Giorgio T. Posted November 29, 2017 Share Posted November 29, 2017 In the hyperfine structure of the nucleus the magnetic moment is tied to the spin and the spin is described as angular momentum. Does this indicate that there is inertial kinetic energy in the nucleus? Link to comment Share on other sites More sharing options...
Giorgio T. Posted November 29, 2017 Author Share Posted November 29, 2017 4 hours ago, Dubbelosix said: Yes. Anything in nature that has a motion is translated into, say linear or rotational kinetic energies. Thanks for your quick answer, Does the kinetic energy of the nucleus then affect the hyperfine frequency and is it a component in the rate stability of atomic clocks ? Link to comment Share on other sites More sharing options...
Giorgio T. Posted November 30, 2017 Author Share Posted November 30, 2017 Dear Dubbelosix and swansont, Thanks for your replies, You guys are obviously the experts. I am sorry but I have almost no knowledge in your field. Consider the question coming from and old fashioned “Natural Philosopher”. Let me pose the question in a different way with a little thought experiment: I have invented an inertial ray gun and I can aim it at things and control mass. I aim it at an old watch with a balance wheel and turn the inertial mass of the parts to one half - the watch goes twice as fast because the balance wheel loses mass and the spring stays the same. What happens if I do the same with an atomic clock? Will it speed up? Regards Link to comment Share on other sites More sharing options...
Giorgio T. Posted December 1, 2017 Author Share Posted December 1, 2017 One of Clark's laws: "For every expert, there is an equal and opposite expert,"[ Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted December 4, 2017 Share Posted December 4, 2017 ! Moderator Note Dubbelosix, Despite the air of superiority you seem to post with, it seems clear based on the number of people with actual degrees in physics that disagree with you that the confusion here is with you, not them. I would ask that you stop polluting this thread with your (as yet) unsupported ideas, as the question posed by the OP was asked in good faith seeking answers offered by mainstream, accepted science. If you wish to argue your points further, I suggest opening another thread. However, if you intend to do this, you must keep the forum rules in mind. That is, we will not accept you constantly shifting the burden of evidence / proof to other people when it is you who is coming up with the stuff not supported by accepted physics. We also will not accept you constantly shifting the goal posts. DO NOT RESPOND TO THIS MOD NOTE WITHIN THIS THREAD. Please report the post or PM a member of staff if you disagree. Any further posts in this thread that do not actively seek to answer the OP within the bounds of what physics currently holds to be true in this context will be removed, and you will receive an official warning. ! Moderator Note On reflection, it didn't seem fair that the OP should have several posts essentially ignored while others engaged in debate. While on topic, the OP has stated they are not an expert, and IMO this makes the thread quite confusing from their perspective. As such, I have split every single post not by the OP into a new thread to give this one an opportunity to provide some actual clarity for them. You may direct the previous conversation to the new thread. Link to comment Share on other sites More sharing options...
swansont Posted December 4, 2017 Share Posted December 4, 2017 On December 1, 2017 at 11:57 AM, Giorgio T. said: One of Clark's laws: "For every expert, there is an equal and opposite expert,"[ We've confirmed that Dubbelosix was a sockpuppet who has been banned from here several times. Crackpot, not expert. Link to comment Share on other sites More sharing options...
Vmedvil Posted December 24, 2017 Share Posted December 24, 2017 (edited) Well, Ya, I agree with dubblesiox there is a V term on the Watson Hamiltonian for Nuclei Angular Momentum and molecular binding for potential kinetic energy of the nucleus. The quarks within the Nucleons of the nuclei are spinning so why would the Nucleus not be? https://courses.lumenlearning.com/suny-physics/chapter/33-5-quarks-is-that-all-there-is/ Edited December 24, 2017 by Vmedvil Link to comment Share on other sites More sharing options...
swansont Posted December 24, 2017 Share Posted December 24, 2017 Then how about addressing the objection that I had to this assertion? The quarks have spin 1/2, and with one anti-aligned, you end up with spin 1/2 for the proton. How do you account for the alleged additional angular momentum? Link to comment Share on other sites More sharing options...
Vmedvil Posted December 25, 2017 Share Posted December 25, 2017 (edited) 12 hours ago, swansont said: Then how about addressing the objection that I had to this assertion? The quarks have spin 1/2, and with one anti-aligned, you end up with spin 1/2 for the proton. How do you account for the alleged additional angular momentum? U = 1/2 , D = -1/2 . The Proton is UUD, So 1 - 1/2 = 1/2 or Neutron DDU = 1-1/2 = 1/2 So the Nucleus has a 1/2 spin as all the nucleons have a 1/2 spin. It is accounted for by the Watson Hamiltonian. And is also the only Equivalent in any of the equations. V ≈ (1/2)∑3N-6s=1 fsqs2 Edited December 25, 2017 by Vmedvil Link to comment Share on other sites More sharing options...
