are talking about from one moment to the next. ;-) 73, Jim AC6XG |
Where Does the Power Go?
On Mon, 09 Oct 2006 15:53:23 -0700, Jim Kelley
wrote: Richard, In the interest of fairness, Cecil said he would be taking a motorcycle trip over the long weekend. At 3 MHz I get 144 micro Kelvin by multiplying frequency by Planck's constant and dividing by Boltzman's. My daughter had a similar problem in her High School freshman Earth Science class a few weeks ago. It might not be entirely fair to translate a lack of response to a lack of ability. 73, Jim AC6XG Hi Jim, Fairness counts, to be sure. Reputation informs us all otherwise. I am not responsible for Cecil's reputation, so impugning an opponent is a matter of local custom. 114 micro Kelvin certainly falls within the parameters of the question offered. I dare say Cecil would have been silent on the specific matter, irrespective of his recreational activity. As for the specific difference between you and your daughter's computation, I use the Wien Displacement Law. It, too, employs the method you describe (albeit with Boltzmann's constant divided by Planck's constant instead as it is frequency not wavelength descriptive), and with an additional constant of multiplication (2.8214). By this method, your 144 micro Kelvins represents 8.466 MHz for the peak wavelength. There are other factors as well, among them being lack of even the slightest amount of interest The nature of posting to the group satisfies that quite simply: folks move on or become part of the thread. That has been adequately demonstrated here. and difficulty deciphering what the heck you are talking about from one moment to the next. ;-) Barring calls for specific explanation, I always treat such whining for what it is. Clearly you are neither whining, nor ignorant/disinterested in the topic. I take it for granted there are a multitude of others who choose to remain silent, but not uninformed. Most will take notice I did not open the door of this side-thread, but having crossed the threshold, I command the topic. 73's Richard Clark, KB7QHC |
Where Does the Power Go?
Richard Clark wrote:
114 micro Kelvin certainly falls within the parameters of the question offered. I dare say Cecil would have been silent on the specific matter, irrespective of his recreational activity. As for the specific difference between you and your daughter's computation, I use the Wien Displacement Law. It, too, employs the method you describe (albeit with Boltzmann's constant divided by Planck's constant instead as it is frequency not wavelength descriptive), and with an additional constant of multiplication (2.8214). By this method, your 144 micro Kelvins represents 8.466 MHz for the peak wavelength. Richard, To be a bit fussy, the temperature of a photon is not defined. Only a distribution of photon energies can be defined with a temperature (sometimes). Assuming a standard blackbody model, your answer is correct of course. Perhaps Cecil was trying to recall the formula for the temperature of a single photon. He might be looking for a while. 73, Gene W4SZ |
Where Does the Power Go?
On Tue, 10 Oct 2006 00:21:58 GMT, Gene Fuller
wrote: To be a bit fussy, the temperature of a photon is not defined. Only a distribution of photon energies can be defined with a temperature (sometimes). Assuming a standard blackbody model, your answer is correct of course. Hi Gene, Fussy is the name of the game here in this forum; and fussy will ultimately dominate; hence I defer to your amplification. I suggested a black body model late in the game certainly, and only through allusion to the frequency of the peak wavelength. Perhaps Cecil was trying to recall the formula for the temperature of a single photon. He might be looking for a while. Ah! Fussiness has been replaced with abundant generosity. Myself, I consider it misplaced. I would think he could summon up the dominant wavelength of a Xerox lamp and correlate it to toner responsivity. 73's Richard Clark, KB7QHC |
Where Does the Power Go?
Richard Clark wrote:
On Tue, 10 Oct 2006 00:21:58 GMT, Gene Fuller wrote: To be a bit fussy, the temperature of a photon is not defined. Only a distribution of photon energies can be defined with a temperature (sometimes). Assuming a standard blackbody model, your answer is correct of course. Hi Gene, Fussy is the name of the game here in this forum; and fussy will ultimately dominate; hence I defer to your amplification. I suggested a black body model late in the game certainly, and only through allusion to the frequency of the peak wavelength. And I assumed we were talking the energy of the photon. By the way Richard, Wein's Law? We've long since learned that energy is quantized. E=hv That's how much energy is in a single photon at frequency v. (At least, that's what Einstein thought.) No fudging needed! 73, Jim AC6XG |
Where Does the Power Go?
