New Science Posted June 27, 2008 Posted June 27, 2008 Creation of Photons Photons, that are a form of light energy, are created when an electron returns to its original orbital state after being 'bumped' to a higher (outer orbit) state by another photon. It then returns to its original state to rediate a photon and this is called the 'emission' state . In this way, it acts as a relay for transmission of light from the central region of the stars to the outer regions of space, since it absorbs a photon before it emits one. So, how do these photons develop? Well, it is the 'magnetic field' component of the EM radiations that creates the photons. The electric component is 'omni-directional' in the sense that it radiates its force equally in all directions that reduces in strength according to the inverse square law, relative to the distance from the particle. So, it is not directional. It is also, a non-variable force that never changes its magnitude but remains constant. Its electric field acts as a 'carrier' of the photon pulses, so in this sense, it does not transmit any intelligence as the magnetic component does. The magnetic component is 'directional' and varies from zero to maximum, relative to the observer and the velocities of the electrons. Naturally, the electrons are always in constant motions, so a magnetic force will always be present. When the electrons make a transition back to their original energy level, these variations in velocity and curvature from a slower velocity to a higher one and from a larger orbit to a smaller one, creates the photon pulse. This type of transition creates a 'black body' type of pulse because of the variations in the electrons increasing velocity and reducing orbital curvature. In this case, the omni-directional electric field varies very little in strength relative to the surrounding space while the magnetic field undergoes a large change during these transitions. In the stars, the electrons trace out a SIGN wave pulse because in these plasmas, the trajectories of the electrons are OPEN orbital passages. The MF also radiates in 'one' direction only to an observer because of the changing electrons movement and direction relative to the observer. The magnetic component is zero relative to the observer when the electron is approaching or receding in the opposite direction away from the observer. The radiation from the MF is at maximum to the observer when the electron is moving laterally to the observer. So, these changes in the MF affect the surrounding electric field charged particles (EFCP) known as 'virtual charged particles'. Although these particles are called 'virtual', they are real because these surrounding fields are 'real'. This is proven because of their 'action at a distance'. The MF's are also real as everyone should know because of the magnets that are common to everyone’s knowledge. So, what happens to the EFCP that constitute and surround the electrons when an electron makes its transition and the MF influences these EFCP to become photons? The MF causes these EFCP to CLUMP or condense together to form a congregate of 'compressed particles' that transmit their momentum by pushing against the field particles in front of them. They then transmit this condensed momentum through the EF as a line of standing dominos will transmit its falls to the next dominoes in front of them. In other words, these field particles will just 'wobble' but remain in their positions rather than moving through the field as individual particles. Of course, this momentum will be transmitted at the 'velocity of light'. This is how I visualize the creation of light as a photon based on the Bohr Planetary Model of the HA.. Originally, light was presumed to be continuous waves but Planck's research and formula has changed it to a 'pulse' known as a quanta or the commonly known photon. However, in the ‘ground’ state, the HA radiates a continuous EF sign wave with a wavelength of about ‘one’ angstrom known as a ‘standing’ wave. Due to the uniform electron velocity around the proton, the MF also is constant in magnitude with a sign wave pattern. New Science
doG Posted June 27, 2008 Posted June 27, 2008 Do you have links to any data that support your mathless proclamations?
New Science Posted June 27, 2008 Author Posted June 27, 2008 doG Math is just a SUB science derived from experiments and observations. It is also just a language. Pictures are worth a thousand words as the saying goes. That is why I rely on VISUALIZATIONS based on the knowledge derived from the experiments and observations. NS
Klaynos Posted June 27, 2008 Posted June 27, 2008 doG Math is just a SUB science derived from experiments and observations. It is also just a language. Pictures are worth a thousand words as the saying goes. That is why I rely on VISUALIZATIONS based on the knowledge derived from the experiments and observations. NS Maths isn't 'just' a language it is THE language of physics used for describing the universe. Without it your idea is pretty meaningless, and it relies on an incorrect model of an atom.
swansont Posted June 27, 2008 Posted June 27, 2008 This is how I visualize the creation of light as a photon based on the Bohr Planetary Model of the HA.. Since the Bohr model is demonstrably wrong, what implications does this have on your model?
