Re: [Vo]:ref QED
On Mar 25, 2009, at 6:10 PM, mix...@bigpond.com wrote: In reply to Horace Heffner's message of Wed, 25 Mar 2009 11:46:46 -0800: Hi, [snip] On Mar 23, 2009, at 1:05 PM, mix...@bigpond.com wrote: I think the electron doesn't spiral into the nucleus because it doesn't have enough angular momentum to create a photon, hence it can't radiate, which means it can't lose energy. This argument must not be true. There is a finite probability of finding the electron near or within the nucleus. If viewed as a point particle and not a wave function, the acceleration can thus become arbitrarily high, as can the kinetic energy and angular momentum, In standard QM the angular momentum of any given orbital is a constant. For the 1S orbital there is no orbital angular momentum, only spin angular momentum. In theory therefore, the electron could radiate by flipping it's spin back and forth, and dropping into ...what? This apparently doesn't happen. In my model, the spin angular momentum of the electron is not a magical quantum property at all, and can in fact be less than h_stripe / 2. Each successive sub_ground state orbital has less angular momentum, and the difference between that and the angular momentum of higher orbitals (up to the ground state) is always less than h_stripe, making photon emission impossible, and neatly explaining the stability of the ground state to normal radiation. You are of course assuming radiation can only come from spin flipping. thus guaranteeing Larmor radiation. In fact, some molecular orbitals exist in a figure 8 configuration, where the center of the 8 is the nucleus, thus guaranteeing constant nuclear traverses. I think ordinary Newtonian point particle models just can't explain the lack of radiation. I think that if you calculate the angular momentum of an electron in/near the nucleus, you may get a surprise. The r part of mvr is very small. Yes, but as r-0 we have energy-inf and the energy available to Larmor radiation goes to infinity. Stable orbital conditions no longer apply. If a spin zero electron enters a nucleus then all bets are off regarding angular momentum and the Coulomb force interaction of the components. Once within the nucleus the spin zero means nothing. It is not a direct hit on any specific charged particle because the nucleus has multiple charged particles, including quarks. Any collision off center has an inherent angular momentum, and the initial spin zero is irrelevant. It is not possible to be on center to multiple targets. I suspect that Larmor radiation is only possible when the electron comes from outside the atom. That then is an assumption outside the boundaries of my statement which explicitly assumed ordinary Newtonian point particle models (i.e. implying your model appears to assume this), and thus proves my point. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:ref QED
In reply to Horace Heffner's message of Fri, 27 Mar 2009 10:50:56 -0800: Hi, [snip] In my model, the spin angular momentum of the electron is not a magical quantum property at all, and can in fact be less than h_stripe / 2. Each successive sub_ground state orbital has less angular momentum, and the difference between that and the angular momentum of higher orbitals (up to the ground state) is always less than h_stripe, making photon emission impossible, and neatly explaining the stability of the ground state to normal radiation. You are of course assuming radiation can only come from spin flipping. Since in standard QM the photon carries h_stripe of angular momentum, and a 1S orbital (all S orbitals?) electron only *has* spin angular momentum, that's all that is available. IOW only a spin flip supplies the required amount. In my model, there is essentially no spin property, just orbital angular momentum, and the change in the value thereof below the ground state is less than that required by a photon. One might however argue that the angular momentum can change from e.g. a small positive value to a large negative value, thus allowing for an h_stripe difference. In my model, the actual allowed values are smaller and smaller fractions, hence the differences are never a full h_stripe. It is actually this fact rather than absolute quantity which prevents radiation. thus guaranteeing Larmor radiation. In fact, some molecular orbitals exist in a figure 8 configuration, where the center of the 8 is the nucleus, thus guaranteeing constant nuclear traverses. I think ordinary Newtonian point particle models just can't explain the lack of radiation. I think that if you calculate the angular momentum of an electron in/near the nucleus, you may get a surprise. The r part of mvr is very small. Yes, but as r-0 we have energy-inf and the energy available to Larmor radiation goes to infinity. Stable