Dovada Posted October 6, 2011 Share Posted October 6, 2011 For a lifetime now (over 40 years) I have been looking to resolve the connection between the physics of the current atomic model and the physics of the all powerful cosmic forces that move all the atoms everywhere within the cosmos. As you can imagine it is quite a massive undertaking. What helped me was I discovered many errors in today's atomic physics models which are still being used today and continue to block our progress in our understanding of how the atom and the cosmic universe interacts. If you are interested in hearing more on this subject please give your best answers to all or any of the following questions: 1. Do orbits exist? 2. Does the cosmic motion of atoms effect their internal energy structures? 3. Why do atoms in the United States have the same characteristics as atoms in Africa and Australia? 4. Why does the Bohr radius exist? 5. What is gravity? Link to comment Share on other sites More sharing options...
DrRocket Posted October 7, 2011 Share Posted October 7, 2011 For a lifetime now (over 40 years) I have been looking to resolve the connection between the physics of the current atomic model and the physics of the all powerful cosmic forces that move all the atoms everywhere within the cosmos. As you can imagine it is quite a massive undertaking. What helped me was I discovered many errors in today's atomic physics models which are still being used today and continue to block our progress in our understanding of how the atom and the cosmic universe interacts. If you are interested in hearing more on this subject please give your best answers to all or any of the following questions: 1. Do orbits exist? 2. Does the cosmic motion of atoms effect their internal energy structures? 3. Why do atoms in the United States have the same characteristics as atoms in Africa and Australia? 4. Why does the Bohr radius exist? 5. What is gravity? What questions do we need to answer to terminate this ? Link to comment Share on other sites More sharing options...
Dovada Posted October 8, 2011 Author Share Posted October 8, 2011 None - Sorry to offend you. It seems strange that Universities and schools are still teaching our new physics students who are travelling at incredible velocities on a moving planet, the wonders of a stationary 3 dimensional quantum atomic model. Link to comment Share on other sites More sharing options...
Aristarchus in Exile Posted October 18, 2011 Share Posted October 18, 2011 (edited) Who is to judge what is better? The consensus which fails so terribly so many times? http://en.wikipedia....aniel_Shechtman "Should a good physical theory predict phenomena that happen always? Yes, of course." One problem to Consensus thought is, there seems ALWAYS to be exceptions to the rule, but this is only a problem if that word "always" is iron bound in the phrase "that happen always." Edited October 18, 2011 by Aristarchus in Exile Link to comment Share on other sites More sharing options...
Klaynos Posted October 18, 2011 Share Posted October 18, 2011 1. Do orbits exist? Planitary or atomic? Atomic orbits do not exists, orbitals do, and have a significant amount of experimental evidence. 2. Does the cosmic motion of atoms effect their internal energy structures? Relativity obviously holds. Having said that, there is no universal rest frame, so an atom that is at rest with the lab is in as good a rest frame as any other frame you choose to use. 3. Why do atoms in the United States have the same characteristics as atoms in Africa and Australia? Because as far as we know and can tell the laws of physics are consistent throughout the universe. 4. Why does the Bohr radius exist? It is a poor approximation. The Bohr model is wrong. 5. What is gravity? A fundamental force that acts between massive objects. 1 Link to comment Share on other sites More sharing options...
Dovada Posted October 20, 2011 Author Share Posted October 20, 2011 Planitary or atomic? Atomic orbits do not exists, orbitals do, and have a significant amount of experimental evidence According to Wikipedia the definition of an orbit in physics is: "In physics, an orbit is the gravitationally curved path of an object around a point in space, for example the orbit of a planet around the center of a star system". If the point in space is constantly moving should the term spiral be used instead? Relativity obviously holds. Having said that, there is no universal rest frame, so an atom that is at rest with the lab is in as good a rest frame as any other frame you choose to use. I don't see this as answering the question. Because as far as we know and can tell the laws of physics are consistent throughout the universe. The main thing that is generally consistent throughout the universe is the galactic motion we have. Does this galactic motion control the uniformity of atomic structure? It is a poor approximation. The Bohr model is wrong. If a simple hydrogen atom which contains one proton and one electron moves in the same direction in the form of an atom within space, can the particles be considered particles in motion? If yes can they be considered the same as parallel currents? If yes the resultant magnetic fields will be opposing, so will the magnetic fields cause the particles to repel each other? Gravity is related to cosmic motion. The cosmic motion drives electromagnetic forces that allow neutral matter to attempt to share induced common magnetic fields. Link to comment Share on other sites More sharing options...
