Dubbelosix Posted May 2, 2021 Report Share Posted May 2, 2021 Anyway, I've fixed it all, you'll see why it has been rewritten as it is, it's much more clearer, except for the fact it's and paste - this will be one if a few threads that discuss the spinor, gravity and torsion coupled to electromagnetism; F_N/F_0 = 1/(Gεµ) nh/(m^2c) By understanding that we have F = Nh/(εµ M_2c) (M/R)^2 by empirical dimensional analysis where relativity as the unifying idea is the same as saying it applies in a general sense the upper limit of the Gravitational force. (m/R)^2 ≡ c^4/G We remind ourselves that the rotational velocity inverse units of time is also a fundamental relation when seen to preserve torsion, it highlights the importance of Newton's G when prepping a theory that obeys the full Poincare group ω = - Ω/2 =GL/2c^2R^3 One thing you can naturally do is plug I'm this torsion equation straight into the covariant derivative where the spin-coupling occurs, Dψ(X,t) = (∂ - i/(4π)ω σ(a,b)) ψ(X,t) The motivation for this is by inviting a correction term that is suitable for a more realistic model for calculations that can predict small corrections that could answer several problems in quantum theory. The modified derivative is D ψ(X,t) = (∂ + i/(4π)Ω/2 σ(a,b)) ψ(X,t) = (∂ - i/(4π)GL/2c^2R^3σ(a,b)) ψ(X,t) In our next session I will take us through the force equations that featured from the beginning F_N/F_0 = 1/(Gεµ) nh/(m^2c) F_N = nh/(εµ M_2c) (M/R)^2 If we inspect the dimensions of these equation, where the first says the gravitational force is caused by all the interesting variables on the left. The second is required to derive the first. By inspecting the dimensions we find an set of interesting solutions under Bohrs incredible theory of the atom However, what I didn't do before was log in an extra factor if c in the denominator sinceω requires one such factor to have dimensions of inverse length, so it has to be modified properly Dψ(X,t) = (∂ - i/(4πc)ω σ(a,b)) ψ(X,t) D ψ(X,t) = (∂ + i/(4π)Ω/2c σ(a,b)) ψ(X,t) = (∂ - i/(4π)GL/2c^3R^3σ(a,b)) ψ(X,t) Now the dimensions are spot on for further work. Bohr obtained two major objects of importance, the Bohr radius and the Bohr inverse mass. He derived the inverse mass from the known classical laws 1/m = mv^2/m^2v^2 ≡(4π ^2Be^2)/h and his radius formula which when cubed is 1/R^3 = (12π^6B^3e^6 m^3)/h^6 these are standard equations from his model which is still considered accurate for a nuclear charge equal to 1, but we will be inviting wave functions soon. First we identify the mass in my following formula F_N/F_0 = 1/(Gεµ) nh/(m^2c) In which we have highlighted because of not only being a dimensionless (and therefore real) observable just so happens to have the mass squared term in the denominator. Plugging in Bohrs inverse mass term after squaring it yields after we simplify by staying hc = e^2 So we cancel these terms out F_N/F_0 = nh/(Gεµ) (16π ^4B^2e^2) and rearrange F_N/F_0 = (16π^4 B^2e^2)/(Gεµ) So how did we arrive at this master equation which helped derive further equations... G ≡ nh/p • B(4π ^2e^2R)/h^2 • 1/εµ = B(4π ^2e^2R)/h^2 • λ/εµ We recognise that the relationship nh/p = λ Is debroglies wave related to matter in motion. The allowed debroglie wave length in an orbital follows from 2πR = nh From here we recognise that Bohrs relationship to inverse mass is 1/m = B(4π ^2e^2R)/h^2 The product of εµ is not just arelated aspect of electromagnetic theory to the metric of space. Itself is s fundamental origin for the speed of light as a parameter in the usual medium of space, c = √1/εµ So while it might look like just a bunch of variables that just so happen to work out in the master equation from dimensional analysis, it wasn't formulated in any ad hoc way. We'll continue later. Now, we should investigate this spin space under the bivector model I invented for analogy to the one which already exists for the EM field. The bivector theorems I published were as essays sent to the gravity research foundation. The premises are self explanatory when concerning the important derivatives required for describing curvature in spacetime. ∇D =∂⋅D + iσ⋅(∇∧D) Where ∧is the wedge product. Thisrsms it can be rewritten as ∇D =∂⋅D + iσ⋅(Γ × D) Where this time × is the cross product and without any silly agents about whether torsion vanishes in general relatobity, we understand it is a crucial non vanishing component when gravity I'd looked at through the spectacles of bivector. This is because the last term on the previous equations defined a crucial relationship with torsion as -(Γ × D) = ∂Ω/∂t And the energy required to make an object spin I'm curved torsion background requires K = L × Ω which is the angular momentum cross product with torsion. A master equation is a terminology we use if all your work is based on and on occasions, itself be an equation with unifying prospects. I explained, hopefully eloquently enough how it was arrived at: G ≡ nh/p • B(4π ^2e^2R)/h^2 • 1/εµ = B(4π ^2e^2R)/h^2 • λ/εµ Bohrs relationship to inverse mass is crucial for the dimensionless force equation and when squared gives 1/m^2 = B^2(16π^4e^4R^2)/h^4 and his radius formula which when cubed is 1/R^3 = (64π^6B^3e^8m^3)/h^8 these are standard equations from his model which is still considered accurate for a nuclear charge equal to 1, but we will be inviting wave functions soon. First we