orbital conditions no longer apply. If a spin zero electron enters a nucleus then all bets are off regarding angular momentum and the Coulomb force interaction of the components. Once within the nucleus the spin zero means nothing. It is not a direct hit on any specific charged particle because the nucleus has multiple charged particles, including quarks. Any collision off center has an inherent angular momentum, and the initial spin zero is irrelevant. It is not possible to be on center to multiple targets. Try actually calculating the maximum off-center angular momentum. This occurs when the distance from the nucleus, or any particle within it, is perpendicular to the path followed by the electron, so you can set r = distance from center of charge, which can also be used to calculate the potential energy, and hence the maximum velocity that the electron can have. I get the formula sqrt((Z*q^2*r*me)/(2*Pi*epsilon_0)). Since r appears in the numerator, and all other values are constants, it shrinks with r (rather than getting larger as one might intuitively expect from increasing energy). A more central collision has even less angular momentum, and a central collision has no angular momentum at all. I suspect that Larmor radiation is only possible when the electron comes from outside the atom. That then is an assumption outside the boundaries of my statement which explicitly assumed ordinary Newtonian point particle models (i.e. implying your model appears to assume this), and thus proves my point. Say what??? Whether or not an electron comes from outside an atom, has no relevance to whether or not it is a point particle, and hence has nothing to do with the boundaries of your argument. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/ Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/Project.html
Re: [Vo]:ref QED
On Mar 23, 2009, at 1:05 PM, mix...@bigpond.com wrote: I think the electron doesn't spiral into the nucleus because it doesn't have enough angular momentum to create a photon, hence it can't radiate, which means it can't lose energy. This argument must not be true. There is a finite probability of finding the electron near or within the nucleus. If viewed as a point particle and not a wave function, the acceleration can thus become arbitrarily high, as can the kinetic energy and angular momentum, thus guaranteeing Larmor radiation. In fact, some molecular orbitals exist in a figure 8 configuration, where the center of the 8 is the nucleus, thus guaranteeing constant nuclear traverses. I think ordinary Newtonian point particle models just can't explain the lack of radiation. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
RE: [Vo]:ref QED
-Original Message- From: Horace Heffner In fact, some molecular orbitals exist in a figure 8 configuration, where the center of the 8 is the nucleus, thus guaranteeing constant nuclear traverses. [JB:] Understanding the figure 8 configuration is perhaps the most important basic challenge in QM and it can also be said that most theorists are simply not up to the task. However, I would add that guaranteeing constant nuclear traverses is a bit of a reach in this regard The following observation may or may not help with that understanding of 2p orbitals- but - if you think about a capacitor as being able to pass AC current, and current consists of, well it consists at least partly of electrons - then the nuclear-cap can be a decent analogy (i.e with the nucleus as a kind of capacitor)... but of course there is the argument that it is not the same electron... However, in Dirac's universe - is not the electron the epitome of fungibility? http://en.wikipedia.org/wiki/Fungibility When I wrote that word fungibility I was almost certain that it would not be in Wiki, but heck, it is getting pretty damn hard to come up with anything that is not in Wiki these daze... Jones BTW here is some interesting commentary on 2p orbitals that adds some credence to the idea that indeed it is not the same electron that crosses the nucleus ... http://winter.group.shef.ac.uk/orbitron/AOs/2p/index.html
Re: [Vo]:ref QED
Instead of asking why the electron doesn't spiral into the nucleus, perhaps one should question the law(s) that lead one to expect it should spiral into nucleus. The mathematical strategy of quantizing energy saves the law(s) and avoids the latter question. Harry
Re: [Vo]:ref QED
