doG Posted June 21, 2008 Posted June 21, 2008 Will you stop trying to make things up and ducking the questions? Is it too late for me to nominate Mike for the Most Unsupported Proclamations Award or something similar?
D H Posted June 21, 2008 Posted June 21, 2008 I substitute 'exact' quantities for my components. h for E, electron mass for mass and Photon for c or light. E = m sub e x photon wavelenth^2. Can you understand that simple explanation? No, I cannot, for three reasons The left hand side has units of energy. The base units of energy are mass*length2/time2. The right hand side has units of mass*length2. The expression makes no sense. It is not even well-formed. What photon wavelength? Photons can have any wavelength. Once again, the expression makes no sense. What makes it valid to combine the mass of an electron with the wavelength of a photon? You can't just toss numbers together willy-nilly and without justification. The expression makes no sense yet again. And you're claiming that joule·seconds divided by mass is a length?[/quote']Square root of, yes. Light is reduced to just ONE single component that is a photon. No! Joule·seconds divided by mass has units of length2/time. You cannot take the square root of a value that has such units.
Reaper Posted June 21, 2008 Posted June 21, 2008 (edited) Is it too late for me to nominate Mike for the Most Unsupported Proclamations Award or something similar? Nope, the Awards have just begun . Although, I would wait before I nominate this guy, as he hasn't been around long enough yet (although his Baez level is higher than Farsight's, so....) And, Mike's crackpottery might even be strong enough to rival Zarkov . Edited June 21, 2008 by Reaper multiple post merged
Sayonara Posted June 21, 2008 Posted June 21, 2008 No, I don't think so. Zarkov didn't even attempt maths, so he is clearly in the lead.
Reaper Posted June 21, 2008 Posted June 21, 2008 (edited) Well, which do you think is worse, not doing the math, or butchering it ?? Edited June 21, 2008 by Reaper multiple post merged
Sayonara Posted June 21, 2008 Posted June 21, 2008 Not doing it, because then you can claim that you would do it, if only you thought your audience were good enough for you to condescend to writing it down for them. As opposed to revealing your mental state by committing it to type from the word go.
Reaper Posted June 21, 2008 Posted June 21, 2008 Ah, ok then. I personally don't consider numerology math, but I concede the point otherwise. But, he is the biggest crackpot to come along since Farsight, having a Baez index even higher then him (my previous experience and skirmishes with him tells me that anyways). Certainly strong enough to play a good round of Crackpot Bingo
doG Posted June 22, 2008 Posted June 22, 2008 Ahem. Let's keep it civil here. Nothing derogatory intended. I was only looking for a term to describe someone which heroically strays from and evades the truth
New Science Posted June 24, 2008 Author Posted June 24, 2008 Why? How? What formulas did you use/derive. What is the physical basis for them. We want to see them' date=' not just hear about it. And, preferably show us algebra or calculus formulas (e.g. functions). [/quote'] Did you read the article page? The answers you ask here are answered in that artcle. You just contradicted yourself. And, again, a photon is both a particle and a wave, all particles have a de Broglie wavelength. The photons are NOT particles. Particles have mass. Photons do not have mass. Photons are both a particle and a wave. And, you do not get frequency by dividing wavelength into the speed of light; it's the other way around: f=c/(lambda). That is what I said. Maybe I should have expressed it mathemathecally. Why? How? What formulas did you use/derive. What is the physical basis for them. We want to see them, not just hear about it. And, preferably show us algebra or calculus formulas (e.g. functions). See 1st question above. That's true for any wave.... So how can the wavelength vary and the frequency be 1? Because the photon has a physical dimension depending on which 'energy' level it was created in. And as I said on another reply' date=' photons are just one energy quanta (photon) pulses. Only their dimentional widths are variable. Really? What physical reason is there for modifying Einstein's formula? E'sF misleads the idea that mass is directly changed into energy. It