devin553344 8 Posted August 26 Report Share Posted August 26 (edited) I have a new unification theory that may be better than my older ideas. The older ideas calculated, but were somewhat not probably. This new theory is more probable and matches all the measured data for the strong force, gravitation, electromagnetic unification with Planck's constant and therefore matter. It keeps the wave-particle system together. Here's the PDF: 20200828WaveCurvatures.pdf I'll post the equations in a while in the thread. Edited September 14 by devin553344 Quote Link to post Share on other sites

MitkoGorgiev 5 Posted August 26 Report Share Posted August 26 Can you tell us in words what the theory is?I don't give a toss about empty mathematics. Quote Link to post Share on other sites

devin553344 8 Posted August 26 Author Report Share Posted August 26 (edited) I explained most of it in the pdf, but basically, there are many concepts in the theory, spelled out in the sentences before and after the equations. When I post the equations in the thread I'll explain the ideas behind the mathematics. Thanks for your interest. I'm excited about this new approach as it calculates accurately a lot of different measured results. The first concept is this, that the charge radius represents a deformation of a wave, such that in any two pi traverse of a wave, four units of linear length are passed. Therefore, for a wave of two pi to exist it must deform 4 units of space, such that: rC = (4c)/(2πf) = 8:412E-16 meters Where rC is the charge radius of the proton, c is the speed of light, f is the frequency of the proton which is (2.26873181532754E+23) hertz. For the charge radius of the proton see the proton page of wikipedia (https://en.wikipedia.org/wiki/Proton). They list it as 8.414E-16 meters. Edited August 26 by devin553344 Quote Link to post Share on other sites

devin553344 8 Posted August 26 Author Report Share Posted August 26 (edited) The second concept in the theory is the there has been a deformation of space for the basic particles, the proton and electron. This deformation is from the charge radius out to the charge radius plus wavelength. Such that the basic form is shown by Deuteron, the Deuteron charge radius is then (Deuteron is the prime example to model nuclear physics since it's the electron orbiting a single proton and neutron): rC + rp = rD = 2.162E-15 meters. Where rC is the charge radius of the proton at 8.412E-16 meters, rp is the Compton wavelength of the proton at 1.321E-15 meters. For the charge radius of Deuteron see: (https://en.wikipedia.org/wiki/Charge_radius). This deformation is a deformation between charge and mass, such that one is expanded beyond the other. Mathematically the charge and mass are demonstrated by something similar to strain energy but for charge: 1/g = 1/2 * ((rC + rp)/rC)^2 = 3.302 Where 1/g is the electromagnetic coupling constant at a value of 3.302 which is dimensionless. g is given by: g = ((8π^2Ke^2)/(hc)) Where K is the electric constant, e is the elementary charge, h is Planck's constant, c is the speed of light. Edited August 26 by devin553344 Quote Link to post Share on other sites

devin553344 8 Posted August 26 Author Report Share Posted August 26 The strong force then is a simple concept of strain and electromagnetic concepts. The strong force is given by: U = Ke^2/r * (1/g)^((rC + rp)/r) Where U is the binding energy, r is the radius between nucleons or general distance of the field, I've already spelled out the previous constants in the last posts here. For Deuteron and its charge radius it calculates to 3.525E-13 Joules. And for large nuclei which are demonstrated at the proton wavelength distance between nucleons they calculate to 1.235E-12 Joules per nucleon. This all agrees with empirical data measured in nuclear physics. The strong force is basically a mixture of electromagnetic energy and wave energy, and creates a force from the push and pull of pressure and vacuum fields. Quote Link to post Share on other sites

