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Natural Science Forum / Physics / Relativity / January 2005Time before the big bang
The general theory is that before the big bang time did not exist, so nothing could be experienced so reality "as we know it didnt exist. Is it just me or does anyone else have a real problem with that. > The general theory is that before the big bang time did not exist, Where did you get that idea from? See http://www.pbs.org/wnet/hawking/mysteries/html/uns_guth_1.html. > so > nothing could be experienced so reality "as we know it didnt exist. > > Is it just me or does anyone else have a real problem with that. If time did not actually exist in what the big bang developed from then I have no problem with that. But the false vacuum of inflation looks like it includes time to me. Thanks Bill > The general theory is that before the big bang time did not exist, so > nothing could be experienced so reality "as we know it didnt exist. > > Is it just me or does anyone else have a real problem with that. Of course. Relativists are simply ducking the issue. Nobody has yet shown any evidence of a big bang anyway. Hubble observed red shift from distant galaxies. He ASSUMED the galaxies are moving away. If we look at Planck's equation, E = hf, what we are really seeing is a drop in energy. Well, we expect to, don't we? A shell of light expands away from its source in a huge sphere, that is a finite quantity of energy being spread over an ever-increasing area. We look at a small slice of that area, and it has less energy than if we'd looked at a similar area from closer to the source. The further away, the less energy. So I expect to find a lower frequency from distant objects, and we do, so why should anyone imagine the object is moving away? And if it isn't moving away, why should anyone imagine a big bang? Androcles. > > The general theory is that before the big bang time did not exist, so > > nothing could be experienced so reality "as we know it didnt exist. [quoted text clipped - 13 lines] > area. > We look at a small slice of that area, and it has less energy than if > we'd looked > at a similar area from closer to the source. > The further away, the less energy. This wave interpretation of light does not account properly for the photoelectric effect. In the photoelectric effect, diminishing intensity does NOT lower the energy delivered to the electrons in the metal. Your interpretation would say that it would. Counter to experimental results. PD > So I expect to find a lower frequency from distant objects, and we do, > so why should anyone imagine the object is moving away? > And if it isn't moving away, why should anyone imagine a big bang? > > Androcles. >> > The general theory is that before the big bang time did not exist, > so [quoted text clipped - 25 lines] > metal. Your interpretation would say that it would. Counter to > experimental results. Fire one photon per second at the metal and measure the current. You'll get at most one electron per second. What do you imagine the intensity of a photon is, anyway? Androcles. > PD > [quoted text clipped - 4 lines] >> >> Androcles. On 1/25/05 2:45 PM, in article KlxJd.24154$2b6.4812@fe3.news.blueyonder.co.uk, "Androcles" <dummy@dummy.net> wrote: >>>> The general theory is that before the big bang time did not exist, >> so [quoted text clipped - 31 lines] > What do you imagine the intensity of a photon is, anyway? > Androcles. And you've missed the point. If E (and f) were attenuated by distance, then attenuating the light would reduce the energy delivered to the electron in the metal, lowering the escape KE of the photoelectrons until some point where there would be finite light impinging on the surface but no photoelectrons emitted. That behavior is NOT seen. The number of photoelectrons is reduced with attenuation but the escape KE is not, and there is no nonzero intensity where photoelectrons are not emitted. Oh, and in case you forgot this, this was another one of Einstein's 1905 papers. PD >> PD >> [quoted text clipped - 4 lines] >>> >>> Androcles. > On 1/25/05 2:45 PM, in article > KlxJd.24154$2b6.4812@fe3.news.blueyonder.co.uk, "Androcles" [quoted text clipped - 40 lines] > > And you've missed the point. No, YOU have missed the point. > If E (and f) were attenuated by distance, E certainly is. It's called the inverse square law. You've probably never heard of it. There is no "If" about it. The enrgy per unit area is reduced as a function of distance. > then > attenuating the light would reduce the energy delivered to the [quoted text clipped - 3 lines] > where there would be finite light impinging on the surface but no > photoelectrons emitted. Absolute nonsense. Spraying bullets from a machine gun mounted on the moon will still kill people and animals on Earth, but your probability of survival will be better than if the machine gun is mounted on the ISS. Line all the dead ones up, pass them through the gates of heaven and start counting, because that is what you are doing when you measure the current in the photo-electric effect. Reducing the energy of the machine gun means slowing down its rate of fire, not taking cordite out of the cartridges. Of course if you do take the cordite out, you can INCREASE the rate of fire and still not kill people. > That behavior is NOT seen. Of course it isn't. You are imagining the bullets are fired using less cordite. In the expanding universe model, only a small area is examined. The number of bullets that kill people in New York City is far less from a machine gun mounted on the moon than from a machine gun mounted on the ISS, but the KE of all bullets doesn't change. You ducked my question, too, which is f.cking typical of a moron like you. What do you imagine