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Jan
31

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Only for the static LNIFs not for the co-moving LIFs

Third try

Think only in terms of emit/absorb or send/receive not present/past or present/future

OK I found my last late night error.

Then for co-moving LIFs

1 + z = f(emit)/f(absorb) = a(absorb)/a(emit) = (absorbed wavelength)/(emitted wavelength)

this works for both past and future light cones (invariance along the time or scale factor axis)

If we send to our future horizon now a(now emit) = 1 to a(horizon absorb) = 2

1 + z(LIF) = 2 = f(emit)/f(absorb) FINITE REDSHIFT for the comoving LIFs.or

However, for the static LNIFs at the future horizon g00 = 0

1 + z(LNIF) = (gtt(absorb)/gtt(emit))^1/2 ---> 0 at the future horizon

= f(emit)/f(absorb) = INFINITE BLUE SHIFT

I think Nick made the false statement that the photon has infinite redshift when it leaves us along our future light cone and hits our future horizon. See below.

Actually Z, the effect I am talking about does not require far-field propagation of on-mass-shell blackbody photons to infinity.

"What is more controversial is whether or not this implies that the system actually radiates."

I don't need it to actually radiate at all. All I need is for the Unruh temperature TU at the g00 = 0 horizon (black hole or de Sitter) to obey

TU > 2mc^2

That's necessary but not sufficient. I also need

probability 1 that every photon at every frequency at r = 0 gets absorbed at r = /^-1/2

in order to get the Wheeler-Feynman total absorber condition.

Not obvious this can happen.

Now what about the static LNIFs?

1 + z = [gtt(future)/gtt(present)]^1/2

at our future horizon this is

1 + z = 0 = f(present)/f(future)

INFINITE BLUE SHIFT

When I was at Birkbeck College, University of London in 1971, David Bohm once said to me that "General Relativity has a complete theory of measurement, while quantum theory doesn't." I think he meant the Wheeler approach that is detector-centric. If you cannot formulate the GR in terms of small detectors in a gedankexperiment you quickly get lost in the opium fog of pure mathematics leading you off the true path to Paradiso in Dante's sense.

*Midway upon the journey of our life I found myself within a forest dark, For the straightforward pathway had been lost.*

On Jan 30, 2011, at 4:49 PM, Paul Zielinski wrote:

ok

Tricky question because GCD assumes a unique vacuum classically, which is not true quantum mechanically. Look at Susskind's horizon complementarity. The real on-mass-shell particle content is a function of the covariant tensor acceleration of the detector. One must use Bohr type approach in GR as well as in QM i.e. "total experimental arrangement" as soon as quantum effects like Unruh effect become important. There is too much formal math in GR without any real anchoring in experiment. Not enough Bridgman operationalism. The Wheeler-Thorne school is the exception to this of course.

On Jan 30, 2011, at 4:43 PM, Paul Zielinski wrote:*Well of course I think *you* still don't get it. The effect is reciprocal. :-) Of course the Unruh phenomenon depends on the acceleration of the detector. The question is whether such effects are objective and can be thus separated from frame artifacts.*

The effect is objective. In my model, which may not work BTW, the virtual electron-positron pair is the accelerating detector because its stuck at the event horizon, whether objective or observer-dependent won't matter.

In the de Sitter observer-dependent case the photon emitted from the observer who is always at r = 0 when g00 = 1 - / ^2 sees a stuck e+-e- pair (static LNIF) at distance H from its (the photon's) horizon.

http://www.npl.washington.edu/npl/int_rep/tiqm/TI_toc.html

past emitter is at r = 0, future absorber is at tiny H from r = /^-1/2 *Your MTW "LNIF" is a chimera, since it bundles objective acceleration-dependent effects (represented by tensor invariants) with frame artifacts (represented by the *components* of 4-tensors). *

