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# A Look Back At The Semi-Classical Friedmann Equation

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The entropy production equation (which is) a modified Friedmann equation for non-conserved pseudo-deSitter space is formed here in a semi-classical fashion by using additional terms to take into account the fluctuations of the zero point fields:

$\frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2}) = \frac{8 \pi G}{3} \int\ [\dot{N} + N\dot{S}_k + N\dot{S}_{ik}] (\frac{8 \pi \hbar \nu^3}{c^3} \frac{1}{\frac{\hbar \nu}{k_BT} - 1} + \frac{8 \pi \hbar \nu^3}{2c^3})\ d\nu$

Where the entropy is defined in dimensionless form:

$S = \frac{4}{3}\frac{U}{k_BT} = (\frac{4 \pi^2 k^3_B}{45c^3\hbar^3})\ VT^3$

In which

$u = \frac{8 \pi \hbar \nu^3}{c^3} \frac{1}{\frac{\hbar \nu}{k_BT} - 1}$

has units of energy per unit density per unit frequency (or inverse time) since

$\int\ \frac{U}{V} = \int\ u(T)\ d\nu = \int\ \frac{U}{V \nu} d\nu = \mathbf{U}\ d\log_{\nu}$

Also take into consideration that the dimensionless particle number is:

$N = \frac{16}{2\pi^2}(\frac{k^3_B \zeta(3)}{\hbar^3c^3}VT^3)$

With \zeta being the Riemann zeta function. It is generally assumed that N remains constant, but this would not be so for particle production insisted by the third derivative, implying N can in fact vary, even if for finite intervals. To form the opening equation, I have taken into respect the Clausius-Clapeyron equation, which takes into respect both reversible and irreversible dynamics when introducing the entropy production - the main idea is that the creation of particles in the universe expresses itself as an irreversible process.

https://en.wikipedia.org/wiki/Entropy_production

https://theory.physics.manchester.ac.uk/~judith/stat_therm/node44.html#1_11bCCderv

For reference to the entropy, energy density per unit frequency and particle number values you can follow

https://en.wikipedia.org/wiki/Photon_gas

The idea is that you can model the early universe in such a way that we can approximate it like a photon condensate - and the result of the thermodynamic change allows it to change from a condensate to a gas (the gas translates into the ordinary background temperatures).

The internal energy density of the photon gas

$\mathbf{U} = \frac{8\pi}{15}(\frac{k^4_B}{\hbar^3c^3}T^4)$

and so we use this in conjunction with replacing for the particle number as well:

$\frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2}) = \frac{8 \pi G}{3}\ \int\ ([\frac{d}{dt} + \dot{S}_k + \dot{S}_{ik}] \frac{16}{2\pi^2}(\frac{k^3_B \zeta(3)}{\hbar^3c^3}VT^3)\ \frac{8\pi}{15}\frac{k^4_B}{\hbar^3c^3}T^4 + \frac{16}{2\pi^2}\frac{k^3_B \zeta(3)}{\hbar^3c^3}\frac{dV}{dt}T^3 \frac{8\pi}{15}(\frac{k^4_B}{\hbar^3c^3}T^4))$

In the past, we combined all the numerical terms, this time I will establish

$\kappa = \frac{8 \pi G}{3}$

And then combining the remaining numerical terms gives

$\frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2}) = \kappa\ \int\ [\frac{d}{dt} + \dot{S}_k + \dot{S}_{ik}] \frac{128}{ 30 \pi}(\frac{k^3_B \zeta(3)}{\hbar^3c^3}VT^3)\ \frac{k^4_B}{\hbar^3c^3}T^4 + \frac{k^3_B \zeta(3)}{\hbar^3c^3}\frac{dV}{dt}T^3 (\frac{k^4_B}{\hbar^3c^3}T^4))$