On Mon, 09 Oct 2006 18:36:22 -0700, Jim Kelley
wrote: By the way Richard, Wein's Law? Hi Jim, And yet he won the 1911 Nobel Prize for the law. He also discovered what was to be called the Proton. How quickly fame fades.... We've long since learned that energy is quantized. E=hv That's how much energy is in a single photon at frequency v. Yes, Wien's law was inappropriate for low frequency application (meaning your daughter's computation is closer, by the factor I offered, if not exact). His law was based on observational data for very much shorter wavelengths. My studies generally confine themselves well above 0°K, probably 2 or 3 degrees to a few hundred. That is why I was initially satisfied with order of magnitude accuracy. (At least, that's what Einstein thought.) No fudging needed! Actually, Planck's explanation anticipated Einstein's photons by five years. Further, he also corrected the massive errors of frequency vs. power in what is called the Ultraviolet catastrophe. This was the presumption that a Black body radiator emits energy with a proportionality to frequency - a classical solution that yields astronomic photonic power output at short wavelengths. I thought I had dodged that bullet with my Wien's Displacement law spread sheet. My focus is more oriented towards Phonon interactions. To return to Gene's comment about the underlying presumption of Black body radiation (perfectly correct), Planck's solution to the Ultraviolet catastrophe was to describe Black body radiation as a composite emission of many resonating cavities (which returns us to single source Photons). 73's Richard Clark, KB7QHC |
Where Does the Power Go?
Tom Ring wrote:
EVERYTHING is a subset of physics. Even biology. If politics is a subset of physics and only Jim Kelley's definitions are allowed, then President Bush doesn't have any power. -- 73, Cecil http://www.w5dxp.com |
Where Does the Power Go?
Jim Kelley wrote:
Every once in a while, when I'm reading the interesting technical prose that Cecil writes here and elsewhere, Cliff Clavin comes to mind for some reason. Was he like some people who post ad hominem attacks devoid of any technical content? :-) -- 73, Cecil http://www.w5dxp.com |
Where Does the Power Go?
Jim Kelley wrote:
It might not be entirely fair to translate a lack of response to a lack of ability. There are other factors as well, among them being lack of even the slightest amount of interest, and difficulty deciphering what the heck you are talking about from one moment to the next. ;-) True, and I also ignore postings that are an obvious diversion away from the topic. Certain people think they can win arguments by taking people on A Wild Goose Chase Down A Primrose Path. All it does is waste time. From now on, I will label such attempts as AWGCDAPP so the perp will know that I am not ignoring him. -- 73, Cecil http://www.w5dxp.com |
Where Does the Power Go?
Cecil Moore wrote: If politics is a subset of physics and only Jim Kelley's definitions are allowed, then President Bush doesn't have any power. Welcome back Cliffy! :-) ac6xg |
Where Does the Power Go?
On Wed, 11 Oct 2006 10:43:12 -0700, Jim Kelley
wrote: Cecil Moore wrote: If politics is a subset of physics and only Jim Kelley's definitions are allowed, then President Bush doesn't have any power. Welcome back Cliffy! :-) ac6xg Ah! the proof of reputation informing us all. :-0 |
Where Does the Power Go?
Jim Kelley wrote:
Cecil Moore wrote: If politics is a subset of physics and only Jim Kelley's definitions are allowed, then President Bush doesn't have any power. Welcome back Cliffy! Had a good time playing with my grand niece and grand nephew in TN whose IQ's are impossible to measure in public schools. It runs in the family. :-) Back to the present thread. I infer that you believe that Hecht's total irradiance equations are in error? Before you reply, let me remind you that Dr. Best was the first one, to the best of my knowledge, to publish the irradiance equations in an amateur radio publication, QEX. In the following fixed font diagram, IR is the Index of Refraction. air | 1/4WL thin-film | Glass 1W Laser---IR=1.0---|----IR=1.222-----|--IR=1.493---... Ifor=1W | Ifor=1.0101W | Ifor=1W Iref=0.01W | Iref=0.0101W | Iref=0 Note that I is "irradiance", not current. Given: The irradiance reflection coefficient is 0.01 at both interfaces. The irradiance transmission coefficient is 0.99 at both interfaces. Please describe your theory of the wave cancellation process occurring at the air to thin- film interface without using the superposition and interference principles that I have been using to which you object. -- 73, Cecil http://www.w5dxp.com |
Where Does the Power Go?