also is using values that are not precise in their meanings like my idea of using single component values/quantities. You just contradicted yourself. And, again, a photon is both a particle and a wave, all particles have a de Broglie wavelength. deBroglie used the Bohr model to create his electron matter wave formula. He also created a 'light' formula. E = hv, v=frequecy. No! Joule·seconds divided by mass has units of length2/time. You cannot take the square root of a value that has such units. What you describe above is Einsteins c^2 component. c is a dimension quantity and a time of one second. Since my formula was devoted to reducing everything to just one value, then c^2 or Lambda^2 can be reduced to the single value by taking the square root of the value. I used Plancks quanta that is a single value and electron mass. So why should not I also reduce the photon to a single value of one photon? NS
Klaynos Posted June 24, 2008 Posted June 24, 2008 Did you read the article page?The answers you ask here are answered in that artcle. I still don't think it answers the physical basis question. The photons are NOT particles. Particles have mass. Photons do not have mass. Photons are indeed particles, this is quite trivial to show with photoelectric effects, varying the intensity and wavelength. That is what I said. Maybe I should have expressed it mathemathecally. But then you go f = 1, which means either c has to be changing (not experimentally seen) or that the wavelength is constant. See 1st question above. See first response. Because the photon has a physical dimension depending on which 'energy' level it was created in.And as I said on another reply, photons are just one energy quanta(photon) pulses. Only their dimentional widths are variable. But you agreed that c = frequency * wavelength. Which is it frequency = 1 and a static wavelength or both varying. Experiments show us c is constant. E'sF misleads the idea that mass is directly changed into energy. Mass is changed into energy, look up binding energy in nuclear physics. It also is using values that are not precise in their meanings like my idea of using single component values/quantities. The values are very precise if you do the maths rigorously. deBroglie used the Bohr model to create his electron matter wave formula. That's not how I've seen it described, he did work with Bohr but Bohr knew his model was wrong, and worked extensively on QM post the model. He also created a 'light' formula. E = hv, v=frequecy. Really, so? What you describe above is Einsteins c^2 component.c is a dimension quantity and a time of one second. c has the dimension of length/time Since my formula was devoted to reducing everything to just one value, then c^2 or Lambda^2 can be reduced to the single value by taking the square root of the value.I used Plancks quanta that is a single value and electron mass. So why should not I also reduce the photon to a single value of one photon? NS So you're making stuff up so it fits what you want?
D H Posted June 24, 2008 Posted June 24, 2008 The photons are NOT particles. Particles have mass. Photons do not have mass. Not all particles have mass. Photons, for example, are massless particles. That is why photons go at the speed of light. No! Joule·seconds divided by mass has units of length2/time. You cannot take the square root of a value that has such units.[/quote']What you describe above is Einsteins c^2 component. I am not talking about "Einstein's c2 component". I am talking about Dimensional Analysis: 1 joule·second / 1 kilogram = 1 meter2/second. This expression does not have dimensions of length2. You cannot take the square root of the numerical value and call it a length. This is very elementary stuff. If you cannot understand this you have absolutely no business delving into advanced physics. Klaynos gave a good reference in this post: New Science please read: http://en.wikipedia.org/wiki/Dimensional_analysis To understand what we mean when we say your equation does not satisfy dimensional analysis! c is a dimension quantity and a time of one second. That makes no sense. c is a velocity, and like all other velocities, it has dimensions of length/time. In the metric system, c is exactly 299,792,458 meters/second. In natural units, c has a numerical value of 1 but it still has dimensions of length/time. c is not a dimensionless constant in any reference system.
