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Posts posted by Dubbelosix

  1. 1 minute ago, swansont said:

    Nobody is denying that a microscopic can rotate. But the topic here is limited to a system of some nucleus, with nucleons that have spin, and your contention that this means there is rotational KE.

    please stop changing the topic.

    I'm not changing the subject. You are ignoring everything that is being said to you, in obvious ways. You are fixated on me being wrong some how, when all the material I have provided you says otherwise. You are actually wasting my time. 

  2. 2 minutes ago, swansont said:

    A hydrogen atom is not. It has a size, and therefore (according to you) should have rotational KE. How can the angular momentum be zero?


    What do you mean, how can the angular momentum be zero? It's angular momentum is not zero, if my memory serves. 

  3. 13 minutes ago, Strange said:

    The spin of a nucleus. Therefore not classical. 

    Yeah, right. 


    If the nucleus makes real rotations in space (which it does) then the spin is classical. Are you having problems with terminology as well?

    Note though, it depends on whether the nucleus is perfectly spherically symmetric. 


    I think I have realised both your problems.


    Just because an atom is a microscopic object, does not mean it cannot classically rotate. Classical rotation, as opposed to an intrinsic case (like we might expect for an electron) is still subject to nuclei of atoms, and to atoms themselves, even to diatomic molecules. 

  4. 9 hours ago, Mordred said:


     I don't have any doubt that you and I will agree on further tests are required.


    Indeed, more stringent tests need to be done. Perhaps torsion coupling can only be seen with more significant contribution of mass, like I expect around black holes.

  5. You have a problem with terminology, or me?


    Classical rotation is just that, a classical rotation in space. Intrinsic means something else. If you are so profoundly confused by the use of spin in this context, I don't think science is your area.

    And regardless, I have also said the mathematics of classical spin and that associated to quantum mechanics, has no formal difference.

    6 hours ago, swansont said:

    A hydrogen atoms has a measurable radius, and yet the orbital angular momentum of the ground state electron is zero. Same for helium. (true of any s-state electrons)



    So? You still don't seem to be getting the point, the point is the electron is the only pointlike system (as far as we can tell in existence). It's a total hit and miss with you.

    8 hours ago, Strange said:

    Then why is it quantised and determined by the number of nucleons and how they pair-up?

    What is quantized exactly?



    Are you also aware there is no formal difference between the mathematics that describes classical spin and the ''intrinsic case'' only appropriate for electrons?

    There are some serious issues, all extending from classical physics, relativity and quantum concerning pointlike systems. They are a problem in physics because they are actually a closely related to the divergence problems of relativity.


    In classical physics, a pointlike system will possess infinite energies.

    For a pointlike system in relativity, the curvature tensor vanishes, implying a zero radius and a singularity in the curvature.

    Finally, pointlike systems do not make sense in phase space - early on, it was found a point particle is smeared in the phase space so points are physically meaningless, according to Von Neumann. 

  6. I was reading the replies to a thread on another site:




    One posters states that torsion vanishes from Riemann geometry and general relativity ''by construction.'' This is still a common myth, even by experienced posters online. Torsion does not ''vanish'' in spacetime as some intrinsic part of relativity, in fact, such a statement has no empirical support.




    It has been noted by some authors, that general relativity without torsion is actually, the most boring kind of model you can work with.The torsion is of course part of the Poincare spacetime symmetries from properties found in rotation.Torsion, if we would accept for a moment could have solid foundation in reality, could offer solutions to standing problems in QM and even Cosmology.


    ''The Einstein–Cartan theory eliminates the general-relativistic problem of the unphysical singularity at the Big Bang.[9] The minimal coupling between torsion and Dirac spinors generates an effective nonlinear spin-spin self-interaction, which becomes significant inside fermionic matter at extremely high densities. Such an interaction replaces the singular Big Bang with a cusp-like Big Bounce at a minimum but finite scale factor, before which the observable universe was contracting. This scenario also explains why the present Universe at largest scales appears spatially flat, homogeneous and isotropic, providing a physical alternative to cosmic inflation.[8]

    Torsion allows fermions to be spatially extended[11] instead of "pointlike", which helps to avoid the formation of singularities such as black holes and removes the ultraviolet divergence in quantum field theory. According to general relativity, the gravitational collapse of a sufficiently compact mass forms a singular black hole. In the Einstein–Cartan theory, instead, the collapse reaches a bounce and forms a regular Einstein–Rosen bridge (wormhole) to a new, growing universe on the other side of the event horizon.''