Nonsense Z the acceleration I use is completely covariant with a tensor invariant. Your statement is wrong because*(The minus sign on σ0 ensures that is future pointing.) This is the frame that models the experience of static observers who use rocket engines to "hover" over the massive object. The thrust they require to maintain their position is given by the magnitude of the acceleration vectorThis is radially outward pointing, since the observers need to accelerate away from the object to avoid falling toward it. On the other hand, the spatially projected Fermi derivatives of the spatial basis vectors (with respect to ) vanish, so this is a nonspinning frame.*

http://en.wikipedia.org/wiki/Tetrad_(general_relativity)

The corresponding formula in the de Sitter case is

[c^2(/^1/2 + / ^2) (1 - / ^2)^-1/2)e1

note that the /^1/2 term is like the hf/2 zero point term in the quantum oscillator - call it the horizon surface gravity Hawking radiation term missing from Einstein's classical calculation.

Similarly in the black hole case we must use

c^2/rs + rs/2r^2 in the Newtonian coefficient

in these units rs = 2m

On Jan 30, 2011, at 8:45 PM, Paul Zielinski wrote:

There are a number of papers arguing that there is no actual Unruh radiation, and that the Unruh temperature seen by an accelerating detector is an effective temperature, but otherwise the physical situation is exactly the same as that experienced by a non-accelerating observer at the elevated temperature.

For example:

http://arxiv.org/abs/quant-ph/0509151

What impact if any would that have on your Wheeler-Feynman model for cosmological horizons?

It would destroy it obviously. That is a good thing. My model is Popper falsifiable, unlike many archive papers. There is another reason that my model may not work, but I am not sure yet.

Jan
30

Tagged in:

However, it looks as though we need H ~ Gm/c^2 ~ 10^-56 cm to get T > 2mc^2 and this is problematical since it's much smaller than the Planck length Lp ~ 10^-33 cm.

It depends how close you are to it as a static LNIF.

Here is one important idea that Nick got hung up on.

The Newtonian surface gravity is

gNewton = c^2rs/r^2 ---> c^2/rs

rs = 2GM/c^2

However, that is not correct for GR we need the time-dilation factor g00^-1/2

g(r) = (c^2/rs)(1 - rs/r)^-1/2 ----> infinity at the black hole horizon for static LNIFs hovering outside it.

we see c^2/rs as rs/r ---> 0

The temperature of the Hawking thermal radiation that the static LNIF detects T = hc/rskB = hc^3/2GMkB

that Unruh cites is what we see far from the black hole where the static LNIF merge to LIFs as rs/r ---> 0

Similarly for the observer-dependent cosmic horizon

g(r) = c^2/^1/2(1 - / ^2)^-1/2

we are at r = 0, so we see T = hc/^1/2/kB

but a static LNIF distance H << /^-1/2 from the horizon sees

T ~ hc/^1/4H^-1/2kB^-1

Of course a geodesic LIF falling through the horizon sees T = 0 because its covariant acceleration is zero.

The photon is a null geodesic LIF, but the virtual electron-photon pairs clamped to the relative horizon of that retarded photon emitted from r = 0 are static LNIF with covariant acceleration c^2/^1/4H^-1/2.

On Jan 29, 2011, at 11:09 PM, JACK SARFATTI wrote:

On Jan 29, 2011, at 10:49 PM, nick herbert wrote:

====================================

Dear Abbie--

To whom should I turn for reliable information

concerning horizon radiation?

eager to learn

Nick in Boulder Creek

=====================================

======================================================

Dear Nick in Boulder Creek--

You could start with Bill Unruh's clever, amusing, detailed, intelligent

investigations about what might happen at horizons. ...

Take a look at: "Dumb Holes: analogues for Black Holes" by Bill Unruh

http://rsta.royalsocietypublishing.org/content/366/1877/2905.full.pdf

Happy Valentine's Day, eager to learn

Abbie

========================================================

Hey Z, please copy and paste the specific Greene text.

"How little, mark! that portion of the ball,

Where, faint at best, the beams of Science fall.

