User talk:FyzixFighter/Archive 2

Latest comment: 14 years ago by FyzixFighter in topic Quantum teleportation

Quantum teleportation edit

I take it that writing about a fringe theory is alright, but that writing a fringe theory about a fringe theory is a no-no. I'm wondering if you know of any processes that involve a discontinuity of the position line of a particle in an S versus T Spacetime chart.WFPM (talk) 17:34, 22 April 2010 (UTC)Reply

Quantum teleportation is hardly a fringe theory - it is a widely accepted and investigated theory in modern physics. However, it appears from the edit I reverted and from your comment above that you misunderstand somewhat the nature of quantum teleportation. Quantum teleportation does not mean that a particle disappears and suddenly appears somewhere else such that the two events (the disappearance and re-appearance) are space-like. Rather it involves a transport of a quantum state from one quantum system to another. The system is not transported, only the quantum information. Therefore I don't see how quantum teleportation involves a "discontinuity of the position line of a particle in an S versus T Spacetime chart". Also, Feynmann diagrams are not spacetime diagrams. --FyzixFighter (talk) 18:36, 22 April 2010 (UTC)Reply

When you mention Quantum I think Quantum. That's a package of energy/matter? First it's here, and then it's over there. And you're calling that translation of distance "teleportation". Maybe in this peculiar situation you don't think energy/matter? but instead an entity called "information", which I, of course have no concept of but I still think that it involves something being moved from here to there. But you seem to think that something that is here can affect something that's over there without any means of connection or translation of anything, kind of like instant mental telepathy; which I consider to be a fringe science. And The creating of a logical deduction process related to how the system works doesn't impress me. And when you use the word "Quantum", which I interpret to be an amount of something, even information, I hope I can be excused for thinking it is something that has to be transported.WFPM (talk) 22:36, 22 April 2010 (UTC)Reply

Like I said, there is a misunderstanding of what "Quantum Teleportation" means. The opening paragraph of the article itself addresses some of the confusion:

Quantum teleportation, or entanglement-assisted teleportation, is a technique used to transfer quantum information from one quantum system to another. It does not transport the system itself, nor does it allow communication of information at superluminal (faster than light) speed. Neither does it concern rearranging the particles of a macroscopic object to copy the form of another object. Its distinguishing feature is that it can transmit the information present in a quantum superposition, useful for quantum communication and computation.

The only "superluminal" stuff going on with this is the quantum non-local connection between the two entangled states, ie the measurement of the state on one of the quantum systems of an entangled pair instantly forces the other system to be measured in the complementary state. This is an accepted outcome of quantum mechanics, first demonstrated experimentally by Alain Aspect in 1981 and 1982 (and more recently in 2008 where. However, do note that, as the intro paragraph states, it doesn't allow for superluminal communication. The alternate name, as indicated in the intro, is probably the more descriptive of what is going on. It's not called "quantum" teleportation because a quantum of whatever is being teleported, but because quantum mechanics effects are being used (entanglement) to teleport the quantum state from one system/particle/quantum fiddly-bit to another. --FyzixFighter (talk) 00:11, 23 April 2010 (UTC)Reply

I wish you would read a story by Isaac Asimov named "The Endochronic Properties of Resublimated Thalitimoline", about the transfer of information. I use it as a standard reference regarding the subject of the movement of anything with, of course, the next question which is Why Not? You're arguing that it is alright for the same idea to occur at two different places at the same time as a means of synchronizing activities? I'd call that a coincidence and a doubtful means of communication.WFPM (talk) 02:21, 23 April 2010 (UTC)Reply

I have read that short story - in High School and a couple of times since. I fail to see how it relates since your responses continue to show a misunderstanding/misreading of the article and of what I've said above. Neither the article nor I state that the non-local collapse of the waveform of the entangled particles at two different places at the same time due to measurement at one place is a means of synchronizing activities. Quantum teleportation#A summary provides a somewhat layman's description of the quantum protocol used, which includes no synchronizing (as I understand the word). Quantum teleportation#The result gives a little more in-depth quantum mechanical description using Dirac notation. This is not just theory, but has been demonstrated experimentally with both photons and atoms, and is a major element in attempts at quantum computing. --FyzixFighter (talk) 14:41, 23 April 2010 (UTC)Reply

Well I'll go with the definition of the word Quantum contained in Wikipedia. If you think that Information can be teleported or that the Quantum has a "Contingency" property that can influence it physically I wont argue the point. But you eleminated my point that it is doubtful that a real Quantum, what ever it is, can be teleported, and that's not a fringe concept.WFPM (talk) 17:24, 23 April 2010 (UTC)Reply

I will also agree with you that the diagrams I am talking about are not Feynman diagrams, but rather S versus T Space-time diagrams, where the position versus time location of particles is simple enough for me to understand, and I still can't understand how there could be a discontinuity in the line related to the motion of a real physical particle.WFPM (talk) 18:38, 23 April 2010 (UTC)Reply

(after EC)I removed your edit for 3 main reasons:
  1. You misrepresented what is meant by "quantum teleportation" by implying that it refers to a particle ceasing to exist in one place and then suddenly showing up in another. The article itself goes to great lengths to clarify that it is the quantum state that is being teleported and not the individual particles/quanta/system. Why make a point that something (that a real Quanum can be teleported) is doubtful when neither the article nor the mainstream meaning of the phrase "quantum teleportation" argue in favor of such a thing?
  2. You misrepresented Feynmann diagrams as spacetime diagrams (you do acknowledge the point above).
  3. You synthesized a wholly original argument about Feynmann diagrams and quantum teleportation based on these misunderstandings. The best solution to reinstate the point you want to make is to find reliable source that makes such a point.
In the end, no one is arguing that "quantum teleportation" involves a discontinuity in the line related to the motion of a real physical particle. All the particles continue to exist, but their states have changed (C's state being teleported to B). You, not me, not the article, and not any of the physics articles on the subject, are the only one who has ever brought up the idea of a quantum teleportation involves a particle disappearing in one place and instantly appearing in another. --FyzixFighter (talk) 18:56, 23 April 2010 (UTC)Reply

No but you are arguing that a "contingency" property of a real physical entity can disappear from one place and popping up in the other, and I fail to make the distinction, like in the Thalitimolitine article. I believe that the Quantum is a real physical entity, or rather a package of real physical entities, such that you can catch one or more and keep them or throw them away. I don't see how somebody can influence them from a distance without some transmission to me. You're evidently thinking that you can send them to me and then throw a switch that you have that changes their identity. We call that "Remote Control" which involves us in systems for sending things like light quanta which you know about. And I have read with interest stories about how people could transmit their perceptions of letters etc by a thought process, but I didn't believe them like I cant understand and/or believe this.WFPM (talk) 22:39, 23 April 2010 (UTC) And I didn't know that a contribution to the talk section required a reliable source of an idea or opinion and I thought I was permitted to dream it up by myself.WFPM (talk) 22:54, 23 April 2010 (UTC)Reply

Ah, so it's the whole quantum entanglement and non-local connection thing that you don't like. Well, Einstein did call it spooky action at a distance. But experiments can be done that show that it really is happening and not a case of hidden variables. If we produce two photons such that they must have random but opposite polarizations, then the instant I measure the polarization of one photon, the other photon instantly assumes the opposite polarization. Before any measurement both photons are in a superposition of states, but after measurement on just one photon, both photons assume a single state. However, I can't control which state they collapse to. The influence does happen at a distance, instantaneously without a transmission but no one can control what the influence is.
For teleportation, if A and B are entangled, then measuring the A and C pair in a certain way puts B into 1 of 4 states related to C's initial state (at least, that's what the math says) because of this effect. But the holder of B can't know which one until the holder of A and C tells him the results of the AC measurement.
And your edit, which I reverted, was not on the talk page but on the article itself. --FyzixFighter (talk) 01:24, 24 April 2010 (UTC)Reply

You're right and I lose on that technicality. I mistakenly contributed to the article. My apologies. But of course my argument stands. And you're an EE like I am so I thought I'd engage you on it. Cheers!WFPM (talk) 03:35, 24 April 2010 (UTC)Reply

I am an EE now but my undergraduate degree is in physics, so I straddle both worlds. If you're interested in a somewhat layman explanation of quantum entanglement and the non-local wavefunction collapse (essentially the EPR paradox), I would suggest this video. --FyzixFighter (talk) 06:45, 24 April 2010 (UTC)Reply

