A Naive Look Into Possible Quantization Of The Modified Raychuanduri Equation

[Dubbelosix]

[math]\Theta(\frac{\ddot{R}}{R}  + \frac{kc^2}{a^2}) =  \frac{8 \pi G}{3}(\frac{\rho + \frac{3P}{c^2}}{n})n \Theta[/math]

 

The left hand side is second time derivative (actually third) but we will impose the fluid expansion is only quantizable in the limit of [math]a/a = R/R \rightarrow L_P[/math]

 

[math][p_R, R] = pR - Rp = -i\hbar[/math]

 

[math]p = m \frac{dR}{dt}[/math] 

 

This is satisfied through the momentum operator found in second quantization

 

[math]p = -i \hbar \frac{d}{dR}[/math]

 

[math]MR \frac{dR}{dt} = -i\hbar[/math]

 

(action has dimensions of [math]mvR = \hbar[/math]) since [math]\frac{dR}{dt}[/math] just gives a velocity term, then the relation [math]MvR[/math] is satisfied.

 

[math]MR \frac{dR}{dt} = -i \hbar[/math]

 

[math]\frac{dR}{dt} = -i \frac{\hbar}{MR}[/math]

 

Rearrange the main equation (without fluid expansion coefficient)

 

[math]\ddot{R}   =  \frac{8 \pi G R}{3}(\frac{\rho + \frac{3P}{c^2}}{n})n -\frac{kc^2}{a^2}R [/math]

 

working on the LHS only we have

 

[math]\ddot{R} = \frac{dR}{dt} \frac{d}{dt}[/math]

 

[math]\ddot{R}\psi = -i \frac{\hbar}{MR} \frac{d\psi}{dt}[/math]

 

 

I said this was naive, probably based on much more complicated quantization models working in mini-superspace. It's also a quick naive derivation because the operators tend to use partial derivatives instead of the total derivatives, but with some extra calculations and whatnot, I am quite sure it can be fixed. It's also naive for a third reason, we have only quantized one component of this equation, when there are actually several I need to quantize. 

Edited April 6, 2017 by Dubbelosix

[OceanBreeze]

 

 

I said this was naive, probably based on much more complicated quantization models working in mini-superspace. It's also a quick naive derivation because the operators tend to use partial derivatives instead of the total derivatives, but with some extra calculations and whatnot, I am quite sure it can be fixed. It's also naive for a third reason, we have only quantized one component of this equation, when there are actually several I need to quantize. 

 

Er, so what good is all this?

 

You could have gone diretly from this:

 

[math]\frac{dR}{dt} = -i \frac{\hbar}{MR}[/math]

 

to this:

 

[math]\ddot{R}\psi = -i \frac{\hbar}{MR} \frac{d\psi}{dt}[/math]

 

In one easy step by taking the second derivative of  R with respect to ψ and be done with it.

 

I am beginning to suspect you are only here to practice or showcase your LaTex prowess, perhaps in the hope of earning some of this  but all you are getting is a lot of  

[OceanBreeze]

I'm very meticulous, you admitted this yourself did you not? 

 

Either way, this isn't ''any attempt'' of showing off. And I am sorry you have that impression.

 

OK then. Maybe I am just suffering from LaTex envy or something.

 

I do think it would be a big help to those of us who are trying to follow your math, to know what it is you are trying to show.

 

For instance, what is the point of a "Naïve Look Into Possible Quantization Of The Modified Raychuandhuri Equation"?

 

What does this tell us?

[OceanBreeze]

So far not very much, but you won't know how difficult my early attempts of this was. The differentiation leading to non-conservation was (I'll show how you differentiate for anyone who might not know), a common term from the Friedmann equation is,

 

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

 

Differentiation is relatively easy, we lower the exponent, and drag a coefficient of 2, what we get is

 

[math]2\dot{R} \ddot{R}[/math]

 

In short, the factor of 2 will eventually be calculated out of the full equation, so what we really deal with for a non-conserved form:

 

[math]\dot{R} \ddot{R}[/math]

I became focused on this because I came to realize from my own studies, that this would be notoriously difficult to quantize - in fact to the point my friend Matti Pikannen (The creator of the topological geometrodynamics TGD) even said, ''its impossible!''

 

Now of course, he is probably right about such an assertion, so it was a real brain storm to realize you could not only derive the components of the Friedmann equation in such a way that each term  can have a coefficient

 

[math]\frac{\dot{R}}{R}[/math]

 

which can be directly seen also as an equivalent

 

[math]\frac{\dot{a}}{a}[/math]

 

This coefficient is nothing more than a fluid expansion term; if all components shared this, like they seem to able to, they can answer for an easier solution to the quantization I had in mind above. So even though we quantized in a naive way the term [math]\ddot{R}[/math] the non-conservation arises from the coefficient itself

 

[math]\ddot{R} \Theta[/math]

 

This may not seem like a surprising result, but I could see a physical meaning in this. In relativity, there is no global time, so generally speaking, a universes total energy content probably is not a conserved quantity; no such quantity exists in relativity. Friedmann assumed the universe remains constant, and I actually show how you can circumvent that in my original post that you first met me at. The assumption, was completely unfounded, there is actually no reason a universe should in theory retain the same energy as (it expands and where curvature and irreversible particle production happens within the context of field theory).

 

Basically, the fluid expansion creates the non-conservation because as the universe expands, new energy has fill the space being created. This is one interpretation but the most basic that makes sense. Secondly, as the universe expands from a dense structure of curvature, non-conservation happens through the production of particles in curved space, a feature of field theory that has been known for a while. 

 

So even though I have not extrapolated much from interpreting the quantization, I hope at least, this has shown some history into my realizations about the early complications I came to. 

 

Thanks for that explanation. It does look like you know what you are doing, but I admit I find the math daunting and I don't think I have either the qualifications or the time to really get into it. Hopefully someone else will come along who can work with you.

 

I do have one question though, what is the difference between your "fluid expansion term" and an integrating factor? They seem to be the same thing to me.