devin553344 8 Posted August 26 Author Report Share Posted August 26 (edited) The gravitation of the particles is generated from the strong force reaching the event horizon energy per meter of a black hole, which splits space into positive and negative space via relativity when the value becomes negative as a square of two values: y^2 = 1 - 2GM/(Rc^2) Where y is the relativistic factor, G is the gravitational constant, M is the mass of the body, R is the radius to the center of mass, c is the speed of light. The positive space just continues into the black hole but the negative space and the event horizons zero distance projects outward to infinite distance with an inverse square falloff. Therefor the gravity of the basic particles (I will demonstrate the proton) is: Gmp^2 = Ke^2 * (1/g)^(rC/-r) Where mp is the proton mass. Once we calculate the radius from the above equation we can verify that this matches an electromagnetic quantum black hole via the following equation: mPc^2/(4πlP) = Ke^2/r^2 * (1/g)^(rC/r) Where mP is the reduced Planck mass: mP = (hc/(16π^2G)^1/2 And lP is the Planck length: lP = (Gh/(2πc^3))^1/2 Note: you have to be mindful of the positive and negative r values produced in the above equations, one uses positive r and the other negative r. In order to get the calculations to work correctly! Cheers. Edited August 27 by devin553344 Quote Link to post Share on other sites

devin553344 8 Posted August 26 Author Report Share Posted August 26 (edited) So to note, the above equations can be used to calculate the electron or proton gravity fields. The proton is basically the same form as the electron, but the pressure and vacuum fields are reverse, such that the proton has positive charge and the electron negative charge which reflects the reverse physics of the matter fields and therefore the strong force is repulsive for the electron and would simulate electric charge. This allows me to define both particles with a wave particle duality. The strong force extends to infinity in the example I gave and turns into standard electrics at a distance beyond. Note: The electron is more precise in the gravitational calculations, the proton is so close (1 digit accurate at the exponential expanse which is pretty precise) that it is considered precise for the idea. Edited August 26 by devin553344 Quote Link to post Share on other sites

devin553344 8 Posted August 26 Author Report Share Posted August 26 I found the electromagnetic logarithmic strain energy for the proton and electron. It is a simple strain from the Deuteron charge radius down to the proton charge radius: 1/g^2 = 1/2 * ln^2(4/3 * π *(((rC + rp)/rC)^3 / (8/3 * π^2)) Where ln is the natural logarithm. Quote Link to post Share on other sites

devin553344 8 Posted August 26 Author Report Share Posted August 26 (edited) The proton, neutron and electron can be defined from a median mass then, such that the electron and proton have the same physics, and are different because of their definition and opposite nature. The median mass is: mm = (mp * me)^1/2 Where mm is the median mass, mp is the proton mass and me is the electron mass. Then the electron and proton definition follow an opposites defintion (each have opposite charges so why not total opposites): me = mm * g^(2 * rp/rC) mp = mm * g^(-2 * rp/rC) The median mass is apparent in the neutron definition from the proton regarding electron capture (https://en.wikipedia.org/wiki/Electron_capture): mn = mp + mm * (rC/(rp + rC))^3 And that unifies the forces in a common language mathematically that can be more easily understood. Edited August 26 by devin553344 Quote Link to post Share on other sites

devin553344 8 Posted August 27 Author Report Share Posted August 27 (edited) I've adjusted the equations in post #6 to reflect a new idea that there are two sides to the particle, one for Planck (mass) and the other for charge. The gravitation then uses the charge radius for its benchmark. It is on the other side than the strong force. I've updated the PDF file in the OP. Edited August 27 by devin553344 Quote Link to post Share on other sites

devin553344 8 Posted August 28 Author Report Share Posted August 28 I'm changing the definition of the electromagnetic coupling constant to be: g^2/4 = (rC/(rC+rp))^4 this complies with a pressure system. I've also changed the definition of the proton and neutron to be: mpc^2 = mnc^2 = 3/2 * mec^2 * ln^2(Pp/Pe) Where mp is the mass of the proton, mn is the mass of the neutron, me is the mass of the electron Pp is the pressure of the proton, and Pe is the pressure of the electron, which are: Pp = hc/rp^4 Where h is Planck's constant c is the speed of light, and Pe = Ke^2/rCe^4 Where rCe is the charge radius of the electron which is 4c/(2πve) Where ve is the frequency of the electron. Quote Link to post Share on other sites