New Science Posted June 28, 2008 Author Posted June 28, 2008 Since the Bohr model is demonstrably wrong, what implications does this have on your model? None. The Schroedinger Eq's did not explain the HA spectrum. It proved that Bohr;s model for the energy levels was correct. All the physics books explain the Bohr model. Isn't that enough for you? NS Do you have links to any data that support your mathless proclamations? Yes. The physics books have the math for Bohr's model that I referred to. I use visualization because, like I said, pictures are worth a thousand words. NS Maths isn't 'just' a language it is THE language of physics used for describing the universe. Without it your idea is pretty meaningless, and it relies on an incorrect model of an atom. You overlook the preliminary Slipher, Hubble and humason observations and all the NASA and other current observations to reveal the universe. Like I said, math is NOT pictures. NS
Klaynos Posted June 28, 2008 Posted June 28, 2008 None. The Schroedinger Eq's did not explain the HA spectrum. It proved that Bohr;s model for the energy levels was correct. All the physics books explain the Bohr model. Isn't that enough for you? NS Yes. The physics books have the math for Bohr's model that I referred to. I use visualization because, like I said, pictures are worth a thousand words. NS Bohr's model is wrong, Bohr new it was wrong, it isn't included in advanced books because it's wrong. There are many reasons why it's wrong, ignoring the fact it doesn't correctly explain the emission spectra. Accelerating charges radiate, and electrons in the Bohr model are accelerating! You overlook the preliminary Slipher, Hubble and humason observations and all the NASA and other current observations to reveal the universe. Like I said, math is NOT pictures. NS They don't just take photos and go "pretty" they do spectral analysis of them (hey look maths) and compare them to models (OH **** more maths) what a shocker! Please go away and find out what real modern physics is. It's about modelling the universe, you CAN'T do that accurately without maths, and any new theory needs to be more accurate than the old ones....
swansont Posted June 28, 2008 Posted June 28, 2008 None. The Schroedinger Eq's did not explain the HA spectrum. It explains a lot more than the Bohr model does. It proved that Bohr;s model for the energy levels was correct. All the physics books explain the Bohr model. Isn't that enough for you? Energy levels are about all it explains correctly. The Bohr model does not account for different orbital angular momentum states, in fact, it gets the ground state wrong — the Bohr model has [math]L=n\hbar[/math] and yet the n=1 state in Hydrogen has no angular momentum. And what of fine structure, hyperfine structure and Zeeman splitting? There's a whole bunch of structure tot he atoms that isn't explained by the model. It's wrong That it gets the energy right is not enough for me, and shouldn't be enough for anyone.
New Science Posted June 29, 2008 Author Posted June 29, 2008 Bohr's model is wrong, Bohr new it was wrong, it isn't included in advanced books because it's wrong. There are many reasons why it's wrong, ignoring the fact it doesn't correctly explain the emission spectra. Accelerating charges radiate, and electrons in the Bohr model are accelerating! They don't just take photos and go "pretty" they do spectral analysis of them (hey look maths) and compare them to models (OH **** more maths) what a shocker! Please go away and find out what real modern physics is. It's about modelling the universe, you CAN'T do that accurately without maths, and any new theory needs to be more accurate than the old ones.... I will give you some maths regarding the BBT. The expansion of the space is additive. So the Hubble expansion of space is 75 kms/megsparsec/second. The velocity of light is 300,000 kms/s Divide 'c' by the Hubble constant and you get 4000 seconds of added expansion that equals 'c' So after 4000 seconds of expansion, the universe is expanding at more than 'c', so it would be invisible after 4000 seonds of expansion. The HDF's see more than 25 billion light years deep. that is more than 4000 seconds and there are still visible objects at that distance. I can provide more such ludicrous maths regarding the BBT. NS It explains a lot more than the Bohr model does. Energy levels are about all it explains correctly. The Bohr model does not account for different orbital angular momentum states, in fact, it gets the ground state wrong — the Bohr model has [math]L=n\hbar[/math] and yet the n=1 state in Hydrogen has no angular momentum. And what of fine structure, hyperfine structure and Zeeman splitting? There's a whole bunch of structure tot he atoms that isn't explained by the model. It's wrong That it gets the energy right is not enough for me, and shouldn't be enough for anyone. The Sch'r orbitals apply to chemistry more than to the light of the universe. In close proximity, the electrons do not remain in a stable orbital plane but keep changing their positions constantly. But in the gases of space and the Sun, they do maintain stable orbits except in the plasmas deeper in the Sun where they shift to OPEN orbital passages. So the Sch'r orbs do not apply t o the Universe radiations. NS
swansont Posted June 30, 2008 Posted June 30, 2008 The Sch'r orbitals apply to chemistry more than to the light of the universe. In a word, no. Absorption spectra that identify the "light of the universe" follow from atomic structure that is explained by quantum mechanics.
