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Temperature, Drag And Gravity


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#1 Dubbelosix

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Posted 23 June 2019 - 06:22 AM

The drag coefficient and drag force has been vital instruments in my own approach to the unification of gravity and cosmology with hopeful progress in quantum mechanics under semi-classical approaches. The drag coefficient was to refresh our memory directly proportional to the drag force:


[math]f = \frac{2F}{\rho v^2\ A} = \frac{A_b}{A_f} \frac{B}{R^2_e}[/math]


and the drag force itself is


[math]F = \frac{1}{2} \rho v^2\ f\ A[/math]


Now, since temperature and pressure are fundamentally related to gravity under the unification of Friedmann's equations obtained from Einstein's field equations, we should explore the physics of the drag force and its dimensionless proportionality [math]f[/math] with it. Certain equations from previous work will be important for such progress, a recent one is that the cosmological constant is more like a impetus thrust rather than associated to strictly a vacuum energy associated to zero point fields:


[math]- \frac{1}{f}\frac{A_f}{A_b}(\frac{mv^2}{n^2x^3} + \frac{mv^2}{n^2y^3} + \frac{mv^2}{n^2z^3}) = \frac{\Lambda c^2}{8 \pi G} = \rho_{vac} v^2 [/math]


Which relates the cosmological constant as a ''thrust.'' And the second replacement for the stress energy:


[math]- \frac{1}{f}\frac{A_f}{A_b}(\frac{mv^2}{n^2x^3} + \frac{mv^2}{n^2y^3} + \frac{mv^2}{n^2z^3}) = - g^{00}\ \frac{\Lambda c^4}{8 \pi G} = T^{00} [/math]


And of course, if we want to entertain the drag force relationship with thermodynamics, we also need to understand it within how the temperature relates to classical gravity:


[math]F = m\frac{2 \pi c\ k_Bt}{\hbar} = m\frac{4 \pi c}{\hbar} \frac{E}{N} = m\frac{4 \pi c^2}{\hbar} \frac{M}{N} = \frac{4 \pi GMm}{A} = G \frac{Mm}{r^2}[/math]

 

[math]N = \frac{A}{\ell^2} = \frac{Ac^3}{\hbar G}T = \frac{\hbar a}{2 \pi c\ k_B}[/math]


So, to find the drag force, we must find it as an application to all the physics underlying the equation above: Doing that is easy, we just plug its definition in so that in a loose way, all the terms are equal to each other


[math]F = \frac{1}{2} \rho v^2\ f\ A = m\frac{2 \pi c\ k_BT}{\hbar} = m\frac{4 \pi c}{\hbar} \frac{E}{N} = m\frac{4 \pi c^2}{\hbar} \frac{M}{N} = \frac{4 \pi GMm}{A} = G\frac{Mm}{r^2}[/math]

 

[math]N = \frac{A}{\ell^2} = \frac{Ac^3}{\hbar G}[/math]

In which again, the temperature is

[math]T = \frac{\hbar a}{2 \pi c\ k_B}[/math]


This will take us into ''new territory'' but first we have to justify the terms in such a way that it is consistent. One immediate consequence is the mass flow of a collection of particles, in which we collect the identity of a heat transfer which fortuitously it was already investigated in my paper yet to be properly updated and properly proof read:


An elementary form of the continuity equation for mass in hydrodynamics is (which is strictly for dynamics involving the cross section only, just as found in the entropic gravity relationships):


[math]\rho_1 v_1 \cdot A_1 = \rho_2 v_2 \cdot A_2[/math]


Such that a mass flow rate is given as


[math]\dot{m} = \rho \cdot \dot{V} = \rho v \cdot A = \mathbf{j} \cdot A[/math]


with [math]\dot{V}[/math] is the volume flow rate and [math]\mathbf{j}[/math] is the mass flux.


This equation though is only true for a surface distribution of the flow. Only dimensionless equations are physically important in physics, which is why Bernoulli's equation is important for fluid mechanics for cosmological purposes -


[math](\frac{v^2}{2c^2} + \frac{\Psi}{\psi} + \frac{P}{\rho})E = C_{vac}[/math]


And we will work on this later. But first of all, we can also identify for the work, a dynamic pressure [math]q = \frac{1}{2} \rho v^2[/math]. The force equation derived will present this quantity:


[math]F = \frac{A_b B}{Re^2_L} \cdot q\ = \frac{1}{2n^2}(x + y + z - \sigma_{x,t})^2\ T_{00}[/math]


(where we treat time as a true operator and observable in contradiction to quantum mechanics as an avenue to why the relativity theories and those of quantum mechanics seem at odds with each other).


This means it related the fluid pressure weighted by the back area with the dynamic pressure as proportional to the Bejan number and inversely related to the Reynolds number as related to the stress energy:


[math]P = \frac{F}{A_b} = \frac{B}{Re^2_L} \cdot q\ = \frac{1}{2n^2\lambda}(x + y + z - \sigma_{x,t})\ T_{00}[/math]


While there is so much to explore, the last obvious relationship that popped up to me was the mass flow identification which relates to the back area:


[math]F = \int\ \frac{1}{2}\frac{A_b B}{Re^2_L} \cdot \rho v\ = \frac{1}{2n^2}(x + y + z - \sigma_{x,t})^2\ T_{00}[/math]


In which we refresh our memory on the following relationships -


[math]\dot{m} = \rho \cdot \dot{V} = \rho v \cdot A = \mathbf{j} \cdot A[/math]


If we just concentrate on the second equality of the equation, you will notice my derivations lead to:


[math]F = \frac{A_bB}{Re^2_L} \cdot q = \int\ \frac{1}{2}\frac{B}{Re^2_L} \cdot \rho v \cdot A_b = \int\ \frac{1}{2}\frac{B}{Re^2_L} \cdot \rho \dot{V} = \int\ \frac{1}{2}\frac{ B}{Re^2_L} \cdot \mathbf{j} \cdot A_b[/math]


The standard definition of the mass flow follows the limit where a velocity term has been set to unity for simplicity


[math]\dot{m} = \lim_{\Delta t \rightarrow 0} \frac{\Delta m}{\Delta t} = \frac{dm}{dt}[/math]


