We will use the mass of the W and Z boson to calculate the Higgs mass. Using the equation below we can discount distances for the sake of simplification. \( (Dw)^2 + (Dz)^2 = {D^{2}H}^2 80.4^2 + 91.2^2 = H^2 /sqrt{14781.6}=121.57 Gev/c^2\) and my table has it at 125.3 Gev/c^2. I think that is pretty close. http://en.wikipedia.org/wiki/List_of_particles#Bosons
I guess it's just coincidental. Why don't you use this method and see how many masses you can calculate?
If you want to sere then please identify your symbols , otherwise it is to impress the reader how good you are .
Units? Just pick some. As long as they are the same units for distance represented by D and the same units for x,y,z we should find the "right" answer. Though if you have three "conditions" and are looking for the fourth you have to solve for zero I believe. Conditions can be anything like heat, pressure, mass, volume of inscribed spheres, area of inscribed triangles, or even time. Conditions are represented by x,y,z I have no valid basis for any of these claims as I just posted the equation without checking any of the math, then went back and relized... Hey! It worked. The formal original equation is \( Dx^2 + Dy^2 = (D^{2}z)^2\) The units represented in the example above are masses of subatomic particles in gigaelectron volts and distance. Tri-chi-square test could be the unofficial name after further review.
That would interesting if that was true considering the two particles as traveling in different lines of force vectors. I was researching about the Higgs Boson, and found a good description of the Higgs Mechanism that describes it as a collision of the W and Z bosons. I think that electroweak symmetry breaking could have explicit symmetry breaking globally in the fact that photons can only come from electrically charged particles, and this leads to the possible existence of other types of Goldstone Bosons. "In particle physics, the Higgs mechanism is a kind of mass generation mechanism, a process that gives mass to elementary particles. According to this theory, particles gain mass by interacting with the Higgs field that permeates all space. More precisely, the Higgs mechanism endows gauge bosons in a gauge theory with mass through absorption of Nambu–Goldstone bosons arising in spontaneous symmetry breaking. The simplest implementation of the mechanism adds an extra Higgs field to the gauge theory. The spontaneous symmetry breaking of the underlying local symmetry triggers conversion of components of this Higgs field to Goldstone bosons which interact with (at least some of) the other fields in the theory, so as to produce mass terms for (at least some of) the gauge bosons. This mechanism may also leave behind elementary scalar (spin-0) particles, known as Higgs bosons. In the Standard Model, the phrase "Higgs mechanism" refers specifically to the generation of masses for the W±, and Z weak gauge bosons through electroweak symmetry breaking.[1] The Large Hadron Collider at CERN announced results consistent with the Higgs particle on July 4, 2012 but stressed that further testing is needed to confirm the Standard Model. The mechanism was proposed in 1962 by Philip Warren Anderson. The relativistic model was developed in 1964 by three independent groups: Robert Brout and Francois Englert; Peter Higgs; and Gerald Guralnik, C. R. Hagen, and Tom Kibble."
I went to Convert GeV to newton-meter then I plugged in the Values of the W, Z, and Higgs Bosons for what they would be in Newton-meters. I found that the distance the Higgs Boson would take is to 0.1 x 10^-8 m of what it should be, that could be considered to be even an accurate margarine of error. The Higgs Boson would only travel about 1.9 x 10^-8 m, when according to its mass alone it would only travel 2.0 x 10^-8 m. That would only be a difference of about 0.1 x 10^-8 m, from the measurement of the mass of the Higgs Boson and a vector of a collision between the masses of a W and Z boson. I don't even know if they measure particle paths with such accuracy. It would seem like their masses could be directly related to each other through the Pythagorean Theorem as a product of the W and Z bosons masses collision would have on a new vector's total distance. W 1.2881505492 x 10^-8 m z 1.4611856976 x 10^-8 m Higgs 2.0075281569 x 10^-8 m W,Z mass vector 1.9479208095 x 10^-8 m That would be very close to the distance a Higgs Boson should travel according to its mass just from straight converting GeV to Newton Meters, and then running it through the Pythagorean Theorem.
My brain Just exploded! Where X, Y, Z are three separate objects we should be able to make a three dimensional spherical matrix with three variables respecting every law of physics which lies between two infinitesimally separated points. \(ds^{4}_{(x,y,z)} = \sum_{i,j,k}^n g_{i,j,k} dx_{i} dx_{j} dx_{k} + \sum_{i,j,k}^n g_{i,j,k} dy_{i} dy_{j} dy_{k} + \sum_{i,j,k}^n g_{i,j,k} dz_{i} dz_{j} dz_{k} \) I could easily be wrong somewhere tho."proof" Differential_geometry
Excellent! Sorry I didn't help to answer, but here you what other people of good will give you a satisfactory reply.
\(dHx_{126\frac{Gev}{C^{2}}}^{5}_{(W,Z,G,Y)} = Wx_{(1,+-1, 80.4\frac{Gev}{c^{2}})}\sum_{(spin,charge,mass)}^{n} g_{(1/2,0,80.4\frac{Gev}{c^{2}})} dx 2.2\frac{ev}{c^{2}}_{(0)} dx .17\frac{Mev}{c^{2}}_{(1/2)} dx 15.5\frac{Mev}{c^{2}}_{(Ve+Vu+Vt)} + Zx_{(1,0, 91.2\frac{Gev}{c^{2}})}\sum_{(s,c,m)}^{n} g_{(1/2,-1,91.2\frac{Gev}{c^{2}})} dx .511\frac{Mev}{c^{2}}_{(1/2)} dx 105.7\frac{Mev}{c^{2}}_{(-1)} dx 1.777\frac{Gev}{c^{2}}_{(e+u+t)} + Gx_{(1,0,0)}\sum_{(s,c,m)}^{n} g_{(1/2,-1/3,0)} dx 4.8\frac{Mev}{c^{2}}_{(1/2)} dx 104\frac{Mev}{c^{2}}_{(-1/3)} dx4.2\frac{Gev}{c^{2}}_{(Down+Strange+Bottom)} + Yx_{(1,0,0)}\sum_{(s,c,m)}^{n} g_{(1/2,2/3,0)} dx 2.4\frac{Mev}{c^{2}}_{(1/2)} dx 1.27\frac{Gev}{c^{2}}_{(2/3)} dx 171.2\frac{Gev}{c^{2}}_{(Up+Charm+Top)} \) That should yield one hydrogen atom if x=1. That's odd 5 dimensions. Time for a cigarette Please Register or Log in to view the hidden image!
