Independent Term

[Dubbelosix]

If an acceleration is defined in the normal way:

 

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

 

The subscript ''i' stands for inertial. Inertial basically means that there is a uniform motion. Let's take a few steps back to the velocity equation to understand how it arises:

 

[math]v \equiv \frac{dR}{dt}[/math]

 

The time derivatives of [math]R(t)[/math] in some rotating reference has two components. 

 

[math]v_i \equiv (\frac{dR}{dt})_r + \omega \times R = v_r + \omega \times R[/math]

 

Again, ''i'' stands for inertial reference frame, so v is unchanging and uniform. But this is generally not the case of a dynamic universe! The universe is not only accelerating (or decelerating), there is a non-uniform motion in the rotation of a universe and the linear expansion rate. For a universe where rotation gives rise to a centrifugal force, it will decay as well, so the rotation is non-uniform as well. 

 

The centrifugal force arises from a time derivative of the velocity equation

 

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

 

If the last term here varies

 

[math] \omega^2 \times R[/math]

 

Then for a universe where an exponentially decaying rotation expands the universe then neither frame is actually uniform. 

 

What is interesting from here is that an expansion of the acceleration equation may give us a candidate for a Friedmann acceleration equation. We calculate that in the following way -

 

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

 

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

 

It can yield a solution in the rotating frame of reference 

 

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

 

There were other ways to construct this without the confusing subscripts,

 

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

 

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

 

expanding like we did before, we actually find an equation of the form

 

[math]a + a_{eul} + a_{cor} + a_{cent} = \frac{d^2R}{dt^2}[/math]

 

Flipping the equation gives

 

[math]\frac{d^2R}{dt^2} = a + a_{eul} + a_{cor} + a_{cent}[/math]

 

What is the first acceleration term written as ''a''? This is not entirely clear to me, but it is independent of rotation! The only independent factor in a Friedmann equation of rotation is [math]\frac{8 \pi GR}{3}\rho[/math]. If we replaced then the acceleration with this term

 

[math]a \rightarrow \frac{8 \pi GR}{3}\rho[/math]

 

we would get

 

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

 

From there you plug in the classical fields for the full equation ie.

 

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

 

[math]a_{cor} = -2 \omega \times \frac{dR}{dt}[/math]

 

[math]a_{cent}= -\omega \times (\omega \times R)[/math]

 

 

 

Conclusions

 

In the two frame of the velocity we picked up an extra term and as we went into the acceleration equations, we expanded the relevant terms and we got an acceleration equation fitting the form

 

[math]\frac{d^2R}{dt^2} = a + a_{eul} + a_{cor} + a_{cent}[/math]

 

You can solve to find the independent factor of ''a'' to be in an inertial or non-inertial reference frame, so the choice to replace the only independent factor 

 

[math]a \rightarrow \frac{8 \pi GR}{3}\rho[/math]

 

Seems logical, if we are to construct an equation of motion for the universe based on the classical equations of motion and such an equation was easily constructed now

 

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

 

This has been shown to a number of people without a true objection. Whether this falls apart mathematically because I have been so loose with terms is uncertain, but generally looking at this last equation, I really don't see any reason why it could not be true, nothing has appeared in the equation in a strange way, everything looks ok in the context of a Friedmann equation. 

 

Edited April 26, 2017 by Dubbelosix