Dubbelosix Posted April 6, 2017 Report Share Posted April 6, 2017 (edited) [math]J = \frac{Gm^2}{v}[/math] angular momentum also [math]J = m \omega R^2[/math] Centrifugal force [math]F = m \omega \times (\omega \times R) = m\omega^2 R[/math] So centrifugal force is also in terms of angular momentum [math]\frac{J}{R}\frac{d\theta}{dt} = m \omega^2 R[/math] where [math]\omega = \frac{d\theta}{dt}[/math] and the relationships also satisfy [math]\omega \times R = \frac{dR}{dt} = \dot{R}[/math] The Friedmann equation with rotation is [math](\frac{\dot{R}}{R})^2 = \frac{8 \pi G}{3}\rho + \omega^2[/math] Similar to how I derive the Friedmann Langrangian, we have after some rearranging [math]mR\ddot{R} = \frac{8 \pi GmR^2}{3}\rho + m\omega^2R^2[/math] Rearranging again we get, we can retrieve the exact form for the centrifugal force arising from the angular velocity [math]m\ddot{R} = \frac{8 \pi GmR}{3}\rho + m\omega \times (\omega \times R)[/math] The time derivatives of any vector function leads to, using the unit vectors [math]\frac{df}{dt}i + \frac{df}{dt}j + \frac{df}{dt}k[/math] Therefore, the time derivative of the radius vector actually arises as [math]\frac{dR}{dt} = (\frac{dR}{dt}i + \frac{dR}{dt}j + \frac{dR}{dt}k) + \omega \times (Ri + Rj + Rk)[/math] and so [math]\ddot{R} = (\frac{d^2R}{dt^2}i + \frac{d^2R}{dt^2}j + \frac{d^2R}{dt^2}k) + \omega^2 \times R = (\frac{d^2R}{dt^2})_k + \omega^2 R(t)[/math] Going back to the derivation of the centrifugal term in the modified Friedmann equation for angular velocity, to understand the acceleration of the universe, we simply omit the mass terms (and omitting the cross products for now) and using the proper terms from derivation and principle, we have for the final equation for cosmology [math]\ddot{R} = (\frac{d^2R}{dt^2})_r + \omega^2 R(t) = \frac{8 \pi GR}{3}\rho + a_{DE}[/math] The last term is a simplification, there is not just a dark energy (centrifugal term) but the Euler and Coriolis. It's really the same thing as seeing expansion in terms of two reference frames, the radius frame and the angular frame, neither are stationary. The radius of course, is a very important frame for an expanding universe, the rotational properties arguably generate the non-stationary aspect of the universes radius (the rotation pushes the universe outwards). The relationship between the two accelerating frames is: [math]v \equiv \frac{dR}{dt} = (\frac{dR}{dt})_k + \omega \times R[/math] then acceleration has to appear like [math]a_{DE} \equiv \frac{d^2R}{dt^2} = (\frac{d^2R}{dt^2})_k + \omega^2 \times R[/math] * the full expansion of the acceleration will yield four acceleration terms in which only one is independent of rotation and so make a substitution [math]a \rightarrow \frac{8 \pi GR}{3}\rho[/math]. This term: [math]\omega^2 R[/math] is the acceleration which imitates dark energy! This equation is more correct than the versions I have seen (see references), where the acceleration term has only been concluded as [math]\omega^2R[/math] when really it should feature the full components given for the reasons above. Alone without the corrective terms, the acceleration of the universe comes out to be [math]\omega^2 R = 10^{-7}cms^{-2}[/math]. Ref: http://prints.iiap.res.in/bitstream/2248/5767/1/Primordial%20Rotation%20of%20the%20Universe,%20Hydrodynamics.pdf https://en.wikipedia.org/wiki/Centrifugal_force Edited April 26, 2017 by Dubbelosix Maine farmer 1 Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.