Cecil Moore wrote:
The irradiance reflection coefficient is 0.01 at both interfaces. "It's a little known fact that cows were domesticated in Mesopotamia and were also used in China as guard animals for the forbidden city." :-) ac6xg |
Where Does the Power Go?
Jim Kelley wrote:
Cecil Moore wrote: The irradiance reflection coefficient is 0.01 at both interfaces. "It's a little known fact that cows were domesticated in Mesopotamia and were also used in China as guard animals for the forbidden city." Perhaps, should I have used the term, "reflectance"? Allow me to refer you to page 120, "Optics", by Hecht, 4th edition, equation for reflectance, R, "ratio of the reflected power to the incident power". (References are a bitch to deal with, huh?) Note that reflectance, irradiance reflection coefficient, and RF engineering's power reflection coefficient are all the same thing. -- 73, Cecil http://www.w5dxp.com |
Where Does the Power Go?
Jim Kelley wrote:
Cecil Moore wrote: The irradiance reflection coefficient is 0.01 at both interfaces. "It's a little known fact that cows were domesticated in Mesopotamia and were also used in China as guard animals for the forbidden city." air | 1/4WL thin-film | Glass 1W Laser---IR=1.0---|----IR=1.222-----|--IR=1.493---... Ifor=1W | Ifor=1.0101W | Ifor=1W Iref=0.01W | Iref=0.0101W | Iref=0 The reflectance (irradiance reflection coefficient) at the air to thin-film interface is [(1.222 - 1.0)/(1.222 + 1.0)]^2 = 0.01 The reflectance at the thin-film to glass interface is [(1.493 - 1.222)/(1.493 + 1.222)]^2 = 0.01 1% of the irradiance is reflected at either surface and 99% of the irradiance is transmitted through the surfaces. Would you like to engage in a technical discussion of how non-reflective glass works or would you like to keep avoiding that technical discussion? -- 73, Cecil, http://www.qsl.net/w5dxp |
Where Does the Power Go?
Cecil Moore wrote: Would you like to engage in a technical discussion of how non-reflective glass works or would you like to keep avoiding that technical discussion? Your memory must be failing you, Cecil, as it was I who originally explained to you how anti-reflective coatings work and provided you with the Melles-Griot web site, the irradiance equations, and several references on the subject. I know how these coatings work. I have the optics bible, "Principles of Optics" by Born and Wolf sitting on my desk right here in front of me, and have referred to it frequently during our many previous technical discussions on the subject. Your pretense here is ridiculous. Please cite a page number where your 1 watt laser problem can be found. Thanks. 73, Jim AC6XG |
Where Does the Power Go?
Jim Kelley wrote:
Please cite a page number where your 1 watt laser problem can be found. It can be found in my posting. Why are you afraid to discuss it? -- 73, Cecil, http://www.qsl.net/w5dxp |
Where Does the Power Go?
On Thu, 12 Oct 2006 07:50:37 -0700, Jim Kelley
wrote: Please cite a page number where your 1 watt laser problem can be found. Hi Jim, Is that sad example being trotted out again? 73's Richard Clark, KB7QHC |
Where Does the Power Go?
Richard Clark wrote: On Thu, 12 Oct 2006 07:50:37 -0700, Jim Kelley wrote: Please cite a page number where your 1 watt laser problem can be found. Hi Jim, Is that sad example being trotted out again? I hadn't seen the one with a 'forward laser power' reading, as if he was reading it from an SWR meter before. Seems suspicious. tnx de ac6xg |
Where Does the Power Go?
Jim Kelley wrote:
I hadn't seen the one with a 'forward laser power' reading, as if he was reading it from an SWR meter before. Seems suspicious. I have presented that example before. You chose not to discuss it back then. The output of the laser is given to be one watt. What's so suspicious about that? Whatever laser one buys comes with a power output rating. If we discuss this example, step by step, you can show me, once and for all, the error of my ways. -- 73, Cecil, http://www.qsl.net/w5dxp |
Where Does the Power Go?