The ratio

$\frac{128}{\pi 30}$

by cancelling the common factor 2, reveals

$= \frac{64}{15 \pi} = 1.35812...$

And so simplifies again to

$\frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2}) = \kappa\ \int\ [\frac{d}{dt} + \dot{S}_k + \dot{S}_{ik}] \frac{64}{15 \pi}((\frac{k^3_B \zeta(3)}{\hbar^3c^3}VT^3\ \frac{k^4_B}{\hbar^3c^3}T^4 + \frac{k^3_B \zeta(3)}{\hbar^3c^3}\frac{dV}{dt}T^3 \frac{k^4_B}{\hbar^3c^3}T^4))$

Now that we combined those numerical values, we can bring back now the definition of $\kappa$ numerically, we get

$\frac{8 \pi}{3} \cdot \frac{64}{15 \pi} = \frac{512}{45}$

It's simple to do this, we simply multiply fractions:

$\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}$

Which when setting Newtons constant into natural units $G = 1$ we get a final result

$\frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2}) = \frac{512}{45}\ \int\ [\frac{d}{dt} + \dot{S}_k + \dot{S}_{ik}] ((\frac{k^3_B \zeta(3)}{\hbar^3c^3}VT^3)\ \frac{k^4_B}{\hbar^3c^3}T^4 + (\frac{k^3_B \zeta(3)}{\hbar^3c^3}\frac{dV}{dt}T^3 )\frac{k^4_B}{\hbar^3c^3}T^4)$

Which is a different take from previous calculations, quite a big difference, so I am wondering if my previous calculations were wrong... not sure until I check previous calculations.

The earliest conditions, possibly the pre-big bang phase, need not be a photon condensate, it could have consisted of an all-matter degenerate state of condensed particles. These alternative views are yet to be studied thoroughly by me.

From wiki:

''Planck's law arises as a limit of the Bose–Einstein distribution, the energy distribution describing non-interactive bosons in thermodynamic equilibrium. In the case of massless bosons such as photons and gluons, the chemical potential is zero and the Bose-Einstein distribution reduces to the Planck distribution. There is another fundamental equilibrium energy distribution: the Fermi–Dirac distribution, which describes fermions, such as electrons, in thermal equilibrium. The two distributions differ because multiple bosons can occupy the same quantum state, while multiple fermions cannot. At low densities, the number of available quantum states per particle is large, and this difference becomes irrelevant. In the low density limit, the Bose-Einstein and the Fermi-Dirac distribution each reduce to the Maxwell–Boltzmann distribution.''

Edited by Dubbelosix
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So I had a check back now, it seems that I have done something numerically wrong in my last approach, I redid these calculations again and I am getting the same result.

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Now, in the same special limit in which we care only about the ground state fluctuations for a pre-big bang phase, the entire equation reduces to:

$\frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2}) = \frac{1536}{45}(\frac{k^3_B \zeta(3)}{\hbar^3c^3}\frac{dV}{dt}T^3)\ P = \frac{512}{45}(\frac{k^3_B \zeta(3)}{\hbar^3c^3}\frac{dV}{dt}T^3)\ \mathbf{U}$

The ratio different comes from $\mathbf{U} = 3P$ for the pressure term. Integration of the volume element yields:

$\int\ \frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2})\ dV = \frac{1536}{45}(\frac{k^3_B \zeta(3)}{\hbar^3c^3}\frac{dV}{dt}T^3)\ PV = \frac{512}{45}(\frac{k^3_B \zeta(3)}{\hbar^3c^3}\frac{dV}{dt}T^3)\ \mathbf{U}V \approx \mathbf{C} \frac{\zeta(4)}{\zeta(3)}\dot{N}k_BT$

where we take $C$ as the coefficient without simpification, meaning we can take $\frac{\zeta(4)}{\zeta(3)}$ as a true solution but modifications needed to be done.

Edited by Dubbelosix
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So next thing I will do then upload here, will be a direct calculation, without the final approximation.

