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We see the cosmic microwave blackbody radiation from the big bang aftermath in our past. The dark energy accelerating our universe's expansion however is black body Hawking radiation back from our future horizon.

See the pdf uploaded to Library Resources Cosmology on 12-31-2012
Now I am going to pull the White Rabbit out of the Top Hat for all the Pundits shining the strong light in the wrong part of the dark room.
On Jan 31, 2012, at 2:51 PM, art wagner wrote:

http://xxx.lanl.gov/pdf/1011.1657v2.pdf

"This is an ongoing review on the idea that the phase space information loss at causal horizons
is the key ingredient of major physical laws."
I have been saying this quite independentlly, but Lee does not say it's a retro-causal effect from out future de Sitter horizon. Our past particle horizon is not de Sitter. Specifically, the dark energy density hc/ALp^2 is the cosmological redshift of hc/Lp^4 from our future horizon.
Tamara Davis Fig 1.1c Ph.D.
There is a contradiction in his abstract, the cosmological constant is not zero, in fact it is 1/A where A is the area of our future horizon.
On the Verlinde model I am not sure - maybe.
Look at Lee's Fig 6 and compare with Tamara Davis's Fig 1.1 c - you see Lee's problem? Which horizon is it?
Note that when g00(r) = 1 - / ^2
for STATIC LNIF (ACCELERATING) DETECTORS at fixed r in accelerating expanding curved spacetime with flat Hubble flow spacelike slices at constant CMB temperature with maximal isotropy of the CMB to 10^-5.
g(r) = +2c^2/ (1 - / ^2)^-1/2
we are at r = 0
the cosmological horizon is at g00(horizon) = 0
Using the cosmic redshift
Therefore,
1 + z = (g00(0)/g00(horizon))^1/2 ----> infinity classically without quantum gravity
but we can only see this if we could see advanced Wheeler-Feynman Hawking radiation because our past horizon is not even closely de Sitter.
Suppose the horizon of area A = /^-1 is Lp thick
Then the fuzzy horizon from quantum gravity is
g00(horizon) = 1 - /(/^-1/2 - Lp)^2)^2 ~ 1 - 1 + 2Lp/^1/2 + higher order terms
since Lp/^1/2 ~ 10^-3310^-28 = 10^-61
g00(horizon)^-1/2 = (1/Lp/^1/2)^1/2
g00(0) = 1
Therefore, the advanced dark energy back-from-the-future redshift is
1 + z = 1/g00(horizon)^1/2 ~ Lp^1/2A-^1/4 = T(horizon)/T(us here-now) = hf(horizon)/hf(us here-now)
i.e. the energy per advanced photon we see is 10^-61 smaller than it was when it left our future horizon and moved back along our future light cone. Indeed, that energy here-now is
hc/A^1/2
At the horizon it isĀ  (hc/A^1/2)(1 + z) = hc/A^1/2Lp^1/2A^-1/4 = hc/A^1/4Lp^1/2
i.e. the energy of the Hawking radiation photon at our future horizon is approximately
hc/(Geometric mean of Planck length with Hubble radius)
The Unruh effect for a static LNIF observer is that the Hawking radiation black body temperature is proportional to its acceleration.
The redshifted Unruh temperature of advanced Wheeler-Feynman-Hawking black body radiation from our future horizon that we see at r = 0 here-now where a0 = 1 in the time-dependent form of the metric, is
kBT(receiver) = c^2/Lp^1/2A^1/4)
Planck et-al showed before 1900 that the energy density ~ T^4 and when you stick in the Stefan-Boltzmann coefficient you wind up with the observed dark energy density hc/Lp^2A.
To see that this obviously correct
Let A = R^2
1 + z = 1/(Lp/R)^1/2 = (R/Lp)^1/2
T(receiver)(1 + z) = T(horizon)
(LpR)^-1/2 (R/Lp)^1/2 = 1/Lp
i.e. energy density at the hologram screen horizon is hc/Lp^4 exactly as it should be using quantum field theory.

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