Soon as they dawn, from Hyperborean skies,

Embody'd dark, what clouds of Vandals rise!" (III 75-8)

Dunciad, Alexander Pope

I did not pay attention seeing Nick Herbert just used it as bait to throw Red Herrings at me.

On Jan 29, 2011, at 7:55 PM, Paul Zielinski wrote:

OK we'll let Jack do his little rodomontade here since he has clearly done his homework in this area

and as usual has little patience for novices.

I still maintain that Jack has not so far produced anything that actually refutes Nick's original comments

about Brian Greene's NYT article.

Jan
29

Tagged in:

http://www.futuretheater.com/listen-to-past-shows/january-29-2011.html

On Jan 29, 2011, at 1:20 PM, William J Birnes wrote:

*I was thinking more along the lines of what do you think it man for the world of UFO researchers. But we can avoid that. *

Idle speculation at this point - you need to ask Jacques Vallee, Stanton Friedman and Nick Pope who were there.

*However, I wonder if you have any opinions/thoughts about theories concerning parallel universes and possible portals between them, portals that either open or close by themselves or, possibly, can be opened or closed. *

Yes, I have been very clear on that since the discovery of dark energy repulsive gravity.

The only way real material ETs from distant places and times can reach us is with star gates and warp drive.

Stargates can connect Max Tegmark's Level 1 and Level 2 parallel universes of course and warp drive can connect different Level 1 parallel universes.

http://www.relativitycalculator.com/articles/max_tegmark/parallel_universes_max_tegmark.html

Also there is general confusion about parallel universes. Brian Greene himself may be confused I'm not sure. There are at least THREE different meanings to the word obvious from Max's picture above.

On Jan 29, 2011, at 2:14 PM, JACK SARFATTI wrote:

On Jan 29, 2011, at 10:49 AM, William J Birnes wrote:

*I would like to ask you about your views on the Saudi conference. Are they trying to be an alternative Ted or are they onto something?*

Bill, I was not there and have no views as yet because I have no information on what happened. Of course it's significant that they were invited in the first place to talk about the subject.

I don't want to talk about other people's physics ideas. My own are exciting enough.

First I need you to plug http://stardrive.org

Good questions to ask me.

What is dark energy?

What is dark matter?

What is the hologram universe?

What is signal nonlocality?

How does signal nonlocality relate to ordinary consciousness and to the remote viewing experiments of the CIA at SRI?

What about the experiments of Bem at Cornell that seem to confirm Bierman's and Radin's presponse experiments on back from the future messaging?

President Obama just gave Yakir Aharonov a medal for his back-from-the-future quantum theory. How does Yakir's theory relate to what you have been saying since the 1970's?

How does that connect to the sudden rude disinvite by Antony Valentini and Michael Towler to you, Nobel Laureate Brian Josephson, and David Bohm's biographer David Peat because of their politically incorrect interests in consciousness, the paranormal, UFOs, and your unconventional non-academic style of presenting your ideas on the internet bypassing journals - though not entirely.

On Jan 29, 2011, at 1:36 PM, JACK SARFATTI wrote:

On Jan 29, 2011, at 8:27 AM, William J Birnes wrote:

We will call in for the show at 3:30 PM Pacific time

On Jan 29, 2011, at 1:59 AM, JACK SARFATTI wrote:

what time?

On Jan 28, 2011, at 10:17 PM, William J Birnes wrote:

We're go for tomorrow. Will go up sometime tonight or tomorrow.

On Jan 28, 2011, at 8:28 PM, JACK SARFATTI wrote:

i don't see anything on http://www.futuretheater.com/

to post to my 2300 facebook friends & 600 yahoo group members + 400/day at http://stardrive.org

Jan
29

Tagged in:

One can do a simple experiment at the space station to test all this in a new way and detect horizon complementarity - more on this another time.

The Pound-Rebka experiment is between two static LNIFS.