On Thu, 12 Oct 2006 09:56:54 -0700, Jim Kelley
wrote: a 'forward laser power' reading Hi Jim, To quote from an era of scandal: "Say it a'in't so, Joe!" .... and I thought the potential for comedy had already drained away. 73's Richard Clark, KB7QHC |
Where Does the Power Go?
Cecil Moore wrote:
Back to the present thread. I infer that you believe that Hecht's total irradiance equations are in error? Before you reply, let me remind you that Dr. Best was the first one, to the best of my knowledge, to publish the irradiance equations in an amateur radio publication, QEX. In the following fixed font diagram, IR is the Index of Refraction. air | 1/4WL thin-film | Glass 1W Laser---IR=1.0---|----IR=1.222-----|--IR=1.493---... Ifor=1W | Ifor=1.0101W | Ifor=1W Iref=0.01W | Iref=0.0101W | Iref=0 Note that I is "irradiance", not current. Given: The irradiance reflection coefficient is 0.01 at both interfaces. The irradiance transmission coefficient is 0.99 at both interfaces. Please describe your theory of the wave cancellation process occurring at the air to thin- film interface without using the superposition and interference principles that I have been using to which you object. Cecil, I don't have a copy of Hecht. Like Jim, I use Born and Wolf as a "textbook" reference and several other books for practical optical design reference. Does Hecht really say that more power goes into the glass than enters the 1/4WL film? Can we somehow bottle this free energy and save the world? By the way, those irradiance equations go back a couple hundred years. It seems unlikely that Dr. Best was the first to publish them, even considering only amateur-oriented publications. 73, W4SZ Gene |
Where Does the Power Go?
Gene Fuller wrote:
Does Hecht really say that more power goes into the glass than enters the 1/4WL film? No, and neither did I. The 0.01W of power reflected in the diagram undergoes destructive interference. That's why they call it "non-reflective" glass. The 0.01W is known as the external reflection. There is a matching 0.01W (not shown) from the internal reflection that causes wave cancellation. All of this is explained in any good physics book. If anyone is willing to go step by step with me, I will either prove my point or be disproved in the process. What I cannot figure out is why everyone is avoiding a step by step technical discussion and instead engaging in ad hominem comments - hoping to prove what? By the way, those irradiance equations go back a couple hundred years. It seems unlikely that Dr. Best was the first to publish them, even considering only amateur-oriented publications. Well, I said that was to the best of my knowledge. I had never seen Ptot = P1 + P2 + 2*SQRT(P1*P2)cos(A) in an amateur radio publication before and I have been reading them since 1952. -- 73, Cecil, http://www.qsl.net/w5dxp |
Where Does the Power Go?
On Thu, 12 Oct 2006 18:13:08 GMT, Gene Fuller
wrote: Does Hecht really say that more power goes into the glass than enters the 1/4WL film? Can we somehow bottle this free energy and save the world? Hi Gene, Isn't this the point at which all should simply surrender to accepting 69% error as being sufficient for a proof? No one here really expects Cecil's dy-no-mite advantage to be set aside for the sake of accuracy - do they? However, your question illuminates the necessity of Cecil being the one to ask "Where Does the Power Go?" That redeems the thread's entertainment value. 73's Richard Clark, KB7QHC |
Where Does the Power Go?
Gene Fuller wrote:
Does Hecht really say that more power goes into the glass than enters the 1/4WL film? Can we somehow bottle this free energy and save the world? Gene, here's a complete fixed font irradiance diagram of the air to thin-film interface including all components. Reflectance = 0.01, Transmittance = 0.99 Forward irradiance component Ifor=1W --| |-- I1=0.99W I3=0.01W--| air | thin-film Reflected irradiance component |-- Iref=0.0101W I4=0.01W--| |-- I2=0.0001W I1 + I2 + 2*SQRT(I1*I2) = 1.0101W = Ifor in the thin-film This is "total constructive interference" per Hecht. I3 + I4 - 2*SQRT(I3*I4) = 0W = Iref in the air This is "total destructive interference" per Hecht. Those are the irradiance equations from "Optics", by Hecht. As Hecht asserts, the destructive interference equals the constructive interference and the reflections toward the source are canceled. -- 73, Cecil, http://www.qsl.net/w5dxp |
Where Does the Power Go?