Okay so I'm a determinist. But I agree with Bohr's original orbit theories as I point out in my talk section about the details of the Whirlpool Galaxy. And the mass of a Planck's constant energy carrying particle calculates to be 10 to the - 47 gms as I pointed out in Talk: Photon. And you're not helping any by calling it an indeterminate particle. And the Heisenberg Uncertainty principle says that you affect the motion of an electron by trying to measure it. So what? The electron has a function related to the physical creation of the atom, probably as a means of getting rid of excess angular momentum. If you're interested in a physics problem, you might consider my desire to do a 3 dimensional Michelson-Morley experiment. That's where you reflect light from all 3 orthogonal directions instead of just 2. And we need a 4th direction relationship to the other 3 directions to do that. And there aren't 4 orthogonal directions. But note that the cube has a diagonal direction that is equal angled to the other 3 directions. And if we sent a light from the corner of a cube to mirrors in the 3 adjacent corners and analyzed them like in the 2 dimensional test it would be the same. But in the 3 dimensional test case the direction of time would be congruent with the diagonal line, and the Lorentz space-time calculation would be carried out about motion in the area plane where the light paths are at 120 degree angles with each other and I don't think the mathematics is the same, and I need a mathematician to figure it out.WFPM (talk) 16:07, 24 April 2010 (UTC) I also like Boolean algebra and that it usually gives a yes or no answer and very few maybees. And I don"t think you could sell this idea to a computer program dedicated to real physical entity calculations.WFPM (talk) 16:56, 24 April 2010 (UTC) And they're playing around with predetermined spin quantities in their discussion about identity association. I had to consider that when I was making my models in Talk:Nuclear model. And I noticed that the idea of spin is related to an idea of direction. So if you say an entity has a positive spin you have to add that's looking down at the top. Because if I look at it from the bottom it has a negative spin. So in isolation what could I say about an item's spin that would be significant to some other frame of reference?WFPM (talk) 17:31, 24 April 2010 (UTC)Reply

You cut a really broad swath but I'll try and hit each point but no guarantees (in fact, I think I'll do this in subheadings to try and localize the discussion by topic so the dialog can be easier to follow). On a few points I don't understand your use of terminology or the connection with previous discussion:

Re: Bohr orbit model edit

Your argument for Bohr's original orbit model is circular (no pun intended), ie saying it doesn't radiate and that it's total energy doesn't change is saying the same thing, but the conundrum remains for why it still doesn't radiate. Classically, all accelerating charges radiate (eg, only when the current is varied will an antenna radiate), and any orbit, be it circular or elliptical involves a radial-inward acceleration.--FyzixFighter (talk) 23:41, 27 April 2010 (UTC

Can't agree. Don't think Earth radiates energy due to acceleration in it's path around the sun. Things radiate energy when they lose it. And and electron's circular path in an electrostatic field would be a constant electrostatic energy level path.WFPM (talk) 06:04, 28 April 2010 (UTC)Reply

Then you need to review basic electrodynamics (like the Larmor formula) and antenna theory (constant currents don't radiate in an antennae, only oscillating currents do). Acceleration=radiation is a well known fact in any text on electrodynamics. The earth has very little, if any, charge so it won't radiate electromagnetically, but according to general relativity, it should radiate gravitational waves. There are macroscopic examples that contradict your argument, like wigglers and undulators and synchrotrons, where a charged particle radiates (usually called synchrotron radiation) because it follows a curved path (ie, undergoes centripetal acceleration) due to magnetic fields but no change in electrostatic potential (in the case of wigglers, the electric field is zero, so no change in electric potential and magnetic fields don't have a potential energy field for charged particles). --FyzixFighter (talk) 06:40, 28 April 2010 (UTC)Reply

I'll buy that as an admission that the constant velocity motion of an electron in an electrostatic field doesn't result in emissions as long as the field has the same static strength.WFPM (talk) 21:42, 28 April 2010 (UTC) But there's a saying that "if you shake an electron you'll get radiation" implying that it can translate without radiating, but not turn or even acelerate. And when the electron gets to the top end of it's path through a Marconi antenna and has to stop, I guess it's the cause of the radiant energy radiated away from the antenna. But in ac currents, the electrons don't radiate, just the constituents of the field.Reply

I've never argued anything differently, I've only ever argued for emissions from charged particles undergoing acceleration. Constant velocity motion never produces radiation because in those cases acceleration is zero. But orbits, even perfectly circular orbits, are not cases of constant velocity motion. Speed is constant in those scenarios, but not direction therefore acceleration is non-zero. Even electrons in a perfectly circular synchrotron with no external field (therefore traveling along a path that is equipotential) will generate synchrotron radiation. Again, this is accepted classical electrodynamics that was known before the advent of both relativity and quantum mechanics. --FyzixFighter (talk) 03:14, 29 April 2010 (UTC)Reply

Please!! A circular orbit certainly involves a constant velocity motion along the tangent line to the circle. I don't see how you can question that. And in gravitationally induced circular orbits, the lost free energy level value of the orbit remains constant and no orbital energy is lost. Then if electrons radiate energy as the result of this motion, and I thought that their radius of motion remained constant for a constant velocity, we're going to have to dream up some other reason for the occurrence.WFPM (talk) 14:31, 29 April 2010 (UTC)Reply

Circular orbits involve constant speed motion. Do you understand what the difference is between speed and velocity? Velocity is a vector - it has both magnitude and direction. If either (or both) of those change, then it cannot be constant velocity motion. Constant velocity implies zero net force and straight line motion (Newton's first law). Circular orbits satisfy neither of these--there is a net force inward towards the center of the orbit (gravitational in the case of planets, and electrostatic for Bohr orbits), and the motion is not straight line. Perhaps you're confusing the radiation theory for atoms with general electrodynamic radiation. In atoms, electrons will radiate when they transition between energy levels, but this categorically not a classical concept. Radiation classically is not a way for particles to get rid of excess energy when it moves across a non-uniform potential. Radiation classically, according to Maxwell, Hertz, Larmor, and all the pre-quantum mechanics/relativity physicists and engineers, is due to charges undergoing acceleration. Any basic physics/engineer textbook will attest to these two classical concepts (circular orbits are not constant velocity motion & accelerating charges radiate). Are you saying the textbooks are wrong? --FyzixFighter (talk) 15:12, 29 April 2010 (UTC)Reply

I'm thinking that velocity is Ds/Dt and in a circular orbit is not changing. And why it is changing in direction is a different matter. Maybe we can adopt Pauli's concept and say there has to be emissions perpendicular to the tangent of motion so that the net directional momentum remains constant. And maybe Yukawa can argue that his meson has the power to convince the nucleons that they should throw material away from a center of "attraction" so that they can move closer together. And in space time considerations I think about the implications of graphs on the time versus distance TS plane. And if there's a force involved in the consideration, I visualize it as a line vector sticking up out of the paper plane. So stationary is a perpendicular line, and straight line motion is a slanted line. And accelerated motion is a curved line. And an Impulse is an area equal to the area of the time vector times the force vector. And the force vector times the lateral distance is the Force times time or the work area. So as a force chases a freely moving particle it is giving it kinetic energy of motion at an increasing rate. But if is truly a freely moving particle, I dont see why it should be doing anything other than storing up the acquired kinetic energy given it by the force vector. And if you apply a side directed force on the particle as they do in mass determination equipment, the side force also adds kinetic energy of side motion to the previous linear motion of the particle as per the calculations. So nature's ability to retain the kinetic energy of velocity through a circular orbit has made me cautious about the ideas about when it can radiate energy. And the elliptical orbits are the same except that each orbit involves the presumably 100% efficient conversion of potential energy to kinetic energy and back again.WFPM (talk) 20:11, 29 April 2010 (UTC) And I think that the textbooks are maybe wrong about the rest mass of the Plank particle. But since they don't have a concept thereon it is not to worry.WFPM (talk) 23:14, 1 May 2010 (UTC) And if I'm pressed, I might do like Tombe and argue that the Vsquared value could go up faster than the mass value goes down and so a particle's kinetic energy could exist at some very low value of mass. WFPM (talk) 01:07, 2 May 2010 (UTC)Reply

Circular orbits and acceleration edit

To make this easier to focus on a single topic of disagreement I'm going to decouple the whole circular orbits and acceleration question from the the general radiation question and create this new sub-subheading. Let's ignore the radiation question for a minute and just look at orbits. First of all I need to ask you what you think the difference is between speed and velocity? You're nearly correct in how you think about velocity, it is Ds(vector)/Dt (the time derivative of the position vector). I generally don't like working in a TS plane for classical mechanics, but we can and by your own argument then an orbit is undergoing accelerated motion. Think about the usual TS representation of the earth orbit - it's a corkscrew spiral, which is in fact a curved line. A force only increases the kinetic energy of a particle if it acts in the same direction as the spatial displacement (see Work (physics)). In the special case of circular orbits, the force is constantly parallel to the displacement so the kinetic energy doesn't change. But since the central force (gravity for the earth, electrostatic for an electron in Bohr's orbit model) is the only force acting on the electron, then according to Newton's 2nd law the non-zero net force produces an acceleration. So to sum up some the important points and the points where I think we disagree:

  1. Velocity is the derivative of the position vector, is a vector itself, and is different than speed.
  2. Acceleration happens when velocity changes either in magnitude and/or in direction. (see 1st paragraph of Acceleration)
  3. The net force in a circular orbit is always radially inward, so the acceleration is also radially inward (Newton's 2nd law).
  4. The acceleration of a particle in a circular orbit like we're discussing is non-zero.
  5. An applied force will only increase the kinetic energy if it has some component that is parallel to the displacement/velocity.
  6. An applied force that is perpendicular to the displacement/velocity will not change the kinetic energy.
  7. The net force in circular orbits is always perpendicular to the displacement/velocity.

