How Might Planck Star Physics Integrate Into The Equations Of Motion For A Universe?

[Dubbelosix]

It was suggested by the original authors that gravitational quantum pressure can explain why a universe may not collapse to a point.

 

The only reason why repulsive quantum gravitational effects where not considered in my recent approach to the Friedmann equation, is because torsion is capable of avoiding singularities. In this work, we'll consider a Friedmann cosmology without it and consider Planck Star physics within its framework. For me, this is only a fun exercise, I don't really see this being the solution avoided singularities, in my own work, torsion does this. 

 

The red shift is related to the scale factor ratio as

 

[math]1 + z = \frac{a(t_0)}{a(t_e)} = \frac{dt_0}{dt_e}[/math]

 

The Schwarzschild relationship of proper time is

 

[math]\Delta \tau = (\frac{Gm}{c^2 r})^{\frac{1}{2}} \Delta t[/math]

 

The red shift is related to the scale factor ratio as

 

[math]1 + z = \frac{a(t_0)}{a(t_e)} = \frac{dt_0}{dt_e} = \frac{\lambda(t_0)}{\lambda(t_e)} = \frac{2Gm}{c^2R}[/math]

 

The quantum gravitational repuslive force arises from a correction on the redshift as follows, in the natural units now

 

[math]1 - \frac{2mR^2}{R^3 + \alpha^2 m}[/math]

 

Using the procedure by Carlo Rovelli, you expand in [math]\frac{1}{R}[/math], 

 

[math]\frac{2m}{R} + \frac{4 \alpha^2 m^2}{r^4}[/math]

 

The simple version of a Friedmann equation is

 

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

 

The scale factor ratio is equal to the radius ratio

 

[math]\frac{a(t_0)}{a(t_e)} = \frac{R(t_0)}{R(t_e)}[/math]

 

So you can also write the left hand side as

 

[math](\frac{\dot{a}(t_0)}{a(t_e)})^2 = \frac{8 \pi G}{3}\rho + \frac{\Lambda c^2}{3}[/math]

 

Using 

 

[math]\dot{a}^2 = a\ddot{a}[/math]

 

we get

 

[math]\frac{a(t_0)}{a(t_e)} \frac{\ddot{a}(t_0)}{a(t_e)} = \frac{8 \pi G}{3}\rho + \frac{\Lambda c^2}{3}[/math]

 

Replacing now for our definition of the redshift we get

 

[math](1 + z) \frac{\ddot{a}(t_0)}{a(t_e)} = \frac{8 \pi G}{3}\rho + \frac{\Lambda c^2}{3}[/math]

 

alternatively it can be just written as (ignoring the Cosmological Constant),

 

[math]\frac{2Gm}{c^2R} \frac{\ddot{a}(t_0)}{a(t_e)} = \frac{8 \pi G}{3}\rho[/math]

 

Using now natural units with [math]G = c = 1[/math] we get

 

[math]\frac{2m}{R} \frac{\ddot{a}(t_0)}{a(t_e)} = \frac{\rho}{3}[/math]

 

Expanding the same terms, 

 

[math](\frac{2m}{R} + \frac{4 \alpha^2 m^2}{r^4})\frac{\ddot{a}(t_0)}{a(t_e)} = \frac{\rho}{3}[/math]

 

 

[math]\frac{\ddot{a}(t_0)}{a(t_e)} = \frac{\rho}{3}\frac{1}{(\frac{2m}{R} + \frac{4 \alpha^2 m^2}{r^4})} = \frac{\rho}{3(\frac{2m}{R} + \frac{4 \alpha^2 m^2}{r^4})}[/math]

 

Working in natural units is supposed to be the scientific method for many papers, for me, it just makes an equation look bare. The extra dynamics suppressed are (with CC)

 

 

[math]\frac{\ddot{a}(t_0)}{a(t_e)} = \frac{8 \pi G}{3}\frac{\rho}{(\frac{2Gm}{c^2R} + \frac{4 \alpha^2 G^2m^2}{c^4r^4})} + \frac{\Lambda c^2}{3}[/math]

Edited April 5, 2017 by Dubbelosix

[exchemist]

 

It was suggested by the original authors that gravitational quantum pressure can explain why a universe may not collapse to a point.

