A Simple Derivation Of The Full Classical Universe In Poincare Symmetry (Torsion)

[Dubbelosix]

I'll assume I don't need to derive the Friedmann equation for this discussion, only the general idea's behind the full equations of motion for the classical universe following all the rotational symmetries of spacetime. 

 

 

The full acceleration equation is, with in order, the coriolis force, the centrifugal force and the Euler force. 

 

 

[math]a_B = a_A - 2\omega \times v_B - \omega \times (\omega \times R_B ) - \frac{d\omega}{dt} \times R_B[/math]

 

 

[math]a_{cor} = -2\omega \times v[/math]

 

(In two reference frames for centrifugal force)

 

[math]a_{cent} = (\frac{d^2R}{dt^2})_r + \omega \times (\omega \times dR)[/math] 

 

[math]a_{euler} = - (\frac{d\omega}{dt}\times R)[/math]

 

 

and we also found the centrifugal force term was also missing an extra term. We can show this quickly. In the two frames of reference we now have, as a correction to the standard way of placing in the rotation term, the acceleration can be written in terms of the inertial frame of reference

 

 

[math]a_i \equiv (\frac{d^2R}{dt^2})_i = [(\frac{d}{dt})_r + \omega \times][(\frac{d}{dt})_r + \omega \times R][/math]

 

 

subscripts ''i and r'' stand for inertial and rotating frames. Carrying differentiations and rearranging yields the acceleration in the rotating frame of reference. We can see then soon from this extension, under this preliminary look, how the Euler force arises from the equations of motion. But first of all, the correct centrifugal term in two frames is:

 

 

[math]a_{cent} \equiv \frac{d^2R}{dt^2} = (\frac{d^2R}{dt^2})_r + \omega^2 \times R[/math]

 

 

*however

 

 

the acceleration in the two frames is

 

 

[math]a_r = a_i - 2 \omega \times v_r - \omega \times (\omega \times R) - \frac{d \omega}{dt} \times R[/math]

 

 

So this brings question over whether the correction should even appear in the final form, is this was a simple derivation from the two frames in the context of my notes for the transport theorem below, then that's fine. 

 

 

And we came to understand how the rotation entered - by a construction of it from the Friedmann equation with rotation (see Sivaram and Arun) for that kind of application. So we ended up with an equation of motion that now featured a centrifugal pseudo force field term ~

 

 

[math]\ddot{R} = \frac{8 \pi GR}{3}\rho + (\frac{d^2R}{dt^2})_r + \omega^2 R(t)  = \frac{8 \pi GR}{3}\rho + a_{cent}[/math]

 

 

*note which is almost correct, clearly we need to extend the full equation for the acceleration which we will soon see how to. It's only for derivatives of velocity can we get the extra term attached (ie. [math](\frac{d^2R}{dt^2})_r[/math]), but for derivatives leading to acceleration, you need the full classical equation. The extra term only appeared because we started viewing things in two frames of reference, but really that extra term later appears to be a functional part of the coriolis force. 

 

 

It was only clear then the full equation of motion with the extra acceleration terms are

 

 

[math]\ddot{R} = \frac{8 \pi GR}{3}\rho - a_{cor} - a_{cent} - a_{eul}[/math]

 

 

note* for frame dragging, the representation of this equation would have Lens-Thirring coefficients probably like so:

 

 

[math]\ddot{R} = \frac{8 \pi GR}{3}\rho - d_1a_{cor} - d_2(a_{cent} - a_{eul})[/math]

 

 

and so, the full acceleration equation would enter like

 

 

[math]\ddot{R} = \frac{8 \pi GR}{3}\rho - 2\omega \times v - \omega \times (\omega \times R ) - \frac{d\omega}{dt} \times R[/math]

 

 

If it has a rotation, Poincare symmetry enforces the torsion to appear in the equation and Prof. Sabbata derived how it might enter in theory, as he shows a torsion term enters the effective density part of a Poisson equation. So the same principle should apply in theory and so torsion would enter like so:

 

 

[math]\ddot{R} =  \frac{8 \pi GR}{3}(\rho - \mathbf{k} \sigma^2) - 2\omega \times v - \omega \times (\omega \times R) - \frac{d\omega}{dt} \times R[/math]

 

 

Before, we differentiated with an extra factor of R on the LHS and so we picked up a factor of 2, we can do without

 

 

[math]\dddot{R} =  \frac{8 \pi GR}{3}(\dot{\rho} - 2\mathbf{k} \sigma\dot{\sigma}) - 2\omega \times \frac{dv}{dt} - \omega \times (\omega \times \frac{dR}{dt}) - \frac{d\omega}{dt} \times \frac{dR}{dt}[/math]

 

 

I'll reside this new post for discussion and work on the classical extension and implications of the classical forces at work in the universe. A few surprises have came along the way. One example is that the Euler force does not just contribute a force in the direction of dark flow, but because the radius of the universe increases with centrifugal expansion, if rotation suddenly stopped, there would be a linear Euler force pushing systems as well. Just like in a car, you are riding along and suddenly the car will stop, you will experience the force pushing you forwards, this is what would happen in the universe as the end stage of a universal rotation (which is manifest of the first law of Newton). The rotation generates the expansion (linear growth) and the flow of rotation (suspected to be dark flow) is the residue of the primordial rotation. The Hubble radius is therefore perpendicular to the Euler acceleration in the rotating reference frame. 

 

 

 

 

notes

 

The absolute acceleration is 

 

[math]a = \frac{d^2R}{dt^2} = \frac{d}{dt}\frac{dR}{dt} = \frac{d}{dt}([\frac{dR}{dt}] + \omega \times R)[/math]

 

[math]= [\frac{d^2R}{dt^2}] + \omega \times [\frac{dR}{dt}] + \frac{d\omega}{dt} \times R + \omega \times \frac{dR}{dt}[/math]

 

[math]= [\frac{d^2R}{dt}] + \frac{d\omega}{dt} \times R + 2\omega \times [\frac{dR}{dt}] + \omega \times (\omega \times R)[/math]

 

From this, we can see something like the coriolis, the Euler and the centrifugal terms arising. 

 

The acceleration equation (standard form) is

 

[math]\frac{d^2R}{dt^2} = -\frac{d\omega}{dt} \times R - 2\omega \times \frac{dR}{dt} - \omega \times (\omega \times R)[/math]

 

The two derivatives for the two reference frames gives us the general form of the transport theorem

 

[math]\frac{d}{dt} f = [(\frac{d}{dt})_r + \omega \times]f[/math]

 

If [math]f[/math] is just some vector field (Hubble radius) what we have for the centrifugal acceleration in the two frame is not:

 

[math]\frac{d^2R}{dt^2}  = (\frac{d^2R}{dt^2})_r + \omega \times (\omega \times R)[/math]

 

The full equation of motion for acceleration is larger than this. 

 

Why is it important to see it in two frames? The radius is an important frame which is directly affected by the rotation of the universe, since the centrifugal term plays the role of expansion (linear growth). 

Edited April 18, 2017 by Dubbelosix