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I'm going to be honest, the whole thing is just getting too complicated: The equation that is just barely readable is:

The entropy production equation (which is) a modified Friedmann equation for non-conserved pseudo-deSitter space is formed here in a semi-classical fashion by using additional terms to take into account the fluctuations of the zero point fields:

$\frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2}) = \frac{8 \pi G}{3} \int\ [\dot{N} + N\dot{S}_k + N\dot{S}_{ik}] (\frac{8 \pi \hbar \nu^3}{c^3} \frac{1}{\frac{\hbar \nu}{k_BT} - 1} + \frac{8 \pi \hbar \nu^3}{2c^3})\ d\nu$

I need to simplify some terms because this is becoming a bit of a headache, this isn't a simple equation and I keep getting lost in the terms all cuddled together. We will use the notation as found from the first two equations stated now:

$\mathbf{u} = \frac{8 \pi \hbar \nu^3}{c^3} \frac{1}{\frac{\hbar \nu}{k_BT} - 1}$

Integrating over the frequency yields the energy density of the photon gas

$\mathbf{U} = (\frac{8 \pi^5k^4_B}{15 \hbar c^3}) T^4$

This is why the integration on the right hand side yields the density due to photon energy content of the universe

$\frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2}) = \frac{8 \pi G}{3} \int\ [\dot{N} + N\dot{S}_k + N\dot{S}_{ik}] (\frac{8 \pi \hbar \nu^3}{c^3} \frac{1}{\frac{\hbar \nu}{k_BT} - 1} + \frac{8 \pi \hbar \nu^3}{2c^3})\ d\nu$

We can symbolize all energy states (including those of zero point fields) under a single term for simplicity

$\frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2}) = \frac{8 \pi G}{3} [\dot{N} + N\dot{S}_k + N\dot{S}_{ik}]\ \mathbf{U}$

The entropy is still taken here under a dimensionless form:

$S = \frac{4}{3}\frac{U}{k_BT} = (\frac{4 \pi^2 k^3_B}{45 \hbar^3 c^3}) T^4$

Since pressure is related to relativistic photon gas as

$\mathbf{U} \approx P$

then we can also write

$\frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2}) = \frac{8 \pi G}{3} [\dot{N} + N\dot{S}_k + N\dot{S}_{ik}]\ \mathbf{U} = \frac{8 \pi G}{3} [\dot{N} + N\dot{S}_k + N\dot{S}_{ik}]\ P$

The integration of a volume element reads

$\int \frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2})\ dV = \frac{8 \pi G}{3} [\dot{N} + N\dot{S}_k + N\dot{S}_{ik}]\ U = \frac{8 \pi G}{3} [\dot{N} + N\dot{S}_k + N\dot{S}_{ik}]\ PV$

Remember

$\mathbf{u} = \frac{8 \pi \hbar \nu^3}{c^3} \frac{1}{\frac{\hbar \nu}{k_BT} - 1}$

has units of energy per unit density per unit frequency (or inverse time) since

$\int\ \frac{U}{V} = \int\ u(T)\ d\nu = \int\ \frac{U}{V \nu} d\nu = \mathbf{U}\ d\log_{\nu}$

Combining the following equations

$\mathbf{U} = (\frac{8 \pi^5k^4_B}{15 \hbar c^3}) T^4$

$N = \frac{16 \pi k^3_B \zeta(3)}{\pi \hbar^3c^3}VT^3$

Gives a solution for the equation of state as

$PV= \frac{\zeta(4)}{\zeta(3)}\ Nk_BT$

Edited by Dubbelosix
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Ok I am following this better under this simplified case as I hope everyone else can in some way. We recall we ended up getting:

$\int \frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2})\ dV = \frac{8 \pi G}{3} [\dot{N} + N\dot{S}_k + N\dot{S}_{ik}]\ U = \frac{8 \pi G}{3} [\dot{N} + N\dot{S}_k + N\dot{S}_{ik}]\ PV$

The formula that can describe both reversible and irreversible entropy is

$\frac{dS}{dt} = \sum_k (\frac{\dot{Q}_k}{T_k} + \dot{S}_k + \dot{S}_{ik})$

With

$\dot{S}_{ik} \geq 0$

So we can simplify again and we get

$\int \frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2})\ dV = \frac{8 \pi G}{3} \frac{dS}{dt}\ U = \frac{8 \pi G}{3} \frac{dS}{dt}\ PV$