If you disagree with any of these statements then you do not believe in classical Newtonian mechanics. In which case, I would say go back and read your college mechanics textbooks and review the basic kinematic and dynamic concepts of mechanics. --FyzixFighter (talk) 22:52, 29 April 2010 (UTC)Reply

I just talked about acceleration in my above discussion. If you're on the TS plane plane and pushing with a force in a straight line to cause acceleration, you are adding something to the particle that is called inertia, and has the same units as kinetic energy of motion. So it first changes it's time rate of change of the velocity, which is called acceleration, and as the velocity increases it continues to accumulate the inertia and the amount of displacement. But why should it emit radiant energy? There's no other force. And it's accumulating inertia as fast as it's coming in, I think. Do you know of any reason why it should radiate energy? And our electrostatic theory of sideways deflection of motion of a particle says that the sideways force adds a sideways motion to the particle proportional to the amount of sideways direction inertia acquired, also I think. And in that case it would involve a change of velocity due to the increased inertia that it has in it's new path.( At least the action is called an acceleration in my Kaplan's Nuclear Physics book, and of course Newton was the one who developed the theory that orthogonal motions functioned independently of each other). But in electromagnetic and light energy deflection situations we're able to get around that by assuming that for some reason or angle of force direction the force is always such as to change the direction while maintaining the standard velocity. I hadn"t heard that summary of dynamic motion principles and I'll have to think about them some more and look in my Robeson again. But that's enough for now.WFPM (talk) 00:00, 30 April 2010 (UTC) Think I cant agree with #6 based on Kaplan.Reply

Which Kaplan book are you talking about? Is it Irving Kaplan's "Nuclear Physics"? What statement in Kaplan do you feel contradicts #6? If that is the same book, I'd also invite you to look at the example Kaplan gives on the bottom of page 84 (this is in the 1962 edition) and continues on the top half 85, which addresses the original point of dispute:
"Assume that the electron revolves about the nucleus in a circular orbit of radius a, and that the velocity of the electron is such that the attractive force between the electron and the nucleus provides just the centripetal force required. The system, nucleus and electron, should then be mechanically stable. The electron, however, is subject to a constant acceleration toward the nucleus, and, according to electromagnetic theory, the electron should radiate energy. The energy of the system should then decrease; the electron should gradually spiral in toward the nucleus, emitting radiation of constantly increasing frequency, and should eventually fall into the nucleus."
Kaplan does talk a little bit about the relationship between force and kinetic energy on pg 124 (equation 6-31), but implicit in this is that F and dx are in the same direction (Kaplan isn't explicit but he does say that F*dx is the work done on the particle, and the full formula for work is F*dx*cos(theta), so theta (the angle between dx and F) in this example has to be 0). What other physics-related texts do you trust and have access to? I've got access to a university library (where I found Kaplan) and I much prefer to rely on texts that we both trust so it's not either of us having to trust in the intellect of the other, and so we can use an agreed upon vocabulary. --FyzixFighter (talk) 03:01, 1 May 2010 (UTC)Reply

I'm glad you have Kaplan 'cause he's my detail man and he's sometimes over my head and I know I can keep learning as I read him. And on page 84 he's talking about the electromagnetic theory, which I mentioned, and on page 124 he's talking about the equivalence of rest mass to kinetic energy such that the "effective" mass of a moving particle can be representative by the sum of the 2 values. And note that he goes on to say that the constituents of a photon cannot have a rest mass because then the equivalence mass of a moving particle would become infinite in accordance with 6-27. And of course I cant agree with that because I think that the energy of light is transported by a stream of planck particles which are boring their way through space and whatever. And now we come to page 200 where he talks about the ability of an electrostatic field to cause an electron to accelerate in the direction of the field. Well consider the implications of that. The particle has a lateral velocity and associated kinetic energy, and now it accelerates at right angles to some increased velocity at an angle, (just like in ballistics), which results in it's having an increased kinetic energy of translation. I think that it was Newton who developed that concept in his Principia. So I don't think your #6 will stand up under scrutiny.WFPM (talk) 15:27, 1 May 2010 (UTC) If you think about this, you can see why Newton stuck to his theory about the particulate nature of light transmission. He probably just couldn,t imagine a small enough particle to do the job. PS I finally found a reference in Wikipedia. Look in Trajectory of a projectile at the velocity calculation where the 2 right angle velocities are added at right angles to each other to get a combined increase in the projectile velocity and resultant kinetic energy of motion.WFPM (talk) 16:39, 1 May 2010 (UTC)Reply

Two points:
1. Comparison to ballistics only works part way. In #6 I specifically argue that the applied force has to be perpendicular to the velocity vector. This isn't true in a ballistic trajectory unless the projectile is fired off with enough speed to go into a bound orbit. But the math from that article is useful, so let's go through the math and see what happens. In cartesian coordinates, the kinetic energy (T) is written:
(1)  
Now let's take the time derivative of this to see what are the condition for the kinetic energy to change
(2)  
By Newton's 2nd law F=ma along the axis.
(3)  
If the velocity is always perpendicular to the velocity, this sum will always be zero, and the kinetic energy will be constant despite an applied force and induced acceleration. The math doesn't lie.
2. It seems more and more like a case of WP:IDIDNTHEARTHAT, or maybe I'm misunderstanding what the original dispute was. It seems to me that Kaplan, in the text from page 85 I quoted above, explicitly makes a statement that contradicts your original argument. Kaplan states that according to classical mechanics and electromagnetics, an electron in a circular orbit, while mechanically stable, would still radiate and spiral in. Mechanically stable means that if the electron didn't have a charge, it would be a stable orbit (but then there would be no electrostatic force keeping it in orbit). So, even Kaplan says that the Rutherford-Bohr orbit model is not classically stable and will radiate. Am I misreading Kaplan (if so, how?), or are you saying that Kaplan is wrong? --FyzixFighter (talk) 19:51, 1 May 2010 (UTC)Reply

The formula you're using says the initial lateral velocity is conserved and the side velocity added by the side force is added at the applied angle to create a new + or - new velocity and associated kinetic energy of motion, and I agree. And that violates #6WFPM (talk) 20:53, 1 May 2010 (UTC) At least the perpendicular part of the side velocity violates #6 and I'm being more general, and I could see how you might have a force that maintains the translational velocity of the particle but it wouldn't be a perpendicular force. And most level balistical trajectories follow a parabolic path into the ground and thus deliver the initial kinetic energy of the level velocity plus that acquired in the falling process.WFPM (talk) 21:08, 1 May 2010 (UTC)Reply

Your intuition is wrong, you don't understand calculus, and you are dodging the math. The equation I give above (I've labeled it eq-3) clearly shows that if the applied force is perpendicular to the instantaneous velocity, the kinetic energy is constant, which supports my statement #6. So here are two questions. 1) Where do you disagree with this math? 2) What is the net force acting on a particle in a circular orbit and, if it's not zero, what direction does it point? --FyzixFighter (talk) 22:41, 1 May 2010 (UTC)Reply
Still would love a response from you to these two questions. --FyzixFighter (talk) 13:38, 7 May 2010 (UTC)Reply

Re: Planck mass and the photon edit

I'm not sure how you did your calculations. In order to get from the Planck constant to energy you need a time scale or a delta-t of some sort. This is usually obtained by combining it with other physical constants as seen in Planck mass. Additionally, the mass of a photon has been experimentally shown to be less than 1e-18 eV, which comes out to be (by my calculations) about 1.7e-51 grams, which is a few orders of magnitude smaller than the number you give.--FyzixFighter (talk) 23:41, 27 April 2010 (UTC)Reply