In reply to Horace Heffner's message of Wed, 25 Mar 2009 11:46:46 -0800: Hi, [snip] On Mar 23, 2009, at 1:05 PM, mix...@bigpond.com wrote: I think the electron doesn't spiral into the nucleus because it doesn't have enough angular momentum to create a photon, hence it can't radiate, which means it can't lose energy. This argument must not be true. There is a finite probability of finding the electron near or within the nucleus. If viewed as a point particle and not a wave function, the acceleration can thus become arbitrarily high, as can the kinetic energy and angular momentum, In standard QM the angular momentum of any given orbital is a constant. For the 1S orbital there is no orbital angular momentum, only spin angular momentum. In theory therefore, the electron could radiate by flipping it's spin back and forth, and dropping into ...what? This apparently doesn't happen. In my model, the spin angular momentum of the electron is not a magical quantum property at all, and can in fact be less than h_stripe / 2. Each successive sub_ground state orbital has less angular momentum, and the difference between that and the angular momentum of higher orbitals (up to the ground state) is always less than h_stripe, making photon emission impossible, and neatly explaining the stability of the ground state to normal radiation. thus guaranteeing Larmor radiation. In fact, some molecular orbitals exist in a figure 8 configuration, where the center of the 8 is the nucleus, thus guaranteeing constant nuclear traverses. I think ordinary Newtonian point particle models just can't explain the lack of radiation. I think that if you calculate the angular momentum of an electron in/near the nucleus, you may get a surprise. The r part of mvr is very small. I suspect that Larmor radiation is only possible when the electron comes from outside the atom. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/Project.html
Re: RE: [Vo]:ref QED
- Original Message - From: Jones Beene jone...@pacbell.net Date: Wednesday, March 25, 2009 4:50 pm Subject: RE: [Vo]:ref QED -Original Message- From: Horace HeffnerIn fact, some molecular orbitals exist in a figure 8 configuration, where the center of the 8 is the nucleus, thus guaranteeing constant nuclear traverses. [JB:] Understanding the figure 8 configuration is perhaps the most importantbasic challenge in QM and it can also be said that most theorists are simply not "up to the task". speaking of figure eights... http://www.facebook.com/photo.php?pid=1277715l=6e4f24e6e2id=676517267 Harry However, I would add that "guaranteeing constant nuclear traverses" is a bit of a reach in this regard The following observation may or may not help with that understanding of 2p orbitals- but - if you think about a capacitor as being able to "pass" AC current, and current consists of, well it consists at least partly of electrons - then the nuclear-cap can be a decent analogy (i.e with the nucleus as a kind of capacitor)... but of course there is the argument that it is "not the same electron"... However, in Dirac's universe - is not the electron the epitome of "fungibility"? http://en.wikipedia.org/wiki/Fungibility When I wrote that word "fungibility" I was almost certain that it would not be in Wiki, but heck, it is getting pretty damn hard to come up with anything that is not in Wiki these daze... Jones BTW here is some interesting commentary on 2p orbitals that adds some credence to the idea that indeed it is "not the same electron" that crosses the nucleus ... http://winter.group.shef.ac.uk/orbitron/AOs/2p/index.html
Re: [Vo]:ref QED
Jones Beene wrote: -Original Message- From: Horace Heffner In fact, some molecular orbitals exist in a figure 8 configuration, where the center of the 8 is the nucleus, thus guaranteeing constant nuclear traverses. [JB:] Understanding the figure 8 configuration is perhaps the most important basic challenge in QM and it can also be said that most theorists are simply not up to the task. I've seen a model of the atom which has the nucleus as a torroid and the electrons going through the center of it. --- Get FREE High Speed Internet from USFamily.Net! -- http://www.usfamily.net/mkt-freepromo.html ---
Re: [Vo]:ref QED
In reply to fznidar...@aol.com's message of Mon, 23 Mar 2009 08:43:37 -0400: Hi Frank, [snip] The reason that the electron does not spiral into the nucleus has been a fundamental mystery. It is currently accepted that the angular momentum cannot be lees that h/2pie.? This has been accepted because the result agrees with experimental spectra.? Puthoff came up with some ideas about the emission and absorption of zero point energy.? I believe that I have solved this problem.? The orbits of the atoms exist as point of electromagnetic and gravitomagnetic accessibility.? The ground state orbit is a point where the transitional frequency ( 1.094 megahertz-meters ) equals the natural frequency of the electron. [snip] I think the electron doesn't spiral into the nucleus because it doesn't have enough angular momentum to create a photon, hence it can't radiate, which means it can't lose energy. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/Project.html