Jefferson laboratory at Harvard University. The experiment occurred in the left "tower". The attic was later extended in 2004.

http://en.wikipedia.org/wiki/Pound–Rebka_experiment

Einstein's gravity redshift from the Sun uses

dt = ds(emit)/(1 - rs/r(emit))^1/2 = ds'(absorb)/(1 - rs/r(absorb))^1/2

f(absorb)/(1 - rs/r(emit))^1/2 = f(emit)/(1 - rs/r(absorb))^1/2

f(absorb) = f(emit)(1 - rs/r(emit))^1/2 /(1 - rs/r(absorb))^1/2

rs << r(emit)

rs << r(absorb)

f(absorb) ~ f(emit)(1 - rs/2r(emit)) (1 + rs/2r(absorb))

~ f(emit)(1 + (rs/2)(1/r(absorb) - 1/r(emit)))

~ f(emit)(1 + (rs/2r(emit))(r(emit)/r(absorb) - 1)

However, in the case of observation of the Sun

r(emit)/r(absorb) ~ 10^11/10^13 = 10^-2

therefore with ~ 1% error

f(absorb) ~ f(emit)(1 + (rs/2r(emit))(10^-2 - 1) ~ f(emit)(1 - .99(rs/2r(emit)) ~ f(emit)(1 - .99(10^5/10^13) ~ f(emit)(1 - .99x10^-8)

so we see that the gravity redshift on spectral lines emitted from the surface of the Sun to Earth is only of order 10^-8, i.e. megaHz for visible light.

The Earth is on geodesic orbit in the curvature field of the Sun. So there are two more approximations here.

Even though the Earth is LIF in the Sun's curvature field we approximate it it like a static LNIF. Note that a static LNIF becomes LIF as r ---> infinity. We see that this approximation only makes a 1% error in the coefficient of the major contribution to the shift of 10^-8.

We also neglect the covariant acceleration of the detector on Earth's surface of g ~ 10 meters/sec^2 in the curvature field of the Earth.

Note that the curvature field of the Sun at Earth is of order 10^5/(10^13)^3 cm^-2 ~ 10^-34cm^-2 i.e. curvature radius of 10^17 cm.

The curvature field of the Earth at the surface of Earth is of order 1/(10^9)^3 ~ 10^-27 cm^-2, i.e. curvature radius of ~ 10^13 cm.

One can do a simple experiment at the space station to test all this in a new way and detect horizon complementarity - more on this another time.

On Jan 28, 2011, at 7:31 PM, JACK SARFATTI wrote:

Your problem Z is that you simply do not look at the diagram.

Pick any vertical (constant co-moving coord) world line you like. The Copernican principle is obeyed.

Its past particle horizon is its future light cone at a(0) = 0. Retarded light in our past cone from the past particle horizon is infinitely redshifted.

Light rays at 45deg & 135 deg on this conformal diagram.

Its future event horizon is its past light cone at a(infinity) = infinity. Advanced light from this horizon back to us at r = 0 is infinitely redshifted, but retarded light from r = 0 to it is infinitely blue shifted (classically).

Here t = proper metric time not the conformal time tau.

1) assume only retarded light signals.

Therefore, we only detect at a point stuff on the past light cone of that point.

2) Now we have a choice.

2a) the future horizon is not a total absorber, so we can see retarded signals from Galaxy X even when we are beyond Galaxy X's future event horizon.

2b) the future horizon is a Wheeler-Feynman total absorber from the real electron-positron plasma therefore no light rays from r = 0 make it past its future horizon. In that case, and only that case is Nick right in one of his statements. Only photons emitted from r = 0 see the plasma at their horizon r = /\^-1/2. LIFs do not see it. The blue shifted photon relative to the LNIF virtual electron-positron pairs clamped to the the r = /\^-1/2 horizon (of thickness Lp) provides the energy needed to pull them out of the vacuum in accord with the Unruh temperature T(r) = hg(r)/ckB.

Again that's horizon complementarity.

