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## Miscellaneous calculations, Friedmann equation

### Miscellaneous calculations, Friedmann equation

Some miscellaneous equations on the Friedmann cosmology.

I investigated the phase transition of the energy in Friedmann cosmology arising as a power equation:

$m \dot{R}\ddot{R} - \frac{8 \pi Gm R^2}{6}\dot{\rho} = \mathbf{P}$

Non-conservation it seemed, just appeared from a time derivative of the Friedmann equation and from there the continuity equation needs a new definition. The continuity has relationships to the first law of thermodynamics:

$dQ = dE + \rho dV = k_BTdS$

Variations in density can be achieved by density perturbations in early cosmology symbolized by

$\delta(x, t) = \frac{\rho(x,t) - \rho_{dia}}{\rho_{dia}}$

Moving on, the variations in density we will use will be how they vary with time including the energy content of the universe, we will explore possible ways (solutions) to the non-conserving model. The curvature part of the Friedmann equation can be written as:

$kc^2 = -\frac{2U}{mx^2}$

In which the Friedmann equation itself is given as

$(\frac{\dot{R}}{R})^2 = \frac{8 \pi G}{3}\rho - \frac{kc^2}{a^2}$

replacing we get (by using $R = ax(t)$

$(\frac{\dot{R}}{R})^2 = \frac{8 \pi G}{3}\rho + \frac{2U}{mR^2}$

Multiplying through by $mR^2$ we can see how the Friedmann equation at the very core is just a statements about the energy of the universe:

$m\dot{R}^2 = \frac{8 \pi Gm}{3}\rho + 2U$

To achieve the non-conserved form, we just take the derivative on both sides:

$m\dot{R}\ddot{R} = \frac{8 \pi Gm}{3}\dot{\rho} + \dot{U}$

Divide through by $mR^2$ retrieves a form close to the original Freidmann equation, except it is for non-conservation

$\frac{\dot{R}}{R}\frac{\ddot{R}}{R} = \frac{8 \pi G}{3}\dot{\rho} + \frac{\dot{U}}{mR^2}$

It's helpful to keep in mind the following relationships:

$\frac{\ddot{R}}{R}= \dot{H} + H^2$

$H = \frac{\dot{R}}{R}$

Another version, differentiating:

$(\frac{\dot{R}}{R})^2 = \frac{8 \pi G}{3}\rho + \frac{2U}{mR^2}$

We can give it as

$2 \frac{\dot{R}}{R}(\frac{R \ddot{R} - \dot{R}^2}{R^2}) = \frac{8 \pi G}{3} \dot{\rho} + \frac{2\dot{U}}{mR^2}$

With variations in the density term and the energy term in the last expression will both meaure how density and energy varies in the evolution of the universe. Going back to this form now

$\frac{\dot{R}}{R}\frac{\ddot{R}}{R} = \frac{8 \pi G}{3}\dot{\rho} + \frac{\dot{U}}{mR^2}$

inverting through $\frac{R}{\dot{R}}$ and using the following fluid equation (which describes how density changes in the universe)

$\frac{\ddot{R}}{R} = \frac{8 \pi G}{3}\dot{\rho}\frac{R}{\dot{R}} + \frac{\dot{U}}{mR^2}\frac{R}{\dot{R}}$

$\dot{\rho}\frac{R}{\dot{R}} = ( \rho + 3P)$

we have

$\frac{\ddot{R}}{R} = \frac{8 \pi G}{3}( \rho + \frac{3P}{c^2}) + \frac{\dot{U}}{mR^2}\frac{R}{\dot{R}}$

(Note in this form, it seems to be the last term that contributes to the variation of energy in the universe).

