The derivation to get to the following equation is very difficult, but we start with the Spin $$\omega$$ And the torsion $$/omega$$ as the starting point
$$\omega = - \frac{\Omega}{2} = \frac{\pi^6}{6} \frac{G L B^3e^4m^2}{c^2}$$
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
$$ \frac{mv^2}{R} = B \frac{e^2}{R^2}$$
I took an equation that I derived two years back when trying to unify electromagnetism with gravity, the resulting equation was
$$ \frac{F_N}{F_G} = \frac{1}{G \epsilon \mu} \cdot \frac{nh}{m^2c} $$
I plugged in the Bohr mass before squaring it
$$ \frac{1}{m} = B \frac{e^2}{R} \cdot (\frac{\lambda }{h})^2 $$
Which gave
$$ \frac{F_N}{F_G} = \frac{1}{G \epsilon \mu} \cdot \frac{nh}{c} \cdot (\frac{4 \pi^2 B e^2 R}{h})^2$$
By noticing that the orbital radius cubed from the Bohr model Is
$$ \frac{1}{R^3} = \frac{12 \pi^6 B^3 e^6 m^3}{h^6}$$
And by using the following dimensional analysis by cancelling unwanted terms,
$$hc = Gm^2 =e^2$$
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
$$\omega = - \frac{\Omega}{2} = \frac{G L }{2c^2R^3}$$
And it gave
$$\omega = - \frac{\Omega}{2} = \frac{\pi^6}{6} \frac{G L B^3e^4m^2}{c^2}$$
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
$$D = \partial - \frac{i}{4} \omega \sigma$$
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
$$\omega = - \frac{\Omega}{2} = \frac{\pi^6}{6} \frac{G L B^3e^4m^2}{c^2}$$
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
$$ \frac{mv^2}{R} = B \frac{e^2}{R^2}$$
I took an equation that I derived two years back when trying to unify electromagnetism with gravity, the resulting equation was
$$ \frac{F_N}{F_G} = \frac{1}{G \epsilon \mu} \cdot \frac{nh}{m^2c} $$
I plugged in the Bohr mass before squaring it
$$ \frac{1}{m} = B \frac{e^2}{R} \cdot (\frac{\lambda }{h})^2 $$
Which gave
$$ \frac{F_N}{F_G} = \frac{1}{G \epsilon \mu} \cdot \frac{nh}{c} \cdot (\frac{4 \pi^2 B e^2 R}{h})^2$$
By noticing that the orbital radius cubed from the Bohr model Is
$$ \frac{1}{R^3} = \frac{12 \pi^6 B^3 e^6 m^3}{h^6}$$
And by using the following dimensional analysis by cancelling unwanted terms,
$$hc = Gm^2 =e^2$$
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
$$\omega = - \frac{\Omega}{2} = \frac{G L }{2c^2R^3}$$
And it gave
$$\omega = - \frac{\Omega}{2} = \frac{\pi^6}{6} \frac{G L B^3e^4m^2}{c^2}$$
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
$$D = \partial - \frac{i}{4} \omega \sigma$$
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
$$D = \partial - i \frac{\pi^6}{24} \frac{G L B^3e^4m^2}{c^2} \sigma$$
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.
$$\omega = - \frac{\Omega}{2} = \frac{\pi^6}{6} \frac{G L B^3e^4m^2}{c^2}$$
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
$$ \frac{mv^2}{R} = B \frac{e^2}{R^2}$$
I took an equation that I derived two years back when trying to unify electromagnetism with gravity, the resulting equation was
$$ \frac{F_N}{F_G} = \frac{1}{G \epsilon \mu} \cdot \frac{nh}{m^2c} $$
I plugged in the Bohr mass before squaring it
$$ \frac{1}{m} = B \frac{e^2}{R} \cdot (\frac{\lambda }{h})^2 $$
Which gave
$$ \frac{F_N}{F_G} = \frac{1}{G \epsilon \mu} \cdot \frac{nh}{c} \cdot (\frac{4 \pi^2 B e^2 R}{h})^2$$
By noticing that the orbital radius cubed from the Bohr model Is
$$ \frac{1}{R^3} = \frac{12 \pi^6 B^3 e^6 m^3}{h^6}$$
And by using the following dimensional analysis by cancelling unwanted terms,
$$hc = Gm^2 =e^2$$
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
$$\omega = - \frac{\Omega}{2} = \frac{G L }{2c^2R^3}$$
And it gave
$$\omega = - \frac{\Omega}{2} = \frac{\pi^6}{6} \frac{G L B^3e^4m^2}{c^2}$$
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
$$D = \partial - \frac{i}{4} \omega \sigma$$
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
$$\omega = - \frac{\Omega}{2} = \frac{\pi^6}{6} \frac{G L B^3e^4m^2}{c^2}$$
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
$$ \frac{mv^2}{R} = B \frac{e^2}{R^2}$$
I took an equation that I derived two years back when trying to unify electromagnetism with gravity, the resulting equation was
$$ \frac{F_N}{F_G} = \frac{1}{G \epsilon \mu} \cdot \frac{nh}{m^2c} $$
I plugged in the Bohr mass before squaring it
$$ \frac{1}{m} = B \frac{e^2}{R} \cdot (\frac{\lambda }{h})^2 $$
Which gave
$$ \frac{F_N}{F_G} = \frac{1}{G \epsilon \mu} \cdot \frac{nh}{c} \cdot (\frac{4 \pi^2 B e^2 R}{h})^2$$
By noticing that the orbital radius cubed from the Bohr model Is
$$ \frac{1}{R^3} = \frac{12 \pi^6 B^3 e^6 m^3}{h^6}$$
And by using the following dimensional analysis by cancelling unwanted terms,
$$hc = Gm^2 =e^2$$
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
$$\omega = - \frac{\Omega}{2} = \frac{G L }{2c^2R^3}$$
And it gave
$$\omega = - \frac{\Omega}{2} = \frac{\pi^6}{6} \frac{G L B^3e^4m^2}{c^2}$$
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
$$D = \partial - \frac{i}{4} \omega \sigma$$
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
$$D = \partial - i \frac{\pi^6}{24} \frac{G L B^3e^4m^2}{c^2} \sigma$$
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.
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