the intensity of a photon is? > The number of > photoelectrons is reduced with attenuation but the escape KE is not, [quoted text clipped - 4 lines] > 1905 > papers. Yeah, I know. That idiot didn't know if his arsehole was punched, bored, or drilled and countersunk. He got the photoelectric effect right by real experiment, but then got into his stupid thought experiments and made a right cockup of it. Let's see. I'll tell everyone the speed of photons is independent of the speed of the source and constant in empty space, since "I am enough of an artist to draw freely upon my imagination." They want to believe it, so that's the easy part. Now, having read H.G. Wells' "Time Machine", (the secret to creativity is knowing how to hide your sources) let's see if we can treat time as a vector and make it happen. It should at least make interesting reading, even if it is as fictional as H.G. Wells, and anyway, Lorentz is getting away with that crap about the Earth stretching and shrinking, and "the only source of knowledge is experience." I'm sure I can do better than that, the rest of the world are a bunch of idiots; "only two things are infinite, the universe and human stupidity, and I'm not sure about the former", they'll swallow any crap I give them if I wrap it up nicely in mathematics, "as far as the laws of mathematics refer to reality, they are not certain; as far as they are certain, they do not refer to reality". "Do not worry about your problems with mathematics, I assure you mine are far greater" and "if we knew what it was we were doing, it would not be called research, would it?" The theory says the universe is expanding from a big bang. Therefore it is a fact. "If the facts don't fit the theory, change the facts." -Einstein. Androcles. > > On 1/25/05 2:45 PM, in article > > KlxJd.24154$2b6.4812@fe3.news.blueyonder.co.uk, "Androcles" [quoted text clipped - 29 lines] > >>> photoelectric effect. In the photoelectric effect, diminishing > >>> intensity does NOT lower the energy delivered to the electrons in > >>> the > >>> metal. Your interpretation would say that it would. Counter to [quoted text clipped - 26 lines] > > Absolute nonsense. Spraying bullets from a machine gun mounted on the > moon will still kill people and animals on Earth, but your probability > of survival will be better than if the machine gun is mounted on the [quoted text clipped - 5 lines] > of the cartridges. Of course if you do take the cordite out, you can > INCREASE the rate of fire and still not kill people. You are correct in this analogy to the photon model of the photoelectric effect. > > That behavior is NOT seen. > [quoted text clipped - 4 lines] > machine gun mounted > on the moon than from a machine gun mounted on the ISS, but the KE of > all bullets doesn't change. Precisely, which is why the wave model of light does not account for the photoelectric effect, which was my point, which you missed. > You ducked my question, too, which is f.cking typical of a moron like > you. > What do you imagine the intensity of a photon is? I didn't duck it. I didn't answer it, because it is a poorly posed question, comparable to asking about the pressure due to one gas molecule hitting the wall of a container. The intensity of light is a quantity with units energy/time/area. For monochromatic light, we can write this for a stream of photons as I = Nhf, where N is the number rate density (number of photons per unit area per unit time). [Here, I'm taking N to be an expectation value, after interference among the photon states has been taken into account.] With this clear definition, it's easy to see what the intensity of one photon per unit area per unit time is, though it is meaningless to ask what the intensity of a single photon is. The inverse-square law reduces N, not f. Your first contribution to this topic (see above) was that the inverse square law accounted for the Hubble-observed red shift, which means a reduction of f, not N. That is precisely what I objected to, and for which you have yet to admit error. Moreover, you later in the string explained how the inverse-square law reduces N, not f, contradicting your own earlier claim. So not only were you wrong, but you apparently didn't even know what you were saying. I hope this clarifies your understanding of light and the inverse square law. PD > > The number of > > photoelectrons is reduced with attenuation but the escape KE is not, > > and > > there is no nonzero intensity where photoelectrons are not emitted. > > > > Oh, and in case you forgot this, this was another one of Einstein's > > 1905 > > papers. > > Yeah, I know. That idiot didn't know if his arsehole was punched, bored, > or drilled and countersunk. He got the photoelectric effect right by > real experiment, but then got into his stupid thought experiments and > made a right cockup of it. Note that Einstein did not perform any experiment. He posited the existence of a photon to explain someone else's experimental results -- much the same as he did with another 1905 paper. [snip rant about the wrong 1905 paper] > Androcles. >> > On 1/25/05 2:45 PM, in article >> > KlxJd.24154$2b6.4812@fe3.news.blueyonder.co.uk, "Androcles" [quoted text clipped - 81 lines] > You are correct in this analogy to the photon model of the > photoelectric effect. Of course I am, but thanks for confirming it. >> > That behavior is NOT seen. >> [quoted text clipped - 11 lines] > Precisely, which is why the wave model of light does not account for > the photoelectric effect, which was my point, which you missed. Ah, but it does. Photons are waves, so you do not have a point to miss. YOUR wave model of light may not account for the phenomenon, by MY wave model of light does. You are making the assumption there is only