Bignose Posted June 30, 2008 Posted June 30, 2008 I will give you some maths regarding the BBT. The expansion of the space is additive. So the Hubble expansion of space is 75 kms/megsparsec/second. The velocity of light is 300,000 kms/s Divide 'c' by the Hubble constant and you get 4000 seconds of added expansion that equals 'c' So after 4000 seconds of expansion, the universe is expanding at more than 'c', so it would be invisible after 4000 seonds of expansion. The HDF's see more than 25 billion light years deep. that is more than 4000 seconds and there are still visible objects at that distance. I can provide more such ludicrous maths regarding the BBT. Not that I expected you to learn from the previous discussions of dimensions, but your math is "ludicrous" H0 = hubbles constant = 70.1 km/s/Mpc (I took the number from wikipedia, there is a margin or error, but the exact number itself is unimportant) Note that since km and Mpc are both units of length, this can also be expressed in units of inverse seconds (1/s or s^-1) and is about 2.5*10^-18 s^-1 c = speed of light = 300 000 km/s Now, in your own words here "Divide 'c' by the Hubble constant" c/H0 = 300 000 km/s / 2.5*10^-18 s^-1 = 1.2*10^23 km It has units of kilometers. And, I have no idea what that number may mean. But the point is it does NOT NOT NOT have units of time!!! Wherever you got 4000 seconds from, it certainly wasn't from the division you said because the division you talk about only has units of length in it. The units of time cancel. Learning how to ensure that your calculations have the right units is day 1 Introduction to Physics stuff. Until you get these things right, I don't know how you can think your "calculations" can be taken seriously.
New Science Posted June 30, 2008 Author Posted June 30, 2008 In a word, no. Absorption spectra that identify the "light of the universe" follow from atomic structure that is explained by quantum mechanics. I have mentioned here a number of times that an electron ABSORBS a photon before it radiates a photon. This is obvious in measuring the absorption lines for determining the redshifts that is more precise that using the emission lines. That is my opinion. NS Not that I expected you to learn from the previous discussions of dimensions, but your math is "ludicrous" H0 = hubbles constant = 70.1 km/s/Mpc (I took the number from wikipedia, there is a margin or error, but the exact number itself is unimportant) Note that since km and Mpc are both units of length, this can also be expressed in units of inverse seconds (1/s or s^-1) and is about 2.5*10^-18 s^-1 c = speed of light = 300 000 km/s Now, in your own words here "Divide 'c' by the Hubble constant" c/H0 = 300 000 km/s / 2.5*10^-18 s^-1 = 1.2*10^23 km It has units of kilometers. And, I have no idea what that number may mean. But the point is it does NOT NOT NOT have units of time!!! Wherever you got 4000 seconds from, it certainly wasn't from the division you said because the division you talk about only has units of length in it. The units of time cancel. Learning how to ensure that your calculations have the right units is day 1 Introduction to Physics stuff. Until you get these things right, I don't know how you can think your "calculations" can be taken seriously. The Hubble Constant has a given distance in seconds also. For every expansion of the HC, a seconf is involved. So the constant represents both distance and TIME. Incidentally, how would you incorporate the mpc into determining the size of the current BBU?
swansont Posted June 30, 2008 Posted June 30, 2008 I have mentioned here a number of times that an electron ABSORBS a photon before it radiates a photon. This is obvious in measuring the absorption lines for determining the redshifts that is more precise that using the emission lines. That is my opinion. NS And you also seem to be saying that the absorption/emission is adequately explained by the Bohr model. That's wrong. Thermal (i.e. blackbody) spectra come from acceleration of charges, which gives you a continuum.
John Cuthber Posted June 30, 2008 Posted June 30, 2008 "That is why I rely on VISUALIZATIONS" http://en.wikipedia.org/wiki/Optical_illusion That's why I don't.
Bignose Posted June 30, 2008 Posted June 30, 2008 The Hubble Constant has a given distance in seconds also. For every expansion of the HC, a seconf is involved. So the constant represents both distance and TIME. Incidentally, how would you incorporate the mpc into determining the size of the current BBU? I don't know what Hubble's constant you could be talking about. http://scienceworld.wolfram.com/physics/HubbleConstant.html http://en.wikipedia.org/wiki/Hubble's_law http://science.nasa.gov/newhome/headlines/ast25may99_2.htm http://www.thefreedictionary.com/Hubble's+constant Every single one of these sources (and many others) states that Hubble's constant is a velocity (km/s typically) divided by a length (parsecs or megaparsecs, typically) which would have units of inverse time. Is there any way you could post a source that lists Hubble's constant in any other units? Such as this distance you seem to think exists?