[math]F = \int\ \frac{1}{2}\frac{ B}{Re^2_L} \cdot \dot{m} = \int\ \frac{1}{2}\frac{ B}{Re^2_L} \cdot \lim_{\Delta t \rightarrow 0} \frac{\Delta m}{\Delta t}[/math]


The total pressure is defined as


[math]P + q = P_0[/math]


In such a case, [math]P[/math] is defined as the static pressure. Recall also that [math]\mathbf{j}[/math] is the flux of [math]q[/math] and is the dynamic component of the pressure. Retrieving the dynamic component, by subtracting the static component from both sides gives:


[math]q = P_0 – P[/math]


Let's just cover very quickly the dimensions in which we have a set a velocity term to unity for [math]v = c = 1[/math] :


[math]F \cdot c \cdot t = mc^2[/math]


Since mass cannot move at light speed, we now deduct that [math]v<c[/math] for the first terms:


[math]F \cdot v \cdot t = mc^2[/math]


Which is already an established equation in relativity. Plugging in the terms is pretty simple, it just yields for the mass flow:


[math]F \cdot v = \int\ \frac{1}{2}\frac{ B}{Re^2_L} \cdot \dot{m}c^2 = c^2 \int\ \frac{1}{2}\frac{ B}{Re^2_L} \cdot \lim_{\Delta t \rightarrow 0} \frac{\Delta m}{\Delta t}[/math]


With this in mind, we now have an argument for the drag under the entropic gravity case, in which the flow of mass will be translated as a heat flow or flux of heat and has already been stated, its form satisfied the cross section:


[math]F = \frac{1}{2} \rho v^2\ f\ A = m\frac{2 \pi c\ k_Bt}{\hbar} = m\frac{4 \pi c}{\hbar} \frac{E}{N} = m\frac{4 \pi c^2}{\hbar} \frac{M}{N} = \frac{4 \pi GMm}{A} = G \frac{Mm}{r^2}[/math]


[math]N = \frac{A}{\ell^2} = \frac{Ac^3}{\hbar G}[/math]


[math]T = \frac{\hbar a}{2 \pi c\ k_B}[/math]


From here we can do loads of ''new'' stuff - we can rewrite the mass with the derivative directly allowing us to define the mass flux. We can even divide off area terms for the fluid mechanics in which we find the specific ratio for back and front areas as defined from the dimensionless coefficient:


[math]f = \frac{2F}{\rho v^2 A} = \frac{A_b}{A_f}\frac{B}{R^2_e}[/math]


From it we find the first area term related to the front area:


[math]F = \frac{1}{2} \rho v^2\ f\ A_f = m\frac{2 \pi c\ k_Bt}{\hbar} = m\frac{4 \pi c}{\hbar} \frac{E}{N} = m\frac{4 \pi c^2}{\hbar} \frac{M}{N} = \frac{4 \pi GMm}{A} = G \frac{Mm}{r^2}[/math]


Which means the pressure is found through a simple division of the front area giving us also the important ratio from fluid mechanics:


[math]P = \frac{1}{2} \rho v^2\ f = \rho_A \frac{2 \pi c\ k_BT}{\hbar} = \rho_A \frac{4 \pi c}{\hbar} \frac{E}{N} = \rho_A \frac{4 \pi c^2}{\hbar} \frac{M}{N} = 4 \pi \frac{A_b}{A_f}\ \frac{ GMm}{r^4} = \rho_A\ \frac{GM}{r^2}[/math]


With [math]\rho_A[/math] being the area density term. Notice also, the configuration of the dimensionless particle number can also be expressed using the same physics we have been applying:

 

[math]N = \frac{A_b}{A_f}\frac{1}{\ell^2} = \frac{A_b}{A_f}\frac{c^3}{\hbar G}[/math]

 

 

Now let's implicate the mass flux. To do that we go back to the force equation and take the derivative with respect to time :


[math]\dot{F} = \frac{1}{2} \rho v^2\ f\ A_f = \dot{m}\frac{2 \pi c\ k_BT}{\hbar} = \dot{m}\frac{4 \pi c}{\hbar} \frac{E}{N} = \dot{m}\frac{4 \pi c^2}{\hbar} \frac{M}{N} = \dot{m}\frac{A_b}{A_f}\frac{4 \pi GM}{r^2} = \dot{m}\frac{GM}{r^2}[/math]

 

 

The acceleration term arises from the equation, but we will come to that in the next post.

Force over time, is just how the force changes over time - but it can viewed a number of different ways, for instance, the impulse is

[math]I = F \Delta t[/math]


Solving for the force we get


[math]F = \frac{I}{(t_1 - t_2)}[/math]


and so it's derivative looks like


[math]\dot{F} = \frac{\dot{I}}{(t_1 - t_2)} = \frac{d^2I}{d(t^2_1 - t^2_2)}[/math]


Alternatively we can view it as being related to the relativistic expression [math]Fvt[/math] by derivation:


[math]F \cdot v \cdot vt = F \cdot v \cdot vt = mc^2 \cdot v[/math]


Dividing through by [math]v^2[/math] gives an inverse Doppler shift [math](\frac{v^2}{c^2})^{-1}[/math] as:


[math]F \cdot t = F \cdot t = (\frac{c^2}{v^2})\ m \cdot v[/math]


And twice the derivative yields:


[math]\dot{F} = (\frac{c^2}{v^2})\ \dot{m} \cdot \frac{dv}{dt} = (\frac{v^2}{c^2})^{-1}\ \dot{m}a [/math]


As far as I know, this equation is unique.


Edited by Dubbelosix, 23 June 2019 - 10:55 AM.


#2 Dubbelosix

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Posted 23 June 2019 - 08:59 AM

How acceleration enters the definition is explained by noticing that the strength of the acceleration [math]g[/math] is described under Newtons law:

 

[math]F = gm =\frac{GMm}{r^2}[/math]

 

We divide through by the small mass term,

 

[math]g = \frac{F}{m} =\frac{GM}{r^2}[/math]

 

And in a nutshell, it shows that the acceleration is dependent on the mass [math]M[/math] of the system.

