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Mordred last won the day on January 2

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About Mordred

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    University of the Caribou
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    cosmology and particle physics

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  1. No they are charge neutral bosons which can be it's own antiparticle another is the Z boson. Higgs boson as well.
  2. The Higgs boson mass can be derived through those mixing angles. There is advantages to using normalized units hence the mixing angles to establish the correlation. The factor that the above also accounts for is helicity of particles under the right hand rule. Example photon has two polarity states. As well as being its own antiparticle.
  3. The above isn't quite accurate what your referring to is the numerous indpendent unitary triangles of the CKM and PMNS mass mixing matrixes https://en.m.wikipedia.org/wiki/Cabibbo–Kobayashi–Maskawa_matrix There are six independent triangles in the CKM. Through those mixing we also correlate particle generations. Yes you get a correlation to the Higgs mass value by the bosons you mentioned however I needed to show you what is meant by triangle. Your descriptive above lacked that clarity for other readers./posters. Ie needs that detail so everyone recognizes your describing a unitary triangle.
  4. I can name several gauge bosons with mass. You might want to consider the w+ and w- gauge bosons. There is no criteria that all gauge bosons must be massless. It is literally the neutrino mass terms that required the SM model to become extended. The mass term also correlates the range of a force. Ever wonder why the EM field is infinite in extent while the weak or strong force are not ? To answer that you need the gauge boson mass terms.
  5. No it the inner product between Lambda and g. Your dealing with vectors. Here is cross product. https://www.mathsisfun.com/algebra/vectors-cross-product.html I don't know if your familiar with the right hand rule in EM theory https://en.m.wikipedia.org/wiki/Right-hand_rule You can from those links how the cross product equates to the right hand rule. The dot product returns a scalar however the cross product returns a vector. So here is a question how can one claim that the EM field and the cosmological constant are related when the latter has no cross product of two vectors and is described strictly via the inner or dot product ?
  6. Ok so can you quarantee particles will have the same mass terms beyond the Event horizon of a black hole ? The theory we have can only handle a given range of mass values. That range is only a small portion of the possible range of mass. Think of the Golden rule the laws of physics must be the same in all reference frames. When answering the above. Next question with symmetry breaking one can know when different fields unify or becomes in thermal equilibrium. This is done by running of the coupling constants. At what temperature does gravity unify with the electroweak force ? If you don't know the answer then can you unify gravity with the other forces.
  7. It contributes to the mass terms but doesn't account for all the mass terms. Yes it is still a working progress but the Higgs field only gives the mass to certain particles such as quarks neutrinos and leptons. However that doesn't account for all the mass of every particle. For example the equation below has all 18 coupling constants involved however the relativity section is non renormalizable. [latex] \mathcal{L}=\underbrace{\mathbb{R}}_{GR}-\overbrace{\underbrace{\frac{1}{4}F_{\mu\nu}F^{\mu\nu}}_{Yang-Mills}}^{Maxwell}+\underbrace{i\overline{\psi}\gamma^\mu D_\mu \psi}_{Dirac}+\underbrace{|D_\mu h|^2-V(|h|)}_{Higgs}+\underbrace{h\overline{\psi}\psi}_{Yukawa}[/latex] As you can see we have the Yukawa and Dirac couplings along with the Higgs couplings.
  8. The Higgs field doesn't account for all the mass terms of all particles. It is simply one of many contributors. It only describes two out of the eighteen SM model coupling constants.
  9. Conjure it might help to understand QFT better before drawing conclusions. The issue lies in the renormalization of a group or set. QFT does set boundary conditions those are the IR and UV boundary conditions. However this doesn't solve renormalization. You can still end up with Infinities in the effective degrees of freedom in regards to the one loop integrals of the Feymann diagrams. For example what is the upper boundary for mass ? I certainly don't know, yet if you quantize gravity where each graviton is described by a one loop interaction then you have no limit to the number of gravitons (just apply that term to a quanta unit of gravity) Ie the graviton propogator. As the gravitational field strength increases the number density of graviton would also increase. However more importantly you cannot limit the number of one loop corrections. https://en.m.wikipedia.org/wiki/Renormalization This is where the real problem lies with unifying gravity. We have successfully unified the other forces. We have effective renormalization groups for the strong, weak and EM field and we can describe those fields at any given energy scale. While most importantly the mass terms of every particle involved do not vary (invariant or rest mass) at any energy scale. Not so with trying to quantize gravity.
  10. Think of it this way our observable universe is a sphere. The volume around a star or planet obviously isn't flat or curved. As gravitational potential falls off at r^2. If you were to draw field lines you would end up with the gravitational potential from a centre of mass being spherical for any given value. So what is really meant by curved or flat spacetime. Obviously we are not describing the potential gravity distribution nor are we talking about a physical shape for lack of a better description. In essence we are talking about the geodesics and particle paths due to spacetime. Take the statement mass tells spacetime how to curve. Mass is resistance to inertia change. This tells us were dealing with the energy momentum relations under coordinates {ct,x,y,z}. Now massless particles follow a particular type of geodesic {null}. In flat spacetime if you took several parallel beams of light those beams will neither diverge or converge. They will remain parallel. Spacetime is flat in this case, however if you took those same beams in curved spacetime the will either converge or diverge. Positive curvature being the former. Now for a massive falling object. If you take two objects in flat spacetime with equivalent momentum and parallel direction. In flat spacetime those paths will remain parallel. Now let's use a curved example. Take two objects and drop them from some radius from Earth. As they approach the Centre of mass the distance between the two dropped objects will converge. You have in this example positive curvature. So it is the paths taken by objects with momentum that is either straight or curved. (Follows the principle of least action). The geodesics that the object or particle follows is what is described by spacetime curvature. The more curved a path an object must follow the longer the interval will be to arrive at the receiver. This also applies to the Feymann path integrals. The paths chosen will affect reaction and interaction times. (If you think about that one can better understand aging and how time itself varies). Granted you must include all particle interactions. Helps to remember we give time units of length by describing time as an interval using (ct) the length contraction itself allows c to remain constant while x is the particle vector direction. The length between individual units of x will be determined by how the interval is affected by the gamma factor of the Lorentz transforms.
  11. At 4.244 light years distance he would have to have a working FTL drive lol.
  12. When you decide to come back hopefully you can provide testable proof and not imagined conjecture.
  13. I'm suppose to take your word you saw a test of proper acceleration between Earth and Alpha Centauri ?
  14. Have you even tried any of your claimed experiments ? I have mentioned actual tested and applied experiments that have already been used to test relativity not imagined and never applied experiments.
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