New Science Posted July 1, 2008 Author Posted July 1, 2008 I don't know what Hubble's constant you could be talking about. http://scienceworld.wolfram.com/physics/HubbleConstant.html http://en.wikipedia.org/wiki/Hubble's_law http://science.nasa.gov/newhome/headlines/ast25may99_2.htm http://www.thefreedictionary.com/Hubble's+constant Every single one of these sources (and many others) states that Hubble's constant is a velocity (km/s typically) divided by a length (parsecs or megaparsecs, typically) which would have units of inverse time. Is there any way you could post a source that lists Hubble's constant in any other units? Such as this distance you seem to think exists? The Hubble Constant is not an actual constant but one measured by a variety of methods. The consensus, as far as I know, is 72 kms/second. NS I don't know what Hubble's constant you could be talking about. http://scienceworld.wolfram.com/physics/HubbleConstant.html http://en.wikipedia.org/wiki/Hubble's_law http://science.nasa.gov/newhome/headlines/ast25may99_2.htm http://www.thefreedictionary.com/Hubble's+constant Every single one of these sources (and many others) states that Hubble's constant is a velocity (km/s typically) divided by a length (parsecs or megaparsecs, typically) which would have units of inverse time. Is there any way you could post a source that lists Hubble's constant in any other units? Such as this distance you seem to think exists? Originally, the universe was split between two factions . One for 50 Kms/mpc/sec while the other was promoting a 100 Kms/mpc/s rate. So this established the age of the universe as being from 20 billion years old to 10 billion years old. So this was comprimised at 75 kms/mpc/s and an age of 15 bioon years. So the latest figures from WMAP is that the universe is 14 billion years old and the Hubble valus as 70.1. Since I believe in a Flat Soace SSU, I do not give any credibility to these findings and previous accepted values for our universe. I just follow the current opinions to keep in touch with the latest news. NS
D H Posted July 1, 2008 Posted July 1, 2008 I will give you some maths regarding the BBT. The expansion of the space is additive. So the Hubble expansion of space is 75 kms/megsparsec/second. The velocity of light is 300,000 kms/s Divide 'c' by the Hubble constant and you get 4000 seconds of added expansion that equals 'c' That is not math, it is gobblygook. It isn't even dimensionally correct, as Bignose has already noted. Not that I expected you to learn from the previous discussions of dimensions, but your math is "ludicrous" Now, in your own words here "Divide 'c' by the Hubble constant" c/H0 = 300 000 km/s / 2.5*10^-18 s^-1 = 1.2*10^23 km It has units of kilometers. And, I have no idea what that number may mean. There's an easier way to examine at this mighty big number. Since Hubble's constant has units of inverse time, the inverse of Hubble's constant has units of time. Not surprisingly, the inverse of Hubble's constant is called Hubble time. Assuming a value of 72 km/sec/megaparsec, 1/H0 = 13.6 billion years. Multiplying by the speed of light yields 13.6 billion light years. What does this value mean? Answer: Nothing. Multiplying the speed of light times the age of the universe yields a value that has units of distance but also has no meaning. Since I believe in a Flat Soace SSU, I do not give any credibility to these findings and previous accepted values for our universe. It certainly is downright mean when some old-school scientist conducts an experiment that shows one's cherished beliefs are wrong. It is best to "jusr ignore" those "teensy distractions". I can provide more such ludicrous maths regarding the BBT. I will not deny that you can provide ludicrous mathematics. I will close with Bignose's words. Learning how to ensure that your calculations have the right units is day 1 Introduction to Physics stuff. Until you get these things right, I don't know how you can think your "calculations" can be taken seriously.
swansont Posted July 1, 2008 Posted July 1, 2008 Since I believe in X, I do not give any credibility to these findings and previous accepted values for Y. I just follow the current opinions to keep in touch with the latest news. Wow. That's almost as unscientific as you can get. Fill in your favorites for the variables.