 

Notice in the derivations before, we also have such a term:

 

[math]P = \frac{1}{2} \rho v^2\ f = \rho_A \frac{2 \pi c\ k_BT}{\hbar} = \rho_A \frac{4 \pi c}{\hbar} \frac{E}{N} = \rho_A \frac{4 \pi c^2}{\hbar} \frac{M}{N} = 4 \pi \frac{A_b}{A_f}\ \frac{GMm}{r^4} = \rho_A\ \frac{GM}{r^2}[/math]

 

The Christoffel symbol can be 'loosely' thought of as being analogous to a force in Newtons equations ~

[math]\Gamma = g_{00}\ G\ \frac{m}{2}\frac{ M}{ x}[/math]

Newtonian formulation of this acceleration is

[math]F = -\frac{\phi}{x}[/math]

 

Where

 

[math]\phi = \frac{Gm}{r}[/math]

 

known as the gravitational potential.

However as mentioned, the gravitational force is not actually a true definition of a force as we come to expect say, in the proposed fundamental fields of nature, which are inherently complex (when quantum gravity is not) and that require quantization of field particles acting as mediators of the force (something which gravity is expected to use to form the unification theory in the opinions of many scientists) but has led to the divergence problems indicating that fundamentally-speaking, gravity cannot be quantized in the similar fashion as we do with the other forces of nature. The pressure equation will naturally introduce the physics of curvature and acceleration described through the Christoffel symbol by using the identities above:

 

[math]P = \frac{1}{2} \rho v^2\ f = g_{00}\ \frac{1}{2}G\ \rho_A\ \cdot \frac{M}{r^2} = \rho_A\ \cdot \Gamma_{00}[/math]

 

With the Christoffel symbol having units of acceleration (as it should be).


Edited by Dubbelosix, 23 June 2019 - 03:01 PM.


#3 Dubbelosix

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Posted 23 June 2019 - 10:50 AM

Now let's move on. The pressure formula is:


[math]P = \frac{1}{2} \rho v^2\ f = \rho_A \frac{2 \pi c\ k_BT}{\hbar} = \rho_A \frac{4 \pi c}{\hbar} \frac{E}{N} = \rho_A \frac{4 \pi c^2}{\hbar} \frac{M}{N} = 4 \pi \frac{A_b}{A_f}\ \frac{ GMm}{r^4} = \rho_A\ \frac{GM}{r^2}[/math]

 

There are some relatinoships/equations we will focus on - one of them being is the particle number as a dimensionless quantity also takes the form:

 

[math]N = \frac{A_b}{A_f}\frac{A_p}{\ell^2} = \frac{A_b}{A_f} \frac{A_p c^3}{\hbar G}[/math]

 

With [math]A_p[/math] being the Planck area which is just the square of the Planck length. Since the particle number in the first equation is inverse, we can simply invert the equation and plug in the definition

 

[math]N^{-1} = \frac{A_f}{A_b}\frac{\ell^2}{A_p} = \frac{A_f}{A_b} \frac{\hbar G}{A_p c^3}[/math]

 

The Planck area by definition is:

 

[math]A_p = \frac{\hbar G}{c^3}[/math]

 

And the Planck pressure is

 

[math]P = \frac{F}{\ell^2_p} = \frac{c^7}{\hbar G}[/math]

 

And a Planck energy as

 

[math]E = \frac{\hbar}{t_p}[/math]

 

Let's first focus on the pressure relationship to the energy term weighted by the particle number (and replace those) for the Planck definitions - this will yield after simplification of one term of the Planck constant and one term of the speed of light as:

 

[math]P = 4 \pi \rho_A\  \frac{A_f}{A_b} \frac{\hbar G}{A_p c^2 t_p} = 4 \pi \rho_A\ \frac{1}{c} \frac{A_f}{A_b} \frac{\hbar G}{V_p} [/math]

 

With [math]V_p[/math] denoting the Planck volume.

 

since [math]k_BT[/math] is a thermodynamic energy term we can also replace this with a Planck temperature and/or the Planck energy, but doing the latter requires a correction coefficient... but we can look into that later as it is related to equipartition. The first equation will be defined under Planck temperature ~

 

[math]P = \frac{1}{2} \rho v^2\ f = 2 \pi \rho_A \frac{mc^3}{\hbar}[/math]

 

With the Planck temperature defined as

 

[math]T_p = \frac{mc^2}{k_B}[/math]

 

Note also the reduced Compton wavelength is

 

[math]\frac{\hbar}{mc}[/math]

 

So we can pull this very important quantity out and distribute the remaining squared factor of the speed of light and redefine as

 

[math]P = \frac{1}{2} \rho v^2\ f = 2 \pi \rho_Ac^2\ \frac{mc}{\hbar} = 2 \pi \rho_Ac^2\ \lambda^{-1}[/math]

 

In which the areal density is a mass over area, so we obtained an areal energy density and the inverse of the wavelength gives the correct dimensions of an energy density associated to the pressure.

 

Or we can also entertain that

 

[math]\Delta E \Delta t \geq \frac{1}{2}\hbar[/math]

 

In which case we would also have for an energy divided by the Planck constant as:

 

[math]\frac{2\Delta E}{\hbar}  \approx \frac{1}{\Delta t}[/math]

 

Which now we define under a Planck energy retrieves the acceleration. Under normal convention, the velocity over time gives the average of the acceleration, but for unification purposes we will stick with the Christoffel symbol ~

 

[math]P = \frac{1}{2} \rho v^2\ f = 2 \pi \rho_A\ (\frac{ c}{ t_p}) = 2 \pi \rho_A \cdot \Gamma[/math]

 

But I will not lie, all this mind work and latex is draining me now, so I will continue this later.


Edited by Dubbelosix, 23 June 2019 - 10:51 AM.


