Dubbelosix Posted September 22, 2017 Report Posted September 22, 2017 It was suggested by Arun (et al.) That rotation enters the Friedmann equation like [math](\frac{\ddot{R}}{R})^2 = \frac{8 \pi G}{3}\rho + \omega^2[/math] [1]. see references It's proposed the correct derivation is not only longer, but in this form, should have a sign change for the triple cross product. It is also apparent, no one has offered in their work a derivation other than the one implied through the Godel model. Really what is in implied by the centrifugal term is a triple cross product, is the use of the cross product terms [math]\ddot{R} = \frac{8 \pi G R}{3}\rho + \omega \times (\omega \times R)[/math] Even though the centrifugal force is written with cross products [math]\omega \times (\omega \times R)[/math] it is not impossible to show it in a similar form by using the triple cross product rule [math]a \times (b \times c) = b(a \cdot c) - c(a \cdot b )[/math] Using [math]\omega \cdot R = 0[/math] because of orthogonality we get [math]\ddot{R} = \frac{8 \pi G R}{3}\rho - \omega^2R[/math] which justifies this form of writing it as well. If the last term is the centrifugal acceleration then the acceleration in the two frame is just, while retaining the cross product definition with positive sign, [math]a_i \equiv \frac{d^2R}{dt^2}_i = (\frac{d^2R}{dt^2})_r + \omega^2 \times R[/math] Expanding you can obtain the asbolute acceleration [math]\ddot{R} = (\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}_r + \frac{d\omega}{dt} \times R + 2\omega \times \frac{dR}{dt} + \omega \times (\omega \times R)[/math] Which is the four-component equation of motion which describes the pseudoforces. This gives us an equation of motion, with using [math]\ddot{R} = \frac{d^2R}{dt^2}_r \righarrow frac{8 \pi G R}{3}\rho[/math], [math]\ddot{R}_i = \frac{8 \pi G R}{3}\rho + \frac{d^2R}{dt^2}_r + \frac{d\omega}{dt} \times R + 2\omega \times \frac{dR}{dt} + \omega \times (\omega \times R)[/math] Or simply as [math]\ddot{R}_i = \frac{8 \pi G R}{3}\rho + a_r + \frac{d\omega}{dt} \times R + 2\omega \times \frac{dR}{dt} + \omega \times (\omega \times R)[/math] In the rotating frame we have [math]\ddot{R}_r = \frac{8 \pi G R}{3}\rho + a_i - \frac{d\omega}{dt} \times R - 2\omega \times \frac{dR}{dt} - \omega \times (\omega \times R)[/math] Since inertial systems are only a local approximation, the inertial frame of reference here may been seen to go to zero leaving us the general equation of motion for a universe [math]\ddot{R} = \frac{8 \pi G R}{3}\rho - \frac{d\omega}{dt} \times R - 2\omega \times \frac{dR}{dt} - \omega \times (\omega \times R)[/math] Using the three standard identities ~ [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] we get [math]\ddot{R} = \frac{8 \pi GR}{3}\rho + a_{eul} + a_{cor} + a_{cen}[/math] The concept of rotation in a universe was initially explored by Godel, but we have made some progress since his day and his simple non-expanding metric. The idea the universe has a rotation was also explored by Hawking who admitted they very well could be the kind of model we associate to reality, except the rotation has to be remarkably slow. In the discovery of dark flow, it seems this could be the perfect candidate of a residual primordial rotation. It was proven by Hoyle and Narlikar that any primordial rotation would exponentially decay in an expanding universe. This is an important realization to understand how rotation in the primordial stages was allowed to be large and decayed as the scale factor of a universe increases, leaving behind presumably, something like dark flow in the observable motion of all the systems in the universe - hopefully this will be supported with further mapping of dark flow over larger quantities of systems. Even though technically speaking, the axis could not be discernible in the Godel universe, it becomes a non-problem in a late universe which experiences a decay in rotation. Arguably, dark flow is way too slow to discern any axis of evil. For exact values on how rotation mimics dark energy (or expansion energy) please read the link below. The rotation will give rise to the classical centrifugal force, arising in a universe and pushing systems away. A good question that puzzled me for a while is, if rotation has in fact almost decayed, why is the universe still speeding up in expansion? I have come to realize the universe is exponentially many times the size it is today, so we must reconcile the tug of gravity has been overwhelmed by the acceleration of the universe, which just like Newtons law of inertia, will continue to expand, or continue to accelerate, unless hindered by something. Dark energy, if the substance exists, is believed only to become significant when a universe gets large enough. This means dark energy does not explain how a dense universe was capable of expanding out of the Planck era. Rotation does explain this, very easily and may offer solutions in which vacuum energy can be stolen by the rotational property of the universe - I call the latter, a Bulk to Rotation process, similar to Bulk to Horizon energy transfers in cosmological models. references [1].https://www.researchgate.net/publication/51956326_Primordial_Rotation_of_the_Universe_and_Angular_Momentum_of_a_wide_rangeof_Celestial_Objects https://en.wikipedia.org/wiki/Centrifugal_force Quote