Nick is mistaken that a retarded photon from r = 0 is infinity redshifted. In fact, its the opposite, its infinitely blue shifted for a static LNIF detector clamped to the horizon. Not so for a co-moving detector because of horizon complementarity. Basically the co-moving detector in de Sitter space has acceleration 0, i.e. LIF, but a static LNIF detector has acceleration g(r) = 2c^2/\r(1 - /\r^2)^-1/2 ---> 0 at r = 0 and ---> infinity (classically) at r = /\^-1/2.

Also the frequency shift is

f(r) = f(0)(1 - /\r^2)^-1/2

The Pound-Rebka experiment is between two static LNIFS.

Jefferson laboratory at Harvard University. The experiment occurred in the left "tower". The attic was later extended in 2004.

http://en.wikipedia.org/wiki/Pound–Rebka_experiment

Einstein's gravity redshift from the Sun uses

dt = ds(emit)/(1 - rs/r(emit))^1/2 = ds'(absorb)/(1 - rs/r(absorb))^1/2

f(absorb)/(1 - rs/r(emit))^1/2 = f(emit)/(1 - rs/r(absorb))^1/2

f(absorb) = f(emit)(1 - rs/r(emit))^1/2 /(1 - rs/r(absorb))^1/2

rs << r(emit)

rs << r(absorb)

f(absorb) ~ f(emit)(1 - rs/2r(emit)) (1 + rs/2r(absorb))

~ f(emit)(1 + (rs/2)(1/r(absorb) - 1/r(emit)))

~ f(emit)(1 + (rs/2r(emit))(r(emit)/r(absorb) - 1)

However, in the case of observation of the Sun

r(emit)/r(absorb) ~ 10^11/10^13 = 10^-2

therefore with ~ 1% error

f(absorb) ~ f(emit)(1 + (rs/2r(emit))(10^-2 - 1) ~ f(emit)(1 - .99(rs/2r(emit)) ~ f(emit)(1 - .99(10^5/10^13) ~ f(emit)(1 - .99x10^-8)

so we see that the gravity redshift on spectral lines emitted from the surface of the Sun to Earth is only of order 10^-8, i.e. megaHz for visible light.

The Earth is on geodesic orbit in the curvature field of the Sun. So there are two more approximations here.

Even though the Earth is LIF in the Sun's curvature field we approximate it it like a static LNIF. Note that a static LNIF becomes LIF as r ---> infinity. We see that this approximation only makes a 1% error in the coefficient of the major contribution to the shift of 10^-8.

We also neglect the covariant acceleration of the detector on Earth's surface of g ~ 10 meters/sec^2 in the curvature field of the Earth.

Note that the curvature field of the Sun at Earth is of order 10^5/(10^13)^3 cm^-2 ~ 10^-34cm^-2 i.e. curvature radius of 10^17 cm.

The curvature field of the Earth at the surface of Earth is of order 1/(10^9)^3 ~ 10^-27 cm^-2, i.e. curvature radius of ~ 10^13 cm.

One can do a simple experiment at the space station to test all this in a new way and detect horizon complementarity - more on this another time.

On Jan 28, 2011, at 7:31 PM, JACK SARFATTI wrote:

Your problem Z is that you simply do not look at the diagram.

Pick any vertical (constant co-moving coord) world line you like. The Copernican principle is obeyed.

Its past particle horizon is its future light cone at a(0) = 0. Retarded light in our past cone from the past particle horizon is infinitely redshifted.

Light rays at 45deg & 135 deg on this conformal diagram.

Its future event horizon is its past light cone at a(infinity) = infinity. Advanced light from this horizon back to us at r = 0 is infinitely redshifted, but retarded light from r = 0 to it is infinitely blue shifted (classically).

Here t = proper metric time not the conformal time tau.

1) assume only retarded light signals.

Therefore, we only detect at a point stuff on the past light cone of that point.

2) Now we have a choice.

2a) the future horizon is not a total absorber, so we can see retarded signals from Galaxy X even when we are beyond Galaxy X's future event horizon.