I'm talking about a hypothetical mass particle whose kinetic energy (mvsquared)/2 would be equal to planck's constant. I got that to be about 10e-47 grams. =((6.6xe-27)/10e+21)/2.WFPM (talk) 06:19, 28 April 2010 (UTC)Reply

I still don't follow your calculations exactly. So the mass of your particle will then be m=2*E/v^2, right? How did you choose E and v? Was it just the Planck constant and the speed of light, respectively? Since the Planck constant is not an amount of energy (it doesn't have units of energy), how did you get the energy equivalent of it? What are the units and sources for the numbers (6.6e-27 and 10e+21) you use? I ask because I think if you keep track of the units attached with the numbers you want to use, it won't come out to be units of grams. --FyzixFighter (talk) 07:00, 28 April 2010 (UTC)Reply

Planck's theory is that radiant energy is quantized. His formula says the quantity is equal to a number of Planck's value unit of action (erg-seconds). But the theory is that the action amount is acquired during a very short time period. And action is energy times time, or lets say the integral of energy accumulated during a time period. So lets say that the Photon is not 1 seconds worth of energy particles, but rather an equivalent amount of energy accumulated in a short time in a few particles, with each particle having a planck's constant value of energy. So we have a planck's constant =6.6 times 10e-27 erg-sec = 6.6 times 10e-27 dyne cm squared/sec squared all accumulated at once, or 6.6 times 10e-27 dyne cm squared. And if it was kinetic energy acquired the v squared value is 10e+21 cm squared/second squared. So the m/2 value would be 6.6 times 10 e -27 -21 which equals 6.6 times e-48, So m = 1.32 times 10e-47 grams. I hope!!! This higher mathematics gets me down. And we would have the photon being the energy contained in the wavelength's frequency value of planck particles, except that that value would be contained within the length of 1 wavelength. And you can see why I would rather have the computer do the dirty work and just give me the answer.WFPM (talk) 09:19, 28 April 2010 (UTC)Reply

Yes, Planck's theory was that radiant energy is quantized, but quantized based on the frequency. In other words, give me any energy and I can create a photon with that energy as long as I pick the appropriate frequency. I don't see in the theory where it says that the action amount is acquired during a very short time period. Where do you see this in the theory? As for units, you're playing a little loose with units, ie an erg-sec is not equal to a dyne-cm squared/sec squared but a dyne-cm-sec (an erg on it's own is a dyne-cm, right?). You can't just change units at will without some kind of conversion factor. If we want the energy of the action to be accumulated in a very short amount of time, we'd actually have to divide the action by that amount of time (action=energy * time duration, ergo energy=action/time). So shorter amounts of time leads to larger energy values, and an all at once instance leads to an infinite amount of energy. Does that make sense? Also, if v is meant to be the speed of light, then v^2 is about 10e20 (or 1e21) cm squared/second squared (I think this is a confusion over what the ##e## notation means. The "e" here means "times 10^". The Planck constant is 6.6e-27 erg-sec, also written as 6.6 times 10^-27 erg-sec). --FyzixFighter (talk) 14:37, 28 April 2010 (UTC)Reply

An erg on it's is 1dyne cmSquared per secondsquared. That's (F times S squared)/ T squared.WFPM (talk) 14:08, 29 April 2010 (UTC)Reply

No, see erg, specifically the line where it says 1 erg = 1 dyn cm. A dyne is a unit of Force, an erg a unit of work or energy. If we want to follow a little dimensional analysis for what an erg (work or energy) consists of we can use some standard physical relationships: kinetic energy=mv^2/2 => M(mass, not force)*(S squared)/(T squared), gravitational potential energy=mgh => M*(S/T squared)*S, or work done by a force=F*d => F*S. It appears in your dimensional analysis that you've confused force and mass - perhaps this is a confusion on the difference between weight (a force) and mass? You do understand that weight and mass are not the same thing, correct? --FyzixFighter (talk) 14:58, 29 April 2010 (UTC)Reply

I just changed the above, so look at it again.70.244.238.120 (talk) 18:25, 28 April 2010 (UTC) I occasionally drop out of being logged in while contributing and then Wiki gives me this contributer number. Why is that?WFPM (talk) 18:42, 28 April 2010 (UTC)Reply

It still doesn't scan. If a photon is made up smaller planck particles, then frequency can only occur in discrete, integer values, which it doesn't. Maybe there's something in QFT or QED that quantizes frequency in any way, but AFAIK there is no reason for the allowable frequency values to be a continuous, non-discretized spectrum. That a photon is made up of smaller particles is not consistent IMO with Planck's original statistical argument for quantization. Nor am I aware of any evidence that a photon is made up smaller pieces. --FyzixFighter (talk) 03:43, 29 April 2010 (UTC)Reply

Re: electron edit

I'm sorry but I don't really follow or see what you're trying to argue with your comments about the electron, the Heisenberg Uncertainty, and the creation of an atom. What indeterminacy that apparently I've talked about are are you referring to here? How does your argument hold for electrons created with no corresponding created atom (like a muon or when a photon creates electron-positron pairs)?--FyzixFighter (talk) 23:41, 27 April 2010 (UTC)Reply

I'm talking about the idea that the uncertaincy of measurements of the location and/or momentum of an electron might mean that it doesn't exist. And if it exists, then whatever that creates it must also exist. See Lucretius. So when we create these force carrying hyopthetical particles, that pop into and out of space time we tend to dematerialize the whole process. And I particularly don't see how you send out a force of attraction from the attraction center. That sound like a bank shot problem to me.WFPM (talk) 06:46, 28 April 2010 (UTC)Reply

Re: 3D michelson-morley experiment edit

I also don't completely understand your desire to do a 3D michelson-morley experiment. What kind of information are you hoping to get that combining the results from two standard 2D M-M experiments, one vertical and one horizontal, would give you? Doing 3D would be a bit difficult - you couldn't use a half-silvered mirror for a beamsplitter. I imagine you could use a volume grating of some kind that would diffract an incident beam into the three directions. When you say that the diagonal of the cube is congruent to the "direction of time", I think you're confusing the vertical time axis of a spacetime diagram with an actual spatial axis. What do you mean when you say that the direction of time is congruent to the diagonal of the cube? Could you also be more clear about what you mean by "Lorentz space-time calculations"? I think I have an idea of where you're going, but I'm not certain.--FyzixFighter (talk) 23:41, 27 April 2010 (UTC)Reply

The Michelson-Morley experiment experiment showed some results, just not enough to be significant. And the Lorentz-Fitzgerald contraction calculation is calculated related to the time of orthogonal motion on the measurement plane. But with a 3 orthogonal direction test within a cubic volume of space, and using the cube diagonal as the time ordinate, the calculation would be made about motion within the plane containing the point of origen and the location of the 3 light instant locations and would involve their motion at a 120 degree angle of separation from each other. This would change the math of the Lorentz=Fitzgerald calculation, and I haven't been able to figure out how much, and it's an interesting physics/mathematical problem. And I live in a 3 dimensional universe and am not positive that my point of view and/or time line of existence is perpendicular to any particular 2 dimensions of space. So how's that for complicated?WFPM (talk) 07:08, 28 April 2010 (UTC)Reply

Ok, so I follow about half of your argument. I don't think anyone has done a Michelson-Morley experiment that showed a significant result larger than their apparatus' experimental error. Most modern attempts at the experiment having further driven down the limits from what the initial experiment's limits were and still no significant result. I don't see how the cube diagonal is the time ordinate. In the Lorentz-Fitzgerald contraction calculations, the only thing that ties the time ordinate to a specific direction is the vector v, which in Lorentz's initial formulation was the system's velocity relative to the ether. If you want v to be along the cube's diagonal, I think what you're looking for would be found at Lorentz transformation#Matrix form where it gives the matrix form for the Lorentz equations with a boost in an arbitrary direction. For a boost along the cube diagonal, Beta_x=Beta_y=Beta_z=Beta/sqrt(3). I think that this would be compatible with Lorentz's derivation of his transformation. But I still think you could very easily look at the results from two three 2D M-M experiments (one for each combination of two axes) to determine the direction of the vector v if it isn't aligned along one of the axes chosen. --FyzixFighter (talk) 07:50, 28 April 2010 (UTC)Reply

But if you use the diagonal as the time line and a plane carrying the origin and equaldistance points as the Lorentz-contraction calculating area, (and they are orthognal to each other), you note that the light position lines move out at 120 degree angles to each other. And the contraction calculation becomes the comparison of a lateral line alongside the river? to a 60 degree oblique line into the river and back. And I've been unable to make this calculation and would like to have some help.WFPM (talk) 22:36, 29 April 2010 (UTC)Reply

