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Re: [Phys-l] Holographic Dark Energy in Induced Gravity Trying Again

Bob,

your note reminds me of this separate question:
As the universe is expanding at an accelerating pace, so that the
"furthest" light we can see is from 13 Bly or thereabouts,
and more distant light never can reach us,
is there an accounting for the mass and dark energy at these
larger separations?

Brian W

Spinozalens@aol.com wrote:
Sorry for some reason the corrections I made weren't saved, Trying again. Holographic Dark Energy in Induced Gravity Ever since the discovery of Dark Energy Physicists have been confronted by two problems, the fine tuning problem and the coincidence problem. The fine tuning problem is simply the need to understand why the observed vacuum energy is 1E-60 times smaller than it's predicted value based on summing over the Zero point energy of all the Quantum fields up to the SUSY breaking scale. The coincidence problem is the need to explain why the total dark energy in the Universe is the same order of magnitude as the density of matter. They evolve very differently over the range of the scale factor, why are they not at very different values in the Universe we live in? A very elegant solution is suggested by the Holographic principle. In the Holographic model Entropy scales not with the volume of space time but with its surface area. This would allow us to scale the Dark Energy at the cosmological scale. Looking at the Einstein equation we would have; G_mu,nu= R_mu,nu -(1/2)*R*g_mu,nu = 8*pi*G*T_mu,nu/c^4 - Lambda*g_mu,nu Lambda is proportional to l^(-2) a scale at the same order as the Cosmological scale. If the vacuum energy were proportional to the inverse Hubble scale the fine tuning problem would be solved which also solves the coincidence problem given the Holographic bound on total energy in any O region of the Universe. Unfortunately this attractive idea produces an equation of state value which predicts no acceleration of the expansion rate. Rho(a) = rho(0)*(a/a(0))^-3*(1+w) 1/a^3= a^-3(1+w) -2*lna=-3*(1+w)*lna w= -1/3 Given d^2a/dt^2= - 4*pi*G * ( rho+ 3*P/c^2)*a w= P/rho*c^2 d^2a/dt^2= - 4*pi*G * ( rho+ (3/c^2)*w*rho*c^2)*a d^2a/dt^2= - 4*pi*G * ( rho-rho)*a =0 Another Holographic Dark Energy approach is to scale the ZPE with the future Horizon scale. However, this would require backward causation and a non local effect. In addition this involves obvious circular reasoning. In order to interpret the cosmic acceleration the Holographic dark energy model has presumed the Acceleration. A different approach that avoids these pitfalls is the idea of Induced Gravity Dark Energy proposed by Zu-Yao Sun and You-Gen Shen. Induced Gravity models can be used as an effective theories of Gravity in the Super String model or a Pre Geometric canonical Quantum theory of Gravity. Shen and Sun, by applying this model to the ZPE of Quantum fields, offer a possible Induced Gravity Dark Energy model. This is closely related to the various proposals which incorporate the extended set of solutions of the Relativistic equations to the ZPE calculation, the models proposed by Klauber, THooft, Moffat, Sundrum, Randel and others. Based on the above we can write the equation T_mu,nu= zeta*{ chi_a*int Dw L(+) + chi^a*int DwL(-)}*g_mu,nu Where Zeta is a constant, chi_a and chi^a are Killing vector fields and the integration is summing over the positive and negative energy vacuum fluctuations. The action for the Induced Gravity model is S= int d^4*sqrt[-g]{ -(1/2)*zeta*phi^2*R - (1/2)*g^mu,nu*pd_mu*pd_nu- V(phi) +L_sm} Where R is the scalar curvature, phi is an effective scalar field and L_sm is the standard model Lagrangian. We get the metric ds^2= chi^a*chi_a dt^2c^2- ((chi^a*chi_a)^-1)*( dx^2 +dy^2 +dz^2) In terms of electromagnetic property of the vacuum we have mu= mu(0)*{ chi^a*chi_a)^(-1/2) and epsilon = epsilon(0)* { chi^a*chi_a)^(-1/2) Based on the Induced Gravity version of the Holographic dark energy the length scale is given by the average radius of the of the Ricci scalar curvature, R^(-1/2) Looking at the Ricci curvature tensor we have R_00=- 3*[ dH/dt + H^2+k/a^2 } R_11 =R_22 = R_33 = -[dH/dt+*H^2] Where H is the Hubble parameter. We assume dH/dt=0 and k= 0 based on current observational evidence. R=TR [R_mu,nu] = 6*H^2 Rho_de = c^2*R/(16*pi*G)= 3*H^2*c^2/(8*pi*G)= 6.4E E -10 J/m^3 Which is on the order of the observed value of vacuum energy density. We can plug this back into the Einstein equation. R_mu,nu -(1/2)R*g_mu,nu + Lambda*g_mu,nu= (8*pi*G/c^4)*T_mu,nu R_mu,nu -(1/2)R*g_mu,nu + 8*pi*G/c^4*rho_de*g_mu,nu= (8*pi*G/c^4)*T_mu,nu R_mu,nu -(1/2)R*g_mu,nu + (8*pi*G/c^4)*c^2/16*pi*G)*R*g_mu,nu= (8*pi*G/c/^4)*T_mu,nu R_mu,nu -(1/2)R*g_mu,nu + (1/2)*c^2R*g_mu,nu= (8*pi*G/c^4)*T_mu,nu R_mu,nu -(1/2)R*g_mu,nu + (mu(0)*epsilon(0)/2)*R*g_mu,nu= (8*pi*G/c^4)*T_mu,nu So here we see that the dark energy term is (mu(0)*epsilon(0)/2)*R*g_mu,nu Using this model we have a formalism which produces the observed vacuum energy density within an order of magnitude of the observed value. In addition, this model equates the vacuum energy density with the electromagnetic parameters of the vacuum using an Induced Gravity model of gravity. Bob Zannelli