REPLYGot it
Went through all the equations and all spot on
Only one tiny aspect to complete
What name are you planning to call this new flavour potato chip?
A gravitomagnetic Spinor Solution to the Dirac Equation
- Thread starter ITisTHY
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Sorry about any typos. My phone is hell when it comes to spell-checker. I have to even educate it in physics terminology. Let's just say it dies my head in Later I'll finish thus thread off from here:
....
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 and by 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)
There's just a little more to cover when we take some of these equations and modify them with some simple substitutions.
Later I'll finish thus thread off from here:....
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 and by 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)
There's just a little more to cover when we take some of these equations and modify them with some simple substitutions.
REPLY
View attachment 4225
I just wouldn't name it after me. Even though the solutions for the spinors are novel, it feels more right to name them by those who did most if the hard work.
To continue where we finished off, we will find out first the funaldamentals that relate Newton's G to the electromagnetic parameters (permeability and permittivity) with that of the usual electromagnetic laws. I was able to find a master equation that led to everthing to work from, it came as
G ≡ nh/p • B(4π ^2e^2R)/h^2 • 1/εµ
= B(4π ^2e^2R)/h^2 • λ/εµ
Keep this in mind, there's some explaining to do. The symbol here is the debroglie wavelength relationship to matter λ - so it unifies a whole bunch of fundamental aspects.
G ≡ nh/p • B(4π ^2e^2R)/h^2 • 1/εµ
= B(4π ^2e^2R)/h^2 • λ/εµ
Keep this in mind, there's some explaining to do. The symbol here is the debroglie wavelength relationship to matter λ - so it unifies a whole bunch of fundamental aspects.
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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.
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.
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 = (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)
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 = (12π^8B^3e^8 m^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 the inverse orbital radius into the following equation?
F_N/F_0 =
16π^4m • B^2(e^4R^3)/h^3
Remember, you need
(12π^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. I'll return later.
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 = (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)
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 = (12π^8B^3e^8 m^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 the inverse orbital radius into the following equation?
F_N/F_0 =
16π^4m • B^2(e^4R^3)/h^3
Remember, you need
(12π^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. I'll return later.
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No need for a summary of the last post. But 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.
Sorry one small error,
1/r^3 = 64π^6B^3e^8 m^3)/h^8
Because
r = h^2/4 π^2Be^2m
So you'll see why it has the exact numerical coefficients because
4^3 = 64
And any squared number to the power of 3 is simply 8. It's no good leaving an excersise with the wrong coefficient, that would be misleading.
1/r^3 = 64π^6B^3e^8 m^3)/h^8
Because
r = h^2/4 π^2Be^2m
So you'll see why it has the exact numerical coefficients because
4^3 = 64
And any squared number to the power of 3 is simply 8. It's no good leaving an excersise with the wrong coefficient, that would be misleading.
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 = h^8/12π^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.
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 = h^8/12π^6B^3e^8 m^3)/h^8
And plugging it into
F_N/F_0 =
16π^4m • B^2(e^4R^3)/h^3
Gives
F_N/F_0 =
16π^4m • B^2(e^4R^3)/h^3
Since the R^3 exists in the numerator on the LHS we take the inverse
(12π^6B^3e^8 m^3)/h^8
By noticing
16 *12 = 192
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 =
192π^(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.
R^3 = h^8/12π^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.
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 = h^8/12π^6B^3e^8 m^3)/h^8
And plugging it into
F_N/F_0 =
16π^4m • B^2(e^4R^3)/h^3
Gives
F_N/F_0 =
16π^4m • B^2(e^4R^3)/h^3
Since the R^3 exists in the numerator on the LHS we take the inverse
(12π^6B^3e^8 m^3)/h^8
By noticing
16 *12 = 192
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 =
192π^(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.
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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.
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.
Decided I'd just fix it mow. If deletion of almost duplicate of post can be done to avoid confusion. Id appreciate it. What you really want to read is this, this time with the right coefficient
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 = h^8/12π^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
(64π^6B^3e^8 m^3)/h^8
We notice how to simplify terms
16 *64 = 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 me. We will also explore other variations to interpret what the equation could mean.
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 = h^8/12π^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
(64π^6B^3e^8 m^3)/h^8
We notice how to simplify terms
16 *64 = 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 me. We will also explore other variations to interpret what the equation could mean.
Ah so you're like the Borg Queen, speaking for a collective order? Very interesting. Must make you feel very powerful to think you can speak for everyone on what appears to be a mindset of only a few. Little do you know, or maybe you do, that a lot of people follow what I say, but I don't need to defend myself on this occasion. The fact you think you're on top of the pecking order is good enough for me.
I know that's true
Found a graphic showing your followers
https://m.imdb.com/title/tt0024451/mediaviewer/rm1660032768/?ref_=ext_shr_lnk
The master equation was so fundamental to a lot of my research, it featured strongly in this very long essay about gravitomagnetism. It covers aspects of almost everything I've spoke about in the forum posts. The only thing it didn't cover was Diracs modified spinor for the full Poincare group, which is why I am posting a link here, because it features so much more interesting things I covered.
https://blackholeradiation.quora.co...ck-Hole-Thermodynamics-and-Linearized-Gravity
https://blackholeradiation.quora.co...ck-Hole-Thermodynamics-and-Linearized-Gravity
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