one wave model. >> You ducked my question, too, which is f.cking typical of a moron like > [quoted text clipped - 4 lines] > question, comparable to asking about the pressure due to one gas > molecule hitting the wall of a container. That IS ducking the question, ducky. You are the one stating that intensity will not change the current and the (intrinsic) frequency of a photon will. > The intensity of light is a quantity with units energy/time/area. So the current increases proportionally to the intensity of the light. You contradict yourself. For > monochromatic light, we can write this for a stream of photons as > I = Nhf, where N is the number rate density (number of photons per [quoted text clipped - 6 lines] > > The inverse-square law reduces N, not f. (Smile) Here's your problem: 4 discrete photons hit a unit square eyeball at distance d. We make a hole in thecntre of the unit square eyeball to let them go through. At distance 2d, 4 discrete photons hit 4 square eyeballs, one photon per unit area each. We make 4 holes in the area of 4 square eyeballs to let them go through, but the little buggers wont play by our rules. Instead, they each decide to go through all four holes at the same time. You'll have to change the game to play by THEIR rules, not yours. > Your first contribution to > this topic (see above) was that the inverse square law accounted for > the Hubble-observed red shift, which means a reduction of f, not N. Yep. > That is precisely what I objected to, and for which you have yet to > admit error. My error? Good grief no. No, no no. You tell the photons THEY are in error for not going through the hole you you wanted them to. I have yet to make an error. The red shift is measured with a diffraction grating, or holes in the unit area. If they obeyed your rules, you wouldn't even be able to measure any shift. If they were not waves, again you wouldn't be able to measure any shift. So now YOU can admit your error. Repeat after me, "Oh... Really?...Oh. I see I was confused. OK, I get it now." Well, you can leave out the "OK, I get it now." because you don't. I doubt you ever will, since you are unwilling to learn. But if you were honest, you'd admit you are confused. But anyway, say the words and I'll explain to you what a photon is and how to create one. You won't, of course. You are too proud for that (and too stupid). > Moreover, you later in the string explained how the inverse-square law > reduces N, not f, contradicting your own earlier claim. Yes, it reduces N, but then N photons refuse to go through one hole when there M holes available. So I'm not contradicting myself, I'm contradicting your photon model as the only explanation for the photo-electric effect. If you've got a source of photons shining on a metal plate, put a diffraction grating in the path and move your source further away (about 100 million light years should do it). Now your good little obedient photons should go through one hole in the grating and make the same current, right? > So not only > were you wrong, but you apparently didn't even know what you were > saying. Ah, but I do know what I'm saying, and you are hopelessly confused. > I hope this clarifies your understanding of light and the inverse > square law. I've got a much clearer understanding than you ever will, but we can hope. Androcles > PD > [quoted text clipped - 23 lines] > >> Androcles. > >> > On 1/25/05 2:45 PM, in article > >> > KlxJd.24154$2b6.4812@fe3.news.blueyonder.co.uk, "Androcles" [quoted text clipped - 104 lines] > model of light does. You are making the assumption there is only one > wave model. Ah! A surprise. Now the proper thing to do would be for you to explain YOUR wave model of light that explains all the observed behavior of the photoelectric effect. > >> You ducked my question, too, which is f.cking typical of a moron like > > [quoted text clipped - 8 lines] > intensity will not change the current and the (intrinsic) frequency of a > photon will. No, I'm not. At all. Show me where I said that. > > The intensity of light is a quantity with units energy/time/area. > > So the current increases proportionally to the intensity of the light. > You contradict yourself. No, I don't. Show me my contradictory statements. > For > > monochromatic light, we can write this for a stream of photons as [quoted text clipped - 22 lines] > decide to go through all four holes at the same time. > You'll have to change the game to play by THEIR rules, not yours. What rules are you attributing to me? How does quantum interference have any bearing on the intensity at any distance? Or to be absolutely rock-solid on this: What is the intensity at 2d assuming no quantum interference? What is the intensity at 2d assuming quantum interference? Explain yourself by answering these two questions. > > Your first contribution to > > this topic (see above) was that the inverse square law accounted for [quoted text clipped - 7 lines] > My error? Good grief no. No, no no. > You tell the photons THEY are in error for not going through the hole > you > you wanted them to. I have yet to make an error. > The red shift is measured with a diffraction grating, or holes in the > unit area. > If they obeyed your rules, you wouldn't even be able to measure any > shift. > If they were not waves, again you wouldn't be able to measure any shift. What rules would that be? Or rather, what rules exactly are you attributing to me, and what are the implications of those rules? > So now YOU can admit your error. Repeat after me, "Oh... Really?...Oh. I > see I was confused. OK, I get it now." [quoted text clipped - 60 lines] > > > >> Androcles. >> >> > On 1/25/05 2:45 PM, in article >> >> > KlxJd.24154$2b6.4812@fe3.news.blueyonder.co.uk, "Androcles" [quoted text clipped - 128 lines] > the > photoelectric effect. You are not ready for that. First you must learn the basics. Or you could research Google for earlier explanations if you are in a real hurry. >> >> You ducked my question, too, which is f.cking typical of a moron > like [quoted text clipped - 12 lines] > > No, I'm not. At all. Show me where I said that. Ok: "In the photoelectric effect, diminishing intensity does NOT lower the energy delivered to the electrons in the metal." - Draper. Turn the light off and find out the easy way. Of course if you don't think there is frequency dependency involved, then I've given you too much credit and you don't understand the photoelectric effect at all; my mistake. >> > The intensity of light is a quantity with units energy/time/area. >> [quoted text clipped - 3 lines] > > No, I don't. Show me my contradictory statements. >> For >> > monochromatic light, we can write this for a stream of photons as [quoted text clipped - 28 lines] > > What rules are you attributing to me? I = Nhf, of course. You are squirming, Draper > How does quantum interference have any bearing on the intensity at any > distance? > Or to be absolutely rock-solid on this: > What is the intensity at 2d assuming no quantum interference? > What is the intensity at 2d assuming quantum interference? > Explain yourself by answering these two questions. f.ck off and read below. I've given the explanation. >> > Your first contribution to >> > this topic (see above) was that the inverse square law accounted [quoted text clipped - 21 lines] > What rules would that be? Or rather, what rules exactly are you > attributing to me, and what are the implications of those rules? Look: Empirical data. Distant objects show red shift. According to you, there is no change in the frequency of photons UNLESS the object is moving away. We see a change in frequency. Therefore the object is moving away. According to Planck, E = hf. We expect a change in total energy, the sphere of light is expanding. E (total at d) = Nhf. E (total at 2d) = Nhf. E (total at Nd) = Nhf. E (unit area at d) = Nhf/N or one photon per unit area. E (unit area at 2d) = Nhf / 4N. What the f.ck is a quarter of a photon, moron? Androcles. >> So now YOU can admit your error. Repeat after me, "Oh... > Really?...Oh. I [quoted text clipped - 71 lines] >> > >> >> Androcles. > >> >> > On 1/25/05 2:45 PM, in article > >> >> > KlxJd.24154$2b6.4812@fe3.news.blueyonder.co.uk, "Androcles" > >> >> > <dummy@dummy.net> wrote: > >> >> > > >> >> >> "PD" <pdraper@yahoo.com> wrote in message news:1106667982.276684.290180@f14g2000cwb.googlegroups.com... > >> >> >>>>> The general theory is that before the big bang time did not > >> > exist, [quoted text clipped - 122 lines] > > Ah! A surprise. Now the proper thing to do would be for you to explain > > YOUR wave model of light that explains all the observed behavior of > > the > > photoelectric effect. > > You are not ready for that. First you must learn the basics. > Or you could research Google for earlier explanations if you are in a > real hurry. > [quoted text clipped - 18 lines] > intensity does NOT lower the energy delivered to the electrons in > the metal." - Draper. And that's not what you said I said. The energy delivered to each electron does not determine the current. The *number* of electrons liberated determines the current. > Turn the light off and find out the easy way. > Of course if you don't think there is frequency dependency involved, > then I've given you too much credit and you don't understand the > photoelectric effect at all; > my mistake. Diminishing the number of incident photons to diminish the intensity does not change the frequency. > >> > The intensity of light is a quantity with units energy/time/area. > >> [quoted text clipped - 3 lines] > > > > No, I don't. Show me my contradictory statements. Contradiction vaporized. > >> For > >> > monochromatic light, we can write this for a stream of photons as [quoted text clipped - 31 lines] > I = Nhf, of course. > You are squirming, Draper And what do you think is the right expression for the intensity? I'm not squirming at all. I'm asking you to explain what the heck you're talking about. > > How does quantum interference have any bearing on the intensity at any > > distance? [quoted text clipped - 4 lines] > > f.ck off and read below. I've given the explanation. No, you haven't. You've given some talk about how the photons go through all apertures at once. You haven't addressed the above questions at all. Pick a value for N at d, your choice. Now give me numbers or algebraic expressions that are answers to the two questions above. > >> > Your first contribution to > >> > this topic (see above) was that the inverse square law accounted [quoted text clipped - 31 lines] > E (total at 2d) = Nhf. > E (total at Nd) = Nhf. Note that the value of N is *different* in each of these cases. Read my definition. Moroever, you've misidentified I (intensity) as E (total at...). Read the definition. > E (unit area at d) = Nhf/N or one photon per unit area. > > E (unit area at 2d) = Nhf / 4N. No. This is just plain wrong. Your algebra is atrocious. Let's set N = 1 at d (that's one photon per unit area per unit time - note the units) at d. Then at d, I = Nhf = hf. That's the energy per unit area per unit time, not a count of photons. Then at 2d, I = (N/4)hf = hf/4. Again, that's the energy per unit area per unit time, not a count of photons. > What the f.ck is a quarter of a photon, moron? A rate of one photon per area of four per unit time is precisely a quarter of a photon per unit area per unit time. You got a problem with that? > Androcles. > [quoted text clipped - 73 lines] > >> > > >> >> Androcles. >> >> >> > On 1/25/05 2:45 PM, in article >> >> >> > KlxJd.24154$2b6.4812@fe3.news.blueyonder.co.uk, "Androcles" [quoted text clipped - 196 lines] > Diminishing the number of incident photons to diminish the intensity > does not change the frequency. So you say. Want to express that mathematically ? >> >> > The intensity of light is a quantity with units > energy/time/area. [quoted text clipped - 6 lines] > > Contradiction vaporized. So you say. Want to express that mathematically ? >> >> For >> >> > monochromatic light, we can write this for a stream of photons [quoted text clipped - 41 lines] > not squirming at all. I'm asking you to explain what the heck you're > talking about. I'm not the one talking quanta, you are. If at distance d one photon falls per unit area, and at distance 2d one quarter of a photon falls per unit area, what is the energy of a quarter of a photon? >> > How does quantum interference have any bearing on the intensity at > any [quoted text clipped - 10 lines] > questions at all. > Pick a value for N at d, your choice. Ok, 1. Now give me numbers or algebraic > expressions that are answers to the two questions above. E = 1 * hf. "What is the intensity at 2d assuming no quantum interference?" E/4, by the inverse square law. "What is the intensity at 2d assuming quantum interference?" I haven't foggiest idea. You tell me. Androcles. > >> >> >> > On 1/25/05 2:45 PM, in article > >> >> >> > KlxJd.24154$2b6.4812@fe3.news.blueyonder.co.uk, "Androcles" [quoted text clipped - 198 lines] > > So you say. Want to express that mathematically ? Sure. I = Nhf. Keep f constant. Lower N. I lowers proportionally. There is no mathematical requirement that f changes. So suppose, conversely, that I drops. In this case, either N would drop or f would drop, or both. Under the photon model, N drops and f does not. You say otherwise. Prove it. > >> >> > The intensity of light is a quantity with units > > energy/time/area. [quoted text clipped - 8 lines] > > So you say. Want to express that mathematically ? You haven't shown me my contradictory statements. > >> >> For > >> >> > monochromatic light, we can write this for a stream of photons [quoted text clipped - 46 lines] > 2d one quarter of a photon falls per unit area, what is the energy of a > quarter of a photon? You are assuming that something has to fall in every unit of area. That is a mistaken assumption. Suppose I have five strawberries and I throw them on a chessboard with 2 inch squares. The density of strawberries on the chessboard is (5 strawberries)/(256 square inches) = 0.0195 strawberries/square inch. Does that mean I have to try to cut the strawberries so that there are 0.0195 strawberries per unit area? How would you do that? > >> > How does quantum interference have any bearing on the intensity at > > any [quoted text clipped - 17 lines] > > E = 1 * hf. Wrong. I = 1*hf. But we'll make the corrections on the fly. > "What is the intensity at 2d assuming no quantum interference?" > > E/4, by the inverse square law. Wrong. I = hf/4. But your general idea is close enough. > "What is the intensity at 2d assuming quantum interference?" > > I haven't foggiest idea. You tell me. I say it's I = hf/4. But YOU said: > >> >> (Smile) > >> >> > >> >> Here's your problem: [...] > >> >> We make 4 holes in the area of 4 square eyeballs to let them go > >> > through, > >> >> but the little buggers wont play by our rules. Instead, they each > >> >> decide to go through all four holes at the same time. > >> >> You'll have to change the game to play by THEIR rules, not yours. Since this was YOUR correction to my rules, you tell me. PD >> >> >> >> > On 1/25/05 2:45 PM, in article >> >> >> >> > KlxJd.24154$2b6.4812@fe3.news.blueyonder.co.uk, [quoted text clipped - 310 lines] > That > is a mistaken assumption. Oh, you mean if I step sideways, I can't see the star. Guess what? I can. Yes, the light illuminates the entire area, by observation, even if not by your theory. "If the facts don't fit the theory, change the facts." --Albert Einstein In fact, you are one of the master fact changers, in fact, and that's a fact. > Suppose I have five strawberries and I throw them on a chessboard with > 2 inch squares. The density of strawberries on the chessboard is (5 > strawberries)/(256 square inches) = 0.0195 strawberries/square inch. > Does that mean I have to try to cut the strawberries so that there are > 0.0195 strawberries per unit area? How would you do that? What you asking me for? Ask the photons why they go through two holes instead of one. All I know is that they do. Heck, I used to carry a diffraction grating in my wallet, and I KNOW they do. You are the one that knows all the answers, you tell me, know-it-all teacher. You didn't answer my question, what is the intensity of a 1/4 of a photon? So I'll answer it myself. E = Nhf at d, so E/4 = Nh f/4 at 2d because energy is conserved and h is a constant. N is an integer and a quantum number, I've chosen N = 1, so the only thing left to divide is the frequency. So I expect to see distant galaxies red shifted without them having to be moving. And guess what? We do. Good thing too, or we'd have all that nonsense about an imaginary big bang to contend with. sh.t, man, I offered to buy Hawking a beer when he was down at Sussex U, but that bitch of a first wife of his wheeled him away before I could straighten him out. Still, he made a couple of quid with that fairy tale "A Brief History of Time", so I suppose she was doing him a favour. What are you doing talking to me for, anyway? Go write a good fairy story, call it physics and make yourself a few bucks. As with all religous stories and fairy tales, make sure it happens way back in the past so that nobody can check it, and you are on a winner. >> >> > How does quantum interference have any