Unified Field Posted December 25, 2017 Share Posted December 25, 2017 I'm also not an expert, but quantum spin shouldn't be rather associated with physical rotation of a particle. It's a common mistake, as people always try to apply standard mechanics to quantum physics - but it simply won't work, so stop thinking about subatomic particles, as about tiny planets. Quantum spin has only 2 possible orientations (up and down), so we can't treat it as a 3D vector (like a normal spin). Besides, quantum spin remains constantly in superposition with other properties of subatomic particles - it's direction in 3D space is not determined and we can't consider it as a simple rotational motion of a body. Quantum spin is directly connected with the magnetic dipole moment of a particle, what is the source of magnetism in matter - and this is probably the only and actual role of quantum spin. In my humble opinion, momentum of a subatomic particle has nothing to do with it's spin... Link to comment Share on other sites More sharing options...
Vmedvil Posted December 25, 2017 Share Posted December 25, 2017 (edited) 48 minutes ago, Unified Field said: I'm also not an expert, but quantum spin shouldn't be rather associated with physical rotation of a particle. It's a common mistake, as people always try to apply standard mechanics to quantum physics - but it simply won't work, so stop thinking about subatomic particles, as about tiny planets. Quantum spin has only 2 possible orientations (up and down), so we can't treat it as a 3D vector (like a normal spin). Besides, quantum spin remains constantly in superposition with other properties of subatomic particles - it's direction in 3D space is not determined and we can't consider it as a simple rotational motion of a body. Quantum spin is directly connected with the magnetic dipole moment of a particle, what is the source of magnetism in matter - and this is probably the only and actual role of quantum spin. In my humble opinion, momentum of a subatomic particle has nothing to do with it's spin... No that just shows it is possible for it to have angular momentum the angular momentum is defined by the Watson Ham, if S = 0, then it would be impossible for it to have angular momentum, but it is S = 1/2. Otherwise, they are unconnected and spin can have way more states than U and D. If it is connected to magnetism which is a Current then the charge must be moving in this case rotating, If it were 0, there would be not magnetism thus not movement of charge or "Not Spinning" magnetic dipole moment, is like saying Magnetic Moment of interia, which instantly implies if not 0 a moment of inertia or rotation. The Dipole part just states it can be (-,+) Nothing that says Hamiltonian is classical for future reference. If I were speaking classically I would say Magnetic Dipole Moment is like a Faraday Loop in Maxwell's equations, which the spin being a non zero current. http://slideplayer.com/slide/11005662/ Then quote ampere's Law http://slideplayer.com/slide/4902429/ Then just say from Ohm's Law that I = Q/Δt or I = Qf Then Quote all the AP Physics Equations from 101 and 102 https://www.pinterest.com/pin/78109374764657260/ Edited December 25, 2017 by Vmedvil Link to comment Share on other sites More sharing options...
Vmedvil Posted December 25, 2017 Share Posted December 25, 2017 (edited) On then I would add one last definition for ϴ Then add the other forms. http://hyperphysics.phy-astr.gsu.edu/hbase/rotq.html Either way you say it you get the same result. Edited December 25, 2017 by Vmedvil Link to comment Share on other sites More sharing options...
swansont Posted December 25, 2017 Share Posted December 25, 2017 10 hours ago, Vmedvil said: U = 1/2 , D = -1/2 . The Proton is UUD, So 1 - 1/2 = 1/2 or Neutron DDU = 1-1/2 = 1/2 So the Nucleus has a 1/2 spin as all the nucleons have a 1/2 spin. So how can it have rotational KE, without having additional rotational angular momentum? Link to comment Share on other sites More sharing options...
Vmedvil Posted December 25, 2017 Share Posted December 25, 2017 1 minute ago, swansont said: So how can it have rotational KE, without having additional rotational angular momentum? It does, both those terms are the same-thing in two different terminologies. Link to comment Share on other sites More sharing options...
swansont Posted December 25, 2017 Share Posted December 25, 2017 23 minutes ago, Vmedvil said: It does, both those terms are the same-thing in two different terminologies. But we have already established that the angular momentum is fully accounted for by the spin. Spin is intrinsic to the quarks, not physical motion. There is no KE associated with it. Link to comment Share on other sites More sharing options...
Vmedvil Posted December 25, 2017 Share Posted December 25, 2017 (edited) 11 minutes ago, swansont said: But we have already established that the angular momentum is fully accounted for by the spin. Spin is intrinsic to the quarks, not physical motion. There is no KE associated with it. in the simpliest form KE = 1/2 MV2 p = MV They are literally to versions of the same thing. Edited December 25, 2017 by Vmedvil Link to comment Share on other sites More sharing options...
swansont Posted December 25, 2017 Share Posted December 25, 2017 1 minute ago, Vmedvil said: in the simpliest form KE = 1/2 MV2 P = MV They are literally to versions of the same thing. We're talking about rotation, but ... If we have accounted for all of the angular momentum with the spin, which has no physical rotation associated with it, how can there be any kinetic energy without having additional rotation? Link to comment Share on other sites More sharing options...