The energy may also be seen as

$U = \frac{4 \pi}{3}R^3\epsilon c^2$

and

$dU = \frac{4 \pi}{3}R^2\epsilon c^2 \frac{dR}{dt} + \frac{4 \pi}{3}R^3 \frac{d\epsilon}{dt} c^2$

Where epsilon is just some other density term related to the energy. Using the previous form

$\frac{\dot{R}}{R}\frac{\ddot{R}}{R} = \frac{8 \pi G}{3}\dot{\rho} + \frac{\dot{U}}{mR^2}$

It is strange though, that the non-conserved form can be achieved very easily using a form of the fluid equation (which itself is a statement about conservation),

$\frac{\ddot{R}}{R} = \frac{8 \pi G}{3c^2}(\rho + 3P)$

Using

$\dot{\rho}\frac{R}{\dot{R}} = (\rho + 3P)$

We obtain after substitution

$\frac{\ddot{R}}{R} = \frac{8 \pi G}{3c^2}\dot{\rho}\frac{R}{\dot{R}}$

$\frac{\dot{R}}{R}\frac{\ddot{R}}{R} = \frac{8 \pi G}{3c^2}\dot{\rho}$

Which is the non-conserved form if and only if:

$\dot{\rho} + 3(\frac{\dot{R}}{R})( \rho + P) \ne 0$

The fluid equation does indeed describe the evolution of the density parameters in the universe but is only a statement about conservation if it satisfies

$\dot{\rho} + 3(\frac{\dot{R}}{R})( \rho + P) = 0$

The first term in the fluid equation describes how density dilutes over time. Second term describes loss of energy because the pressure has done work as the universes volume increases. The equation is either incomplete, or incompatible as a conservation equation satisfying only the $\ne 0$-solution.
GarethMeredith

Posts: 100
Joined: Tue Aug 02, 2016 4:13 am

### Re: Miscellaneous calculations, Friedmann equation

Non-conservation for a universe is not ad hoc, it is required as a more general assumption about nature than what was available to Newton using the then current model.

miscellaneous-calculations-friedmann-equation-t1275.html

Sean explains, concerning the momentum tensor

$\nabla_{\mu} T^{\mu \nu} = 0$

The details aren’t important, but the meaning of this equation is straightforward enough: energy and momentum evolve in a precisely specified way in response to the behavior of spacetime around them. If that spacetime is standing completely still, the total energy is constant; if it’s evolving, the energy changes in a completely unambiguous way.
GarethMeredith

Posts: 100
Joined: Tue Aug 02, 2016 4:13 am

### Re: Miscellaneous calculations, Friedmann equation

As was explained previously, the first law of thermodynamics is intimately related to the derivation of the continuity equation (aka. equation of state, fluid equation). The fluid equation in the context used in modern theory, uses the adiabatic system. The adiabatic representation of the first law is

$dQ = dW + dU$

$dU = -dW$

The diabatic formulation of this has been given in history as:

$\dot{E} = \dot{W} - \dot{U} + \dot{m}(u_1 + \frac{V^2_1}{2} + gz_1) - \dot{m}(u_1 + \frac{V^2_2}{2} + gz_2)$

It is unlikely this will be the form we will want for cosmology, as you don't need an outside to the universe to actually change the energy content, that comes from field theory and how we measure the metric to change over time.

However, with that said, we can see that to get the variation in the energy $\dot{E}$ we had to assume a third component:

$\dot{m}(u_1 + \frac{V^2_1}{2} + gz_1) - \dot{m}(u_1 + \frac{V^2_2}{2} + gz_2)$

Which explained those dynamics of why energy was changing, we need to something similar in a new formulation for the new wedding of physics with relativity under the examination of non-de Sitter space. We can see though, it is just a power equation in the end of the day, just like the Friedmann power equation derived before.

What this shows is that the first law requires an additional component for the diabatic process, to simply explain it.

http://www.learnengineering.org/2013/03 ... ystem.html
Last edited by GarethMeredith on Mon Mar 20, 2017 5:40 am, edited 1 time in total.
GarethMeredith

Posts: 100
Joined: Tue Aug 02, 2016 4:13 am

### Re: Miscellaneous calculations, Friedmann equation

It can be argued that the rotation term might enter the first law of thermodynamics so that the statement of energy in a universe (may be) the following:

$dE_{total} = dW + dU + dmR^2 \omega^2$

or as

$\dot{E} = \dot{W} + \dot{U} + \dot{U}_R$

$U_R$ being the rotational energy of course.