 

The only reason why repulsive quantum gravitational effects where not considered in my recent approach to the Friedmann equation, is because torsion is capable of avoiding singularities. In this work, we'll consider a Friedmann cosmology without it and consider Planck Star physics within its framework. For me, this is only a fun exercise, I don't really see this being the solution avoided singularities, in my own work, torsion does this. 

 

The red shift is related to the scale factor ratio as

 

[math]1 + z = \frac{a(t_0)}{a(t_e)} = \frac{dt_0}{dt_e}[/math]

 

The Schwarzschild relationship of proper time is

 

[math]\Delta \tau = (\frac{Gm}{c^2 r})^{\frac{1}{2}} \Delta t[/math]

 

The red shift is related to the scale factor ratio as

 

[math]1 + z = \frac{a(t_0)}{a(t_e)} = \frac{dt_0}{dt_e} = \frac{\lambda(t_0)}{\lambda(t_e)} = \frac{2Gm}{c^2R}[/math]

 

The quantum gravitational repuslive force arises from a correction on the redshift as follows, in the natural units now

 

[math]1 - \frac{2mR^2}{R^3 + \alpha^2 m}[/math]

 

Using the procedure by Carlo Rovelli, you expand in [math]\frac{1}{R}[/math], 

 

[math]\frac{2m}{R} + \frac{4 \alpha^2 m^2}{r^4}[/math]

 

The simple version of a Friedmann equation is

 

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

 

The scale factor ratio is equal to the radius ratio

 

[math]\frac{a(t_0)}{a(t_e)} = \frac{R(t_0)}{R(t_e)}[/math]

 

So you can also write the left hand side as

 

[math](\frac{\dot{a}(t_0)}{a(t_e)})^2 = \frac{8 \pi G}{3}\rho + \frac{\Lambda c^2}{3}[/math]

 

Using 

 

[math]\dot{a}^2 = a\ddot{a}[/math]

 

we get

 

[math]\frac{a(t_0)}{a(t_e)} \frac{\ddot{a}(t_0)}{a(t_e)} = \frac{8 \pi G}{3}\rho + \frac{\Lambda c^2}{3}[/math]

 

Replacing now for our definition of the redshift we get

 

[math](1 + z) \frac{\ddot{a}(t_0)}{a(t_e)} = \frac{8 \pi G}{3}\rho + \frac{\Lambda c^2}{3}[/math]

 

alternatively it can be just written as (ignoring the Cosmological Constant),

 

[math]\frac{2Gm}{c^2R} \frac{\ddot{a}(t_0)}{a(t_e)} = \frac{8 \pi G}{3}\rho[/math]

 

Using now natural units with [math]G = c = 1[/math] we get

 

[math]\frac{2m}{R} \frac{\ddot{a}(t_0)}{a(t_e)} = \frac{\rho}{3}[/math]

 

Expanding the same terms, 

 

[math](\frac{2m}{R} + \frac{4 \alpha^2 m^2}{r^4})\frac{\ddot{a}(t_0)}{a(t_e)} = \frac{\rho}{3}[/math]

 

 

[math]\frac{\ddot{a}(t_0)}{a(t_e)} = \frac{\rho}{3}\frac{1}{(\frac{2m}{R} + \frac{4 \alpha^2 m^2}{r^4})} = \frac{\rho}{3(\frac{2m}{R} + \frac{4 \alpha^2 m^2}{r^4})}[/math]

 

Working in natural units is supposed to be the scientific method for many papers, for me, it just makes an equation look bare. The extra dynamics suppressed are (with CC)

 

 

[math]\frac{\ddot{a}(t_0)}{a(t_e)} = \frac{8 \pi G}{3}\frac{\rho}{(\frac{2Gm}{c^2R} + \frac{4 \alpha^2 G^2m^2}{c^4r^4})} + \frac{\Lambda c^2}{3}[/math]

 

Original authors of what?  This reads like a section cut and pasted from some longer article. If you are quoting you should say from whom.