#4 Dubbelosix

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Posted 23 June 2019 - 05:48 PM

Under these new sets of formula's, there are so many avenues to cover that it will take a while to get through them all.  Before we do, you might want refresh the study of gravielectromagnetism I had done from reading Sciama's work on the origin of inertia:

 

https://spinorbit.qu.../Linear-Gravity

 

The pressure formula is:


[math]P = \frac{1}{2} \rho v^2\ f = \rho_A \frac{2 \pi c\ k_BT}{\hbar} = \rho_A \frac{4 \pi c}{\hbar} \frac{E}{N} = \rho_A \frac{4 \pi c^2}{\hbar} \frac{M}{N} = 4 \pi \frac{A_b}{A_f}\ \frac{ GMm}{r^4} = \rho_A\ \frac{GM}{r^2}[/math]

 

we also recall we obtained:

 

[math]P = \frac{1}{2} \rho v^2\ f = 2 \pi \rho_Ac^2\ \frac{mc}{\hbar} = 2 \pi \rho_Ac^2\ \lambda^{-1}[/math]

 

And acceleration enters like:

 

[math]P = \frac{1}{2} \rho v^2\ f = 2 \pi \rho_A\ (\frac{ c}{ t_p}) = 2 \pi \rho_A \cdot \Gamma[/math]

 

When [math]G[/math] is used in standard cgs units under Sciama's definition of gravielectromagnetism,

 

[math]\frac{1}{r}\frac{\partial m}{\partial r} = \frac{\omega^2r}{G} = \frac{a}{G}[/math]

 

[math]P = \rho_A\ \frac{GM}{r^2}[/math]

 

Dividing through by G gives us

 

[math]\frac{P}{G} = \rho_A\ (\frac{M}{r^2}) = \rho_A\ (\frac{\omega^2r}{G}) = \rho_A\ (\frac{a}{G})[/math]

 

The gravimagnetic field defined for magnetic coupling to orbit is, as the master equation we will work from:

 

[math]\mathbf{B} = \frac{1}{mc^2 e} \frac{1}{r} \frac{\partial U}{\partial r} \mathbf{J} = \frac{1}{me}(\frac{\phi}{c^2})\frac{\partial v}{\partial t} \mathbf{J} = -\frac{1}{e}\frac{1}{Gm}\frac{\partial v}{\partial t} \mathbf{J}= -\frac{1}{me}\frac{a}{G} \mathbf{J} = -\frac{1}{me}\frac{\omega^2 r}{G} \mathbf{J} = -\frac{1}{m e} \frac{m}{r^2} \mathbf{J}[/math]

 

where we have used the relationships I derived for clarity:

 

[math]\frac{1}{e}\frac{\partial U}{\partial r} = \frac{c^2}{e} \frac{\partial m}{\partial r}[/math]

 

[math]\frac{1}{r}\frac{\partial m}{\partial r} = \frac{\omega^2r}{G} = \frac{a}{G}[/math]

 

The gravimagnetic field in standard physics is defined as:

 

[math]\mathbf{B} = \Omega[/math]

 

And is known as the torsion field. The definition of the gravimagnetic field then under standard approach is to treat it with the same dimensions as frequency. The frequency field is related to a central potential:

 

[math]\Omega^2 = \frac{1}{mc^2}(\frac{d^2U}{dt^2})[/math]

 

In which case, since the gravimagnetic field is defined by the torsion field

 

The gravimagnetic field is derived this from the ordinary spin coupling equation which has dimensions of:

 

[math]\mathbf{B} = \frac{1}{mc^2 e} \frac{1}{r}\frac{\partial U}{\partial r} \mathbf{J}[/math]

 

With an electric field identified as:

 

[math]\mathbf{E} = \frac{1}{e} \frac{\partial U}{\partial r}[/math]

 

The gravielectric formula is just

 

[math]\nabla \cdot \mathbf{E} = -\frac{\rho}{\epsilon_G} = -4 \pi G \rho[/math]

 

Comparing this with Maxwell’s equation for electric field,

 

[math]\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon}[/math]

 

There is no real formal difference, with

 

[math]\frac{1}{\epsilon_G} = 4 \pi G[/math]

 

[math]\frac{1}{\mu_G} = \frac{c^2}{4 \pi G}[/math]

 

The units though for the potential are the same as the gravitational potential, which we find the relationship:

 

[math]\mathbf{E} = -\nabla \phi = \frac{1}{r} \frac{Gm}{r}[/math]

 

and so units become obvious when:

 

[math]\nabla \cdot \mathbf{E} = -G \frac{m}{r^3} = -G \rho[/math]

 

In our equations, the factor of G will drop out of the definition for the gravielectric field.

 

[math]\frac{P}{G} = \rho_A\ (\frac{M}{r^2}) = \rho_A\ (\frac{\omega^2r}{G)} = \rho_A\ (\frac{a}{G})[/math]

 

The gravimagnetic field defined for magnetic coupling to orbit is, as the master equation we will work from:

 

[math]\mathbf{B} = \frac{1}{mc^2 e} \frac{1}{r} \frac{\partial U}{\partial r} \mathbf{J} = \frac{1}{me}(\frac{\phi}{c^2})\frac{\partial v}{\partial t} \mathbf{J} = -\frac{1}{e}\frac{1}{Gm}\frac{\partial v}{\partial t} \mathbf{J}= -\frac{1}{me}\frac{a}{G} \mathbf{J} = -\frac{1}{me}\frac{\omega^2 r}{G} \mathbf{J} = -\frac{1}{m e} \frac{m}{r^2} \mathbf{J}[/math]

 

From the master equation we can see by a distribution of [math]\frac{1}{m} \frac{\mathbf{J}}{2e}[/math] with [math]\Phi = \frac{\mathbf{J}}{2e}[/math] being the Josephson constant (or magnetic flux) we regain the gravimagnetic field by plugging it into the pressure formula:

 

 

[math]\frac{P}{G} = \rho_A\ \frac{1}{m}(\frac{M}{r^2}) \cdot (\frac{\mathbf{J}}{2e}) [/math]

 

[math]= \rho\ \frac{1}{r} \cdot (\frac{\mathbf{J}}{2e}) = \rho_A\ (\frac{\omega^2r}{Gm}\frac{\mathbf{J}}{2e}) = \rho_A\ (\frac{a}{Gm}\frac{\mathbf{J}}{2e})[/math]

 

In which [math]\rho[/math] is now the mass density. This will do for now, as I will do more later.


Edited by Dubbelosix, 24 June 2019 - 03:38 AM.


