Dubbelosix 152 Posted April 25 Report Share Posted April 25 (edited) Say we have a following formula E - 1/(M_1c^2)^1/2 + E - 1/(M_2c^2)^1/2 = N Where E = energy M_1 and M_2 are rest masses with c^2 (celeritas squared of light speed) N =some number related to Energy Let us substitute a = E - 1/(M_1c^2)^1/2 An b = E - 1/(M_2c^2)^1/2 So we now have a + E - 1/(M_1c^2)^1/2 = N And a + b = x Using some algebra we find (a -b)(a + b) = N ( a - b) The left hand side becomes a difference of squares ie. a^2 - b^2 = N(a - b) This will now give a^2 = E - 1/(M_1c^2)^1/2 And b^2 = E - 1/(M_2c^2)^1/2 These solutions are important and can relationshipthe golden ratio. Edited Sunday at 01:43 PM by Dubbelosix Quote Link to post Share on other sites

HallsofIvy 4 Posted April 25 Report Share Posted April 25 1.678 is NOT the golden ratio. The golden ratio is $\frac{1+ \sqrt{5}}{2}$, an irrational number. Even approximated to three decimal places the golden ratio is 1.618, not 1.678. Dubbelosix 1 Quote Link to post Share on other sites

Dubbelosix 152 Posted April 25 Author Report Share Posted April 25 Sorry, i was trying to work from memory, correction noted. But it still holds. Quote Link to post Share on other sites

Dubbelosix 152 Posted April 27 Author Report Share Posted April 27 (edited) On 4/25/2021 at 1:03 PM, HallsofIvy said: 1.678 is NOT the golden ratio. The golden ratio is $\frac{1+ \sqrt{5}}{2}$, an irrational number. Even approximated to three decimal places the golden ratio is 1.618, not 1.67 Now let's continue this. We can impose an extra relativistic correction to (a^2 -b^2)N = N/N_0 It is now a relativistic operator in the style of the ratio of energies it's a rough approximation without the need of the gravitational gamma function. This is highly unique because, it's said there are no relativistic operators but I have proven one exists. Edited Sunday at 01:44 PM by Dubbelosix Quote Link to post Share on other sites

Dubbelosix 152 Posted April 27 Author Report Share Posted April 27 Operators are used in relativity all the time, here's a good reference to such operators https://en.m.wikipedia.org/wiki/Poisson's_equation If time is quantized from my essays on a similar idea then we must utilise it to find a path for a grand unified theory. Quote Link to post Share on other sites

Dubbelosix 152 Posted April 27 Author Report Share Posted April 27 (edited) 3 hours ago, Dubbelosix said: Now let's continue this. We can impose an extra relativistic correction to (a^2 -b^2)N = N - b = N/N_0 It is now a relativistic operator in the style of the ratio of energies it's a rough approximation without the need of the gravitational function. This is highly unique because it's said there are no relativistic operators but I have proven one exists. The formula I created in this fashion, was motivated with the localisation formula, it's a precursor to measuring the differences found in relativity and is a relativistic operator https://en.m.wikipedia.org/wiki/Newton–Wigner_localization Edited April 27 by Dubbelosix Quote Link to post Share on other sites

Dubbelosix 152 Posted April 27 Author Report Share Posted April 27 (edited) Now, we can even mathematically prove the last derivations can describe an exponential decay law. By taking the derivative of time through the energy N, which is a number itself with units of mass, we get the mass flow and can be written as d/dt(a^2 - b^2) = dN/dNt(e^{ b}) Noticing that_0 we can replace dN/dt = - λN where the italic lambda represents a decay constant. Now N is now a function of time and is related to N_0 as E(t) = E_0 e^ λ which is the decay rate formula. Edited April 29 by Dubbelosix Quote Link to post Share on other sites

Dubbelosix 152 Posted April 27 Author Report Share Posted April 27 (edited) Say the ratio represents a dimensionless set of numbers related to an energy solution E(t)/E_0V= e^ λe ~ TO FINE STRUCTURE Now If a^2 +b^2 = 9 as a number in an atom, or follows a cicular trajectory. If we plug in a^2/9 +b^2/4 = 9 it travels the path in a hyperbolic trajectionary. Yet if N=O we get Bohrs "stable/special trajectory where no radiation is given off: N(a^2/9 - b^2/4) = 0 Edited Sunday at 01:47 PM by Dubbelosix Quote Link to post Share on other sites

Dubbelosix 152 Posted April 27 Author Report Share Posted April 27 (edited) Summary. We have explained the atom without quantum wavefunctions. Edited April 27 by Dubbelosix Quote Link to post Share on other sites

Dubbelosix 152 Posted April 27 Author Report Share Posted April 27 Let's say g is the acceleration amd a new coefficient constant parameter attached to lambda. Then we get to play some more physics at hand. dE/dt = - λ βN β = g/2(R/v)^2 Absorbing the contents we cam simple write it as dE/dt = - η N This is a gravitational flow related to the energy of a free falling body. Quote Link to post Share on other sites

Dubbelosix 152 Posted April 28 Author Report Share Posted April 28 (edited) Using the quadratic difference of squares (a^2 -b^2)N = N - b = N/N_0 We can impose the uncertainty from the localised Wigner operator such that (a^2 -b^2)N = N/N_0 turns into [a^2, b^2]N = B N/N_0 B is related to h snd energy Edited April 29 by Dubbelosix Quote Link to post Share on other sites

Dubbelosix 152 Posted April 28 Author Report Share Posted April 28 (edited) 2 hours ago, Dubbelosix said: Let's say g is the acceleration amd a new coefficient constant parameter attached to lambda. Then we get to play some more physics at hand. dE/dt = - λ βN β = g/2(R/v)^2 Absorbing the contents we cam simple write it as dE/dt = - η N This is a gravitational flow related to the energy of a free falling body. To add. The left hand side is the quadratic uncertainty operator. It may also be relevant to the muon spin anomaly. Edited April 28 by Dubbelosix Quote Link to post Share on other sites

Dubbelosix 152 Posted April 28 Author Report Share Posted April 28 Tomorrow we will substitute the N for energy and second quantize it Quote Link to post Share on other sites

Dubbelosix 152 Posted April 28 Author Report Share Posted April 28 (edited) On 4/28/2021 at 12:54 AM, Dubbelosix said: i[a^2, b^2] ∂h/ ∂t ψ = B N/N_0 Where B is s notation for the energy and Planck constant Edited April 29 by Dubbelosix Quote Link to post Share on other sites

Dubbelosix 152 Posted April 28 Author Report Share Posted April 28 This is now, quantum gravity, without any photons Quote Link to post Share on other sites

Dubbelosix 152 Posted April 28 Author Report Share Posted April 28 (edited) Now. If there are any questions to the said work, please do not hesitate to ask. Edited April 28 by Dubbelosix Quote Link to post Share on other sites

Dubbelosix 152 Posted April 28 Author Report Share Posted April 28 (edited) Now s layman summery of the final gravity equation The left hand side we have the operator. Denoted in the square brackets, so when we integrate it through quadratic relationship we get on the right a correction term, a constant of +C. The symbol that looks like a trident, is a wave function notation. It will translate to a gravitational wave. We plugged in the energy and used the operator version of it and will be an interpretation of the gravitational wave. The N/N_0 measures also energy. Specifically how the energy is effected while in motion of any particle ralated to the particles of the standard model. They are the systems which contribute to the stress energy tensor which causes curvature in spacetime. Edited April 28 by Dubbelosix Quote Link to post Share on other sites

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