I still don't see why the diagonal has to be the time line. In the Lorentz-contraction equations the "time line" is tied to a spatial direction based on the direction of the velocity of the system relative to the ether. If we want the "time line" to be along the diagonal, then we are stating that the system is moving in that direction relative to the ether (or that the ether wind is along the diagonal). So by my math the Lorentz-contraction equations would be:
 
I did this first in the matrix notation and I hope I transferred it right to separate equations. --FyzixFighter (talk) 23:42, 29 April 2010 (UTC)Reply

Re: Boolean logic edit

I'm not certain where Boolean logic enters in exactly, but I'm guessing it's with respect to quantum stuff. I'm also not certain what you mean when you talk about having "a computer program dedicated to real physical entity calculations". Is this in reference to my mention of quantum computing? Boolean logic tends to break down with quantum mechanics. Some examples of this are the double-slit experiment with electrons, the sequential Stern-Gerlach experiment, or sequential non-orthogonal polarization measurements of a photon. Quantum computing relies on the ability to put a bit into both a 1 and a 0 at the same time, but it's only good for solving a few limited problems and not running standard computations. A good book that I enjoyed for discussing quantum computing and its limited applications is "Ultimate zero and one: computing at the quantum frontier" by Williams and Clearwater.--FyzixFighter (talk) 23:41, 27 April 2010 (UTC)Reply

Boolean logic involves the theory of sets. And as far as I can see, all information, all communicated scientific information can fit into either or both of the two categories fact, or opinion. So when I get information, I immediately try to fit it into those 2 category sets. But I'm human and live with uncertaincies, and believe in luck, and have hunches etc, which are not consistent with Boolean logic. And I've dealt with the mathematics of probability and know that if we went through that formalized process every time we had to make a decision we'd never get anything done. So as senator Eagleton said about action in the Senate "We muddle along". But using strict Boolean logic in today's concepts about scientific phenomena wouldn't get us very far. It would bog down at the travelling twins paradox. But if we took a computer and told it "Stick to rigid physical logic and go as far as you can and then call me, rather than jumping into hypothetical nonmaterial alternatives, we might be able to move the level of real physical entity calculations down into the level of detail that it would need to really explain some of these natural events.WFPM (talk) 07:34, 28 April 2010 (UTC)Reply

So what you're arguing is that if we knew more details, knew the physical attributes of the particles involved with more certainty, we wouldn't need quantum mechanics? That the uncertainty isn't inherent in the system, but is just the case of hidden variables because we have measured the state precisely enough? If that is an accurate summary of your argument, then I would respond that an experiment has been devised that can distinguish between these two views of quantum mechanics (it's talked about both in the book "Ultimate Zero and One" I mention below, and in the video I linked to above) and the results don't support the hypothesis that the uncertainty is due to not knowing enough detail. Instead, the results support quantum mechanics' claim that the uncertainty is inherent and outcomes truly unpredictable. --FyzixFighter (talk) 08:14, 28 April 2010 (UTC) Uncertaincy of measurement I'll buy. Uncertaincy of existence I wont. I'll go with Lucretius, except that there's a lot more variables than he could imagine. And likewise with Newton and the moon.WFPM (talk) 23:33, 1 May 2010 (UTC)Reply

My idea concerning this is like considering what can happen if we throw 50 pennies up in the air and ask you how they are going to come down. You get all excited and start making calculations on the distribution of probabilities related to each of the discrete possible occurrences and wind up with a probability summary totaling 100%. Do you think nature goes through that process? of course not! The individual and maybe the collective mass of pennies go through the physical process of doing what they should do under the circumstances and the end result is whatever is the result of the details of that process. And the only reason that a fast computer couldn't do this is that it doesn't have a sufficient description of the details of the process. I remember the programming problem related to the generation of a random number by the computer, but that we wanted to repetitively generate the same series of random numbers for debugging purposes, so we had to seed a number overflow program so we could always begin at the same point for analysis.WFPM (talk) 23:55, 1 May 2010 (UTC)Reply

Sorry but once again no, you're wrong. Did you even watch the video or look at the experiments I talked about above? If you did you would know that there are simple experiments where quantum mechanics and the hidden variable (not enough detail of the process) produce different probability distributions. These experiments have been carried out many, many times and the statistical distribution matches quantum mechanics, confirming that the observed uncertainty is inherent in the state of the particle and not due to not knowing enough about the state to predict how it evolves. I'm more than willing to entertain multiple models, but once an experiment is performed that contradicts a model, it's time to throw out that model. The experimental evidence contradicts your arguments, therefore you're wrong. --FyzixFighter (talk) 16:51, 6 May 2010 (UTC)Reply

I contributed about the Stern-Gerlich Experimemt in the next section and in the talk section thereon. It sounds like a process of organizing two different orientation of spin into two different groups and sending them to different target locations. I haven't read the book. But I have read the story "Skewered" by Isaac Asimov and noted his developed of the multiplicity of time interval events that can occur within a scene the size of control volume of a neutron during the lifetime of the universe, and I've lost interest in other additional possibilities for the occurrence of an event. So events occur as the result of a sequence of causative factors. And as a means of keeping track of this process about an event of interest, we have to monitor the discernible entities of nature, and which as far as I can see involves matter moving around in space. So I don't believe in even the temporary discontinuity of existence of either the space or the matter, although it might be that the matter might get smaller than I can discern, like when it's very small and moving very fast like in electromagnetic energy generation and transmission. And I can't see instantaneous translation in space, because the time value interval related to motion is already from some upper physical limit to practically zero, and I don't know anything in nature that can be subdivided into a number of zero magnitude subentities. And it can be that our 3 dimensional concept physical laws of motion might be wrong due to some quirk of nature, such as a lack of orthogonality of spacial dimensions in our spacial location, and maybe the space plenum volume is expanding with time as is the hypothesis. But the fundamental process is that of mass entities moving within a volume of some kind of space, and as IBM used to say "It's all battery and ground and the rest is details".WFPM (talk) 23:49, 6 May 2010 (UTC)Reply

I've responded to your SG experiment comment below (again, you don't understand the physics of the experiment) and awhile back to your planck particle idea (would require frequency to be quantized, which experimentally it isn't). This has nothing to do with a discontinuity in space or time of a particle, as I've mentioned ad nauseum. If you watched the video where the type of experiment is explained you would understand this. The results of the experiment are random, but since two entangled, quantum mechanical particles are involved, the measurements on the two particles are related. As I've said time and time again, and you continue to not understand apparently, is that the experiment would produce different statistical distribution of the results if quantum mechanics is right (the uncertainty in state is inherent to the particle) or the hidden variable interpretation is right (the uncertainty is only because we don't know enough details about the particle). Experimental evidence does not match up with the hidden variable interpretation, but does match up with quantum mechanics. Therefore we throw out the model that doesn't match experiment, and keep the one that does. In this experiment, as with all other experiments that distinguish between quantum mechanics and your arguments both about this and about the physical model of the nucleus, Quantum mechanics wins out. --FyzixFighter (talk) 13:36, 7 May 2010 (UTC)Reply

Re: Spin and nuclear model edit

I'm not sure what you mean by the discussion of the identity association. Who are "they" in that sentence and what is the "identity association" they are discussing? Inherent when anyone talks about spin-up or spin-down is the type of measurement made which includes the observer's point of view (spin-up and spin-down being the eigenstates of the measurement). See again the Stern-Gerlach experiment. As to your physical nuclear model, the main problem is it doesn't match up with experimental scattering data, as pointed out by User:Sbharris here, ie that the "electron scattering experiments show the charge in He-4 nuclei is a smooth exponential function, maximal in the center and tailing off outward". Quantum mechanical model of the nucleus predict this behavior, your model (as I understand it) does not.--FyzixFighter (talk) 23:41, 27 April 2010 (UTC)Reply

I just can't buy that the protons in an atom can all clump in the same location. We discussed that in Spherical packing. Also if you allow that concept how are you going to explain that Q=Z times e, and is thus divisible into parts. Consider the gravitational attraction problem between that of a 1 gram of gas entity and a 1 gram point particle, that are each enclosed in a 1 cm dia volume of space and in contact with each other. If you sent a particle through between the 2 spheres at the point of contact, the moving particle wouldn't have a net force in either direction. It's only when you analyze the bias of the net force of the concentrated particle with relation to the variance in passage locations than you begin to understand the capability of matter concentration to bias motion in it's direction of existence. So your trying to tell me that a sphere of marble material is the same size as an equal mass volume of accumulated marbles, and I refer you to the spherical packing article.WFPM (talk) 08:06, 28 April 2010 (UTC)Reply