#5 Dubbelosix

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Posted 24 June 2019 - 03:41 AM

Just a small note before we continue:

 

 

[math]\frac{P}{G} = \rho_A\ (\frac{\omega^2r}{G})[/math]

 

 

The centrifugal force appears here naturally as a consequence of the spin-dynamics of the system:

 

 

[math]\frac{P}{G} = \rho_A\ \cdot \omega \times (\frac{\omega \times r}{G})[/math]

 

 

Simplifying by removal of the gravitational constant yields the centrifugal force related to the pressure:

 

 

[math]P = \rho_A\ \cdot \omega \times (\omega \times r)[/math]


Edited by Dubbelosix, 24 June 2019 - 03:52 AM.


#6 Dubbelosix

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Posted 24 June 2019 - 07:31 AM

[math]P = \frac{1}{2} \rho v^2\ f = g_{00}\ \frac{1}{2} G\ \rho_A\ \frac{M}{r^2} = \rho_A\ \cdot \Gamma_{00}[/math]

 

With the Christoffel symbol bounded in units of acceleration ~

 

[math]\Gamma_{00} = g_{00} \frac{GM}{r^2}[/math]

 

Dividing through a pressure term, we can call out a pressure ratio:

 

[math]\frac{P}{\rho_B} = \frac{1}{2} (\frac{\rho}{\rho_B}) v^2\ f = g_{00}\ G\ (\frac{\rho_A}{\rho_B}\ \frac{M}{2r^2}) =  \Gamma_{00}[/math]

 

 

With the numerator and denominator as the same dimensions, which makes [math]\frac{\rho_A}{\rho_B}[/math] a dimensionless ratio, not too different to the ratio of pressure to density found  in Bernoulli's fluid mechanics. The difference between the ratio's become a coefficient to the acceleration term - which itself will modify the mechanics of accelerated motion as something that depends upon it.


Edited by Dubbelosix, 24 June 2019 - 09:58 PM.


#7 Dubbelosix

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Posted 24 June 2019 - 08:45 AM

The same pressure differences can be formulated from the pressure equation:

 

[math]P = \frac{1}{2} \rho v^2\ f = 2 \pi \rho_A\ (\frac{ c}{ t_p}) = 2 \pi \rho_A \cdot \Gamma[/math]

 

Distribution of the factor of 2, to give twice the pressure

 

[math]2P = \rho v^2\ f = 4 \pi \rho_Ac^2\ \frac{mc}{\hbar} = 4 \pi \rho_Ac^2\ \lambda^{-1}[/math]

 

Dividing through by the pressure term with coefficients we get:

 

[math]2 \pi\ (\frac{P}{\rho_A}) = \frac{2P}{4 \pi \rho_A} = \frac{\rho}{4 \pi \rho_A} v^2\ f = c^2\ \frac{mc}{\hbar} = c^2\ \lambda^{-1}[/math]

 

Divide through [math]c^2[/math]  so that we get a Doppler shift term and further simplify the constants

 

[math]\frac{1}{c^2} (\frac{P}{\rho_A}) = \frac{1}{c^2}\frac{2P}{8 \pi^2 \rho_A} = \frac{\rho}{8 \pi^2 \rho_A} (\frac{v^2}{c^2})\ f = \frac{mc}{2 \pi \hbar} =( 2 \pi \lambda)^{-1}[/math]

 

Which leaves the inverse of the Compton wavelength with adjustable parameters.We'll do some stuff later on with the Doppler shift using algebraic techniques/hacks.


Edited by Dubbelosix, 24 June 2019 - 09:51 PM.


#8 Dubbelosix

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Posted 24 June 2019 - 09:28 AM

The pressure formula from the fundamental principles of entropic gravity is continued from the last derivation:

 

[math]\frac{1}{c^2} (\frac{P}{\rho_A}) = \frac{1}{c^2}\frac{2P}{8 \pi^2 \rho_A} = \frac{\rho}{8 \pi^2 \rho_A} (\frac{v^2}{c^2})\ f = \frac{mc}{2 \pi \hbar} = 2 \pi \lambda^{-1}[/math]

 

So what is the significance of the Doppler term [math]\frac{v^2}{c^2}[/math]. Ok, special relativity is inherently tied to the Gamma function, which itself describes the variations of velocity to that of the speed of light. Following algebra hacks I wrote up a short while ago,

 

(https://calculusalge...nd-Useful-Hacks)

 

We find that in a simplified expression, the Doppler effect is

 

[math]\frac{v^2}{c^2} = 1 - \frac{1}{\gamma^2}[/math]

 

When solving for [math]\gamma[/math] we get a very nice set of solutions:

 

[math]\gamma = i\frac{c\sqrt{v^2 - c^2}}{v^2 - c^2}[/math]

 

and

 

[math]\gamma = -i\frac{c\sqrt{v^2 - c^2}}{v^2 - c^2}[/math]

 

Where we notice that for this to be true, [math]v \ne c[/math]. You could go further and state that [math]v \ne - c[/math] but this is where pure algebra reaches a limit because there is no such thing as a negative speed of light. Solving for [math]\gamma^2[/math] we get the usual relationship found from the manipulation of the variables

 

[math]\gamma^2 = -\frac{c^2}{v^2 - c^2}[/math]

 

With all these useful hacks to the side, the first identity simply enters the pressure formula as

 

[math]\frac{1}{c^2} (\frac{P}{\rho_A}) = \frac{1}{c^2}\frac{2P}{8 \pi^2 \rho_A}= \frac{mc}{2 \pi \hbar} = (2 \pi \lambda)^{-1} = \frac{\rho}{8 \pi^2 \rho_A} (1 - \frac{1}{\gamma^2})\ f [/math]


Edited by Dubbelosix, 24 June 2019 - 09:58 PM.


