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Friedmann Equation And The Black Hole Inequality

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#1 Dubbelosix



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Posted 04 September 2019 - 06:34 AM

The black hole inequality is probably a universal relationship to the fundamental systems of particles - so long as it has a charge and an angular momentum to speak of. The inequality is

[math]Q^2 + J^2 \geq m^2[/math]

You can make this inequality part of the Friedmann equations - I noticed this from a dimensional analysis involving the corrections to the inequality for the charge squared on the system (important for Regge trajectories as well, and perhaps welcoming for those who think the universe is literally a black hole):

[math]\frac{4 \pi}{3}(Q^2 + Jc) \geq \frac{4 \pi Gm^2}{3}[/math]

Notice now, with units restored, a Friedmann equation takes the form of

[math]\frac{\ddot{R}}{R} = -\frac{4 \pi G}{3}\rho[/math]


[math]\ddot{R} = -\frac{4 \pi G R}{3}\rho[/math]

this has units of acceleration and distributing one factor of the Hubble radius:

[math]\dot{R}^2 = -\frac{4 \pi G R^2}{3}\rho [/math]

This has units of velocity squared - let's treat the effective density parameters with units of energy:

[math]\dot{R}^2 = -\frac{4 \pi G R^2}{3c^2}\rho[/math]

This has units of velocity squared.

The appearance of the pressure is actually related to continuity, but as shown from previous work, the continuity is obtained from an extra derivative:

[math]2 \dot{R}\ddot{R} = -\frac{4 \pi G R^2}{3c^2}\dot{\rho}[/math]

and the continuity equation is

[math]\dot{\rho} = (\rho + 3P)\frac{\dot{a}}{a}[/math]

in which [math]a[/math] is the scale factor. Plugging it in, we have a non-conserved form of the Friedmann equations, so long as the effective density parameters are not equal to zero - in previous investigations, there were neat ways to do this, such as a non-conserved phase of particle production in early cosmology. Also from previous investigations, we also worked out the Langrangian for the Friedmann equation ~

Without the curvature component just for now, the Friedmann equation is

[math](\frac{\dot{R}}{R})^2 = \frac{8 \pi G}{3}\rho + \frac{\Lambda c^2}{3}[/math]


[math]\dot{R}^2 = \frac{8 \pi GR^2}{3}\rho + \frac{\Lambda c^2}{3}R^2[/math]

Putting all terms on the left hand side, after distributing a mass term, we get a Langrangian,

[math]m\dot{R}^2 - \frac{8 \pi GmR^2}{3}\rho + \frac{\Lambda mc^2}{3}R^2 = \mathcal{L}[/math]

How did I know this was a Langrangian? Simply because there is already a standard Langrangian out there and it has identical terms

[math]\mathcal{L}(R,\dot{R}) = T - U = \frac{1}{2}m\dot{R}^2 + \frac{GM^2}{R} + \frac{mc^2}{6} \Lambda R^2[/math]

To make the numerical part all one needs to do is distribute a factor of 1/2 on the LHS

[math]\frac{1}{2}m\dot{R}^2 - \frac{8 \pi GmR^2}{6}\rho + \frac{\Lambda mc^2}{6}R^2 = \mathcal{L}[/math]

The black hole inequality can also satisfy the parameters as a charge equation:

[math]mR \dot{R} = \frac{4 \pi}{3}(Q^2 + Jc) \geq \frac{4 \pi Gm^2}{3}[/math]

A force equation is (after changing some numerical coefficients is):

[math]F = \frac{8 \pi}{3}(\frac{Q^2}{R^2} + \frac{Jc}{R^2}) \geq \frac{8 \pi Gm^2}{3R^2}[/math]

And we recognize a pressure as well

[math]P = \frac{8 \pi}{3}(\frac{Q^2}{R^4} + \frac{Jc}{R^4}) \geq \frac{8 \pi Gm^2}{3R^4}[/math]

And a simple energy equation is

[math]E = \frac{ 8 \pi R^3}{3}(\frac{Q^2}{R^4} + \frac{Jc}{R^4}) \geq \frac{8 \pi Gm^2}{3R}[/math]

Recognizing that the scalar field is

[math] \phi = \frac{8 \pi G}{3c^2}(\frac{Q^2}{R^2} + \frac{Jc}{R^2} )[/math]

Then we also have an unexpected dimensionless parameter:

[math]\frac{F}{F_P} = \frac{\phi}{c^2} = \frac{8 \pi G}{3c^4} (\frac{Q^2}{R^2} + \frac{Jc}{R^2}) \geq \frac{8 \pi G}{3c^4} (\frac{Gm^2}{R^2})[/math]

With [math]F_P[/math] being the upper limit of the classical gravitational field (Planck force).

We'll leave it here for now, because later I want to get back to the pressure equation

[math]P = \frac{8 \pi}{3}(\frac{Q^2}{R^4} + \frac{Jc}{R^4}) \geq \frac{8 \pi Gm^2}{3R^4}[/math]

And treat it as the effective parameters in the Friedmann equation:

[math]\ddot{R} = -\frac{4 \pi G R}{3}(\rho + \frac{3P}{c^2})[/math]

Edited by Dubbelosix, 24 October 2019 - 01:09 PM.