2b) the future horizon is a Wheeler-Feynman total absorber from the real electron-positron plasma therefore no light rays from r = 0 make it past its future horizon. In that case, and only that case is Nick right in one of his statements. Only photons emitted from r = 0 see the plasma at their horizon r = /\^-1/2. LIFs do not see it. The blue shifted photon relative to the LNIF virtual electron-positron pairs clamped to the the r = /\^-1/2 horizon (of thickness Lp) provides the energy needed to pull them out of the vacuum in accord with the Unruh temperature T(r) = hg(r)/ckB.

Again that's horizon complementarity.

Nick is mistaken that a retarded photon from r = 0 is infinity redshifted. In fact, its the opposite, its infinitely blue shifted for a static LNIF detector clamped to the horizon. Not so for a co-moving detector because of horizon complementarity. Basically the co-moving detector in de Sitter space has acceleration 0, i.e. LIF, but a static LNIF detector has acceleration g(r) = 2c^2/\r(1 - /\r^2)^-1/2 ---> 0 at r = 0 and ---> infinity (classically) at r = /\^-1/2.

Also the frequency shift is

f(r) = f(0)(1 - /\r^2)^-1/2

Jan
28

Jan 28, 2011, at 8:49 AM, nick herbert wrote:

Two questions

1) where did you get the equation of cosmic acceleration as function of H?

Nick this is textbook general relativity

The covariant acceleration of static LNIFS at fixed "r" in the SSS metrics is

g(r) = (Newtonian "force" acceleration)g00^-1/2

in the case of the de Sitter metric / > 0 in this static representation corresponding to the firing rockets in Hawking's picture (static LNIF detectors)

g00 = 1 - / ^2

r = 0 is the observer

i.e. g00 = 0 is the horizon relative to the observer at r = 0.

The Newtonian acceleration is

2c^2/
Therefore the curved spacetime acceleration is

g(r) = 2c^2/ /(1 - / ^2)^1/2

The horizon is at

r = /^-1/2

Let H be a small distance from a horizon from the inside - we are inside our de Sitter horizon

r = /^-1/2 - H

H << /^-1/2

/ = 1/R

then do the algebra, I may have dropped a factor of 2 or so but the basic result is

g(H) ~ c^2/(RH)^1/2

i.e. geometric mean of distance H to the classical horizon surface with square-root of area/entropy of that surface.

2) where does that preposterously small distance L(p) = 10^-33 cm come from?

That comes from Lenny Susskind's "stretched membrane" horizon and basic quantum gravity Heisenberg uncertainty of the metric fluctuation - quantum fuzzy thickness of the classical horizon.

"Three postulates asserting the validity of conventional quantum theory, semi-classical general relativity and the statistical basis for thermodynamics are introduced as a foundation for the study of black hole evolution. We explain how these postulates may be implemented in a “stretched horizon” or membrane description of the black hole, appropriate to a distant observer. The technical analysis is illustrated in the simplified context of 1+1 dimensional dilaton gravity. Our postulates imply that the dissipative properties of the stretched horizon arise from a course graining of microphysical degrees of freedom that the horizon must possess. A principle of black hole complementarity is advocated. The overall viewpoint is similar to that pioneered by ’t Hooft but the detailed implementation is different." http://arxiv.org/pdf/hep-th/9306069v2

Actually Martin may still only need six numbers since the numbers here are derivative as long as he has the Hubble scale as one of the six.

Now one question is what happens to the geodesic observer falling through the black hole? where

g00 = 1 - rs/r

If the observer carries a net surface charge his virtual photons will couple to the virtual electron-positron pairs it seems and he may no longer be a LIF.

We seem to be getting to something like George Chapline's dark star? Maybe nothing can get through the black hole surface and it all smears over the surface as in Lenny Susskind's picture above? Not sure about this.

This does not happen in the subjective cosmological horizon obviously.

Not just SIX NUMBERS but SEVEN?