New Science Posted June 25, 2008 Author Posted June 25, 2008 Not all particles have mass. Photons, for example, are massless particles. That is why photons go at the speed of light. So, I am discussing photons, not the old contiuous wave concept. I am not talking about "Einstein's c2 component". I am talking about Dimensional Analysis: 1 joule·second / 1 kilogram = 1 meter2/second. This expression does not have dimensions of length2. You cannot take the square root of the numerical value and call it a length. This is very elementary stuff. If you cannot understand this you have absolutely no business delving into advanced physics. What do you call the meter? It is a component of your J-S. Using your D A, how would you explain the reason for Einsteins use of C^2? Klaynos gave a good reference in this post: Why use him for an answer? That makes no sense. c is a velocity, and like all other velocities, it has dimensions of length/time. In the metric system, c is exactly 299,792,458 meters/second. In natural units, c has a numerical value of 1 but it still has dimensions of length/time. c is not a dimensionless constant in any reference system. Who said it was? When I reduce a photon to a frequency of one, it still has its own identity with its wavelenght representing different energy levels. But they all still have a 'f' of ione. NS
Klaynos Posted June 25, 2008 Posted June 25, 2008 E = mc2 Joule = Kg m/s m/s Which has dimensions, mass length^2 / time^2 Now let's consider: E = 0.5 mv2 Joule = 0.5 Kg m/s m/s Which has dimensions, mass length^2 / time^2 ODG! it works! This isn't how E=mc2 is derived at all but is a valid test that if it were to fail it would mean the equation is wrong, have you read the link I posted and the D H referred you back to on DA? You've still not addressed my question above, are we having fixed frequency and wavelength, or changeable wavelength and frequency? You've got to pick one...
New Science Posted June 26, 2008 Author Posted June 26, 2008 Klaynos #62 I still don't think it answers the physical basis question. Photons are indeed particles, this is quite trivial to show with photoelectric effects, varying the intensity and wavelength. But then you go f = 1, which means either c has to be changing (not experimentally seen) or that the wavelength is constant. C does not apply to the concept of photons. A photon is just a fraction of C. A photon does have a dimension (wavelenth) and an elepsed time of c / wavelength = 1 / f. But that does not mean it has a frequency other than one because a 'photon' is not a continuous wave.. But you agreed that c = frequency * wavelength. Which is it frequency = 1 and a static wavelength or both varying. Experiments show us c is constant. I never said it wasn't. It can be used though to determine the elapsed time of a photon because photons do move at c velocity. The wavelength of one applies to the photons only. Mass is changed into energy, look up binding energy in nuclear physics. The values are very precise if you do the maths rigorously. When a fission bomb explodes, their is both a mass increase and a lot of energy created. If this energy was a mass conversion, than the mass should have descreased, not increased. The energy actially was there in the form of potential energy that was contained by the strong force. So mass was not even involved in this transformation. That's not how I've seen it described, he did work with Bohr but Bohr knew his model was wrong, and worked extensively on QM post the model. As far as the HA is concerned, Bohr's, planetary model is correct because the S'r equations confirmed his energy levels. Besides, all the physics books contain his model. So you're making stuff up so it fits what you want? My posts are NOT imaginary creations. But I do use a lot of visualizations to give me my new interpretations of science. NS
Klaynos Posted June 26, 2008 Posted June 26, 2008 I like the copy and pastes btw, you argued this before somewhere else? Klaynos #62 C does not apply to the concept of photons. A photon is just a fraction of C. A photon does have a dimension (wavelenth) and an elepsed time of c / wavelength = 1 / f. But that does not mean it has a frequency other than one because a 'photon' is not a continuous wave.. The speed of light does not apply to light, nice, very nice. You seem to be arguing against yourself here, you say that it has a freqency (although are you sure c/wavelenght = 1/f is right and it's not wavelength * frequency = c ?) I never said it wasn't. It can be used thoughto determine the elapsed time of a photon because photons do move at c velocity. The wavelength of one applies to the photons only. Elapsed time? But relativity clearly shows us that photons are timeless. How can that work, if you fire a single photon you can measure it's wavelength/frequency quite easily. When a fission bomb explodes, their is both a mass increase and a lot of energy created. I think you'll find it's a mass decrease, the bonding energy/nuceon moves left on the graph.... If this energy was a mass conversion, than the mass should have descreased, not increased. See my point above.... I've actually studied nuclear fission weapons... so lets check if I'm right... Q = (M(A1,Z1 - M(A2,Z2 - M(A3,Z3) Where 1 is the start product, 2 and 3 are the decays. Fission can only occure if Q > 0, so the mass must be decreaseed. If we put real values into the SEMF we get a nice graph, that matches our experiements quite accurately, and that includes measuring the decayed products. The energy actially was there in the form ofpotential energy that was contained by the strong force. So mass was not even involved in this transformation. But colour does not venture outside of the nucleon! As far as the HA is concerned, Bohr's, planetarymodel is correct because the S'r equations confirmed his energy levels. Besides, all the physics books contain his model. They also say that it's wrong. And certainly not ALL physics books contain it, you've been talked to about why books have it in before. There is massive amounts of evidence that shows the Bohr model of the atom is wrong... Bohr himself KNEW it was wrong. My posts are NOT imaginary creations. But I do use a lot of visualizations to give me my new interpretations of science. NS Still appears you're making stuff up so it fits.
