Dubbelosix 152 Posted May 1 Report Share Posted May 1 (edited) The derivation to get to the following equation is very difficult, but we start off with [tex]\omega = - \frac{\Omega}{2} = \frac{\pi^6}{6} \frac{G L B^3e^4m^2}{c^2}[/tex] Where B is the constant from Bohrs model, I believe it's equivalent to the Boltzmann constant. L is the angular momentum e is the charge and m mass with c the speed of light. The result was obtained from a dimensional argument and some simple algebra from Bohrs model where we have [tex] \frac{mv^2}{R} = B \frac{e^2}{R^2}[/tex] I took an equation that I derived two years back when trying to unify electromagnetism with gravity, the resulting equation was [tex] \frac{F_N}{F_G} = \frac{1}{G \epsilon \mu} \cdot \frac{nh}{m^2c} [/tex] I plugged in the Bohr mass before squaring it [tex] \frac{1}{m} = B \frac{e^2}{R} \cdot (\frac{\lambda }{h})^2 [/tex] Which gave [tex] \frac{F_N}{F_G} = \frac{1}{G \epsilon \mu} \cdot \frac{nh}{c} \cdot (\frac{4 \pi^2 B e^2 R}{h})^2[/tex] By noticing that the orbital radius cubed from the Bohr model Is [tex] \frac{1}{R^3} = \frac{12 \pi^6 B^3 e^6 m^3}{h^6}[/tex] And by using the following dimensional analysis by cancelling unwanted terms, [tex]hc = Gm^2 =e^2[/tex] You can work out the same interesting stuff as my line of enquiry led me, so I simplified the search. I used the Bohr radius formula and plugged it into [tex]\omega = - \frac{\Omega}{2} = \frac{G L }{2c^2R^3}[/tex] And it gave [tex]\omega = - \frac{\Omega}{2} = \frac{\pi^6}{6} \frac{G L B^3e^4m^2}{c^2}[/tex] Now the unification of all important terms from spin to gravity and torsion, to electromagmetism can be plugged into the traceless Covariant Derivative spinor formula for the Dirac equation [tex]D = \partial - \frac{i}{4} \omega \sigma[/tex] We'll put in the necessary subscripts at a later date, I am just writing thus up as a preliminary result after stumbling across it last night. Plugging the innovative solutions we get through substitution of the verbein spin connection The derivation to get to the following equation is very difficult, but the result is [tex]\omega = - \frac{\Omega}{2} = \frac{\pi^6}{6} \frac{G L B^3e^4m^2}{c^2}[/tex] Where B is the constant from Bohrs model, I believe it's equivalent to the Boltzmann constant. L is the angular momentum e is the charge and m mass with c the speed of light. The result was obtained from a dimensional argument and some simple algebra from Bohrs model where we have [tex] \frac{mv^2}{R} = B \frac{e^2}{R^2}[/tex] I took an equation that I derived two years back when trying to unify electromagnetism with gravity, the resulting equation was [tex] \frac{F_N}{F_G} = \frac{1}{G \epsilon \mu} \cdot \frac{nh}{m^2c} [/tex] I plugged in the Bohr mass before squaring it [tex] \frac{1}{m} = B \frac{e^2}{R} \cdot (\frac{\lambda }{h})^2 [/tex] Which gave [tex] \frac{F_N}{F_G} = \frac{1}{G \epsilon \mu} \cdot \frac{nh}{c} \cdot (\frac{4 \pi^2 B e^2 R}{h})^2[/tex] By noticing that the orbital radius cubed from the Bohr model Is [tex] \frac{1}{R^3} = \frac{12 \pi^6 B^3 e^6 m^3}{h^6}[/tex] And by using the following dimensional analysis by cancelling unwanted terms, [tex]hc = Gm^2 =e^2[/tex] You can work out the same interesting stuff as my line of enquiry led me, so I simplified the search. I used the Bohr radius formula and plugged it into [tex]\omega = - \frac{\Omega}{2} = \frac{G L }{2c^2R^3}[/tex] And it gave [tex]\omega = - \frac{\Omega}{2} = \frac{\pi^6}{6} \frac{G L B^3e^4m^2}{c^2}[/tex] Now the unification of all important terms from spin to gravity and torsion, to electromagmetism can be plugged into the traceless Covariant Derivative spinor formula for the Dirac equation [tex]D = \partial - \frac{i}{4} \omega \sigma[/tex] We'll put in the necessary subscripts at a later date, I am just writing thus up as a preliminary result after stumbling across it last night. Plugging the innovative solutions we get through substitution of the verbein spin connection [tex]D = \partial - i \frac{\pi^6}{24} \frac{G L B^3e^4m^2}{c^2} \sigma[/tex] This is a natural thing to do as it unifies at last the spin with gravity and magnetism, coupling it with torsion making it part of the full Poincare group. Edited May 1 by Dubbelosix Quote Link to post Share on other sites

Dubbelosix 152 Posted May 1 Author Report Share Posted May 1 Ok, tried using tex commands but it is still not displaying, but I've updated anyway and when I have more time I'll just cut and paste symbols. Quote Link to post Share on other sites

Dubbelosix 152 Posted May 1 Author Report Share Posted May 1 (edited) It's so difficult without latex presentation so later I'm rewriting the it symbol by symbol and blow by blow. Edited May 1 by Dubbelosix Quote Link to post Share on other sites

Dubbelosix 152 Posted May 2 Author Report Share Posted May 2 12 hours ago, Dubbelosix said: This will be sorted soon because the insights I pieced together has relationships vital from this thread 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 relaxation 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, D ψ(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^3 σ(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 ....session two coming soon Quote Link to post Share on other sites

Dubbelosix 152 Posted May 2 Author Report Share Posted May 2 (edited) However, what I didn't do before was log in an extra factor if c in the denominator since ω requires one such factor, so it has to be modified properly D ψ(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^3 σ(a,b)) ψ(X,t) Now the dimensions are spot on for further work Edited May 2 by Dubbelosix Quote Link to post Share on other sites

Dubbelosix 152 Posted May 2 Author Report Share Posted May 2 (edited) Session 2 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.pluggimg 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) And we'll continue tomorrow as it's really late here now. Edited May 2 by Dubbelosix Quote Link to post Share on other sites

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