Klaynos Posted October 20, 2011 Share Posted October 20, 2011 According to Wikipedia the definition of an orbit in physics is: "In physics, an orbit is the gravitationally curved path of an object around a point in space, for example the orbit of a planet around the center of a star system". If the point in space is constantly moving should the term spiral be used instead? That would be relative. If for example you plot the moons orbit relative to the earth it's a circle, relative to the sun it's a spiral. There is no absolute reference frame. I don't see this as answering the question. That is because the question if flawed. There is no "cosmic" motion as that implies an absolute rest frame. The main thing that is generally consistent throughout the universe is the galactic motion we have. Does this galactic motion control the uniformity of atomic structure? No, the rest frame of the earth is used. Using the rest frame of the galaxy would be somewhat cumbersome and add no physical improvement. If a simple hydrogen atom which contains one proton and one electron moves in the same direction in the form of an atom within space, can the particles be considered particles in motion? The bound atom is in motion relative to the assumed rest frame. Space itself is not a rest frame. If yes can they be considered the same as parallel currents?If yes the resultant magnetic fields will be opposing, so will the magnetic fields cause the particles to repel each other? Atoms are electromagnetically bound particles. It doesn't matter how fast they are moving relative to you, they are still bound. Gravity is related to cosmic motion. The cosmic motion drives electromagnetic forces that allow neutral matter to attempt to share induced common magnetic fields. Cosmic motion is a flawed concept, there is no universal rest frame. Gravity is not an electromagnetic force, it does not behave like one. 1 Link to comment Share on other sites More sharing options...
Dovada Posted October 21, 2011 Author Share Posted October 21, 2011 Klaynos: I would like to have a fair and open discussion regarding some of these points you have answered. I have some strong convictions about why we have failed to make any headway in discovering the workings of areas classified as phenomena (gravity, propagation) etc. It is my belief we have overlooked and/or failed to comprehend fully some of the theory we work with. This is such that we have compounded the issue using questionable concepts to the point whereby we have placed stumbling blocks in our own path. To discuss this we might have to go out on a limb so as to consider certain facts from another perspective. This is a perspective that may appear initially to be contradictory to currently held belief. After all the old paths we have taken in the past, have so far led us down the path of failure. If you are you interested in accepting that we might go out on that limb, I will be willing to continue our discussion. Link to comment Share on other sites More sharing options...
Klaynos Posted October 21, 2011 Share Posted October 21, 2011 Unfortunately you're asking me to ignore a fantastic wealth of experimental evidence. Unless you have counter experimental evidence, I doubt we will agree that we can ignore reality and just run with it. Link to comment Share on other sites More sharing options...
Dovada Posted October 21, 2011 Author Share Posted October 21, 2011 Unfortunately you're asking me to ignore a fantastic wealth of experimental evidence. Unless you have counter experimental evidence, I doubt we will agree that we can ignore reality and just run with it. I would not ask you to do such a thing as ignore a fantastic wealth of experimental evidence. Sometimes I will be looking at things from a different perspective using different logic. This is what I requested from you, that this fantastic wealth of experimental evidence should be used to consider some facts from a different point of view. For example Newton's constant big G = 6.673x10-11 N(m / kg)^2 All atomic structure in our local group of galaxies is traveling at a velocity close to 1/500 times the speed of light, this is a velocity of c/500 or 600,000 meters per second. As a consequence of this velocity we can calculate the big G gravitational constant. where: Galactic velocity = 600,000 meters/sec Galactic U velocity = 9,000 meters/sec Mass proton / mass electron = 1836.151 U is the inward 9,000 meters per second motion of the sun and our solar system toward the galactic center. Big G = ((9,000/ (1836.151))/600,000)^2 = 6.673x10-11 Newton's constant was derived empirically, only later to be measured with any accuracy. Yet interestingly the constant value appears to be related to cosmic velocity. Link to comment Share on other sites More sharing options...
Ophiolite Posted October 21, 2011 Share Posted October 21, 2011 All atomic structure in our local group of galaxies is traveling at a velocity close to 1/500 times the speed of light, this is a velocity of c/500 or 600,000 meters per second. Relative to what? Link to comment Share on other sites More sharing options...