where

$c^2 = R^2 \omega^2$

It should also be noted that $mR^2$ is also the rotational inertia. As shown we needed an extra term in the conservation equation to explain the diabatic physics. Perhaps rotation was such a term, because early in the investigation, it was found that rotation would lower the Friedmann energy levels of a universe: This can be understood in the following way: It costs energy to rotate a thing, if there is no outside to the universe, then there has to be a bulk energy transfer (possibly to the horizon) where rotational energy might be stored. If this is true, there has been paper written to try and explain similar bulk transfer models of cosmology, https://arxiv.org/pdf/1208.2482.pdf

In this paper, there are no initial modifications or extra terms to explain the diabatics. It's only stated that for diabatics, $dQ \ne 0$. Is this enough? I don't think it is, while it is true the diabatic law states that $dQ$ is not zero, it hasn't explained why in thermodynamic equation why this should be imposed. In the open system thermodynamic law case, we have an extra term that explains the Reynolds transport into and out of a system (hence imposing) the condition that internal energy is truly changing. This is why rotation in my model can provide the crucial component in the thermodynamic law if and only if the rotation in the early universe is indeed taking energy from the primordial bulk.

The transfer of the vacuum bulk energy to the rotational properties could explain the cosmological vacuum energy problem. The thinning of the vacuum energy coupled with rotation (acting like inflation but is really a centrifugal force field of the universe) could have been enough to take us out of the dense Planck phase.

http://cbc.arizona.edu/~salzmanr/480b/s ... att02.html

http://auworkshop.autodesk.com/library/ ... pen-system
GarethMeredith

Posts: 100
Joined: Tue Aug 02, 2016 4:13 am

### Re: Miscellaneous calculations, Friedmann equation

Using a new definition of how rotational velocity change effects bulk energy we have

$\Delta \dot{E} = \dot{W} + \dot{U} + \dot{m}R^2(\omega^2_2 - \omega^2_1)$

$R^2(\omega^2_1 - \omega^2_2) = \Delta v^2$

where $v$ is the velocity of angular rotation.

$\dot{E}_{bulk} = \dot{W} + \dot{U} + \dot{m}R^2 \omega^2$

$\Delta \dot{E}_{bulk} = \dot{W} + \dot{U} + (\dot{m}R^2 \omega^2_2 - \dot{m}R^2 \omega^2_1)$

As a modified first temperature law for decaying rotational properties, implying a change in the bulk energy where

$\Delta (\dot{m}R^2 \omega^2) = \Delta(\dot{m} v^2) = \dot{m}R^2 \omega^2_2 - \dot{m}R^2 \omega^2_1$

Extra term, with varying rotational velocity (just as was shown by Hoyle and Narliker, velocity is allowed to change and does exponentially decay with time). Seems natural in this formulation to assume that the change in bulk energy arises also from the dynamics of a rotational velocity which is itself not a conserved quantity.

But even still, it should be noted, this equation:

$\Delta \dot{E}_{bulk} = \dot{W} + \dot{U} + \dot{m}R^2(\omega^2_2 - \omega^2_1)$

Is still not enough because in the limit of $\omega \rightarrow 0$ we end up with simply with the first law again

$\dot{E}_{bulk} = \dot{W} + \dot{U}$

in which again, we have to imply $\dot{E}_{bulk} \ne 0$ because of the effects of general relativity and the expansion of space. So while this is a naive first attempt at a look at how rotation and the bulk energy in a first law set-up effect each other, there is clearly more physics needed to explain dynamic changes after the rotation era.
GarethMeredith

Posts: 100
Joined: Tue Aug 02, 2016 4:13 am

### Re: Miscellaneous calculations, Friedmann equation

$m \dot{R}\ddot{R} - \frac{8 \pi Gm R^2}{6}\dot{\rho} = \mathcal{P}$

$\dot{\rho} = \frac{\dot{R}}{R}(\rho + 3P)$

We obtain after substitution

$m \dot{R}\ddot{R} - \frac{8 \pi Gm R^2}{6}\frac{\dot{R}}{R}(\rho + 3P) = \mathcal{P}$

We only have 6 in the denominator, because of how this specific equation was derived, but since they are adjustable parameters, it's probably ok to give it as

$m \dot{R}\ddot{R} - \frac{8 \pi Gm R^2}{3}\frac{\dot{R}}{R}(\rho + 3P) = \mathcal{P}$

In the model we search for, the expansion of spacetime leads to variations in the metric energy - appearance of new spacetime corresponds to new energy and we might think of it as a type of particle non-conservation. We can describe that in the following way:

