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Everything posted by Vmedvil
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Ya, you are talking to someone way different I know what the Electromagnetic Energy Stress Tensor says do not confuse me with Swan. Maxwell's equations
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Yes, but it is governed by SR and generated by GR.
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GR is not motion it is Gravity, SR is motion polymath.
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Gravitational Fields & Anti-Gravity Propulsion
Vmedvil replied to Unified Field's topic in Speculations
Ya, man I was pulling for you but you said the magic word which makes me hate you Anti-gravity. Doesn't work that way. Define "Anti-gravity" in your version because I have never found a person that ever said the word "Anti Gravity" that knew what they were talking about with gravity. -
Does the spin of a nucleus have kinetic energy?
Vmedvil replied to Giorgio T.'s topic in Quantum Theory
Oh, so it goes toward the rest mass term I was wrong about this which also makes sense for quarks. If that is the case, then nuclei do not have angular momentum or angular kinetic energy in the form I was thinking about. The Rest momentum and Rest Energy is stored as Rest mass. Then here are your Energy Terms for it. To you let say my Fist has a mass of 1 kilogram, and I move it with a velocity of 5 m/s one of my weak punches it will indeed have both. as MV = p and V2M(1/2) = KE , (p = 5 kg*m/s , KE = 12.5 Joules) which are two versions of saying the physical value of the velocity in stopping ability needed to counter it. -
Does the spin of a nucleus have kinetic energy?
Vmedvil replied to Giorgio T.'s topic in Quantum Theory
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.- 51 replies
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Does the spin of a nucleus have kinetic energy?
Vmedvil replied to Giorgio T.'s topic in Quantum Theory
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 -
It is really sad Strange that you do not understand the higher level physics I am speaking about, This is about QCD not my personal views. The Gauge of the Strong Nuclear force or Color Charge and its symmetry not my personal views. https://www.physi.uni-heidelberg.de/~uwer/lectures/ParticlePhysics/Vorlesung/Lect-5a.pdf http://www.t39.ph.tum.de/T39_files/Lectures_files/StrongInteraction2011/QCDkap2.pdf
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Does the spin of a nucleus have kinetic energy?
Vmedvil replied to Giorgio T.'s topic in Quantum Theory
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. -
Does the spin of a nucleus have kinetic energy?
Vmedvil replied to Giorgio T.'s topic in Quantum Theory
in the simpliest form KE = 1/2 MV2 p = MV They are literally to versions of the same thing. -
Does the spin of a nucleus have kinetic energy?
Vmedvil replied to Giorgio T.'s topic in Quantum Theory
It does, both those terms are the same-thing in two different terminologies. -
Did I do a good engineering job on this one? For Anyone that found my Youtube Via another post. My Real life version of the Empire's Death Star in Star Wars. Before you ask it causes Supernova via Rapid Helium Flash. '' http://slideplayer.com/slide/6039599/ Merry Christmas, Any Space-faring civilization would love one of these. My only Type III Civilization technology.
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This is an annoying question. The only way I can answer this in Standard physics is ds2 = dx2 + dy2 + dz2 - (Cdt')2 Which is from a Youtube video but is still the invariant version of 4-D Space-time, but I liked this picture, I want to give this Kid youtuber credit for knowing what he was talking about https://www.youtube.com/watch?v=zScn3tV9YPU , but I want to correct you on something if you notice why your youtube video got a bunch of hits that NO, that is not a new concept and has been a concept of SR for a long time even Quantum Gravity equations are bound to it at some point. In any case, here is a link to Wikipedia with the same equation https://en.wikipedia.org/wiki/Special_relativity#4D_spacetime ds = S http://hyperphysics.phy-astr.gsu.edu/hbase/rotq.html Where dx2 + dy2 + dz2 = ∇ http://asonika.com/models/ then Laplace operator goes to Schrodinger's equation solved for it. http://www.butterflyeffect.ca/Close/Pages/SchrodingersEquation.html Where dt' https://physics.stackexchange.com/questions/110669/is-gravitational-time-dilation-different-from-other-forms-of-time-dilation Where V2 = C2 - VΛ2 , C2 in the Schwarzchild metric which is solved for C2 http://hyperphysics.phy-astr.gsu.edu/hbase/Astro/blkhol.html VΛ = H d(Kiloparsec) Hubble's constant, goes to Friedmann equation. https://web2.ph.utexas.edu/~coker2/index.files/friedmann.htm
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Does the spin of a nucleus have kinetic energy?
Vmedvil replied to Giorgio T.'s topic in Quantum Theory
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. -
Does the spin of a nucleus have kinetic energy?
Vmedvil replied to Giorgio T.'s topic in Quantum Theory
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/ -
That is color charge impossible missing green again. The most likely from looking at this form QCD is p++p−→π+ + π− + π0 (P+,P-) + KE --> π0 + π0 + π- + π+ + π- + π+ would be fine Even p+ + p− → π0 + π0 + π0 Even p+ + p− →π+ + π− + n0 But not anything that isn't (N + 3), (N+1) is missing Blue and Green, (N+2) is missing Green
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Well, this is based on the idea that they are at low energies where kinetic energy is near zero @ higher kinetic energy that would be true. if (P+,P-) + KE ≥ (P+,P-) + UD + DU Which doesn't explain why there is missing Green Color Charge symmetry in. RBG + RBG + KE = RR + BB + GG + RR + BB Or in (P+,P-) + KE ≥ DU + UD +UD + DU + DD Which is missing Green in sensei's model. RBG + RBG + KE = RR + BB + GG + RR + BB All the ones Sensei posted is missing a Green Component being 5 and not 3,6,9,12,15 ..... N + 3) The Strong nuclear force disagrees with that statement and tells you it will not bind this without a green component. Like (P+,P-) + KE --> π0 + π0 + π- + π+ + π- + π+ would be fine (P+,P-) + KE -----> DD + UU + DU + UD + DU + UD Then RBG + RBG + KE = RR +BB + GG + RR + BB + GG Which is not missing a Green component.
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Does the spin of a nucleus have kinetic energy?
Vmedvil replied to Giorgio T.'s topic in Quantum Theory
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