[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]

Rearrange

[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]

Rearrange

[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.**