Phi for All Posted October 21, 2011 Share Posted October 21, 2011 To discuss this we might have to go out on a limb so as to consider certain facts from another perspective. This is a perspective that may appear initially to be contradictory to currently held belief. After all the old paths we have taken in the past, have so far led us down the path of failure. If you are you interested in accepting that we might go out on that limb, I will be willing to continue our discussion. Sorry, but this reminds me of the story where the guy goes to a tailor but can't afford a custom suit, so the tailor has him try on a suit rejected by another customer. The lapel is crooked, the sleeves are different lengths and the trousers are baggy. "If you tuck the lapel under your chin like this," says the tailor, "and hold your arms out crooked like this, and grab the front of the trousers here while walking with your legs stiff, no one will notice the suit doesn't fit." The guy buys the suit and wears it out of the shop. Two other guys see him across the street. One guys says, "Look at that poor man, he must be crippled to have to walk that way." The other guys says, "Yeah, but that sure is a nice suit he has on!" Aren't you basically saying that, if you overlook all its flaws, and ignore what accepted science has to say, your model makes perfect sense? 1 Link to comment Share on other sites More sharing options...
swansont Posted October 21, 2011 Share Posted October 21, 2011 U is the inward 9,000 meters per second motion of the sun and our solar system toward the galactic center. Ignoring the absurdity of atomic structure depending on motion (as it is frame-dependent which Ophiolite has pointed out), why should the structure of atoms not in our solar system depend depend on the motion of our sun? If for some reason it was motion-dependent, why wouldn't atomic structure vary over the course of the day and year as the earth rotates and revolves? Link to comment Share on other sites More sharing options...
Dovada Posted October 23, 2011 Author Share Posted October 23, 2011 Ignoring the absurdity of atomic structure depending on motion (as it is frame-dependent which Ophiolite has pointed out), why should the structure of atoms not in our solar system depend depend on the motion of our sun? If for some reason it was motion-dependent, why wouldn't atomic structure vary over the course of the day and year as the earth rotates and revolves? It does but only slightly, for example as the orbit velocity slows the orbit radius increases etc, so much quietly goes on unnoticed. We need to ask the question in the first place; if our atomic physicists have managed to succeed in predicting the behavior of the atomic functions so accurately, how can their base theory be considered as wrong? This is a common defense presented by physicists. So what is it that has made their atomic predictions so reliable? The answer is they have used mathematical empirical constants! Mathematical empirical constants commonly used in atomic physics are necessary and have been required to compensate for the effect of our helical motion of atomic structure with its charged particles within the cosmic environment. That is to say, atomic physicists have in the past, have neglected the influence of the external cosmic environment which includes cosmic motion from its atomic interaction and as a consequence must use empirically derived constants to make their calculations work in accordance with observed atomic laboratory experimental results as they experimented with atoms and associated particle physics. In this way the physicist was mathematically forced, by the use of these physical constants, to compensate for the cosmic environmental conditions whether he was aware of them or not. A friend of mine who is strong in mathematics, once said to me, if a single constant could be calculated by using the velocity associated the cosmic environment then that could be classified as just a coincidence. If two constants could be calculated by using the velocity associated the cosmic environment then it is not necessarily a coincidence and recommended that it should be investigated further. If on the other hand three or more constants could be calculated by using the velocity associated the cosmic environment then there is no coincidence and it can be concluded that the cosmic environment is playing an important role in the use of the constants. I tend to agree with him so I investigated further and found most empirical constants could be expressed using conditions within the cosmos. For example: Where: Mean solar spiral orbit velocity = 29,770 meters/sec Galactic motion = 600,000 meters/sec galactic spin = 225,000 meters/sec Galactic u velocity = 9,000 meters/sec Sun surface gravity = 274 meters/sec 1836.151 = mass proton/mass electron 464.463 = square-root(225,000 - 9,000 - 274) Planck’s constant = (electron volt x (29,770/464.463))^2 x 2pi = 6.626x10-34 Planck’s h-bar = (electron volt x (29,770/464.463))^2 = 1.054x10-34 Two components control the value necessary to define the Rydberg Constant and these are: 1: The vertical motion of the solar system within the Galaxy = (W Velocity) = 7,000 m/s 2: Earths magnetic frequency energy relative to the sun freq = c/Rse = 0.00200396 Hertz Where: c = speed of light in vacuum = 299,792,458 meters/sec Rse = approx sun/earth radial distance = 149,600,000,000 meters Rydberg constant = (7,000 * Pi)/freq = 10,973,844.534 m Fine structure = 1/(0.5 * 274) = 1/137 = 0.00729 Big G = ((9,000/ (1836.151))/600,000)^2 = 6.673x10-11 There are many more examples but the post get too long. Link to comment Share on other sites More sharing options...
