I compute that black holes have much shorter evaporation times than Hawking et-al first computed. They computed surface vibrations and neglected thickness vibrations due to geometrodynamical field zero point vacuum fluctuations.
On Apr 9, 2014, at 5:02 PM, Paul Zielinski <
On 4/9/2014 4:42 PM, JACK SARFATTI wrote:According to Einstein’s classical geometrodynamics, our future dark energy generated cosmological horizon is as real, as actualized as the cosmic blackbody radiation we measure in WMAP, Planck etc.
But doesn't its location depend on the position of the observer? How "real" is that?
Alice has to be very far away from Bob for their respective de Sitter horizons not to have enormous overlap.
We all have same future horizon here on Earth to incredible accuracy.
What else? Obviously.
No, static LNIF hovering observers have huge proper accelerations at Lp from the horizon with redshifted Unruh temperature T at us
I assume by "dark energy generated" you simply mean that the FRWL metric expansion is due to /, and
/ registers the presence of dark energy.
We have actually measured advanced back-from-the-future Hawking radiation from our future horizon. It’s the anti-gravitating dark energy Einstein cosmological “constant” / accelerating the expansion of space.
OK so the recession of our future horizon produces Hawking-like radiation due to the acceleration of our frame of reference
wrt the horizon?
kBT ~ hc/(A^1/2Lp^1/2)^1/2
use black body law
energy density ~ T^4
to get hc/ALp^2
The static future metric is to good approximation
g00 = (1 - r^2/A)
we are at r = 0
future horizon is g00 = 0
imagine a static LNIF hovering observer at r = A^1/2 - Lp
his proper radial acceleration hovering within a Planck scale of the horizon is
g(r) ~ c^2(1 - r^2/A)^-1/2 (A^1/2 - Lp)/A
= c^2(1 - (A^1/2 - Lp)^2/A)^-1/2(A^1/2 - Lp)/A
= c^2(1 - (1 - 2 Lp/A^1/2 + Lp^2/A )^-1/2(A^-1/2 - Lp/A)
= c^2(2Lp/A^1/2 - Lp^2/A )^-1/2(A^-1/2 - Lp/A)
~ c^2(2Lp^-1/2/A^-1/4 )A^-1/2(1 - Lp/A^1/2)
~ c^2(A^1/4/Lp^1/2)A^-1/2 ~ c^2/(Lp^1/2A^1/4)
f(emit) = c/(Lp^1/2A^1/4)
1 + z = (1 - (A^1’2 - Lp)^2/A)^-1/2 = (A^1/4/Lp^1/2)
f(obs) = f(emit)/(1 + z) = Lp^1/2/A^1/4c/(Lp^1/2A^1/4) = c/A^1/2
OK this is the standard low energy Hawking radiation formula from surface horizon modes
However, there is a second high energy quantum thickness radial mode
f'(emit) ~ c/Lp
f’(obs) = (Lp^1/2/A^1/4)c/Lp = c/(Lp^1/2A^1/4)
This advanced Wheeler-Feynman de Sitter blackbody radiation is probably gravity waves not electromagnetic waves.
You seem to be drawing a direct physical analogy between cosmological horizons and black hole horizons.
This requires the anti-Feynman contour for advanced radiation in quantum field theory.
i.e. mirror image of this
so that w = + 1/3 blackbody advanced radiation anti-gravitates
It’s energy density is ~ hc/Lp^2AA = area of future horizon where the future light cone of the detector intersects it.