bearing on the intensity > at [quoted text clipped - 20 lines] > > Wrong. I = 1*hf. But we'll make the corrections on the fly. Blame Max Planck. It is his equation, not mine. I fail to see what problem you have associating Energy with Intensity, but that's your hangup. >> "What is the intensity at 2d assuming no quantum interference?" >> >> E/4, by the inverse square law. > > Wrong. I = hf/4. But your general idea is close enough. Tell it to Max Planck. It's his equation, not mine. >> "What is the intensity at 2d assuming quantum interference?" >> >> I haven't foggiest idea. You tell me. > > I say it's I = hf/4. Yeah, well, you want to disagree with Planck. Pick a fight with him, not me. > But YOU said: > [quoted text clipped - 11 lines] > > Since this was YOUR correction to my rules, you tell me. What, now that you've snipped it? Kiss my arse. Androcles. > PD > >> >> >> >> > On 1/25/05 2:45 PM, in article > >> >> >> >> > KlxJd.24154$2b6.4812@fe3.news.blueyonder.co.uk, [quoted text clipped - 319 lines] > In fact, you are one of the master fact changers, in fact, and that's a > fact. Not at the rate of a few photons per unit time, light doesn't. And it's plain you haven't observed light at low intensities. I have, in a ring-imaging Cerenkov counter. If you want to see a simulation of how low rate light looks, check out http://www.colorado.edu/physics/2000/applets/twoslitsb.html The label says electrons, but the physics is EXACTLY the same as with photons. If you wait, you will see the 2-slit interference pattern emerge, but you will also see the the particles land on the screen one at a time, even though quantum mechanically each particle goes through both slits. This is indeed reproducible in the lab, with photons, and it has been done. > > Suppose I have five strawberries and I throw them on a chessboard with > > 2 inch squares. The density of strawberries on the chessboard is (5 [quoted text clipped - 5 lines] > instead of one. All I know is that they do. Heck, I used to carry a > diffraction grating in my wallet, and I KNOW they do. You are the one > that knows all the answers, you tell me, know-it-all teacher. > [quoted text clipped - 5 lines] > E/4 = Nh f/4 at 2d because energy is conserved and h is a constant. N is > an integer and a quantum number, I've chosen N = 1, so the only thing > left to divide is the frequency. N is *not* a quantum number and it is NOT an integer, not in the definition of intensity. Read the definition above, instead of redefining it. And that's the difference between E and I, which you don't seem to get. The units are different, just like mass and mass density. Let's go to a region of space where there is one proton every 1000 cubic meters. What is the density of matter in protons/cubic meter? Oh, hell, that's 0.001 protons/cubic meter. Where am I going to get a thousandth of a proton? WAIT... But I suppose that means that if I step to one side, I'm in empty space, when I *know* it's not empty because there's 1 proton every 1000 cubic meters. Ah, it's ok, after all, because the proton goes through all the holes simultaneously, so it's really one proton in every cubic meter, so the density must be really m = N * (m_p), where m_p is the mass of one proton. and since N =1, the density is m_p. Am I getting it? > So I expect to see distant galaxies red shifted without them having to > be moving. > And guess what? We do. > Good thing too, or we'd have all that nonsense about an imaginary big > bang to contend with. > > sh.t, man, I offered to buy Hawking a beer when he was down at Sussex U, > but that bitch of a first wife of his wheeled him away before I could > straighten him out. I'll bet she did. Hope she took out a restraining order, as well. > Still, he made a couple of quid with that fairy tale > "A Brief History of Time", so I suppose she was doing him a favour. [quoted text clipped - 31 lines] > problem you have associating Energy with Intensity, but that's your > hangup. Um... different units? But maybe you're right, we can identify force with pressure, mass with density, energy with power -- no problem, close enough. Prefer one over the other anyway. Let's just use the ones we like. > >> "What is the intensity at 2d assuming no quantum interference?" > >> [quoted text clipped - 3 lines] > > Tell it to Max Planck. It's his equation, not mine. For *intensity*? Don't think so. Look again. Or better yet, cite for proof. > >> "What is the intensity at 2d assuming quantum interference?" > >> [quoted text clipped - 3 lines] > > Yeah, well, you want to disagree with Planck. Pick a fight with him, > not me. No, you are maintaining that Planck said something he did not. Is that slander? > > But YOU said: > > [quoted text clipped - 14 lines] > What, now that you've snipped it? Kiss my arse. > Androcles. Snipped nothing. It's all above, in original context. It's *copied* here below. Not in the mood. > > PD >> >> >> >> >> > On 1/25/05 2:45 PM, in article >> >> >> >> >> > KlxJd.24154$2b6.4812@fe3.news.blueyonder.co.uk, [quoted text clipped - 371 lines] > plain you haven't observed light at low intensities. I have, in a > ring-imaging Cerenkov counter. Then you should know that particles can travel faster than light. Next time, think about the aurora. > If you want to see a simulation of how > low rate light looks, check out > http://www.colorado.edu/physics/2000/applets/twoslitsb.html > The label says electrons, but the physics is EXACTLY the same as with > photons. Oh, electrons travel at the same speed a photons, then ? Actually I seen the real thing, I don't need a simulation. The electrons were bounced off a crystal lattice. You'll never do it with slits, the gap is too large. If you wait, you will see the 2-slit interference pattern > emerge, but you will also