    I've also shown, through investigations into the work of Venzo de Sabbata and C Sivaram, (who have both done extensive work in the past on torsion theories) that the torsion should arise in a Friedmann equaion in the following way, by entering the effective density ~


    [math](\frac{\ddot{R}}{R})^2 = \frac{8 \pi G}{3}(\rho - k\sigma^2)[/math]


    In which the Torsion is related to the Poisson equation


    [math]\nabla^2 \phi = 4 \pi G(\rho - \mathbf{k}\sigma^2)[/math]


    You can calculate the torsional part by the following the dimensional arguments:


    [math]\mathbf{k}\sigma^2 = \frac{Gm^2v^2R^2}{c^4L^6} = \frac{Gm^2}{c^2L^4}[/math]


  7. I didn't avoid anything. Did I or did I not provide you reading material pertinent to the discussion?


    It's you who is avoiding what was given to you. You want a demonstration, I don't have one to give you at the moment, but I have provided plenty material which supports my claim. It's not something new to me, I knew there was a rotational energy associated to the nucleus, in fact, anything that has a measurable radius has internal degree's of freedom and the idiom ''intrinsic spin'' is silly. It's just a spin, around an axis. 

    It's you refusing to engage in a meaningful discussion. You are still failing to admit you objected to my assertions and then tried to say you weren't objecting to anything.

  8. 1 hour ago, swansont said:

    The nucleus is a quantum object, not a classical one.



    Let's start here, first of all, you don't seem to understand what ''classical rotation'' means, as to what a quantum spin is. Though mathematically there is nothing different between the two, the former involves actual rotations in space, whereas the latter means '''instrinsic'' which means it is not a classical rotation, that means, it is not an actual rotation. Now, a nucleus certainly rotates - as it has internal dynamics ie. degrees of freedom.


    Let's see if you can understand what it means now?

    Because we cannot measure the electron, it is often taken to mean that it truly is fundamentally a pointlike object - but over the years it is not talked about enough, but there are problems which exist when thinking the electron has no internal structure/radius. A point like particle experiences a divergence problem in its self-energies. 

  9. I know fine well what it means, it was me who explained it was not a classical property of the electron in our current models. Maybe you should read the conversation properly. He was the one who mentioned the electron and then claimed the links I provided suggested he was correct and I was somehow wrong. 


    Point is, and what is frustrating about this, is even when you [provide] the material, certain posters are still incapable of understanding its content and in what context.

    14 minutes ago, J.C.MacSwell said:

    ...and you get from that he thinks classical spin has no kinetic energy...that's not obtuse?

    That's just wrong - the electron is the only system subject to these ''non-internal degree's of freedom'' it is the only pointlike particle in existence since we cannot measure the radius (yet). We have attempted to measure the shape of electrons which suggests they are in fact spherical. 

    Classically rotating systems always possess a kinetic energy associated to the rotation. This even applies to a nucleus. 

  10. 55 minutes ago, swansont said:

    How can you hear it again, when I never said any such thing? I have not questioned anything classical. Don't be obtuse.



    You never said such a thing, so you are saying you did not disagree with me?


    Let's try again, did you not say


    ''The specific question that was asked was if the spin of a nucleus means it has KE. You said yes. I disagree.'' 


    Since this is a direct quote, this is a no brainer. You disagree with something, so would be nice if you were not obtuse and get to the point. 

  11. 21 hours ago, swansont said:


    Yes, the royal ''we'' have you never seen someone use it in such a context?

    Let me get one thing clear before we continue, I want to hear it again from you. You do not believe rotational kinetic energy exists? You question that a classical rotating object has any rotational kinetic energy?


    I just want to get this clear. And you also think the links I provided shows you are right? Is also what you claim?

