We can define

[math]\frac{\partial S}{\partial U_{\lambda}} = \frac{S}{k_BT}[/math]

multiplying through a factor of temperature, with entropy becoming dimensionless and taking a derivative I get:

[math]\frac{\partial^2 S}{\partial U^2_{\lambda}}\ k_B\dot{T} = \frac{\partial S}{\partial U_{\lambda}} \frac{\dot{T}}{T}[/math]

I wanted to see how that derivative might enter the main equation. The equation we formed was:

[math]\frac{\dot{R}}{R} (\dot{H} + H^2 + \frac{kc^2}{a}) = \frac{8 \pi G}{3}\ \int\ ([\frac{d}{dt} + \dot{S}_k + \dot{S}_{ik} ] \frac{\hbar c}{\frac{\hbar \omega}{k_BT} - 1} + \frac{\hbar c}{2})\ \frac{dk}{dt}\ dk^3[/math]

Letting the entropy take on the dimensions of the Boltzmann constant allows we to write in the following form, after distributing a partial derivative:

[math]\frac{\partial}{\partial U_{\lambda}}\frac{\dot{T}}{T}\ (\dot{H} + H^2 + \frac{\mathbf{k} c^2}{a}) = \frac{8 \pi G}{3}\ ([n\frac{\partial S}{\partial U_{\lambda}} + \frac{\partial \mathbf{S}_k}{\partial U_{\lambda}} + \frac{\partial \mathbf{S}_{ik}}{\partial U_{\lambda}} ] \frac{\dot{T}}{\frac{k_BT}{\hbar \omega} - 1} + \int\ \frac{\partial S}{\partial U_{\lambda}}\frac{\dot{T}}{2}\ dk^3)[/math]

Even though the derivative appears to be of first power on the left hand side, we have also shown that the second power can also be considered

[math]\frac{\partial^2 }{\partial U^2_{\lambda}}\ k_B\dot{T} = \frac{\partial }{\partial U_{\lambda}} \frac{\dot{T}}{T}[/math]

Best not to write it in explicitly but at least we know it is there. The left hand side of the equation

[math]\frac{\partial}{\partial U_{\lambda}}\frac{\dot{T}}{T}\ (\dot{H} + H^2 + \frac{\mathbf{k} c^2}{a}) = \frac{8 \pi G}{3}\ ([n\frac{\partial S}{\partial U_{\lambda}} + \frac{\partial \mathbf{S}_k}{\partial U_{\lambda}} + \frac{\partial \mathbf{S}_{ik}}{\partial U_{\lambda}} ] \frac{\dot{T}}{\frac{k_BT}{\hbar \omega} - 1} + \int\ \frac{\partial S}{\partial U_{\lambda}}\frac{\dot{T}}{2}\ dk^3)[/math]

Looks a bit messy, but keep in mind we can just write the whole thing as:

[math]\frac{\partial }{\partial U_{\lambda}}\frac{\dddot{R}}{R} = \frac{8 \pi G}{3}\ ([n\frac{\partial S}{\partial U_{\lambda}} + \frac{\partial \mathbf{S}_k}{\partial U_{\lambda}} + \frac{\partial \mathbf{S}_{ik}}{\partial U_{\lambda}} ] \frac{\dot{T}}{\frac{k_BT}{\hbar \omega} - 1} + \int\ \frac{\partial S}{\partial U_{\lambda}}\frac{\dot{T}}{2}\ dk^3)[/math]

Another form of the equation I looked at a while back considered the quantum thermal application of the system to form a condensate.

I suggest an approach to how we might come to describe the conditions required for the phase transition. A simple equation of state is

[math]\frac{d}{dt}(NV) = \dot{N}V + N \dot{V} = (\dot{N} + 3 \frac{\dot{R}}{R}N)V = NV\Gamma[/math]

Where [math]N[/math] is the particle number and [math]R[/math] is the radius of a universe and [math]\Gamma[/math] is the particle production which is also related dynamically to the fluid expansion [math]\Theta = 3\frac{\dot{a}}{a}[/math]. To find the required object to describe a condensate, we divide through by [math]N^2\lambda^3[/math] where [math]\lambda[/math] is the thermal wavelength and [math]\lambda^3[/math] will replace the role of the density term,

[math]\frac{d}{dt}(\frac{NV}{N^2 \lambda^3}) = \frac{\dot{N}V}{N^2\lambda^3} + \frac{N \dot{V}}{N^2\lambda^3} = (\frac{V}{N\lambda^3})\Gamma[/math]

In which we can measure the statistics from the interparticle distance where

[math]\frac{V}{N \lambda^3} \leq 1[/math]

In which the interparticle distance is smaller than its thermal wavelength, in which case, the system is then said to follow Bose statistics or Fermi statistics. On the other hand, when it is much larger ie.

[math]\frac{V}{N \lambda^3} >> 1[/math]

Then it will obey the Maxwell Boltzmann statistics. The latter here is classical but the former, the Bose and Fermi statistics describes a situation where classical physics are smeared out by the quantum. In this picture, it may make describe the ability to construct a condensate universe from a supercool region that existed before the big bang (the stage in which the universe began to heat up).

Using the previous information, you can construct a Friedmann equation with the necessary terms, which are attached to the effective density. The equation, in a similar non-conserved form as given in the opening post we have:

[math]\frac{\dot{R}}{R}(\frac{\ddot{R}}{R} + \frac{kc^2}{a}) = \frac{8 \pi G}{3}(\dot{\mathbf{q}}_{rev} + [(\frac{\rho V}{n \lambda^3}) + 3P_{irr}(\frac{V}{n \lambda^3})]n\Gamma)][/math]

The pre big bang phase, if it existed, would imply a supercool region obeying non-classical statistics. As the universe underwent a collapse, the liquid state of the universe changed into the radiation vapor we see today. As the universe got larger the interparticle distances clearly changes and this picture allows us to look at the universe in a completely different light. How is temperature related to motion?

The virial theorem, the potential to be clear, is also related to the thermodynamics of the system.

[math]2<k_BT> =\ - \sum^N_{k=1}\ <\mathbf{F}_k \cdot r_k>\ = n\ <V_{tot}>[/math]

And as stated before, the temperature is roughly related to the kinetic energy of the system

[math]k_B T = \frac{1}{2}\sum^N_{k=1} m_kv^2_k[/math]

and can be related to a mass tensor, with Dirac notation:

[math]= \frac{1}{2}\sum^N_{k=1} <\frac{d\mathbf{q}_k}{dt}|\mathbf{M}| \frac{d\mathbf{q}}{dt}>[/math]

using generalized coordinates. Above we saw that the temperature was twice the magntitude of the potential, it is also true the temperature related to generalized coordinates as:

[math]2k_BT = \mathbf{p} \cdot \dot{\mathbf{q}} = m(\frac{ds}{dt})^2[/math]

with [math]ds[/math] as a metric term and this relationship is well-known under the Maupertuis principle.

**Edited by Dubbelosix, 17 January 2019 - 08:24 PM.**