Vmedvil Posted October 8, 2017 Report Posted October 8, 2017 (edited) It was suggested by Arun (et al.) That rotation enters the Friedmann equation like [math](\frac{\ddot{R}}{R})^2 = \frac{8 \pi G}{3}\rho + \omega^2[/math] [1]. see references It's proposed the correct derivation is not only longer, but in this form, should have a sign change for the triple cross product. It is also apparent, no one has offered in their work a derivation other than the one implied through the Godel model. Really what is in implied by the centrifugal term is a triple cross product, is the use of the cross product terms [math]\ddot{R} = \frac{8 \pi G R}{3}\rho + \omega \times (\omega \times R)[/math] Even though the centrifugal force is written with cross products [math]\omega \times (\omega \times R)[/math] it is not impossible to show it in a similar form by using the triple cross product rule [math]a \times (b \times c) = b(a \cdot c) - c(a \cdot b )[/math] Using [math]\omega \cdot R = 0[/math] because of orthogonality we get [math]\ddot{R} = \frac{8 \pi G R}{3}\rho - \omega^2R[/math] which justifies this form of writing it as well. If the last term is the centrifugal acceleration then the acceleration in the two frame is just, while retaining the cross product definition with positive sign, [math]a_i \equiv \frac{d^2R}{dt^2}_i = (\frac{d^2R}{dt^2})_r + \omega^2 \times R[/math] Expanding you can obtain the asbolute acceleration [math]\ddot{R} = (\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}_r + \frac{d\omega}{dt} \times R + 2\omega \times \frac{dR}{dt} + \omega \times (\omega \times R)[/math] Which is the four-component equation of motion which describes the pseudoforces. This gives us an equation of motion, with using [math]\ddot{R} = \frac{d^2R}{dt^2}_r \righarrow frac{8 \pi G R}{3}\rho[/math], [math]\ddot{R}_i = \frac{8 \pi G R}{3}\rho + \frac{d^2R}{dt^2}_r + \frac{d\omega}{dt} \times R + 2\omega \times \frac{dR}{dt} + \omega \times (\omega \times R)[/math] Or simply as [math]\ddot{R}_i = \frac{8 \pi G R}{3}\rho + a_r + \frac{d\omega}{dt} \times R + 2\omega \times \frac{dR}{dt} + \omega \times (\omega \times R)[/math] In the rotating frame we have [math]\ddot{R}_r = \frac{8 \pi G R}{3}\rho + a_i - \frac{d\omega}{dt} \times R - 2\omega \times \frac{dR}{dt} - \omega \times (\omega \times R)[/math] Since inertial systems are only a local approximation, the inertial frame of reference here may been seen to go to zero leaving us the general equation of motion for a universe [math]\ddot{R} = \frac{8 \pi G R}{3}\rho - \frac{d\omega}{dt} \times R - 2\omega \times \frac{dR}{dt} - \omega \times (\omega \times R)[/math] Using the three standard identities ~ [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] we get [math]\ddot{R} = \frac{8 \pi GR}{3}\rho + a_{eul} + a_{cor} + a_{cen}[/math] The concept of rotation in a universe was initially explored by Godel, but we have made some progress since his day and his simple non-expanding metric. The idea the universe has a rotation was also explored by Hawking who admitted they very well could be the kind of model we associate to reality, except the rotation has to be remarkably slow. In the discovery of dark flow, it seems this could be the perfect candidate of a residual primordial rotation. It was proven by Hoyle and Narlikar that any primordial rotation would exponentially decay in an expanding universe. This is an important realization to understand how rotation in the primordial stages was allowed to be large and decayed as the scale factor of a universe increases, leaving behind presumably, something like dark flow in the observable motion of all the systems in the universe - hopefully this will be supported with further mapping of dark flow over larger quantities of systems. Even though technically speaking, the axis could not be discernible in the Godel universe, it becomes a non-problem in a late universe which experiences a decay in rotation. Arguably, dark flow is way too slow to discern any axis of evil. For exact values on how rotation mimics dark energy (or expansion energy) please read the link below. The rotation will give rise to the classical centrifugal force, arising in a universe and pushing systems away. A good question that puzzled me for a while is, if rotation has in fact almost decayed, why is the universe still speeding up in expansion? I have come to realize the universe is exponentially many times the size it is today, so we must reconcile the tug of gravity has been overwhelmed by the acceleration of the universe, which just like Newtons law of inertia, will continue to expand, or continue to accelerate, unless hindered by something. Dark energy, if the substance exists, is believed only to become significant when a universe gets large enough. This means dark energy does not explain how a dense universe was capable of expanding out of the Planck era. Rotation does explain this, very easily and may offer solutions in which vacuum energy can be stolen by the rotational property of the universe - I call the latter, a Bulk to Rotation process, similar to Bulk to Horizon energy transfers in cosmological models. references [1].https://www.researchgate.net/publication/51956326_Primordial_Rotation_of_the_Universe_and_Angular_Momentum_of_a_wide_rangeof_Celestial_Objects https://en.wikipedia.org/wiki/Centrifugal_force Well, I know that rotation of Black holes puts pressure against the gravitational force pushing inward, if at one point the universe was indeed a blackhole then you would expect rotation that would decay as the universe's density changed via expansion. The more massive the black hole the faster the rotation like super-massive black holes rotate near the speed of light where as normal blackholes a bit slower, You may be on to something. Map that using a rotating black hole with a Schwarzchild radius the size of the universe. See the exact speed of Dark energy flow using the Kerr metric. Kerr in parts. Kerr normal form. Or if you don't want to do that use the ultra simple Medvil's Kerr hack for a less exact approximation which is that modified schwarzchild metric for rotating Black Holes that other one could win a nobel prize, this one not so much. Schwarzchild metric Then here is Spin and Gravity for a BH in my modified Schwarzchild Metric, "Kerr hack" the Laplace operators can be canceled if you want it simpler knowing the total energy of the body along with that spin and energy-mass are homogeneous which Dark Energy is I think. This not being General relativistic like the kerr metric but special relativistic. ∇Eb(x,y,z) = ∇(1/((1-(((2MbG / Rs) - (Isωs2/2Mb))2/C2))1/2))MbC2 Original Laplace Prime Form. ∇'(x',y',z') = ∇(1-(((2MbG / Rs) - (Isωs2/2Mb))2/C2))1/2 For your application rotation of Dark Energy ETotΛ = (1/((1-(((2MΛG / Ru) + (Isωs2/2MΛ))2/C2))1/2))MΛC2 u terms universeΛ terms Dark EnergyS terms of spin rotation Lastly, for your application that middle term should be positive given that your Dark Energy and spin expand the universe instead of compress the Black hole. The ratio is something like this graph below, so you can find the energy-mass content of Dark Energy versus total energy of the universe. Reverse Solve the original Equation. ∇(x,y,z) = ∇'(x',y',z')(1-(((2MbG / Rs) - (Isωs2/2Mb))2/C2))1/2 Fitting Directly into the Schrodinger Equation or vice versa. Edited December 21, 2017 by Vmedvil Maine farmer 1 Quote
Vmedvil Posted October 11, 2017 Report Posted October 11, 2017 (edited) Well... emm... I didn't expect such a big reply with some math for once. I will however read it soon and chew over what you said to see if any of it makes sense. Well, since you are looking at it I will explains the parts of the original equation by breaking it into parts like the Kerr Metric is. Relativistic Energy Form. ∇Eb(x,y,z) = ∇(1/((1-(((2MbG / Rs) - (Isωs2/2Mb))2/C2))1/2))MbC2 (2MbG / Rs) = Schwarzchild Metric First Term being the gravitational pull of gravity at radius = C2 (Isωs2/2Mb) = The rotation of Particles inside the Black Hole or Gravity well that are spinning = C2 C2 - C2 = 0 , or the velocity on the object at Rest without kinetic energy in the (X,Y,Z) directions, so it is the same as saying V in Special Relativity, which makes this only work for none directional moving objects unless you add V back in like below, which still only works in a time slice unless you add that last summation like I did in the previous post over Time2 not considering acceleration without that part. Moving and spinning Black Hole or Gravity Well, relativistic energy. ∇Eb(x,y,z) = ∇(1/((1-(((2MbG / Rs) - (Isωs2/2Mb) + V)2/C2))1/2))MbC2 Velocity or V can be in many forms, I usually use these two. (Isωs2/2Mb) = what you think the kinetic energy of a rotating object or set of objects in the gravity well. Is = Can be in many forms which will be wrote below for calculating the movement as many particle masses with any radius or a Single large mass with the radius in any shape of rs, once you take a shape of the Momentum of inertia the Masses cancel in the above part. Literally, any shape as long as you know the function of it for Is, rs , or Ls This equation is relativistic mass in SR, the original equation this was based off. Then Relativistic kinetic energy for a Rest Mass carrying object is. The Laplace operator adds for every point of space in (X,Y,Z) which goes to (X',Y',Z') or (Xi,Yj,Zk) which is gradient that has gravity then the rest is pretty well explained. It is just a Multi-variable differential manifold with 12 dimensions or less of change for gravity that fits into any equation with the Laplace operator ∇ "Euclidean space" or Geodesics ds2 "Isotropic Coordinates" in physics. So, How I think the Universe handles gravity at the Planck level or "Medvil's Quantum Gravity Hyper-complex" being a Hyper-complex Manifold or quaternionic Manifold, I dunno if you can integrate across (i,j,k) so probably, "Medvil's Quaternionic Quantum Gravity" meaning for people that don't know what that means, it have a bunch of layers stacked. Edited October 11, 2017 by Vmedvil Maine farmer 1 Quote
Maine farmer Posted October 11, 2017 Report Posted October 11, 2017 Well... emm... I didn't expect such a big reply with some math for once. I will however read it soon and chew over what you said to see if any of it makes sense. Yep, Vmedvil always has a good mix of math, text, and cool pictures! A hearty welcome for Vmedvil is in order! Quote
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