see the the particles land on the screen one > at a time, even though quantum mechanically each particle goes through > both slits. This is indeed reproducible in the lab, with photons, and > it has been done. I don't need a lab, sonny. I can buy a quality diffraction grating for $64.00 http://www.edmundoptics.com/onlinecatalog/displayproduct.cfm?productID=1896&srti tem_price=D >> > Suppose I have five strawberries and I throw them on a chessboard > with [quoted text clipped - 24 lines] > > N is *not* a quantum number and it is NOT an integer, N is not an integer? Wow! You sure fooled me with that one. You have a nice day, now. Androcles. not in the > definition of intensity. Read the definition above, instead of > redefining it. And that's the difference between E and I, which you [quoted text clipped - 128 lines] > >> > PD All conversations with you are now terminated. PD > All conversations with you are now terminated. > PD "Androcles, in your case, I will get over my disenchantment. But I want this to be a fruitful exchange between the two of us, so let's agree on some ground rules. We'll go things one little step at a time. When we get to a point of conflict, we'll identify what the error is on either side, and the party in error MUST acknowledge the error and remove the erroneous statement from further discussion." - Draper I knew I'd get you to to eat those words, Draper. How does crow taste, you f.cking imbecile? Androcles. Androcles clearly stated that he consider that 1/x = x for any non null x. >> > On 1/25/05 2:45 PM, in article >> > KlxJd.24154$2b6.4812@fe3.news.blueyonder.co.uk, "Androcles" [quoted text clipped - 53 lines] >> E certainly is. It's called the inverse square law. You've probably >> never heard of it. And you obviously don't understand it. In order for the inverse square law to exist, E must be constant. It is surface area that increases with the square of the radius. Since a fixed amount of energy is distributed over the surface area of a sphere, its surface density must decrease as the radius increases. It is intuitively obvious to the most casual observer that if E also decreased, then the inverse square law would not hold true. >> There is no "If" about it. The enrgy per unit area is reduced as a >> function of distance. exactly as it should for individual photons of constant energy. >> > then >> > attenuating the light would reduce the energy delivered to the [quoted text clipped - 8 lines] >> moon will still kill people and animals on Earth, but your > probability silly - none of the bullets would escape the moon, and if any did they would never make it through the atmosphere. >> of survival will be better than if the machine gun is mounted on the >> ISS. [quoted text clipped - 27 lines] > >> You ducked my question, too, which is f.cking typical of a moron like Pot. Kettle. Black. >> you. >> What do you imagine the intensity of a photon is? [quoted text clipped - 52 lines] > >> Androcles. >>> E certainly is. It's called the inverse square law. You've probably >>> never heard of it. > > And you obviously don't understand it. I thought I was on your kill-file? What the f.ck are you sticking your oar in for now? Oh, sorry, you mean Draper doesn't understand it. That's right, he doesn't. > In order for the inverse square > law to exist, E must be constant. That's right, ducky. Energy is conserved, as if we didn't know. It is, however, spread over an ever- increasing area It is surface area that increases with > the square of the radius. Yes, ducky, with a finite quantity of conserved energy spread over that increasing area. Since a fixed amount of energy is distributed > over the surface area of a sphere, its surface density must decrease > as > the radius increases. Yes ducky, that's what I've been saying. It is intuitively obvious to the most casual observer > that if E also decreased, then the inverse square law would not hold > true. Good for you, ducky. Well done, you are making my point for me. Do carry on. >>> There is no "If" about it. The enrgy per unit area is reduced as a >>> function of distance. [quoted text clipped - 53 lines] > > Pot. Kettle. Black. You've got it, and I'm black. Now laugh. Got any more irrelevant babble to offer? Nope? Well, do carry on with the analysis, you kinda stopped short when you got to the finite energy spreading over an ever-increasing area. Androcles. > The general theory is that before the big bang time did not exist, so > nothing could be experienced so reality "as we know it didnt exist. > > Is it just me or does anyone else have a real problem with that. Time doesn't *exist* except as a intellectual construct. It's just a tool used by scientists. Unfortunately, even most of them mistake it for something physical. Ask yourself, if you were in a space devoid of any objects whatsoever, which of your senses would *feel* time? > > The general theory is that before the big bang time did not exist, so > > nothing could be experienced so reality "as we know it didnt exist. [quoted text clipped - 4 lines] > a tool used by scientists. Unfortunately, even most of them mistake > it for something physical. I'm glad to find out that we finally agree. > Ask > yourself, if you were in a space devoid of any objects whatsoever, > which of your senses would *feel* time? You would feel your heart beat. Dirk Vdm > > > The general theory is that before the big bang time did not exist, so > > > nothing could be experienced so reality "as we know it didnt exist. [quoted text clipped - 6 lines] > > I'm glad to find out that we finally agree. So you must've learned something from me. > > Ask > > yourself, if you were in a space devoid of any objects whatsoever, > > which of your senses would *feel* time? > > You would feel your heart