Vmedvil Posted December 25, 2017 Share Posted December 25, 2017 (edited) 2 minutes ago, swansont said: We're talking about rotation, but ... If we have accounted for all of the angular momentum with the spin, which has no physical rotation associated with it, how can there be any kinetic energy without having additional rotation? Because Angular Momentum and Angular Kinetic Energy describe the same physical interaction. If you say Angular momentum it is like saying Angular Kinetic Energy, the two are interchangeable as long as you use them in the right context. Edited December 25, 2017 by Vmedvil Link to comment Share on other sites More sharing options...
Vmedvil Posted December 25, 2017 Share Posted December 25, 2017 (edited) Just so people know, why I just said the linear ones because they are basically the same as the angular http://hyperphysics.phy-astr.gsu.edu/hbase/rke.html http://hyperphysics.phy-astr.gsu.edu/hbase/amom.html http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html Edited December 25, 2017 by Vmedvil Link to comment Share on other sites More sharing options...
swansont Posted December 25, 2017 Share Posted December 25, 2017 <sigh> As mentioned earlier in the thread, we know that the spin of fundamental particles is intrinsic angular momentum. There is no motion associated with it. The spin is not physical. Since that spin accounts for the spin of a proton or neutron, we start from the state of the spin being COMPLETELY explained by this intrinsic angular momentum, that LACKS MOTION. So how can there be this kinetic energy, when there is no additional angular momentum. You have admitted that you can't have one without the other. IOW, your own argument says there is no KE, because there is no extra angular momentum. Link to comment Share on other sites More sharing options...
Vmedvil Posted December 25, 2017 Share Posted December 25, 2017 (edited) 7 minutes ago, swansont said: <sigh> As mentioned earlier in the thread, we know that the spin of fundamental particles is intrinsic angular momentum. There is no motion associated with it. The spin is not physical. Since that spin accounts for the spin of a proton or neutron, we start from the state of the spin being COMPLETELY explained by this intrinsic angular momentum, that LACKS MOTION. So how can there be this kinetic energy, when there is no additional angular momentum. You have admitted that you can't have one without the other. IOW, your own argument says there is no KE, because there is no extra angular momentum. Swan I have never had the impulse to want to punch someone through the internet before, but here I want to, I will wait for Mordred to get done with Christmas to explain, since you will not listen to me. I will say this one last time, Kinetic Energy and Momentum are the same effect on matter. If my fist were to punch this screen, it would have both kinetic energy and momentum which are the same property. Edited December 25, 2017 by Vmedvil -1 Link to comment Share on other sites More sharing options...
geordief Posted December 25, 2017 Share Posted December 25, 2017 OT but timely https://sciencealert.com/graphene-paired-sheets-diamene-deformation-diamond-property Link to comment Share on other sites More sharing options...
StringJunky Posted December 25, 2017 Share Posted December 25, 2017 54 minutes ago, Vmedvil said: kinetic Energy and Momentum are the same effect on matter. If my fist were to punch this screen, it would have both kinetic energy and momentum which are the same property. How can they be when they are scalar and vector respectively? Punch a screen in a straight path there is no change in momentum or KE but punch the screen with a constant speed in a curved path, the KE stays the same but the momentum changes because of the change in direction. Link to comment Share on other sites More sharing options...
Mordred Posted December 25, 2017 Share Posted December 25, 2017 (edited) The instrinsic spin of a particle has no actual motion involved. It literally involves the symmetry/ antisymmetry relations specifically to the types of equations needed to describe a wavefunction. The spin doesn't add to KE, in point of detail it adds to the rest mass term itself of the particle. All principle number quantum wavefunctions do so. They all form the invariant identity of the particle. Spin is intrinsic to the phase space used to model it. Edited December 25, 2017 by Mordred Link to comment Share on other sites More sharing options...
swansont Posted December 25, 2017 Share Posted December 25, 2017 2 hours ago, Vmedvil said: Swan I have never had the impulse to want to punch someone through the internet before, but here I want to, I will wait for Mordred to get done with Christmas to explain, since you will not listen to me. I will say this one last time, Kinetic Energy and Momentum are the same effect on matter. If my fist were to punch this screen, it would have both kinetic energy and momentum which are the same property. No, they are not the same, and repeating yourself will not magically make you correct. Mordred has now weighed in. So, are you interested in correcting your misconception, or do you have a valid answer to my question? Link to comment Share on other sites More sharing options...
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