#9 Dubbelosix

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Posted 24 June 2019 - 11:06 PM

Under these new sets of formula's, there are so many avenues to cover that it will take a while to get through them all.  Before we do, you might want refresh the study of gravielectromagnetism I had done from reading Sciama's work on the origin of inertia:

 

https://spinorbit.qu.../Linear-Gravity

 

The pressure formula is:


[math]P = \frac{1}{2} \rho v^2\ f = \rho_A \frac{2 \pi c\ k_BT}{\hbar} = \rho_A \frac{4 \pi c}{\hbar} \frac{E}{N} = \rho_A \frac{4 \pi c^2}{\hbar} \frac{M}{N} = 4 \pi \frac{A_b}{A_f}\ \frac{ GMm}{r^4} = \rho_A\ \frac{GM}{r^2}[/math]

 

we also recall we obtained:

 

[math]P = \frac{1}{2} \rho v^2\ f = 2 \pi \rho_Ac^2\ \frac{mc}{\hbar} = 2 \pi \rho_Ac^2\ \lambda^{-1}[/math]

 

And acceleration enters like:

 

[math]P = \frac{1}{2} \rho v^2\ f = 2 \pi \rho_A\ (\frac{ c}{ t_p}) = 2 \pi \rho_A \cdot \Gamma[/math]

 

When [math]G[/math] is used in standard cgs units under Sciama's definition of gravielectromagnetism,

 

[math]\frac{1}{r}\frac{\partial m}{\partial r} = \frac{\omega^2r}{G} = \frac{a}{G}[/math]

 

[math]P = \rho_A\ \frac{GM}{r^2}[/math]

 

Dividing through by G gives us

 

[math]\frac{P}{G} = \rho_A\ (\frac{M}{r^2}) = \rho_A\ (\frac{\omega^2r}{G}) = \rho_A\ (\frac{a}{G})[/math]

 

The gravimagnetic field defined for magnetic coupling to orbit is, as the master equation we will work from:

 

[math]\mathbf{B} = \frac{1}{mc^2 e} \frac{1}{r} \frac{\partial U}{\partial r} \mathbf{J} = \frac{1}{me}(\frac{\phi}{c^2})\frac{\partial v}{\partial t} \mathbf{J} = -\frac{1}{e}\frac{1}{Gm}\frac{\partial v}{\partial t} \mathbf{J}= -\frac{1}{me}\frac{a}{G} \mathbf{J} = -\frac{1}{me}\frac{\omega^2 r}{G} \mathbf{J} = -\frac{1}{m e} \frac{m}{r^2} \mathbf{J}[/math]

 

where we have used the relationships I derived for clarity:

 

[math]\frac{1}{e}\frac{\partial U}{\partial r} = \frac{c^2}{e} \frac{\partial m}{\partial r}[/math]

 

[math]\frac{1}{r}\frac{\partial m}{\partial r} = \frac{\omega^2r}{G} = \frac{a}{G}[/math]

 

The gravimagnetic field in standard physics is defined as:

 

[math]\mathbf{B} = \Omega[/math]

 

And is known as the torsion field. The definition of the gravimagnetic field then under standard approach is to treat it with the same dimensions as frequency. The frequency field is related to a central potential:

 

[math]\Omega^2 = \frac{1}{mc^2}(\frac{d^2U}{dt^2})[/math]

 

In which case, since the gravimagnetic field is defined by the torsion field

 

The gravimagnetic field is derived this from the ordinary spin coupling equation which has dimensions of:

 

[math]\mathbf{B} = \frac{1}{mc^2 e} \frac{1}{r}\frac{\partial U}{\partial r} \mathbf{J}[/math]

 

With an electric field identified as:

 

[math]\mathbf{E} = \frac{1}{e} \frac{\partial U}{\partial r}[/math]

 

The gravielectric formula is just

 

[math]\nabla \cdot \mathbf{E} = -\frac{\rho}{\epsilon_G} = -4 \pi G \rho[/math]

 

Comparing this with Maxwell’s equation for electric field,

 

[math]\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon}[/math]

 

There is no real formal difference, with

 

[math]\frac{1}{\epsilon_G} = 4 \pi G[/math]

 

[math]\frac{1}{\mu_G} = \frac{c^2}{4 \pi G}[/math]

 

The units though for the potential are the same as the gravitational potential, which we find the relationship:

 

[math]\mathbf{E} = -\nabla \phi = \frac{1}{r} \frac{Gm}{r}[/math]

 

and so units become obvious when:

 

[math]\nabla \cdot \mathbf{E} = -G \frac{m}{r^3} = -G \rho[/math]

 

In our equations, the factor of G will drop out of the definition for the gravielectric field.

 

[math]\frac{P}{G} = \rho_A\ (\frac{M}{r^2}) = \rho_A\ (\frac{\omega^2r}{G)} = \rho_A\ (\frac{a}{G})[/math]

 

The gravimagnetic field defined for magnetic coupling to orbit is, as the master equation we will work from:

 

[math]\mathbf{B} = \frac{1}{mc^2 e} \frac{1}{r} \frac{\partial U}{\partial r} \mathbf{J} = \frac{1}{me}(\frac{\phi}{c^2})\frac{\partial v}{\partial t} \mathbf{J} = -\frac{1}{e}\frac{1}{Gm}\frac{\partial v}{\partial t} \mathbf{J}= -\frac{1}{me}\frac{a}{G} \mathbf{J} = -\frac{1}{me}\frac{\omega^2 r}{G} \mathbf{J} = -\frac{1}{m e} \frac{m}{r^2} \mathbf{J}[/math]

 

From the master equation we can see by a distribution of [math]\frac{1}{m} \frac{\mathbf{J}}{2e}[/math] with [math]\Phi = \frac{\mathbf{J}}{2e}[/math] being the Josephson constant (or magnetic flux) we regain the gravimagnetic field by plugging it into the pressure formula:

 

 

[math]\frac{P}{G} = \rho_A\ \frac{1}{m}(\frac{M}{r^2}) \cdot (\frac{\mathbf{J}}{2e}) [/math]

 

[math]= \rho\ \frac{1}{r} \cdot (\frac{\mathbf{J}}{2e}) = \rho_A\ (\frac{\omega^2r}{Gm}\frac{\mathbf{J}}{2e}) = \rho_A\ (\frac{a}{Gm}\frac{\mathbf{J}}{2e})[/math]

 

In which [math]\rho[/math] is now the mass density. This will do for now, as I will do more later.

 

I must elucidate that the unification of the spin-orbit equation to that of the investigations in this post concerning the pressure formula:

 

[math]P = \frac{1}{2} \rho v^2\ f = \rho_A \frac{2 \pi c\ k_BT}{\hbar} = \rho_A \frac{4 \pi c}{\hbar} \frac{E}{N} = \rho_A \frac{4 \pi c^2}{\hbar} \frac{M}{N} = 4 \pi \frac{A_b}{A_f}\ \frac{ GMm}{r^4} = \rho_A\ \frac{GM}{r^2} = \frac{F_d}{A}[/math]

 

In which the drag force must be in some way equivalent to the gravitational force.