Just Six Numbers: The Deep Forces that Shape the Universe

Just six numbers govern the shape, size, and texture of our universe. If their values were only fractionally different, we would not exist.

On Jan 27, 2011, at 10:37 PM, JACK SARFATTI wrote at Nick Herbert's request:

The static LNIF virtual electrons stuck to the r = /\^-1/2 future horizon of Galaxy X located at r = 0 accelerate at distance H from the /\ horizon

H << /\^-1/2

g(r) ~ c^2/\^1/2(1 - /\(/\^-1/2 - H))^2^-1/2 ~ c^2/\^1/2(2H/\^1/2)^-1/2 = c^2/\^1/2/(H/\^1/2)^1/2 = c^2/\^1/4/H^1/2 ---> infinity as H ---> 0

The Unruh temperature is

TUnruh(H) = hg(H)/ckB

/\^1/4 ~ 10^-7cm^-2

If H ~ Lp ~ 10^-33 cm

/\^1/4/H^1/2 ~ 10^9 cm^-1

g(r) ~ 10^30 cm/sec^2

T ~ 10^-2710^3010^16/10^10 ~ 10^4610^-37 ~ 10^9 deg K

mc^2 ~ 10^6 ev ~ 10^6 (1/40)300 ~ 10^9 deg K

Curiously, using Lenny Susskind's Lp thickness from quantum gravity, our FUTURE horizon is just the right size to get an electron-positron plasma at an Unruh temperature of 2mc^2.

Just Six Numbers: The Deep Forces that Shape the Universe

Just six numbers govern the shape, size, and texture of our universe. If their values were only fractionally different, we would not exist.

On Jan 27, 2011, at 10:37 PM, JACK SARFATTI wrote at Nick Herbert's request:

The static LNIF virtual electrons stuck to the r = /\^-1/2 future horizon of Galaxy X located at r = 0 accelerate at distance H from the /\ horizon

H << /\^-1/2

g(r) ~ c^2/\^1/2(1 - /\(/\^-1/2 - H))^2^-1/2 ~ c^2/\^1/2(2H/\^1/2)^-1/2 = c^2/\^1/2/(H/\^1/2)^1/2 = c^2/\^1/4/H^1/2 ---> infinity as H ---> 0

The Unruh temperature is

TUnruh(H) = hg(H)/ckB

/\^1/4 ~ 10^-7cm^-2

If H ~ Lp ~ 10^-33 cm

/\^1/4/H^1/2 ~ 10^9 cm^-1

g(r) ~ 10^30 cm/sec^2

T ~ 10^-2710^3010^16/10^10 ~ 10^4610^-37 ~ 10^9 deg K

mc^2 ~ 10^6 ev ~ 10^6 (1/40)300 ~ 10^9 deg K

Curiously, using Lenny Susskind's Lp thickness from quantum gravity, our FUTURE horizon is just the right size to get an electron-positron plasma at an Unruh temperature of 2mc^2.

Jan
28

Tagged in:

Subject: Score a minor victory for Nick Herbert, but only if I am right that the future horizon is the total Wheeler-Feynman absorber (Dr. Quantum)

On Jan 27, 2011, at 6:10 PM, Paul Zielinski wrote:

Change this to:

"That is not what Nick is talking about. He is saying that if we are located at a remote observer's future horizon, not only will light emitted by that observer's galaxy at some point in the distant past not be able reach us after we cross that observer's horizon,"

Look at Nick's version centered at Galaxy X at z = 0.

Let us be on redshift world line z = 1 relative to Galaxy X's z = 0.

This is symmetrical. You can interchange "us" and "Galaxy X" to get my original diagram.

We cross Galaxy X's future event horizon at a(t) somewhere between 1.5 and 2.