identify the mass in my following formula F_N/F_0 = 1/(Gεµ) nh/(m^2c) By recognising that we have an inverse mass term, by plugging it we get after simplifying (canceling) some direct terms yields F_N/F_0 = 16π^4/(Gεµ) • 1/c • B^2(e^4R^2)/h^3 Since 1/εµ is exactly c^2, we can even rewrite it as F_N/F_0 = 16π^4/G • c^2 • B^2(e^4R^2)/h^3 = 16π^4c^2/G • B^2(e^4R^2)/h^3 This coefficient 16π^4c^2/G Is very special, it is a constant found in general relativity. If it has a name it would by the Schwarzschild constant. Working out its dimensions is fun, you have say the gravitational parameter Gm = r(s)c^2 Hence by rearranging you find Gm/c^2 = r(s) the Schwarzschild radius. Further rearranging we get G/c^2 = r(s)/m So it has dimensions of length over the mass, a relationship we utilised before. It's inverse is naturally c^2/ G= m/r(s) Which is a quick demonstration of why 16π^4c^2/G Is important in any gravitational theory. We find then it's dimensions will always satisfy the following modifications F_N/F_0 = 16π^4/G • c^2 • B^2(e^4R^2)/h^3 = 16π^4c^2/G • B^2(e^4R^2)/h^3 = 16π^4m/R(s) • B^2(e^4R^2)/h^3 We now find we have an equation with an inverse length cubed because F_N/F_0 = 16π^4m • B^2(e^4R^3)/h^3 Why is this important? Simply because all the principles we have worked on is itself based from fundamental relationships within the Bohr model of the atom. The object of particular importance is Bohrs orbital radius equation which when cubed gives 1/R^3 = (64π^6B^3e^8m^3)/h^8 This itself can be plugged into F_N/F_0 = 16π^4m • B^2(e^4R^3)/h^3 What do you get when you plug tje inverse orbital radius into the following equation? F_N/F_0 = 16π^4m • B^2(e^4R^3)/h^3 Remember, you need (64π^6B^3e^8 m^3)/h^8 And I'll leave it here as a fun excersise for people until I return, remember to cancel out necessary terms when you can and there is more than one way to simplify it. Summary: I would like to mention, as I forgot before to highlight, that the muons anomalous spin discrepancy may be capable being fixed by using the modified spinor equations that belong to the Full Poincare group for particles that are Fermions with spin 1/2. Because many scientists often ignore the non trivial aspect of torsion and it's implications for a linearized theory of gravity, perhaps the physics does not require a new force but instead, a tweak of the theory by remembering the first principles of spacetime symmetries. Not just these symmetries, but also crucial questions of the rotating frame of reference inducing a gravitational and magnetic coupling. It's been understood for a while for instance, that the Coriolis effect appears to be a type of gravimagnetism theory. Quote Link to comment Share on other sites More sharing options...

Dubbelosix Posted May 2, 2021 Author Report Share Posted May 2, 2021 (edited) Ok, let's solve this with the instructions given, if you've done it at home, well done. I said we needed Bohrs inverse case of the orbital radius equation , so we just invert it again because the term R^3 lives in the numerator on the RHS of the ratio of forces equation. So, we have now R^3 = (64π^6B^3e^8 m^3)/h^8 And plugging it into F_N/F_0 = 16π^4m • B^2(e^4R^3)/h^3 ... Is not too difficult to solve. The hardest part is keeping all the terms at hand when finishing the simplification. Ok let's solve this, if you've done it at home, well done. I said we needed Bohrs inverse case of the orbital radius equation , so we just invert it again because the term R^3 lives in the numerator on the RHS of the ratio of forces equation. So, we have now R^3 = 64π^6B^3e^8 m^3)/h^8 And plugging it into F_N/F_0 = 16π^4m • B^2(e^4R^3)/h^3 By using (12π^6B^3e^8 m^3)/h^8 We notice how to simplify terms 64 *16 = 1024 π^6*π^4 = π^10 =93648.0474761 Use the exponential rule, it's simple enough for rudimentary calculus, when two quantities multiply side by side, you add the exponents ie. A^n*B^m = A^{n +m} So it holds for all the other quantities: B^2*B^3 = B^5 m*m^3 = m^4 e^8e^4 =e^12 And for the denominator h^3*h^8 = h^11 Plugging the results in we get F_N/F_0 = 1024π^(10) • B^5(m^(4)e^(12))/h^(11) So this is the exact case, if my calculations are right, what you'd get when you plugged in Bohrs orbital radius appropriately. It can be simplified further by dimensional analysis, but you can't just look for one specific way as there are no doubts many ways to crunch it down. It's a strange looking equation on the face of it, which is why simplification, if you can should always be encouraged. Let's finish the night off with at least one juicy interpretation that pops out, all these higher factors of e in the numerator and those in the denominator is crucial for understanding the Einstein slope where V is a "stopping" potential, so we'd have eV = hν - W_0 It's important that the quantity can be dimensionally recognised as related to this, since the master equation involved many concepts surrounding the UV spectrum or divergence of wavelengths. The equation just provided can be rearranged to give V = (h/e) ν - W_0/e Rearranging for h/e gives (V + W_0/e) / ν = (h/e) We'll mess around with this later, it's bed time for ne. We will also explore other variations to interpret what the equation could mean. Edited May 3, 2021 by Dubbelosix Quote Link to comment Share on other sites More sharing options...

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