#2 Dubbelosix



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Posted 07 September 2019 - 08:54 AM

Ok so we want to continue now with this. First I would like readers to entertain the paper: https://arxiv.org/abs/1202.1275 as it may prove useful for any understanding special mathematical parameters of black holes. I will also highlight that the stress energy tensor is related to density and pressure parameters:


[math]T^{00} = g^{00}(\rho + P_0 - P) = (\rho_m + \frac{P_0 - P}{c^2})v^0v^0 = (\rho_m + \frac{q}{c^2})v^0v^0[/math]


And the pressure equation we will use later is:


[math]P = \frac{8 \pi}{3}(\frac{Q^2}{R^4} + \frac{Jc}{R^4}) \geq \frac{8 \pi Gm^2}{3R^4}[/math]


Just to keep in mind, the [math]J[/math] represents the total angular momentum, meaning it is described in two parts:


[math]J = L + S[/math]


In which [math]S[/math] will be reserved for any Poincare symmetries involving cosmological rotation and the [math]L[/math] could be taken as an angular momentum of systems inside of it (such as galaxies). We'll take this back up soon as I do not have much time today to really make a go of this. 

Edited by Dubbelosix, 07 September 2019 - 08:54 AM.

#3 Dubbelosix



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Posted 10 September 2019 - 07:10 AM

[math]m\dot{R}^2 - \frac{8 \pi GmR^2}{3}\rho + \frac{\Lambda mc^2}{3}R^2 = \mathcal{L}[/math]


The Langrangian (which involves a not too descriptive parameter of a cosmological constant), has with it, an additive expression of [math]\frac{Gm^2}{R}[/math] which is an approxmate quantity oberying an inequality rule of the following equation:


[math]\mathcal{L}(R,\dot{R}) = T - U = \frac{1}{2}m\dot{R}^2 + \frac{Gm^2}{R} + \frac{mc^2}{6} \Lambda R^2[/math]


[math]P = \frac{8 \pi}{3}(\frac{Q^2}{R^4} + \frac{Jc}{R^4}) \geq \frac{8 \pi Gm^2}{3R^4}[/math]


For those wondering what charge or even angular momentum has to do as effective pressure parameters isn't so outside the realms of physics since such studies have shown relationships of centrigues with that of pressure. Since charge and mass have undeniable relationships especially within the presence of spin dictating other electromagnetic contributions to mass would indicate that a spinning stationary mass still experiences a magnetic moment, including other electromagnetic charge contributions. The motion of heavy particles in a fluid is related to a pressure known as the kinetic pressure: 


[math]q = \frac{1}{2}\rho v^2[/math]


It should be noted, using [math]q[/math] is a well-used symbol alternatively used when describing pressure: In any equation involving pressure, a relativistic coerrection for the total pressure involves for instance (using a bold notation) 


[math]\mathbf{q} = P_0 - P[/math]


Which features for further example in the stress energy and momentum part of the Friedmann equation:


[math]T^{00} = g^{00}(\rho + P_0 - P) = (\rho_m + \frac{P_0 - P}{c^2})v^0v^0 = (\rho_m + \frac{q}{c^2})v^0v^0[/math]


Concerning [math]J[/math] in:


[math]P = \frac{8 \pi}{3}(\frac{Q^2}{R^4} + \frac{Jc}{R^4}) \geq \frac{8 \pi Gm^2}{3R^4}[/math]


It also, let's remind ourselves consists of two parts:


[math]J = L + S[/math]


From here, I would like to continue a related line of work - while being an alternative to ordinary geometric theory of relativity, it has surpassed it in some ways by unveiling many unseen mathematical consequences under the scrutiny of geometric algebra. My own discovery came under a relatization that the torsion part could not be ''done away'' in any ad hoc manner as if often done under general theory of Einstein's work. If in our cosmological work, [math]L_{galaxies}[/math] refers to the angular momentum of galaxies inside of the universe, then it is argued this is the only observable part of [math]S[/math] (excluding) background axis of evil which is thought to have since dissipated under the consequence of an exponentially decaying primordial spin [math]S[/math]. This real and imaginary part could be serviced under a bivector description of the total angular momentum density under geometric algebra:


[math]\mathbf{J} = (\mathbf{L} + i\mathbf{S})\gamma_0[/math]


I'd also like to take the stress energy 


[math]T^{00} = g^{00}(\rho + P_0 - P) = (\rho_m + \frac{P_0 - P}{c^2})v^0v^0 = (\rho_m + \frac{q}{c^2})v^0v^0[/math]


In the style of the gravitational pseudo-field description of either the hydrodynamic time derivative:


[math]\frac{d}{dt} = \frac{\partial}{\partial t} + v \cdot \nabla[/math]


or possibly weighted by an inverse Planck pseudo- force under the normal field equations


[math]\mathbf{G}_{\mu \nu} = \frac{ 8 \pi G}{c^4} T_{\mu \nu} = \partial_{\mu} \cdot \mathbf{D}_{\nu} + i \sigma \cdot (\Gamma_{\mu} \times \mathbf{D}_{\nu})[/math]