    See, you said the ''first poster agreed with you.'' And actually he didn't if you read him carefully. The only thing he agrees in is that a pointlike particle does not rotate. To do so, it has to rotate something like 720 degrees just to return to its original orientation. He clearly said:


    If you're talking about a single particle, the rest mass is defined to be the total energy when the particle is at rest - there is no way to separately discuss contributions to this energy. Furthermore, "spin" does not represent a degree of freedom - there is no motion associated with it, and hence no kinetic energy.


    This only applies to systems that have no internal degree's of freedom, ie. electrons as we model them


    Some compound particles on the other hand have genuine rotational degrees of freedom. A deformed (non-spherical) nucleus can rotate, and may therefore possess rotational bands: excited states with increasing angular momentum, and associated rotational kinetic energy.

    Reference https://www.physicsforums.com/threads/does-spin-have-rotational-kinetic-energy.540443/

    Now I said the nucleus has rotational energy, you came in here, tried to start an argument because you didn't agree with something. Let's just reflect on the evidence.

  12. We've had this problem about your interpretation of things, including the interpretation of the work of accelerations on different superpositioned energy states. I will come back to this later, as I will chase what this poster said and explain it how I understand it, then we can compare differences afterwards. 

  13. 3 minutes ago, swansont said:

    The specific question that was asked was if the spin of a nucleus means it has KE.

    You said yes. I disagree, and want you to justify your response.

    I didn't start talking about the proton, it was inherent in the OP. If you missed that, then say so. No harm, no foul.

    But if you deny this, then we can use Hydrogen as a system. Spin-1/2 nucleus, which is a single proton. What is its rotational KE?

    If you choose another nucleus, then we must wonder how it can have rotational KE, which requires it have angular momentum, but the angular momentum is the sum of its spins. There is no angular momentum left over to account for any physical rotation of the nucleus.


    So...do you want to change your answer, or do you want to answer my objections to it?


    Let's do one better and change the format of the question. If my assumptions hold for protons, they should hold for molecules as well. You will actually find much more literature on that subject



    Here, the first poster seems to agree with me, compound particles do actually have rotational kinetic energy



  14. 11 minutes ago, swansont said:

    I did what? You claimed the proton spin means it has rotational KE. 

    Where did I initially say this anyway? It was you who started talking about the proton, then you stepped it up a gear and started talking about an electron. The question to me is simple, does it cost energy to rotate a thing, the answer is yes. In physics, the rotational energy is a correction to the total kinetic energy. 


    When you get to subjects like protons and electrons, it just complicates the issue. The poster was asking about the spin of a nucleus. Which is of course, related to the dynamics of the spin of its constituents. Just as a proton is the sum of quarks. 


    It still has rotational energy - the electron is the only particle in the standard model with exception since it has no classical rotation in our current model.

    3 minutes ago, swansont said:

    I don't want to guess. It's your claim. Back it up. (And in terms of the quarks having spin, which accounts for all of the spin of the proton.)

    How much rotational KE is in a proton? A link should suffice. If this is standard physics, surely it's been measured or at least calculated.

    Yeah, again, when was this actually my claim? You were the one who brought up the discussion of protons and electrons. 

  15. 3 minutes ago, swansont said:

    If it holds for a proton, it should hold for an electron.

    The electron is evidence that QM spin is NOT the same as classical spin, which is why I brought it up.

    Pointless, I know the differences thank you.

    2 minutes ago, swansont said:

    I did what? You claimed the proton spin means it has rotational KE. How does one calculate it?

    The proton has a spin. It is not a point system. Take a guess how it's done. 

  16. Just now, swansont said:

    How fast is an electron spinning? What is its rotational KE?

    So we have went from the proton to an electron in your question?


    Is this intended or designed to try and trip me up somewhere? The electron is normally classified as a point like particle so probably best to stick with protons. Since protons do have spin, and since quantum mechanical spin is the same as classical spin, you will expect there to be corrections to the total kinetic energy of the system. 

  17. Consider for instance, two particles separated by billions of light years, but are quantum entangled. How does each particle know the state of the other when you come to measure one of them? Classical signalling won't answer for it because the signal would have to travel faster than light. Einstein speculated that maybe there where hidden variables associated with the system. Though local hidden variables have been ruled out, non-local ones are yet to follow the same path. 


    Still, you get the sense there may be more going on than meets the eye. 

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