 

Gave

 

 

[math]\mathbf{B} = \frac{1}{mc^2 e} \frac{1}{r} \frac{\partial U}{\partial r} \mathbf{J}  = \frac{1}{2} \frac{1}{G}\frac{1}{me} (\frac{P}{\rho_A}) \frac{\partial v}{\partial t} \mathbf{J} = \frac{1}{2} \frac{1}{G}\frac{1}{me} (\frac{\phi}{c^2}) \frac{\partial v}{\partial t} \mathbf{J} = -\frac{1}{m}\frac{a}{G} \frac{\mathbf{J}}{2e}[/math]

 

[math] = -\frac{1}{m} (\frac{m}{r^2}) \frac{\mathbf{J}}{2e}  = -2 \pi \cdot \frac{1}{Gm} (\frac{\mathbf{J}}{2e})\Gamma =- 2 \pi \frac{1}{Gm} (\frac{c}{ t_p}) \frac{\mathbf{J}}{2e} = -2 \pi \frac{1}{m} \cdot \omega \times (\frac{\omega \times r}{G}) \frac{\mathbf{J}}{2e} = -2 \pi \frac{1}{\mu} (\frac{\mathbf{J}}{2e}) \Gamma[/math]

 

We also note for dimensional purposes from the pressure equation:

 

[math]P = \frac{1}{2} \rho v^2\ f = \rho_A \frac{2 \pi c\ k_BT}{\hbar} = \rho_A \frac{4 \pi c}{\hbar} \frac{E}{N} = \rho_A \frac{4 \pi c^2}{\hbar} \frac{M}{N} = 4 \pi \frac{A_b}{A_f}\ \frac{ GMm}{r^4} = \rho_A\ \frac{GM}{r^2}[/math]

 

That the last term [math]\rho_A\ \frac{GM}{r^2}[/math] [is] a gravitational charge density weighted by an extra component of length which can be explained through the nodes of virtual particles in theory:

 

[math]P = \frac{Gm^2}{r^4} = \frac{\hbar c}{r^4} = \frac{e^2}{r^4}[/math]

 

Those nodes have been investigated by Sakharov as a correction term to gravity.



#10 Dubbelosix

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Posted 24 June 2019 - 11:46 PM

So this brings me to the next big discovery (aside from viewing the cosmological constant as a thrust), we now know a proper origin theory for inertia. Contrary to what Mach believed, it seems that the local gravitational dynamics play the significant role in the inertial properties of a system, which is ''the resistance to accelerate'' from a position of rest. The generic formula for gravitational mass as linked through weak equivalence with that of inertial mass has always been seen as:

 

[math]m_g = m_i[/math]

 

From the fluid dynamical equations, we learn a correction to this by viewing it proportional to the drag force

 

[math]m_d = m_g = m_i[/math]

 

This is quite a significant realization because it explains that Einstein was right and also Mach, but for alternative reasons. Einstein knew the inertial mass was somehow connected to the inertia of the system, but rejected Mach's hypothesis, even though his stress energy tensor could be formulated for the entire collection of objects in the universe - instead, it will be only the local dynamics which play the significant role in which gravity is an aether, and things move inside the aether and as a consequence, inertia is a drag.



#11 Dubbelosix

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Posted 25 June 2019 - 12:02 AM

A simple thought experiment can prove this assertion to good accuracy. It is also direction dependent when using the frame of reference of the Earth - for instance, it takes more energy to leave the gravitational field of the planet than to simply move across its surface. The escape velocity is due to the ''thickness'' of the gravitational medium.



#12 Dubbelosix

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Posted 25 June 2019 - 01:22 AM

Moving on, say I would like to take the inverse of the equation,

 

[math]\frac{1}{c^2} (\frac{P}{\rho_A}) = \frac{1}{c^2}\frac{2P}{8 \pi^2 \rho_A} = \frac{\rho}{8 \pi^2 \rho_A} (\frac{v^2}{c^2})\ f = \frac{mc}{2 \pi \hbar} =( 2 \pi \lambda)^{-1}[/math]

 

This would give using the following formula:

 

[math](\frac{v^2}{c^2} = 1 - \frac{1}{\gamma^2})^{-1} = \frac{1}{(-\frac{1}{\gamma^2} + 1 = \frac{v^2}{c^2})} = \frac{1}{(\frac{1}{\gamma^2} - 1)}[/math]

 

[math]\frac{\rho_A c^2}{P} = \frac{8 \pi^2 \rho_A c^2}{2P} = \frac{8 \pi^2 \rho_A}{\rho} \frac{1}{(\frac{1}{\gamma^2} - 1)} f = \frac{2 \pi \hbar}{mc} = 2 \pi \lambda [/math]

 

This is a nice equation, but we can further simplify for the Compton wavelength:

 

[math]\frac{\rho_A c^2}{2 \pi P} = \frac{2 \pi \rho_A c^2}{P} = 4 \pi\ (\frac{\rho_A}{\rho}) \frac{1}{(\frac{1}{\gamma^2} - 1)} f = \frac{ \hbar}{mc} =  \lambda [/math]


Edited by Dubbelosix, 25 June 2019 - 01:31 AM.