Cecil Moore wrote:
Gene Fuller wrote: Does Hecht really say that more power goes into the glass than enters the 1/4WL film? Can we somehow bottle this free energy and save the world? Gene, here's a complete fixed font irradiance diagram of the air to thin-film interface including all components. Reflectance = 0.01, Transmittance = 0.99 Forward irradiance component Ifor=1W --| |-- I1=0.99W I3=0.01W--| air | thin-film Reflected irradiance component |-- Iref=0.0101W I4=0.01W--| |-- I2=0.0001W I1 + I2 + 2*SQRT(I1*I2) = 1.0101W = Ifor in the thin-film This is "total constructive interference" per Hecht. Not exactly as per Hecht. Note to the casual reader: please be advised that unlike Cecil, Eugene Hecht does not claim that power is equal to irradiance. Nor does he imply that scaler quantities can be treated mathematically in the same way as vector quantities, and he therefore does not substitute power for irradiance in any of his textbook equations as Cecil has want to do. These equations do in fact give correct results macroscopically. However, it is inaccurate to infer from interference equations that a given electromagnetic wave is 100% reflected from any partially reflecting boundary. It is only after multiple partial reflections from the inner boundaries of the intermediate medium that the total energy from any given wave in conveyed from source to load. 73, ac6xg |
Where Does the Power Go?
Jim Kelley wrote:
Not exactly as per Hecht. Note to the casual reader: please be advised that unlike Cecil, Eugene Hecht does not claim that power is equal to irradiance. It's only fair to tell everyone reading your posting that you do not accept the definition of power in the IEEE Dictionary nor the definition of power used by the average RF engineer. Irradiance, Poynting vectors, and power flow vectors are valid concepts no matter what esoteric definition of power that you choose to assert. Hecht gives the dimensions of irradiance which are the same as the Poynting vector (power flow vector) in RF engineering. That you don't consider watts to be power is no reason to question the validity of irradiance or Poynting vectors. I don't have "Optics" with me at work so I cannot do an exact Hecht quote. Nor does he imply that scaler quantities can be treated mathematically in the same way as vector quantities, and he therefore does not substitute power for irradiance in any of his textbook equations as Cecil has want to do. I am using Hecht's irradiance equations. Whatever Hecht implied by those equations is exactly what I am doing. Any angle used in the irradiance equations is the angle between the electric fields of the two superposed waves. It was not me, but Dr. Best, who first substituted power for irradiance in his series of QEX articles. But Poynting vectors can obviously be substituted for irradiance since they represent the same thing and have identical dimensions. Most RF engineers would assert that Poynting vectors and power flow vectors represent power. It is your narrow definition of "power" that is the culprit here, not anything I have said. If it will make you feel better, forget the power equations and call them the power flow vector equations. I will try to remember to do that in the future. These equations do in fact give correct results macroscopically. However, it is inaccurate to infer from interference equations that a given electromagnetic wave is 100% reflected from any partially reflecting boundary. It is only after multiple partial reflections from the inner boundaries of the intermediate medium that the total energy from any given wave in conveyed from source to load. All the values are average *steady-state* values, the same values that would be read by a Bird directional wattmeter. All multiple reflections are rolled into the average Iref value displayed by the Bird. All multiple forward components are rolled into the average Ifor value displayed by the Bird. -- 73, Cecil, http://www.qsl.net/w5dxp |
Where Does the Power Go?
Cecil Moore wrote:
Gene Fuller wrote: Does Hecht really say that more power goes into the glass than enters the 1/4WL film? Can we somehow bottle this free energy and save the world? Gene, here's a complete fixed font irradiance diagram of the air to thin-film interface including all components. Reflectance = 0.01, Transmittance = 0.99 Forward irradiance component Ifor=1W --| |-- I1=0.99W I3=0.01W--| air | thin-film Reflected irradiance component |-- Iref=0.0101W I4=0.01W--| |-- I2=0.0001W I1 + I2 + 2*SQRT(I1*I2) = 1.0101W = Ifor in the thin-film This is "total constructive interference" per Hecht. I3 + I4 - 2*SQRT(I3*I4) = 0W = Iref in the air This is "total destructive interference" per Hecht. Those are the irradiance equations from "Optics", by Hecht. As Hecht asserts, the destructive interference equals the constructive interference and the reflections toward the source are canceled. Cecil, Don't bother. I understand the physics quite well, thank you. It is your message that causes grief. On the other hand, I don't really care to try to educate you any further, so the grief is quite small. 73, Gene W4SZ |
Where Does the Power Go?