Using the first law, we can state

$dE = dQ - PdV + (\frac{\rho + P}{n}) dN$

and

$\dot{E} = \dot{Q} - P\dot{V} + (\frac{\rho + P}{n}) \dot{N}$

$n$ is the number density $\frac{N}{V}$ and $\rho$ is an energy density. This also induces the diabatic form nature of the universe. We don't want the equation above to be devoid of the density meaning, so we just insert $n$ in replace of $N$ and this retrieves the density component and assuming I have done this right, inserting a factor of $c^2$ in the denominator of the second term to account that we are dealing with an energy density, I get:

$m \dot{R}\ddot{R} - \frac{8 \pi Gm R^2}{6c^2}(\frac{\rho + 3P}{n})\dot{n} = \mathcal{P}$

By introducing the heat per unit particle $d\bar{q} = \frac{dQ}{dN}$ the above component reduces to a Gibbs expression from the Gibbs equation. The process of particle creation in the sense above can be thought of being induced gravitationally (gravitational particle production) and in fact, the particle number coupled to the gravitational field looks like;

$N^{\mu}_{:\mu} \equiv n \sigma \Gamma \equiv s \Gamma \ne 0$

https://arxiv.org/pdf/1512.03100.pdf
GarethMeredith

Posts: 100
Joined: Tue Aug 02, 2016 4:13 am

### Re: Miscellaneous calculations, Friedmann equation

GarethMeredith wrote:.... his right, inserting a factor of $c^2$ in the denominator of the second term to account that we are dealing with an energy density, I get:

$m \dot{R}\ddot{R} - \frac{8 \pi Gm R^2}{6c^2}(\frac{\rho + 3P}{n})\dot{n} = \mathcal{P}$

By introducing the heat per unit particle $d\bar{q} = \frac{dQ}{dN}$ the above component reduces to a Gibbs expression from the Gibbs equation. The process of particle creation in the sense above can be thought of being induced gravitationally (gravitational particle production) and in fact, the particle number coupled to the gravitational field looks like;

$N^{\mu}_{:\mu} \equiv n \sigma \Gamma \equiv s \Gamma \ne 0$

The $\Gamma$ symbol here is actually known as the particle production rate and for a non-conserved universe, $n\Gamma$ has a non-zero value. For the adiabatic universe, particle production rate is simply

$\dot{n} + \Theta n = 0$

In our case we have

$\dot{n} + \Theta n = n\Gamma$

and Theta represents

$\Theta^{\mu}_{:\mu} = 3H = 3 \frac{\dot{a}}{a}$

denotes the fluid expansion

https://arxiv.org/pdf/1512.03100.pdf

http://cds.cern.ch/record/405307/files/9910483.pdf
GarethMeredith

Posts: 100
Joined: Tue Aug 02, 2016 4:13 am

### Re: Miscellaneous calculations, Friedmann equation

$\frac{\dot{R}}{R}\frac{\ddot{R}}{R} - \frac{8 \pi G}{3c^2}(\frac{\rho + 3P}{n})\dot{n} = \frac{2\dot{U}}{mR^2}$

and using

$\frac{kc^2}{a} = - \frac{2U}{mR^2}$

You can get back the recognizable form of the non-conserved Friedmann equation,

$\frac{\dot{R}}{R}\frac{\ddot{R}}{R} - \frac{8 \pi G}{3c^2}(\frac{\rho + 3P}{n})\dot{n} = - \frac{kc^2}{a^2}\frac{\dot{R}}{R}$

rearrange

$\frac{\dot{R}}{R}\frac{\ddot{R}}{R} + \frac{kc^2}{a^2}\frac{\dot{R}}{R} = \frac{8 \pi G}{3c^2}(\frac{\rho + 3P}{n})\dot{n}$

and of course

$\frac{\dot{R}}{R}(\frac{\ddot{R}}{R} + \frac{kc^2}{a^2}) = \frac{8 \pi G}{3c^2}(\frac{\rho + 3P}{n})\dot{n}$
GarethMeredith

