Are we missing a rotation term in the Friedmann equation?

Discussion in 'Pseudoscience' started by SimonsCat, Jan 20, 2017.

  1. SimonsCat Registered Member

    Writing the Einstein-Friedmann equation,as an energy equation with rotation and a cosmological constant we have:

    \(E = m\dot{R}^2 = \frac{8 \pi GmR^3}{R}(\rho - mR^2 \omega^2 + mc^2 \Lambda R^2)\)

    In which a Lagrangian can be derivable by setting the equation to zero

    \(\mathcal{L} = m\dot{R}^2 - \frac{8 \pi GmR^3}{R}(\rho - mR^2 \omega^2 + mc^2 \Lambda R^2)\)

    To get a rotation term in the universe may seem like an assumption that the rotation term enters the equation with a negative sign. However, the sign is not unwarranted if we considered gravity as part of the full Poincare group which involves the spin and torsion (Venzo and Sabbata) also see Sivaram, or if we consider that dark flow could be a residual spin that has exponentially decayed since its nucleation. Torsion enters the gravitational Poisson equation as

    \(\nabla^2 \phi = \pi G(\rho - k\sigma^2)\)

    where k is Einsteins conversion factor and

    \(k \sigma^2 = \frac{Gm^2 v^2 r^2}{c^4 L^6} = \frac{Gm^2}{c^2L^4}\)

    In which:

    \(\rho(energy) = \frac{Gm²}{L^4}\)

    \(\rho(mass) = \frac{Gm²}{c²L^4}\)

    Notice though now, the whole point of the torsion term, is not only to explain dark energy or the correct symmetries of spacetime, but it may also explain why there is a small cosmological constant: Notice if the cosmological constant is related to spin, then large torsional energies will arise with an opposite sign to the cosmological constant. Effectively, we would have

    \( - k\sigma^2 + mc^2 \Lambda R^2 = \epsilon\)

    Let's deviate away from de Sitter space, with rotation and non-conservation!

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