Again, all I can say is that experimental evidence does not match your physical nuclear model. Inverse scattering problems do exactly as you suggest, taking into account the bias of the net force on the variance in passage locations. When electrons are scattered off of He-4 nuclei and we extrapolate back to what the charge distribution has to be to give us that kind of scattering distribution, we find that the positive charge is not localized to two separate spheres, but is spherically symmetric around the center of the atom with an exponential drop off with radius. Thus classical images of separate particles fail to model known charge distributions in very small nuclei. You might not be able to buy this, but it is what experiments tell us. --FyzixFighter (talk) 08:47, 28 April 2010 (UTC)Reply

Well, I don't see that I have a conceptual problem. Maybe a communications problem. And you have the conceptual compatibility problem. And my thoughts about the properties of the magnetized models are leading me in what I think is the right direction towards understanding physical reality. And we'll see. And thanks for communicating.WFPM (talk) 09:36, 28 April 2010 (UTC)Reply

The power of any model is its predictive capability. That's why physicists have a love/hate relationship with string "theory" - there are currently no new, realistically testable predictions. AFAIK it makes some predictions, but the experimental conditions required are very much outside of our current capabilities. The best way to prove your model over that of quantum mechanics is for it to explain experimental data that either current quantum mechanics can't or in a manner more elegant (channeling Dirac) than quantum mechanics. Quantum mechanics is nice IMO because it is not specific to a certain scenario but it is this kind of general set of rules that when applied to various scenarios correctly predicts a variety of behavior. Since you said you had a penchant for Asimov (I too read a lot of Asimov as a teenager), I would also recommend another set of books on these topics which have a very Asimov-esque feel to them. They are George Gamow's Mr. Tompkins series of books (the first two which deal with physics can be found together in "Mr Tompkins in Paperback"). --FyzixFighter (talk) 14:52, 28 April 2010 (UTC)Reply

I have Gamow's "The Atom and it's nucleus" and some others and I admire his sense of humor and think he argued in favor of the pairing of the proton and neutron, although I can't find it as a reference.WFPM (talk) 20:40, 28 April 2010 (UTC)Reply

(I hope this does not come across as wiki-stalking/wik-hounding - if so, I apologize) I notice also that you seem to claim here that the rationale for the next block in the ADOMAH PT table being 18 blocks is not apparent with quantum mechanics. On the contrary, quantum mechanics does predict that the next block should be 18 blocks wide. Each expansion part of the ADOMAH table corresponds to the filling of the next orbital subshell type (hence the labels at the top) and is as wide as the number of electrons that will fit in that subshell. The solutions to the Schrondinger equation for a central potential are characterized by three parameters: n, l, m_l (there is a fourth parameter for spin up/spin down, but that will just add in a factor of 2 at the end). The first parameter, n, can have any positive integer value and is the number for the "electron shell". The second parameter, l, can have integer values from 0 to n-1. Each l value corresponds to a different subshell: l=0 is s, l=1 is p, l=2 is d, l=3 is f, and so on. The third parameter, m_l, can have integer values from -l to +l. Therefore for a given subshell, the number of unique solutions within the subshell is 2*l+1. Multiplying by a final factor of 2 for the up/down spins, we get that the total number of electrons that can fit in a subshell is 2*(2*l+1). For s(l=0), this is 2; for p(l=1), it's 6; for d(l=2), it's 10; for f(l=3), it's 14; and for the next orbital g(l=4), it would be 18. So, as I see it, quantum mechanics predicts logically an 18 unit expansion concept. --FyzixFighter (talk) 16:54, 28 April 2010 (UTC)Reply

Boy!!! I'd just rather build an octahedral model of the filled 8 square (32 element) model series with the Z = 120 element, and then note that for additional elements I have to start a new 10 square series that will involve 50 more deuteron additions, starting with 18 around the periphery. And since 2 layers of additions are required to form a base for the next layer I also know how to extend the table if I want to.WFPM (talk) 22:53, 1 May 2010 (UTC)Reply

Your octahedral model might be a nice mathematical crutch, but it cannot be what corresponds to physical reality. The way one model wins out over another is based on its predictive power. What predictions does your model make that quantum mechanics doesn't (or predicts something differently)? Because experimental evidence (electron scattering experiments) shows that the charge distribution in the nucleus is not localized to packed spheres, but "smeared" like quantum mechanics predicts (the nucleons being non-localized), therefore I will gladly take the quantum model over your model. Unless you can explain how your model is consistent with these experimental results, your model will never be more than a mathematical crutch for people who are either incapable of understanding or unwilling to put in the effort to understand quantum mechanics. --FyzixFighter (talk) 17:05, 6 May 2010 (UTC)Reply

When I asked V Smith, bless him, to help me get my image into my atomic model discussion, I argued that a cognitive picture is worth a thousand words in generating interest and understanding, and thus even disagreement about subject matter. And that goes for your ADOMAH PT image, because I don't see how you can sell an 18 unit addition to the structure. And I'm not too in love with the shell theory such that it's parameters control my thinking about the structure of the atom. And I cant understand orbitals without angular momentum, unless you'll consider maybe tethered orbits, which you probably wont. So we lack a way to make sensible comparisons of different concepts, so lets consider them both and let the best one win out. And note that spin involves the existence of 2 positive energy entities in such a way that the direction of motion vectors are opposite to each other. And does that mean that the contained energy of motion will cancel out? Of course not!! It just means that if they intersect without being properly aligned the angular motion properties of the pair will cease to exist as it probably didn't in the first place. It's just compatibility of motion that counts.WFPM (talk) 19:40, 28 April 2010 (UTC) And if you had a particle (like a nucleon?) that was spinning and had an appendage on each end, and you assumed that the free end of the appendage was the "top" of it, then you would have to say that one of them had a positive spin and the other one had a negative spin. And you might think about that.WFPM (talk) 20:14, 28 April 2010 (UTC)Reply

IMO, the structure is only apparent if you start to accept that in quantum mechanics, particles are non-localized until you try to measure their positions. Until then, they act more like fields. The funny thing about angular momentum in quantum mechanics is that no two individual components of quantum angular momentum can be simultaneously specified for a given system, whereas the total angular momentum can be simultaneously specified along with any one of the operator's components. This applies to the angular momentum of orbitals as well as spin. The up/down when we talk about spin isn't in reference to the total angular momentum, but only the z-direction (ie the axis along which we're measuring) components. The angular momentum components along the other two axes become indeterminate when the measurement is made, hence the results of sequential stern-gerlachs.
I don't understand what you mean by 2 positive energy entities. Can clarify what you mean by this? Are you talking about the angular momentum of the orbital and angular momentum of the spin? As to the idea of appendages, it doesn't work so much because while the total angular momentum is quantized, the angular momentum around any axis you measure will also be quantized (but is always less than the total angular momentum). I'm not sure if that addresses some of your statements - I really don't understand what you mean by a two appendage structure. Are you saying the electron has a dumbbell-like structure, and the two ends are free to rotate independent of one another? --FyzixFighter (talk) 04:18, 29 April 2010 (UTC)Reply

I was about to add to the discussion that I dreamed up last night. It's that if I had a spinning radiation emitting particle that emitted a theoretical up spin emitted sub particle from one end and a theoretical down spin emitted sub particle from the other, and I wanted to reconnect those particles, then I could design a simple axial connection device and install them on the particles such that either of the sub particles could be axially connected to either end of the emitting particle. And in doing I would have made the sub particle to become identical to each other such that you could not have distinguished them from each other. Which means that up and down spinning motion along a given axial direction is identical motion except for the end of connection relationship. And I'm saying that the emitting particle would have to emit the particles away from either of it's two axial point locations such that the emitted particle leaves with the same direction of rotation of its components as did the emitting particle.WFPM (talk) 15:04, 29 April 2010 (UTC) Let's call that end emission, as opposed to peripheral emission, which even so might work the same way.WFPM (talk) 15:07, 29 April 2010 (UTC)Reply

Such a structure does not match with experimental observations of electrons. For one thing the total angular momentum of the structure is zero. It would also not have the dipole structure that we observe for electrons, but instead has a higher multipole moment. In experiments like the Stern-Gerlach, electrons look somewhat like bar magnets. If we take a random assortment of free electrons and pass them through a Stern-Gerlach apparatus, they'd split into two groups based on whether north or the south of the "bar magnet" is on top relative to the axis of the Stern-Gerlach. The structure you're proposing would look like two bar magnets attached along the joining axis so that they have the same pole (north or south) in the middle. This kind of structure would behave completely differently in a Stern-Gerlach apparatus. --FyzixFighter (talk) 16:25, 29 April 2010 (UTC)Reply