If we set $\kappa =\frac{8 \pi G}{3}$, then we find the solution again, with the correction coefficient I left out before

$\int \frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2})\ dV = \kappa\ \frac{dS}{dt}\ \frac{\zeta(4)}{\zeta(3)} Nk_BT$

Edited by Dubbelosix
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It's not hard to see why that was all becoming a bit of a strain on the brain, because while we had a relatively easy equation in the form of:

$\int \frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2})\ dV = \frac{8 \pi G}{3} [\dot{N} + N\dot{S}_k + N\dot{S}_{ik}]\ U = \frac{8 \pi G}{3} [\dot{N} + N\dot{S}_k + N\dot{S}_{ik}]\ PV$

We got many complicated expressions to go with it:

Internal Energy

$U = (\frac{8 \pi^2k^4_B}{15\hbar^3c^3})VT^4$

Expected Particle Number

$N = (\frac{16 \pi k^3_B \zeta(3)}{h^3c^3}) VT^3$

Pressure

$P = \frac{1}{3}\frac{U}{V} = (\frac{8 \pi^2 k^4_B}{45\hbar^3c^3})T^4$

Entropy

$S = \frac{4}{3} \frac{U}{k_BT} = (\frac{4 \pi^2k^3_B}{45 \hbar^3c^3})VT^3$

Edited by Dubbelosix
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Just plugging them in shows how complicated it could get...

$\int \frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2})\ dV = \frac{8 \pi G}{3} [\frac{d}{dt} + \dot{S}_k + \dot{S}_{ik}]\ (\frac{16 \pi k^3_B \zeta(3)}{h^3c^3}) VT^3 \cdot (\frac{8 \pi^2k^4_B}{15\hbar^3c^3})VT^4$

And in terms of the effective density parameter in form of pressure and bringing particle number out of parenthesis we get:

$\frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2}) = \frac{8 \pi G}{3} [\frac{d}{dt} + \dot{S}_k + \dot{S}_{ik}]\ (\frac{16 \pi k^3_B \zeta(3)}{\hbar^3c^3}) VT^3 \cdot (\frac{8\pi^2k^4_B}{15\hbar^3c^3})T^4$

But going through it like this has made it a lot easier to follow, even for me. Combining the particle number with that of the pressure density we have:

$\frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2}) = \frac{8 \pi G}{3} [\frac{d}{dt} + \dot{S}_k + \dot{S}_{ik}]\ (\frac{128 \zeta(3) k^7_B}{15 \pi^2 \hbar^6c^6})VT^7$

Edited by Dubbelosix
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$\frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2}) = \frac{8 \pi G}{3} [\frac{d}{dt} + \dot{S}_k + \dot{S}_{ik}]\ (\frac{128 \zeta(3) k^7_B}{15 \pi^2 \hbar^6c^6})VT^7$

Last time I got $\frac{128}{30 \pi^2}$ because I concerned myself with the ground state, which for zero point fluctuations has a kinetic one-half term attached to it, as we see, I ignored it in this approach. Because the ''expected'' particle number has $16 \pi$ in the numerator, where I would have $\pi^3$ in the denominator above, reduces to simply $\pi^2$. We do this with the reduced Planck constant so that $2\pi \hbar = h$ gives back the ordinary Planck constant except, I have yet to take into consideration that there is an extra factor of two for each reduced Planck constant. Since I was concerned with three, this naturally means there is an extra factor of 9 in the denominator. But whether I add this extra factor in is unsure to me at the moment. Even if I did, there would be nothing particularly interesting in doing so. There is one last avenue to look at, we know there are simpler versions for the entropy and the pressure as I have shown above. As shown, we had derived:

$\int \frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2})\ dV = \frac{8 \pi G}{3} [\dot{N} + N\dot{S}_k + N\dot{S}_{ik}]\ PV$

Instead of pulling the particle number out, we can bring out the entropy terms, and take the derivative of the particle number as equally a particle creation term using the same reversible and irreversible dictations within it:

$\int \frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2})\ dV = \frac{8 \pi G}{3} [\dot{N} + \dot{N}_k + \dot{N}_{ik}]\ \frac{4}{3} S \cdot \frac{1}{3} \frac{U}{V} V$

Which reduces to an entropy and an energy, which we may combine:

$\int \frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2})\ dV = \frac{32 \pi G}{27} [\dot{N} + \dot{N}_k + \dot{N}_{ik}]\ S_{energy}$

Where we may define it as an entropy of energy: Entropy can be define in terms of energy

$E = k_B T\ ln(2)N$

With $N$ being the information in ''bits.'' To obtain $ln(2)$ comes from the well-explored relationship for entropy

$S = -\sum \frac{1}{N}\ ln(\frac{1}{2})$

$= -N \frac{1}{N}\ ln(\frac{1}{2})$

$= ln(2)$

So we would have

$\int \frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2})\ dV = \frac{32 \pi G}{27} [\dot{N} + \dot{N}_k + \dot{N}_{ik}]\ k_B T\ ln(2)$

Using the relationship of the particle number to the moles of the gas

$N = nN_A$

Which uses Avogadro's constant with $n$ being the known amount of the substance would be a valid way to go, but there is also no need to write the particle number in because it already functions inside the square parenthesis.

Edited by Dubbelosix
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The Physics of the Condensate

When wavelengths are described by their thermal counterpart, the physics of quantum mechanics can successfully avoid singular solutions by imposing that a particle cannot be confined into lengths smaller than their own wavelengths. I suggest another approach to show how we might come to describe the conditions required to satisfy quantum condensates. A simple equation of state is

$\frac{d}{dt}(NV) = \dot{N}V + N \dot{V} = (\dot{N} + 3 \frac{\dot{R}}{R}N)V = NV\Gamma$

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

$\frac{d}{dt}(\frac{NV}{N^2 \lambda^3}) = \frac{\dot{N}V}{N^2\lambda^3} + \frac{N \dot{V}}{N^2\lambda^3} = (\frac{V}{N\lambda^3})\Gamma$

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

$\frac{V}{N \lambda^3} \leq 1$

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.

$\frac{V}{N \lambda^3} >> 1$

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).

These were my original idea's to an extension to the semi-classical Friedmann equation.

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:

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

The pre-big bang phase, if it existed, would imply a super-cool 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 and equally what we call the background temperatures as a gas (which is distributed not entirely homogeneously since it has an error factor of about 1 in 10,000 in every direction we look). 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.

Note: $P_{irr}$ is the irreversible pressure.

Edited by Dubbelosix
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We do this with the reduced Planck constant so that $2\pi \hbar = h$ gives back the ordinary Planck constant except, I have yet to take into consideration that there is an extra factor of two for each reduced Planck constant. Since I was concerned with three, this naturally means there is an extra factor of 9 in the denominator. But whether I add this extra factor in is unsure to me at the moment. Even if I did, there would be nothing particularly interesting in doing so.

You could just introduce a Compton 'wavelength of preference' and the extra $2 \pi$ could disappear, or not be there in the first place. ;)

Standard Compton wavelength ${\lambda} = \frac {h}{mc}$

Reduced Compton wavelength ƛ = $\frac {\lambda}{2 \pi} = \frac {\hbar}{mc}$

I think it is classic that $\Lambda$, ƛ and $\lambda$ all represent Lambda and the ratio between total m and observed m in the $\Lambda$ model also equals the ratio between $\lambda$ and ƛ.

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You could just introduce a Compton 'wavelength of preference' and the extra $2 \pi$ could disappear, or not be there in the first place. ;)

Standard Compton wavelength ${\lambda} = \frac {h}{mc}$

Reduced Compton wavelength ƛ = $\frac {\lambda}{2 \pi} = \frac {\hbar}{mc}$

I think it is classic that $\Lambda$, ƛ and $\lambda$ all represent Lambda and the ratio between total m and observed m in the $\Lambda$ model also equals the ratio between $\lambda$ and ƛ.

Yes I suppose this is a viable alternative, except I would prefer the wavelength be defined as the ''thermal wavelength'' since we are talking about condensates.

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