devin553344 8 Posted August 30 Author Report Share Posted August 30 (edited) I've solved point particles, the particle is compressed from its wavelength down to zero, a point. The equation that describes this is: mc^2 = mf^2(r-0)^2 Where m is the mass, c is the speed of light, f is the frequency of emitted light during annihilation of the particle, which is one over period, r is the wavelength which is subtracted from zero. This makes it like Hooke's Law (https://en.wikipedia.org/wiki/Hooke%27s_law). And if it is compressed down to zero, then the electron is a point particle and would use a Riemann zeta negative function, I will use a cubic Riemann zeta -3: hc/(120 * (rc+re)^4) = Ke^2/re^4 Here h is Planck's constant, 120 represents Riemann zeta -3, rc is the charge radius of the electron which is wavelength times 4 divided by two pi, re is the wavelength, K is the electric constant, e is the elementary charge. It will not have a strong force like the proton since it's radius is zero, but it does have a wavelength and a strain energy differential which I described above. And this unifies matter and electromagnetic forces. Then the proton, which I already described is formally (using a 4 dimensional n-sphere (https://en.wikipedia.org/wiki/N-sphere)): hc/(2π^2 * (rc+rp)^4) = Ke^2/rc^4 Edited August 30 by devin553344 Quote Link to post Share on other sites

current 6 Posted August 30 Report Share Posted August 30 Can you tell us in words what the theory is?I don't give a toss about empty mathematics.Agreed neither do I . Its the Physical Reality which counts the most . Quote Link to post Share on other sites

devin553344 8 Posted August 30 Author Report Share Posted August 30 (edited) Agreed neither do I . Its the Physical Reality which counts the most .I described it after that post, it's a radius strain theory (https://en.wikipedia.org/wiki/Strain_energy)(https://en.wikipedia.org/wiki/Deformation_(physics)) that describes curvatures and the relationship between electric and Plank's constant and therefor mass. Gravity is a strong force echo from the splitting of space at the event horizon. The strong force is a mixture of electric and Planck's constant, Therefor it's a wave force. I've described composite particles as strain energies and point particles also. The proton is made of two positrons and one electron (which is what the quarks decay into) that are strained into higher energies. It mathematically describes what I've said. So it's both mathematical and mechanical and therefore a candidate for an alternate theory. Working out the math is no easy task. Don't under estimate the importance of math. Edited August 30 by devin553344 Quote Link to post Share on other sites

current 6 Posted August 30 Report Share Posted August 30 I described it after that post, it's a radius strain theory that describes curvatures and the relationship between electric and Plank's constant and therefor mass. Gravity is a strong force echo from the splitting of space at the event horizon. The strong force is a mixture of electric and Planck's constant, Therefor it's a wave force. I've described composite particles as strain energies and point particles also. The proton is made of two positrons and one electron (which is what the quarks decay into) that are strained into higher energies. It mathematically describes what I've said. So it's both mathematical and mechanical and therefore a candidate for an alternate theory. Working out the math is no easy task. Don't under estimate the importance of math.Gravity is not a strong force , its the weakest force of them all . Therefore I Disagree with your reasoning and the resultant logic . Quote Link to post Share on other sites

devin553344 8 Posted August 30 Author Report Share Posted August 30 Gravity is not a strong force , its the weakest force of them all . Therefore I Disagree with your reasoning and the resultant logic .Of course it's the weakest, which I described why it's so weak. What I've done is unify electric, matter, the strong force and gravity. What else would a unification theory do? So are you saying you don't agree with a unification theory? Quote Link to post Share on other sites

current 6 Posted August 30 Report Share Posted August 30 (edited) Of course it's the weakest, which I described why it's so weak. What I've done is unify electric, matter, the strong force and gravity. What else would a unification theory do? So are you saying you don't agree with a unification theory?Why is gravity so weak ? And how do you define gravity , Physically ? Mathematics are important to understanding our Real Physical Reality , In No-Way though does the Mathematics create reality . Without Physical Things there is No mathematics . Edited August 30 by current Quote Link to post Share on other sites

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