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Home Jack Sarfatti's Blog Proof that dark energy is Hawking radiation back from the future

f(0) = g00(r)^1/2f(r)

We here-now are at r = 0.

OK, from quantum gravity the thickness of our future horizon 2D surface is the Planck Length.

Its Hawking temperature from the short waves ~ Lp is

T(horizon) ~ hc/LpkB ~ f(horizon)

However, AND THIS IS PRETTY,

g00(horizon) ~ (1 - (A^1/2 - Lp)^2/A)

where Lp/A^1/2 << 1

Therefore, the Taylor series expansion to lowest order in small quantities is

g00(horizon) ~ 1 - 1 + 2Lp/A^1/2 + O(Lp^2/A)

Therefore, AND THIS IS MY ORIGINAL POINT,

g00(horizon) ~ (2Lp/A^1/2)^1/2

Therefore, the redshifted advanced Hawking radiation temperature we see BACK FROM THE FUTURE HORIZON is

T(0) ~ (2Lp/A^1/2)^1/2hc/LpkB = hc/(LpA^1/2)^1/2kB = hc/Lp^1/2A^1/4kB

The black body radiation density ~ T^4 and when you stick in the Stefan-Boltzmann constant the result is

energy density ~ hc/ALp^2

which is the observed number.

So I say that the dark energy accelerating our universe is retrocausal advanced Wheeler-Feynman thermal radiation from our future de Sitter horizon.

I challenge anyone to objectively refute this interpretation. It won't work if you try retarded radiation from our past particle horizon because of Tamara Davis's horizon area calculation fig 5.1 of her PhD.

Also the metric in our past is not de Sitter.

On Jan 6, 2012, at 5:56 PM, JACK SARFATTI wrote:

On Jan 6, 2012, at 4:32 PM, Paul Zielinski wrote:

On 1/6/2012 3:44 PM, This email address is being protected from spambots. You need JavaScript enabled to view it.

Paul,

The point is that while phi = c^2 is a horizon condition in Bernstein's calculation, as he himself notes, this interpretation is different in GRT.

OK.

How is it different and what significant difference does it make. The problem here is that you are both basically using words and not showing the relevant math.

For example, the purely dark energy de Sitter metric in static LNIF coordinates

g00 = 1 - r^2/A

A = area of the future horizon.

We are at r = 0.

The cosmological horizon is OBSERVER-DEPENDENT.

Horizon means

g00(HORIZON) = 0

OK, ONLY in SSS metrics without any Ai = g0i at all

g00 = 1 - phi/c^2

so for us at r = 0

g00(A^1/2) = 0

in this dS metric example

phi(r)/c^2 = r^2/A

The classical frequency shift is

ds(0)/g00(0)^1/2 = ds(r)/g00(r)^1/2

f(r)/g00(0) = f(0)/g00(r)^1/2

f(0) = g00(r)^1/2f(r)

therefore

f(0) = 0 for radiation reaching us from the horizon at r = A^1/2

Now if the horizon is in our future then that radiation can only be advanced Wheeler-Feynman radiation.

However, this tells us nothing about alleged Jim's Mach Effect.

I doubt that his effect exists.

JFW: In particular, the horizon condition holds everywhen/where in the cosmology because of homogeneity and isotropy making everywhere/when the same as far as the horizon condition interpreted as an energy condition is concerned.

Z: Yes but doesn't this just mean that for any observer, phi = c^2 at that observer's event horizon?

Doesn't that alone satisfy uniformity and isotropy?

Category: MyBlog

Written by Jack Sarfatti

Published on Friday, 06 January 2012 23:52

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