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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 by Dubbelosix
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Posted (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 by Dubbelosix
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Posted (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 by Dubbelosix
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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 by Dubbelosix
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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 by Dubbelosix
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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.

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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 by Dubbelosix
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Posted (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 by Dubbelosix
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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 by Dubbelosix
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