The Stern-Gerlach process sounds like an alignment process to me. You're running a stream of particles through a machine that wants them to be aligned either up or down, say due to a spinning appendage on one end. And after they're aligned you add a deflection process that deflects the ups in one direction and the downs in the other. And in the beginning, they could be identical except for their spin orientation.WFPM (talk) 16:54, 1 May 2010 (UTC)Reply

Sorry again, but no. Because the field is inhomogeneous, there is a net force on the dipole particles along the axis of the S-G apparatus. Going through the physics, this is what you find - If the particles were classical spinning objects (or even objects like what you suggest above), the distribution of their spin angular momentum vectors would be random and continuous. Each particle would then be deflected by a different amount, producing a smooth distribution on the detector screen. Instead, the particles passing through the Stern-Gerlach apparatus are deflected either up or down by a specific amount. This result indicates that spin angular momentum is quantized (i.e., it can only take on discrete values), so that there is not a continuous distribution of possible angular momenta. For spin-1/2 particles, this means there are two groups, and for spin-3/2 there are four, and so on. Again, experimental evidence in favor of quantum mechanics. --FyzixFighter (talk) 13:21, 7 May 2010 (UTC)Reply

I think you're neglecting the primary operation occurring in the apparatus. It evidently wants the spin of the particles passing through to be in one of two opposite directions. And that's an alignment process. Then the magnetic interaction force vector operating on the aligned particles is such as to cause them to avoid a centralized area of avoidance, probably due to a helical path of travel, and such as to result in their impact on the target in accordance with the observed pattern. So they start out randomly oriented, and then they get coerced into two different paths on the way to the target. And this coercion process must be the operating principal of the inhomogeneous magnetic field, and if you didn't have that, you probably wouldn't have the travel path resulting in the area of avoidance.WFPM (talk) 15:48, 7 May 2010 (UTC)Reply

I'm not neglecting anything. I'm simply analyzing the apparatus using simple classical electromagnetism. If the magnetic field were homogeneous, the force on either end of the dipole would be equal and you would simply get a torque on the dipole that would align it to the magnetic field. This would be like placing a small bar magnet between the two poles of large horseshoe magnetic - it wouldn't move toward either pole, but would spin to align itself. However, if the field is inhomogeneous as it is in the SG apparatus, the force on either end of the dipole is not necessarily equal so along with a net torque, you also get a net force in either direction that will be related to the initial orientation of the the dipole. The dipoles which enter with a dipole moment perpendicular to the SG axis would be deflected the most, and those with a dipole moment parallel to the axis would not be deflected at all. This is all based on classical electromagnetism. You're description doesn't rely on a known physics, let alone classical (non-quantum, non-relativity) physics, but instead relies on inventing new, never-before-seen forces between dipoles and magnetic fields to create the "centralized area of avoidance". In addition, your description would not explained the multiple "areas of avoidance" seen when particles with spin greater than 1/2 are run through the apparatus. --FyzixFighter (talk) 20:01, 7 May 2010 (UTC)Reply

Well I'm obviously a novice at Stern-Gerlach experiments and would like to see the target of a 3/2's spin particle with multiple areas of avoidance. But in the absence of a controlling magnetic field I don't understand how translating magnetized dipoles can be expected to still be organized into two different categories of spin direction. But maybe the translated particles' interrelationship is such as to organize the spin polarities or something. Like when I throw 50 pennies into the air in a heads/tails ratio experiment, I don't assume that they have any particular heads/tails distribution while they're in the air, nut that the probability is pretty good that they will have one at the end of the experiment.WFPM (talk) 00:18, 8 May 2010 (UTC)Reply

The only way that the translated dipoles can occur in two different categories is if the component of their dipole moment along the axis of the SG apparatus is quantized. Well, this is exactly what quantum mechanics says should happen, namely that the component of the dipole moment/angular momentum along a measured axis can only have values of -S...+S (times a proportionality factor - I think it's h-bar for angular momentum) in integer steps. The total angular momentum of the particle is sqrt(S(S+1)). A physical analogy to this is that the particle looks like a spinning top that isn't standing perfectly upright so that it precesses about the center axis. The real fun quantum stuff happens when you start putting a series of SG apparati together. --FyzixFighter (talk) 01:48, 8 May 2010 (UTC)Reply

You keep baiting and I keep responding in my limited way. A nodulating spinning top is an illustration of a failed initial process due to an unsustainable initial position and describes the process of recovery to a stable position. And the cause of the initial inadequate position can be explained. And you're telling me that we don't have to have an explanation for the initial failed position because like Feynman says it just has to happen, and probably does every way it can. But I'm thinking that physics is about determining about how things are as opposed to how things potentially could be. See Cotes' preface in the second edition of the Principia. I guess that's the way Engineers think about things. We don't spend much time on unutilizable concepts. But we do try to understand the processes that we utilize. I've worked with PhDs who would analyze a subject to death, while I was just trying to meet a specific application requirement. But I'm in favor of maximum knowledge about a subject matter if we have the time.WFPM (talk) 04:07, 8 May 2010 (UTC)Reply

The only problem with saying that it's an initial failed position is that it doesn't matter what axis you use, you always get two groups. You can even take the same set of particles, run them through an SG apparatus, bring the two groups back together, run them through another SG apparatus at another angle, and so and so forth, and every time you will still get two groups. The problem with such a classical, deterministic view of physics is that there are experiments like this where it just can't explain the results. --FyzixFighter (talk) 15:10, 8 May 2010 (UTC)Reply

Okay, so you run them through the first SG apparatus and they come out organized into two orientation directions, and with an unknown property of orientation stability, and then you run you run them into a second apparatus and repeat the process. Well, if the first SG unit could temporarily organize the orientation property, I don't see why the second and additional units couldn't do the same thing, unless the orientation property is related to the magnetic field of the transient stream of magnetized particles, which wouldn't argue in favor the particle stream being made up of opposed orientation particles. But that organization of opposed orientation magnetic properties in a stream of bar magnets becomes almost a requirement for your proposition of the continued existence of up/down magnetic properties in your test stream of the magnets. So we're involved with the orientation properties of bar magnets at such a minute level of existence that it's hard to know when and/or where to figure out their activity within the system.WFPM (talk) 19:00, 8 May 2010 (UTC)Reply

But if I point out that one of the features of the structure of my models in Talk:Nuclear model is that they are constructed in a manner consistent with way that magnets go together, that information is disregarded as not being relevant to the subject matter. So how's that for a catch 22?

But the fact that I still always get two groups regardless of what axis I use for the SG experiment is incompatible with an initial failed position argument you put forward and the known mechanism for how an inhomogeneous magnetic field interacts with particles with dipole moments. There are other variations of the SG experiment involving blocking one of the exit groups and using orthogonal axes (see Stern-Gerlach experiment#Sequential experiments). It's not so much that the particles are composed of two groups of opposed orientations, but that the results of the measurement are quantized, and that measurements along different axes don't commute. It doesn't matter that we are dealing with bar magnets (or rather magnetic dipoles) at such a minute level, classical electromagnetism still doesn't work. And that's the point, classical physics fails but quantum mechanics can explain it. Quantum mechanics is a simple set of rules that explains and predicts a variety of behavior that classical mechanics (even if we assume a lack of known details in our calculations) can't. That's the measure of any model.
I couldn't care less whether your nuclear model uses the idea of how magnets go together because your model doesn't measure up to what we know about reality. You may be able to get the numbers right, but so does quantum mechanics. However, your model can't be what is physically going on in an atomic nucleus because a host of experimental results disagree with it. The results do agree with the quantum mechanical model of the nucleus. Your models could be a nice mnemonic, but if I needed a physical description of the atom for making something like semiconductors or something that dealt with physics at the atomic level, I'd take quantum mechanics because it matches with measurements of the physical world and it accurately predicts behavior that your model does not. --FyzixFighter (talk) 22:39, 8 May 2010 (UTC)Reply