swansont Posted October 23, 2011 Share Posted October 23, 2011 It does but only slightly, for example as the orbit velocity slows the orbit radius increases etc, so much quietly goes on unnoticed. We need to ask the question in the first place; if our atomic physicists have managed to succeed in predicting the behavior of the atomic functions so accurately, how can their base theory be considered as wrong? This is a common defense presented by physicists. So what is it that has made their atomic predictions so reliable? The answer is they have used mathematical empirical constants! Mathematical empirical constants commonly used in atomic physics are necessary and have been required to compensate for the effect of our helical motion of atomic structure with its charged particles within the cosmic environment. That is to say, atomic physicists have in the past, have neglected the influence of the external cosmic environment which includes cosmic motion from its atomic interaction and as a consequence must use empirically derived constants to make their calculations work in accordance with observed atomic laboratory experimental results as they experimented with atoms and associated particle physics. In this way the physicist was mathematically forced, by the use of these physical constants, to compensate for the cosmic environmental conditions whether he was aware of them or not. Baloney. If atomic structure depended on velocity the variation in structure would be easily seen. Our speed will vary by almost 1 km/s over the course of a day and 60 km/s over the course of a year. We can do spectroscopy to lots better than part in 10^4. Atomic clocks do more than 10 orders of magnitude better than that. They don't measure a timing variation. Clocks on opposite sides of the earth should give an error of a part in 10^6 because of the velocity difference. Not seen. It's not just that this violates relativity, it's that it violates it so badly that the discrepancies would have been noticed a century ago — relativity would have never been accepted in the first place. Evidence contradicts your claims and you have presented none of your own aside from the numerology, which lacks any mechanism to suggest why it might be valid. A friend of mine who is strong in mathematics, once said to me, if a single constant could be calculated by using the velocity associated the cosmic environment then that could be classified as just a coincidence. If two constants could be calculated by using the velocity associated the cosmic environment then it is not necessarily a coincidence and recommended that it should be investigated further. If on the other hand three or more constants could be calculated by using the velocity associated the cosmic environment then there is no coincidence and it can be concluded that the cosmic environment is playing an important role in the use of the constants. Your friend is wrong. Given the large number of constants and velocity values and ways to combine terms in equations, the probability is not small. Which you have shown. Link to comment Share on other sites More sharing options...
John Cuthber Posted October 23, 2011 Share Posted October 23, 2011 "A friend of mine who is strong in mathematics, once said to me, if a single constant could be calculated by using the velocity associated the cosmic environment then that could be classified as just a coincidence. If two constants could be calculated by using the velocity associated the cosmic environment then it is not necessarily a coincidence and recommended that it should be investigated further." Your friend underestimated your bloodymindedness. If you are prepared to spend enough time forming random combinations of constants then you will find that some of them are nearly the same as other constants. So what? And you are continuing to ignore the point made elsewhere Your "constants" change. you should say "Rse = approx sun/earth radial distance = 149,600,000,000 meters at the moment, but it's ill defined because the orbit is elliptical (roughly) and also it's changing continuously over the aeons." So, this week, and for one week only "Big G = ((9,000/ (1836.151))/600,000)^2 = 6.673x10-11" It has not always been true, and it will not be true in the future. It just happens that I live at a time when this coincidence occurs. Do you understand that? Link to comment Share on other sites More sharing options...
Dovada Posted October 25, 2011 Author Share Posted October 25, 2011 So, this week, and for one week only "Big G = ((9,000/ (1836.151))/600,000)^2 = 6.673x10-11"It has not always been true, and it will not be true in the future. It just happens that I live at a time when this coincidence occurs. Why in your opinion, did Newton need to use the numerical big G constant 6.673x10-11 in the first place? Link to comment Share on other sites More sharing options...
Ophiolite Posted October 25, 2011 Share Posted October 25, 2011 Why in your opinion, did Newton need to use the numerical big G constant 6.673x10-11 in the first place? This isn't an opinion. Newton empirically determined that the attractive forces between two masses was proportional to the product of those masses and the inverse of the square of the distance between them. The value of G varies depending upon what units your masses and distance are in. What are challenging in this? Link to comment Share on other sites More sharing options...