Gene Fuller wrote:
Don't bother. I understand the physics quite well, thank you. So, is anything technically wrong with what I posted? It is all copied out of various parts of "Optics", by Hecht. -- 73, Cecil http://www.w5dxp.com |
Where Does the Power Go?
Cecil Moore wrote: Gene Fuller wrote: Don't bother. I understand the physics quite well, thank you. So, is anything technically wrong with what I posted? It is all copied out of various parts of "Optics", by Hecht. From what page in "Optics" by Hecht is the 1 watt laser problem and analysis copied? 73, Jim AC6XG |
Where Does the Power Go?
Jim Kelley wrote:
From what page in "Optics" by Hecht is the 1 watt laser problem and analysis copied? The 1 watt laser mental exercise is my idea. Everything else is directly from "Optics", by Hecht. A 1 watt ideal laser was chosen for its single frequency and coherent characteristics to avoid any more "brighter than the surface of the sun" postings. Why are you afraid to discuss a 1 watt laser? -- 73, Cecil http://www.w5dxp.com |
Where Does the Power Go?
Jim Kelley wrote:
Not exactly as per Hecht. Note to the casual reader: please be advised that unlike Cecil, Eugene Hecht does not claim that power is equal to irradiance. Here's a quote from "Optics", by Hecht concerning irradiance. "... since the *power* cannot be measured instantaneously, the detector must integrate the energy flux over some finite time, T. ... The time-averaged value of the Poynting vector, ..., is a measure of I." - where I is the irradiance and the Poynting vector is the power flow vector. Hecht seems to treat irradiance the same way that RF engineers treat power flow vectors. Hecht also says that integrating energy flux over some finite time is a measure of power. Presumably, that could be a non-destructive mental integration. In "Optics", the definition of irradiance is: "the average energy per unit area per unit time". That's the same definition as the power flow vector from the IEEE Dictionary: "... *power* per unit-area propagating in the wave". Also from the IEEE Dictionary: "Poynting vector - ... The integral of P(t,r) over a surface is the instantaneous electromagnetic *power flow* through the surface." Do you really think Hecht used the Poynting vector while ignorant of its definition? When an astronomer draws a spherical boundary around the sun and calculates the power output of the sun, most of that power is heading out to empty space and doing no work. Maybe you should campaign to have that calculation removed from publication. -- 73, Cecil http://www.w5dxp.com |
Where Does the Power Go?
Cecil Moore wrote:
Gene Fuller wrote: Don't bother. I understand the physics quite well, thank you. So, is anything technically wrong with what I posted? It is all copied out of various parts of "Optics", by Hecht. Cecil, Yes, there was something technically wrong. The message I responded to had a glaring violation of conservation of energy. Your follow-up corrected the problem. 73, Gene W4SZ |
Where Does the Power Go?
Cecil Moore wrote:
Richard Clark wrote: And electric and magnetic fields exist quite independently of ANY photons (invisible or otherwise) ... Richard, I suggest you take time to digest the material before making any more obviously false statements like the above. -- 73, Cecil, w5dxp.com Cecil, Looked at your site.Love your bike.73 Bill KC9IRR |
Where Does the Power Go?
sapper wrote:
Looked at your site. Love your bike. 73, Bill KC9IRR Just put 1500 miles on it over the long weekend. My sister asked: "What if it's a bad day?" (weather-wise) Told her any day on a Harley is a good day and if I died on it, I would die happy. -- 73, Cecil http://www.w5dxp.com |
Where Does the Power Go?
Gene Fuller wrote:
Yes, there was something technically wrong. The message I responded to had a glaring violation of conservation of energy. Your follow-up corrected the problem. Gene, as you know, there is no such thing as a violation of conservation of energy. But the reflectance at the thin-film surface is 0.01 and a reflection is unavoidable. So how does the reflected 0.01 watts/unit-area of irradiance keep from violating the conservation of energy principle? Where does that energy go? My follow-up answered those questions. Two rearward pointing power flow vectors are associated with wave cancellation of the EM fields. That's destructive interference resulting in constructive interference in the opposite direction. -- 73, Cecil http://www.w5dxp.com |
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