Ole!! And I'm glad that you appreciate my models as a mnemonic because then it's good for something. And who cares about what I know about models? Or about what you know. It's just a comment about a subject of Amiable? discussion, And if you disagree that seriously you must have at least looked at it. And maybe you can tell me what a semiconducter hole really is. You'll probably say it's an atom lacking and therefor having am open position for an "orbital" electron and that involves the shuffeling of this "hole" in the upstream (negative) direction towards a barrier next to the negative voltage potential source. But maybe you've got a quantum electrodynamics theory explanation, but I doubt that it's simpler. And I'm with Newton in thinking that simpler is better as long as you can rationally explain the situation. So we'll each look for solutions to problems using our own MO. But I like my models because they tell me where to put Z=119 and 120 in the (Janet) periodic table. And I wish they would do a set of real physical models using the standard tables and then try to explain their rationale.WFPM (talk) 01:28, 9 May 2010 (UTC)Reply

One other question. If I want to have a quantum electrodynamic concept of the atom, how do I start out with a visual concept? Is it like Dr Pauling's "Hypothetical" models that he shows in his General Chemistry (page 94)? Or on Gamow's paperback "the Atom and it's nucleus" With the orbit electron mesh network. And when does this concept change from a sophisticated organization of electron orbitals, (which don't conserve angular momentum, and are therefor probably pseudo), in 18 of the 26 trisected cubic space volumes around the atom, (and with none in the 8 corners, so they must have been thinking about a spherical space volume) to something I can visualize? And how do I visualize the spin of say 4Be9 to be -3/2 and then 5 B10 to be +3. Does QE tell me that?WFPM (talk) 02:12, 9 May 2010 (UTC)Reply

Sorry, I speak Portuguese, not Spanish, so the "Ole" is a bit lost on me :). Semiconductor physics is usually done with a semiclassical approach to quantum mechanics. That is, quantum mechanics and the Schrodinger equation explain the band structure, but the electrons get to be treated as particles moving along the potential within the bands. That means holes can be treated the same way, as essentially positive imaginary particles like you described. Simpler is better, but not always accurate - I'm fine with it if it's just accurate enough for what I'm doing - that's why we can get away with Newtonian mechanics for most day-to-day real life physics without resorting to Relativity - unless you're doing stuff with GPS software (I'm suddenly reminded about a physics joke involving a spherical cow with an even distribution of milk). Only in some cases like in tunneling diodes or spintronics do the electrons have to be treated as quantum mechanical "fuzzy" objects. In these cases, an electron (or a hole) have to be treated as a non-localized entity that can interfere and act like waves spread out across multiple atoms. And to tell the truth, I find quantum mechanics to be completely rational - it's an entirely consistent set of underlying rules and assumptions that make sense to me.
With regards to elements 119 and 120, are you saying you disagree with putting them under the s^1 and s^2 columns? Or do agree with that placement, which is also where quantum mechanics predicts they should be. Since the Janet table is how the valence electrons are filled, the closest thing to a physical model would be the electron cloud images, like those at Atomic orbital#Orbitals table, which are related to the 3D spherical harmonics and arise from the solutions to the Schrodinger wave equation for a central Coulombic potential. I don't see how a trisected cubic space volume (maybe because I don't exactly know what you mean by this) enters into the spherical harmonic solutions. The spherical harmonic solutions form an infinite set so I don't see a problem when the number of electrons in an energy level gets above 18.
Pauling's hypothetical models can't be truly accurate, at least in the case of the helium nucleus. As I've said before scattering experiments show that for He4, the nucleus (and the wavefuction of each of the nucleons) is spherically symmetric and the nucleons are all most likely to be found in the same space, at the center of the nucleus. Additionally the size of the nucleus only approximately matches up with the size needed for a close-packing scheme of spheres (that and some stable molecules are not spherical). Atomic nucleus#Nuclear models has a decent summary of the modern models for the atomic nucleus and how they match up with experiment. I haven't had a chance to look at the Gamow reference yet so I can't comment on that.
And finally the case of 4Be9 and 5B10. Actually, the nuclear shell model does a good job of predicting exactly that the nuclear spin of 4Be9 should be 3/2- and that of 5B10 should be 3+. Note that the sign (+ or -) should come after the number - it refers to the parity of the spin-orbit state and not to a up/down opposing direction description. The spin number, j, is that of the combined angular momenta of the orbital, l, and the intrinsic spin, s, of the nucleon. These two angular momentum can either be parallel, in which case we add s and l to get j, or anti-parallel, in which case we subtract s from l to get j. The parallel case is slightly a lower energy state than the anti-parallel. Even number of protons and neutrons give a nuclear spin of 0+ because protons and neutrons at the same level form spin up/spin down pairs (these are proton-proton/neutron-neutron pairs) such that the their combined angular momenta cancel out. When one (but not both) of the numbers is odd, the pin is determined by the spin-orbit of the extra, unpaired nucleon - in the case of 4Be9, the extra neutron goes into the j=3/2 (l=1, s=1/2 and parallel) of level 1 (which has odd parity), so we get 3/2-. For the odd-odd numbers we can add the j's of the extra proton and neutron - in the case of 5B10, the additional proton also goes into a j=3/2 stat just like the extra neutron. So the total J will be 3/2+3/2=3, and the parity is the product of the two parities -1*-1=+1 (even parity). --FyzixFighter (talk) 07:00, 9 May 2010 (UTC)Reply
Well I've milked a lot of cows and the distribution is not even. And my wife speaks Spanish naturally and also English and French and can understand Portugese pretty well, so consider it as a complimentary commentary. And I agree with your proposed location of 119 and 120 in S1 and S2 and wonder if IUPAC and Norman Holden et al will do that in their proposed new table. Because that's where my models tell me where they should go. And the spin numbers are not actually talking about just the rotation direction of the nucleons, like I could understand visually, and combines that property with some other spin-orbit parity concept that I don't understand and cant visualize, and all this change due to the addition of one neutron to the atom! Note that I just added a commentary on the pseudo characteristic nature of the orbitals to the previous. And the side bonded magnets agree that adjoining magnets have an up/down and accordingly spin orientation relationship. And that means that they interconnect like gears and you can't use a magnet to side bond to both of 2 side bonded magnets. So the side bonding of 2 zero spin 2He4 atoms by either a neutron or a proton is out. And thus the 2He4 atoms must be made up of the bottom 4 and then the top 4 nucleons, and then bonded at the intermediate vertical level by the bonding neutron or proton, with the neutron bond being the more stable. So I'll work on understanding the j value.WFPM (talk) 11:37, 9 May 2010 (UTC)Reply
I don't follow what you mean when you say that the orbitals shells don't conserve angular momentum. Care to elucidate? --FyzixFighter (talk) 16:11, 9 May 2010 (UTC)Reply

Well all conical orbits involve a directional path and velocity of motion such that the angular momentum of the orbiting particle remains constant, as well as the lost free energy. That is the characteristic of conical orbits. And in the "orbital" path concept they gave up on the idea of constant angular momentum and settled on a hypothetical path that has a "constant contained kinetic energy" level value, without regard to the angular momentum property and allows figure 8 and 4 leaf clover paths of movement, and are thus not conical orbit paths.

You might note that the series level organizational system in my models is like the Janet table in that all 8 series levels of the Janet table end with the accumulation of an alpha particle, and which in my models is one that is always added to the top of the atom. It is this feature of the standard table that doesn't recognize the existence of a second accumulation of an alpha particle as a completed series accomplishment that causes the difference in format of the 2 tables. And I consider that to be a "fatal flaw" in the standard table. And so the elements 119 and 120 are the end of the 8th layers of accumulation of deuterons into the basic structure. And the "extra neutrons" are added in a dynamically balanced manner to the surface of the structure as the the extra neutron positions are created. But the extra neutron number lags behind the available location number, due to corner stability and other factors, such that the atomic stability trend lines for the atom have the formula A = 3Z - an even number, with the even number increasing as the atomic nucleus gets larger. In the 79Au to 83B area, the stability trend line formula is A = 3Z - 40 and in the 69Tm to 78Pt area it is A = 3Z - 38. So there is some method in my madness, as I have been trying to point out.WFPM (talk) 17:49, 9 May 2010 (UTC)Reply

Actually they didn't give up the concept of a constant angular momentum the shell model, but what they gave up was the classical idea of a definitive path because in quantum mechanics the particle is not a localized particle until you choose to measure where it is. The "clover leaf" images (for the electron orbitals I linked to) are not descriptions of path, but show where the electron is likely to be found. That's why we use the term "orbital" and not "orbit". Until you measure it, the electron is "smeared" out across a volume surrounding the atom - it's a standing wave within the potential. It's only at specific energies that you get the proper interference for there to be standing wave, which defines the allowed energies levels of an atom. This is the wave-particle duality for particles, which is a pretty well established physics fact. You can't explain the electron double-slit experiment without it. --FyzixFighter (talk) 23:29, 9 May 2010 (UTC)Reply