#13 Dubbelosix

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Posted 25 June 2019 - 05:23 AM

The Compton wavelength, which for an electron we place in its value:

 

[math]\frac{\rho_A c^2}{2 \pi P} = \frac{2 \pi \rho_A c^2}{P} = 4 \pi\ (\frac{\rho_A}{\rho}) \frac{1}{(\frac{1}{\gamma^2} - 1)} f = \frac{ \hbar}{mc} = \lambda =\ 2.426\ 310\ 2367(11) \times 10^{-12}m[/math]

 

Its inverse term plays a role in physics, for instance, the Klein-Gordon equation would use the inverse of the Compton wavelength:

 

[math]\nabla^2\psi - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\psi = (\frac{mc}{\hbar})^2 \psi[/math]

 

Or simply the wave equation by Schrodinger: where normally we would take the energy for its respective operator:

 

[math]\hat{E} = i \hbar \frac{\partial}{\partial t}[/math]

 

Giving

 

[math]i \hbar \frac{\partial}{\partial t} \psi = \hat{E} \psi [/math]

 

We can though concentrate on the Klein Gordon equation in the following form:

 

[math](\frac{\partial}{\partial t^2} + c^2 \nabla^2) \phi = (\frac{mc^2}{\hbar})^2 \psi[/math]

 

Retrieving the wavelength can be achieved by taking the square root, which in turn would take us into the Clifford algebra, but we will avoid calculating it out for now ~

 

[math]\sqrt{\frac{\partial \psi}{\partial t^2} + c^2 \nabla^2} \psi = \frac{mc^2}{\hbar} \psi[/math]

 

and so

 

[math]\frac{1}{c}\sqrt{\frac{\partial \psi}{\partial t^2} + c^2 \nabla^2} \psi = \frac{mc}{\hbar} \psi[/math]

 

This is often seen as an abuse of notation, but it is generally the idea that Dirac had in mind. It is an abuse of notation unless you take into consideration it is of first order following the Clifford algebra which will implement the role of the Pauli matrices:

 

[math](A\frac{\partial}{\partial x} + B\frac{\partial}{\partial y} + C\frac{\partial}{\partial z} + i \frac{1}{c}D\frac{\partial}{\partial t})\psi = \frac{mc}{\hbar}\psi[/math]

 

We notice then that the equation we derived:

 

[math]\frac{1}{c^2} (\frac{P}{\rho_A}) = \frac{1}{c^2}\frac{2P}{8 \pi^2 \rho_A} = \frac{\rho}{8 \pi^2 \rho_A} \frac{1}{(1 - \frac{1}{\gamma^2})}\ f = \frac{mc}{2 \pi \hbar} = \frac{1}{2 \pi} \frac{mc}{\hbar}[/math]

 

 

Is intrinsic to the same mathematical applications.


Edited by Dubbelosix, 26 June 2019 - 03:02 AM.


#14 Dubbelosix

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Posted 25 June 2019 - 07:11 AM

Purely theoretically speaking, it is possible to obtain a front and back pressure term -

 

[math]f = \frac{2F_d}{\rho v^2\ A} = \frac{A_b}{A_f} \frac{B}{R^2_e}[/math]

 

Using the equation above, we should not forget it can be formulated also by using the definition of the drag

 

[math]f = \frac{2F_d}{\rho v^2\ A} = \frac{A_b}{A_f} \frac{B}{R^2_e}[/math]

 

as

 

[math]\frac{1}{c^2} (\frac{P}{\rho_A}) = \frac{\rho}{4 \pi^2 \rho_A} \frac{F_d}{(1 - \frac{1}{\gamma^2})}\ \frac{1}{\rho v^2 A}[/math]

and the drag force itself is related to definition as a pseudo-force:


[math]F_d = \frac{1}{2} \rho v^2\ f\ A [/math]

 

This introduces the possibility of not only having described the dragging force related to front and back area's, but an exact dimensional consequence is that it is also a front and back pressure term and recovering a possible Doppler inverse term:

 

 

 

[math]\frac{1}{c^2} (\frac{P}{\rho_A}) = \frac{\rho}{4 \pi^2 \rho_A} \frac{f}{(1 - \frac{1}{\gamma^2})}\ \frac{\rho_b}{\rho_f} (\frac{c^2}{v^2}) \frac{A_b}{A_f} = \frac{\rho}{4 \pi^2 \rho_A} \frac{1}{(1 - \frac{1}{\gamma^2})}\ \frac{\rho_b}{\rho_f} (\frac{c}{v})^2 \cdot (\frac{A_b}{A_f})\frac{B}{R^2_e}[/math]

 

This may look trivial, but it is not. Rearranging we get, in a natural unit base of [math]c=1[/math] and Doppler shift replaced by the usual notation [math]\beta = \frac{v}{c}[/math]

 

 

[math](\frac{A_b}{A_f}\frac{B}{R^2_e}\frac{P}{\rho_A}) = (\frac{\rho}{4 \pi^2 \rho_A}\frac{\rho_b}{\rho_f})\frac{f}{\beta^2 (1 - \frac{1}{\gamma^2})}\   = (\frac{\rho}{4 \pi^2 \rho_A}\frac{\rho_b}{\rho_f}) \frac{1}{\beta^2 (1 - \frac{1}{\gamma^2})}\ [/math]

 

While [math]\frac{\rho_b}{\rho_f}[/math] is dimensionless, we must keep in mind that [math]\frac{P}{\rho_A}[/math] is not, so it can be replaced by a unique notation [math]\xi[/math],

 

 

[math](\frac{A_b}{A_f}\frac{B}{R^2_e})\xi = (\frac{1}{4 \pi^2}\frac{\rho_b}{\rho_f})\frac{f}{\beta^2 (1 - \frac{1}{\gamma^2})}\ \xi   = (\frac{1}{4 \pi^2}\frac{\rho_b}{\rho_f}) \frac{1}{\beta^2 (1 - \frac{1}{\gamma^2})}\ \xi[/math]

 

And distribution of the dimensionless term gives finally

 

 

[math](\frac{A_b}{A_f} \frac{\rho_b}{\rho_f} \frac{B}{R^2_e})\xi = \frac{1}{4 \pi^2} \frac{f}{\beta^2 (1 - \frac{1}{\gamma^2})}\ \xi [/math]

 

And since [math]\frac{A_b}{A_f} \frac{\rho_b}{\rho_f} \frac{B}{R^2_e}[/math] is completely dimensionless, we can arrive at this term as a dragging coefficient primarily:

 

 

[math]\frac{1}{4 \pi^2} \frac{f}{\beta^2 (1 - \frac{1}{\gamma^2})}\ \xi  = (\frac{A_b}{A_f} \frac{\rho_b}{\rho_f} \frac{B}{R^2_e})\ \xi  [/math]

 

 

and/or

 

 

[math] \frac{f}{\beta^2 (1 - \frac{1}{\gamma^2})}\ \xi = 4 \pi^2\ (\frac{A_b}{A_f} \frac{\rho_b}{\rho_f} \frac{B}{R^2_e})\ \xi  [/math]