Reaper Posted July 1, 2008 Posted July 1, 2008 Since I believe in a Flat Soace SSU, I do not give any credibility to these findings and previous accepted values for our universe. I just follow the current opinions to keep in touch with the latest news. NS That has got to be one of the most classic crackpot quotes I have ever heard. Maybe I will start a new "hall of shame" thread on this...
pioneer Posted July 1, 2008 Posted July 1, 2008 A photon acts as both a particle and a wave. I often wondered where the particle aspect of the photon goes when it is absorbed by the atom. It only appears to transfer a wave affect. But as the electron drops energy level the particle appears again. I am not highjacking but trying to help. If you look at a hydrogen atom absorbing a photon, there is not a 50/50 EM split, conceptually. The higher mass of the hydrogen proton precludes as much velocity and therefore makes it hard to equally share the magnetic aspect of the photon, since magnetism is charge in motion. Only the electric, conceptually, is equally shared. Maybe this is due to the positive charge having three components and negative only one. A positive charge in motion may be able to use all three but positive may have flexibility in terms of the atomic state.
Sayonara Posted July 1, 2008 Posted July 1, 2008 Go for it Pioneer. I don't think there is any remote chance of a successful hijack in this thread.
swansont Posted July 1, 2008 Posted July 1, 2008 A photon acts as both a particle and a wave. I often wondered where the particle aspect of the photon goes when it is absorbed by the atom. It only appears to transfer a wave affect. But as the electron drops energy level the particle appears again. The atoms recoils. That's not a wave behavior. Though one must keep in mind that particle and wave are convenient descriptions of the entity, and not quantities that are conserved.
Bignose Posted July 1, 2008 Posted July 1, 2008 The Hubble Constant is not an actual constant but one measured by a variety of methods.The consensus, as far as I know, is 72 kms/second. NS Originally, the universe was split between two factions . One for 50 Kms/mpc/sec while the other was promoting a 100 Kms/mpc/s rate. So this established the age of the universe as being from 20 billion years old to 10 billion years old. So this was comprimised at 75 kms/mpc/s and an age of 15 bioon years. So the latest figures from WMAP is that the universe is 14 billion years old and the Hubble valus as 70.1. Since I believe in a Flat Soace SSU, I do not give any credibility to these findings and previous accepted values for our universe. I just follow the current opinions to keep in touch with the latest news. NS Actually, the big, giant, overwhelming point that you are completely missing here is that while the specific value of Hubble's constant is not agreed upon, every single source has the same units -- a velocity divided by a length or equivalently inverse time. Is there any possibility that you could cite any source whatsoever that calls a number Hubble's constant in said constant doesn't have the units of inverse time?
New Science Posted July 2, 2008 Author Posted July 2, 2008 Actually, the big, giant, overwhelming point that you are completely missing here is that while the specific value of Hubble's constant is not agreed upon, every single source has the same units -- a velocity divided by a length or equivalently inverse time. Is there any possibility that you could cite any source whatsoever that calls a number Hubble's constant in said constant doesn't have the units of inverse time? Well, the constant is quoted as being an expansion of 72 kms/mpc/s (consensus). So the expansion is for every second. The question here is how do you incorporate the mpc into any formula? NS That has got to be one of the most classic crackpot quotes I have ever heard. Maybe I will start a new "hall of shame" thread on this... My post on the Flat Space universe (former SSU) is based on the LAWS of CONSERVATION, the M-M EXP and ARPS RS Anomaly. I cannot think of anything that is more scientific rather than accepting a CosmoGONY universe. This is a matter of faith in science. NS
Bignose Posted July 2, 2008 Posted July 2, 2008 Well, the constant is quoted as being an expansion of 72 kms/mpc/s(consensus). So the expansion is for every second. The question here is how do you incorporate the mpc into any formula? NS right. A km (kilometer) is a unit of distance. And a mpc (megaparsec) is a unit of distance. Therefore, 72 km/mpc/s has units of inverse time. Period. The math you posted doesn't have the right units and it therefore 100% wrong. If you can't even get the units correct when performing a calculation, any interpretation of that calculation is completely wrong and meaningless and useless. Start getting the units right (on this particular calculation, and in your your thread about the "MAJOR" discovery) and then maybe we can talk slightly intelligently about the interpretation of those calculations. But, without the correct units on your calculations, you are spouting nonsense. I might as well say "I have to drive 47 Kelvin in my car to get to work every day, and my car gets 26 amperes per mole. I get paid 0.0812 deciliters per lumen squared. On the way home, I buy dinner at the grocery store and pay an average of 14.87 hair follicles." Do you see how ridiculous this gets? Without correct units, I might as well be talking in a foreign language.
Recommended Posts