beat. That's an object, you moron. > > "AllYou!" <idaman@conversent.net> wrote in message > news:qLydnbmV_48SomvcRVn-pA@conversent.net... [quoted text clipped - 11 lines] > > So you must've learned something from me. Actually, that comment belonged before your sentence that starts with the word "Unfortunately", but never mind. > > > Ask > > > yourself, if you were in a space devoid of any objects whatsoever, [quoted text clipped - 3 lines] > > That's an object, you moron. So you actually request someone to ask himself, if he were not alive, which of his senses would feel time. Now that is a very interesting question to ask in a science newsgroup. Dirk Vdm > > > "AllYou!" <idaman@conversent.net> wrote in message > > news:qLydnbmV_48SomvcRVn-pA@conversent.net... [quoted text clipped - 25 lines] > So you actually request someone to ask himself, if he were > not alive, which of his senses would feel time. I'm sure you interpreted it that way. > Now that is a very interesting question to ask in a science > newsgroup. I'm sure you feel that way. > > "AllYou!" <idaman@conversent.net> wrote in message > news:HLGdnZkQQqkH3mvcRVn-hA@conversent.net... > > > > > > > "AllYou!" <idaman@conversent.net> wrote in message > > > news:qLydnbmV_48SomvcRVn-pA@conversent.net... [snip] > > > > > Ask > > > > > yourself, if you were in a space devoid of any objects whatsoever, [quoted text clipped - 8 lines] > > I'm sure you interpreted it that way. Can you seriously give another way to interpret it? Dirk Vdm > > > "AllYou!" <idaman@conversent.net> wrote in message > > news:HLGdnZkQQqkH3mvcRVn-hA@conversent.net... [quoted text clipped - 18 lines] > > Can you seriously give another way to interpret it? Yes. > > "AllYou!" <idaman@conversent.net> wrote in message > news:V-ednScj77np1GvcRVn-qQ@conversent.net... [quoted text clipped - 27 lines] > > Yes. How do you interpret it in different way than I do? Dirk Vdm > > > "AllYou!" <idaman@conversent.net> wrote in message > > news:V-ednScj77np1GvcRVn-qQ@conversent.net... [quoted text clipped - 29 lines] > > How do you interpret it in different way than I do? You'd never understand, or even if you did, you'd never admit it. Your transparency is laughable. > > > > [snip] > > > > [quoted text clipped - 19 lines] > You'd never understand, or even if you did, you'd never > admit it. Your transparency is laughable. Good job. http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Transparency.html Thanks :-) Dirk Vdm Dear Steve J: > The general theory is that before the big bang time did not exist, so > nothing could be experienced so reality "as we know it["] didnt exist. Before there was "left" there was "up". Before the mass/energy in this Universe cooked up spacetime, there was spacetime (or timespace) in the Universe that contains ours. Time doesn't exist at the quantum level, so the Unvierse doesn't really require time to "experience reality". And "as we know it" is pretty provincial, since "as we know it" covers the surface of this planet pretty well, but not a whole lot else. David A. Smith > Time doesn't exist at the quantum level, so the Unvierse doesn't really > require time to "experience reality". Examine a lump of radium-216m. Then wait awhile. Then examine it again. Something has changed even on the quantum level.  SignatureThe world is not fair. Observations of Bernard - No 70 Dear Bernardz: >> Time doesn't exist at the quantum level, so the Unvierse doesn't really >> require time to "experience reality". > > Examine a lump of radium-216m. Then wait awhile. Then examine it again. > Something has changed even on the quantum level. Look at the word "lump". This implies a statistical quantity, a large sample. In the large, the second law of thermodynamics applies, which provides a "preferred direction" to the "arrow of time". At the quantum level, it is possible, given the availability of constituents, that any single radium-216 atom can absorb an alpha particle and become thorium-220. Or the radium-216 could decay, then absorb the daughters, and become radium-216 again. You cannot look at any single atom, a quantum particle, and say anything about "half life". Processes are fully reversible (which goes against the second law), and there is no way to mark the passage of time. Even the "orbiting" of the electron cannot be used as the hands of a clock... David A. Smith > Dear Bernardz: > [quoted text clipped - 7 lines] > sample. In the large, the second law of thermodynamics applies, which > provides a "preferred direction" to the "arrow of time". Are you saying that universe is not such a statistical quantity? > At the quantum level, it is possible, given the availability of > constituents, that any single radium-216 atom can absorb an alpha particle [quoted text clipped - 5 lines] > second law), and there is no way to mark the passage of time. Even the > "orbiting" of the electron cannot be used as the hands of a clock... > David A. Smith SignatureIf you give people more money, they can spend it Observations of Bernard - No 71 Dear Bernardz: >> Dear Bernardz: >> [quoted text clipped - 11 lines] > > Are you saying that universe is not such a statistical quantity? It certainly is. But "the Universe" is only quantum at a scale that we are only now trying to understand. And at that scale, the second law of thermodynamics, and time, do not apply. David A. Smith befor the big bang the wall of the balloon of condenced energy particals was 114 atoms thick on the wall of a balloon 1 billion light years wide. The universe is now 34 billion light years wide. Planets we see 14 billion years ago have speed evrything till now has a speed of 0.02 C max X 14 billion is 170 million light years of distance traved by matter. |
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