New Science Posted June 26, 2008 Author Posted June 26, 2008 E = mc2 Joule = Kg m/s m/s Which has dimensions, mass length^2 / time^2 Now let's consider: E = 0.5 mv2 Joule = 0.5 Kg m/s m/s Which has dimensions, mass length^2 / time^2 ODG! it works! This isn't how E=mc2 is derived at all but is a valid test that if it were to fail it would mean the equation is wrong, have you read the link I posted and the D H referred you back to on DA? You've still not addressed my question above, are we having fixed frequency and wavelength, or changeable wavelength and frequency? You've got to pick one... As I have said in defining the photon, once it is radiated, it has a fixed 'f' of one but its wavelength is expanding because of the intrinsic force causing the photon to expand on its journey through space. So it creates the illusion that space is expanding. I do not concern myself with the contiuous standing waves that exist when a HA is NOT energized by another photons bump of the electron. The breif transition of the electron returning to its original state is not continuous in these photon creations. So, photons come in different energy (wavelengths) levels as a breif pulse of 'one'. But the wavelength is expanding and therefore not fixed. I will post an article on the 'Creation of Photons' in the near future to clarify this scenerio. Regarding the math above, mass is not present in creating the photons but only as the carriers of the forces that do the creations. That definition is of a Joule is similar to the definition of a 'Newton' as a force. However, a Joule is defined as identical to the Coulomb charge. NS
D H Posted June 26, 2008 Posted June 26, 2008 As I have said in defining the photon, once it is radiated, it has a fixed 'f' of one Stop spouting nonsense. - One what? (Units, please). - Different photons have different frequencies. That is why we see colors.
swansont Posted June 26, 2008 Posted June 26, 2008 When a fission bomb explodes, their is both a mass increase and a lot of energy created. If this energy was a mass conversion, than the mass should have descreased, not increased. The energy actially was there in the form of potential energy that was contained by the strong force. So mass was not even involved in this transformation. Evidence? It's trivial to show that the mass of U-235 is larger than the fission products you get, so good luck with that. As I have said in defining the photon, once it is radiated, it has a fixed 'f' of one but its wavelength is expanding because of the intrinsic force causing the photon to expand on its journey through space.So it creates the illusion that space is expanding. Sorry, but you have to use the same definitions as everyone else. What a photon is is already defined. It has a frequency, it has an energy, and we observe the effects of them being absorbed by atoms/molecules with well-defined energy differences.
Klaynos Posted June 26, 2008 Posted June 26, 2008 Surely if this was true if I move my detector further from the photon source I would measure a different wavelength/energy?
ajb Posted June 27, 2008 Posted June 27, 2008 ...and all because [math][\hbar]=Energy \cdot Time[/math].
New Science Posted June 27, 2008 Author Posted June 27, 2008 Stop spouting nonsense.- One what? (Units, please). - Different photons have different frequencies. That is why we see colors. Quit acting like a teacher. Can't you grasp the reality that photons are JUST a pulse of ONE? And what is a photon that has an energy level coincidental with its wavelength (pulse), A wavelength of red is 6.56^-7 meters. This would also define the width of the photon. So the different wavelength constitute the different colors. NS
Sayonara Posted June 27, 2008 Posted June 27, 2008 So what you are saying, is that X-rays differ from radio waves because they are different colours?
swansont Posted June 27, 2008 Posted June 27, 2008 Quit acting like a teacher. Can't you grasp the reality that photons are JUST a pulse of ONE? Do not, do not, do not tread down this path. And what is a photon that has an energy level coincidental with its wavelength (pulse), Since the wavelength changes in a medium, how is energy being conserved?
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