Posts: 100
Joined: Tue Aug 02, 2016 4:13 am

### Re: Miscellaneous calculations, Friedmann equation

sorry fixed that last typo, was supposed to be an addition.
GarethMeredith

Posts: 100
Joined: Tue Aug 02, 2016 4:13 am

### Re: Miscellaneous calculations, Friedmann equation

To get back a recognizable form of the Friedmann equation, just multiply through by $mR^2$ and write it in the form, using first,

$kc^2 = - \frac{2U}{mx^2}$

Divide through by $a^2$

$\frac{kc^2}{a^2} = - \frac{2U}{mR^2}$

where $R = ax$

and

$\frac{kc^2}{a^2}\frac{\dot{R}}{R} = - \frac{2\dot{U}}{mR^2}$

and after some more rearranging, we can end up with an equation (which is my preferred form, now with only scale factors since $\frac{\dot{a}}{a} = \frac{\dot{R}}{R}$)

$\frac{\dot{a}}{a}(\frac{\ddot{a}}{a} + \frac{kc^2}{a^2}) = \frac{8 \pi G}{3c^2}(\frac{\rho + 3P}{n})\dot{n}$

and since

$\dot{H} + H^2 = \frac{\ddot{a}}{a} = \frac{\ddot{R}}{R}$

then it can also be written in the following way

$\dot{H}\frac{\dot{a}}{a} + H^2\frac{\dot{a}}{a} + \frac{kc^2}{a^2}\frac{\dot{a}}{a} = \frac{8 \pi G}{3c^2}(\frac{\rho + 3P}{n})\dot{n}$

Now, there is another technical detail that was cancelled out, very early on and the time derivative went to the particle creation, before it disappeared, we had an equation (written with scale factors)

$m \dot{a}\ddot{a} - \frac{8 \pi Gm R^2}{6}\frac{\dot{a}}{a}(\rho + 3P) = \mathcal{P}$

When the effective density is rewritten in terms of the particle production energy, you can in fact keep the definition of $\frac{\dot{a}}{a}$ with the particle creation number, by recognizing the following: The fluid expansion is given as:

$\frac{1}{3}\Theta^{\mu}_{:\mu} = \frac{\dot{a}}{a}$

and

$\dot{n} + \Theta n = \dot{n} + n\frac{\dot{a}}{a} = n \Gamma$

and so

$n\frac{\dot{a}}{a} = -\dot{n} + n \Gamma$

Rewriting the previous equations then, this time without a vanishing $\frac{a}{a}$ factor we have

$\dot{H}\frac{\dot{a}}{a} + H^2\frac{\dot{a}}{a} + \frac{kc^2}{a^2}\frac{\dot{a}}{a} = \frac{8 \pi G}{3c^2}(\frac{\rho + 3P}{n})n\frac{\dot{a}}{a}$

and plugging in the previous definition we get a new representation of the equation

$\dot{H}\frac{\dot{a}}{a} + H^2\frac{\dot{a}}{a} + \frac{kc^2}{a^2}\frac{\dot{a}}{a} = \frac{8 \pi G}{6c^2}(\frac{\rho + 3P}{n})\Theta n$

and

$\frac{\dot{a}}{a}(\frac{\ddot{a}}{a} + \frac{kc^2}{a^2})= \frac{8 \pi G}{6c^2}(\frac{\rho + 3P}{n})\Theta n$

So there is this representation as well, making use of simple relationships. Writing it in this form again.

$\dot{H}\frac{\dot{a}}{a} + H^2\frac{\dot{a}}{a} + \frac{kc^2}{a^2}\frac{\dot{a}}{a} = \frac{8 \pi G}{3c^2}(\frac{\rho + 3P}{n})n\frac{\dot{a}}{a}$

and putting everything to the left hand side we can see how every term shares this fluid expansion coefficient

$\frac{\Theta}{3}(\dot{H} + H^2 + \frac{kc^2}{a^2} - \frac{8 \pi G}{3c^2}(\frac{\rho + 3P}{n})n) = 0$

and

$\frac{\Theta}{3}(\frac{\ddot{a}}{a} + \frac{kc^2}{a^2} - \frac{8 \pi G}{3c^2}(\frac{\rho + 3P}{n}) n) = 0$
GarethMeredith

Posts: 100
Joined: Tue Aug 02, 2016 4:13 am

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