Dovada Posted October 25, 2011 Author Share Posted October 25, 2011 This isn't an opinion. Newton empirically determined that the attractive forces between two masses was proportional to the product of those masses and the inverse of the square of the distance between them. The value of G varies depending upon what units your masses and distance are in. What are challenging in this? F = G ((m1 x m2)/r^2) If m is in (kg) and radius in (meters) then G is referenced as 6.673x10-11 N(m/kg)^2 If m is in (imperial lbs) and radius in (yards) then G is referenced as 6.673x10-11 N(what)? Or does the number 6.673x10-11 remain the same? Link to comment Share on other sites More sharing options...
Bignose Posted October 25, 2011 Share Posted October 25, 2011 Forgive me for being blunt, but if you honestly cannot convert G from one system of units to another, it is hard to believe that any of your other calculations are correct. Unit conversion is a grade-school level skill, and it is usually well-practiced by 1st year undergrads as homework and exam questions thoroughly test the student's ability to convert units and work in one consistent system. Or, if your question was merely rhetorical, please just get to the point and don't be coy about it. Link to comment Share on other sites More sharing options...
Dovada Posted October 25, 2011 Author Share Posted October 25, 2011 (edited) Forgive me for being blunt, but if you honestly cannot convert G from one system of units to another, it is hard to believe that any of your other calculations are correct. Unit conversion is a grade-school level skill, and it is usually well-practiced by 1st year undergrads as homework and exam questions thoroughly test the student's ability to convert units and work in one consistent system. Or, if your question was merely rhetorical, please just get to the point and don't be coy about it. I have a reason for asking (the units are possibly wrongly defined), if it is so simple please explain by example. Edited October 25, 2011 by Dovada Link to comment Share on other sites More sharing options...
Ophiolite Posted October 25, 2011 Share Posted October 25, 2011 F = G ((m1 x m2)/r^2) If m is in (kg) and radius in (meters) then G is referenced as 6.673x10-11 N(m/kg)^2 If m is in (imperial lbs) and radius in (yards) then G is referenced as 6.673x10-11 N(what)? Or does the number 6.673x10-11 remain the same? Of course the number changes? As Bignose has stated either you can't do simple arithmetic operations, or you are being deliberatly obtuse. You claim it is the latter. Please stop it and state clearly in what way you think the units are improperly defined. You are the one making the challenge to established science. It is up to you to make the case, not up to us to teach you basics. (See, if you had come onto the forum and said "I'm not quite sure how this constant business works," then many members would have been happy to explain it, or direct you to good online sources. But you have appeared here with some arm waving and a tired claim that you have made some remarkable discovery. That won't work. 1 Link to comment Share on other sites More sharing options...
Dovada Posted October 25, 2011 Author Share Posted October 25, 2011 I have a reason for asking (the units are possibly wrongly defined) The gravitational constant, denoted G, is an empirical physical constant involved in the calculation of the gravitational attraction between objects with mass. My experience tells me that the formula is required to correct the dynamic physical distance between the mass bodies. This means that while the distance between mass bodies is the radius length in meters, it is not the correct value as seen by the electromagnetic conditions of the cosmos. The cosmos which includes the matter within it, is in continual motion. Based from that perspective the gravitational attractive force needs big G is to correct the dynamic distance (impedance) between the mass bodies. To understand this, consider the radial distance between the sun and the earth. Let us say the distance could be at a particular moment 149,792,529,789.47 meters The galactic velocity the sun and the earth has is measured as 600,000 meters/sec System energy = 149,792,529,789.47 x 600,000 = 8.9875517873682 x10+16 or simply c^2 As an electromagnetic wave, we start to travel from the sun on our journey to the earth only to find that when we get to where the earth should be, we find it has moved its not there anymore. The reality is that we cannot travel in a straight line to the earth as the formula in the equation suggests (m1 x m2)/r^2 where the r (radius) is referenced. We must take into account that at every second the earth is continually moving and so is the sun. This continual motion increases the length for the radius. In effect the path to the earth electromagnetically is curved and this curving must be compensated for. Hence the need for the big G correction constant. This observation could establish the gravitational field having the same velocity as the speed of light and also being electromagnetic in nature. Link to comment Share on other sites More sharing options...
Klaynos Posted October 26, 2011 Share Posted October 26, 2011 But at every moment the earth is stationary and everything else is moving relative to it. Rest frames are arbitrary. I don't get your system energy equation, it appears meaningless. Link to comment Share on other sites More sharing options...
Dovada Posted October 26, 2011 Author Share Posted October 26, 2011 But at every moment the earth is stationary and everything else is moving relative to it. Rest frames are arbitrary. I don't get your system energy equation, it appears meaningless. Just added this to show that the energy area being swept out is c^2 Link to comment Share on other sites More sharing options...
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