At that moment our past light cone crosses Galaxy X's world line at a(t) ~ 0.6

OK Nick has a point only if he accepts my plasma model that all photons from Galaxy X are absorbed at Galaxy X's future horizon by the Unruh temperature effect that we do not measure because our quantum vacuum is unitarily inequivalent to that of Galaxy X's photon at Galaxy X's future horizon. Only then will Galaxy X suddenly disappear from our past light cone because no photons from Galaxy X can tunnel through the Unruh real electron-positron plasma. If Galaxy X's future horizon is not a total Wheeler-Feynman absorber for all photons coming from Galaxy X then we will continue to see Galaxy X in our past light cone even after we cross its future horizon! In general if the future horizon of a source X is a Wheeler-Feynman absorber for Cramer transactions

then and only then will source X suddenly disappear from our past light cone when we pass through source X's future event horizon because no photons from it will penetrate the real electron-positron plasma barrier. We do not see that plasma when we pass through through source X's future horizon all we see are the virtual electron-positron vacuum polarization zero point fluctuations. This is the Horizon Complementarity inherent in the Unruh effect that the real/virtual particle divide is dependent on the intrinsic tensor acceleration of the detector.

"but light emitted by our galaxy *at the same time* in the distant past will not be able to reach the remote observer. So we can locally experience the entire event. The effects are reciprocal."

For sure all the effects are reciprocal. The Copernican Principle is not violated.

Z, I don't understand your final portion because I don't know what you mean by "the remote observer" - draw the diagram. Also I don't know what you mean by "at the same time."

"Davis is right, this really is a confusing subject."

Nick mistakenly thinks of / de Sitter horizons like the black monolith of 2001 sitting there in space the same for every detector. That is not the case. The / horizons from dark energy depend on the quantum vacuum structure and that changes for different detectors with different accelerations because of the Unruh effect. The future / horizon dense hot plasma for Galaxy X is only real for photons emitted by Galaxy X not for us momentarily passing through it at same time and place as the photons. This is different from the black hole case where the horizon is not observer-dependent in the same way.

On Jan 27, 2011, at 9:41 PM, JACK SARFATTI wrote

On Jan 27, 2011, at 4:34 PM, nick herbert wrote:

"I agree, Z. As we pass through its future horizon the light from Galaxy X would, from our point of view, simply vanish (probably infinitely red-shifted). Even tho we are exactly present at the future horizon of Galaxy X we would see no plasma sheaths, no electron-positron pairs, no photon mirrors, nothing at all."

Nick, your error is assuming that everyone must see the same thing. You don't understand horizon complementarity and the inequivalent quantum vacua for detectors with different accelerations of the Unruh effect. Of course we at Galaxy X's future horizon do not see any real plasma as I said many times already. Only the photon from Galaxy X sees that plasma.

Also that photon is infinitely blue shifted not infinitely redshifted because the metric that has the horizon is

g00 = 1 - / ^2

for the photon emitted at r = 0 at frequency f0 has frequency

f(r) = f(0)/(1 - / ^2)^-1/2 ---> infinity at the future horizon for static LNIFs at fixed r, e.g. the virtual electron-positron pairs stuck at r = /^-1/2 where g00 = 0.

i.e.

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

f(r) = 1/ds(r) = f(0)/g00(r)^1/2 ~ kBTUnruh .

Lenny Susskind explains all this.

Here is where you are confused. You know that the black hole horizon is an infinite redshift surface. But that's for retarded photons leaving it arriving at us. The effect is reciprocal. A photon from us arrives at the black hole horizon infinitely blue shifted for the virtual electron pairs stuck to the horizon.

Similarly if we are at r ---> infinity for a black hole in the LNIF representation

g00 = 1 - rs/r

ds(infinity) = ds(r)/(1 - rs/r)^1/2

f(r) = f(infinity)/(1 - rs/r)^1/2 ---> infinity at r = rs

the effect is reciprocal so any retarded photon from rs will have zero frequency at our detectors.

You cannot use the co-moving metric representation where the photon wavelength stretches as the universe expands in the cosmology case for horizon problems. Comoving detectors are not the same as LNIF detectors (virtual particles) clamped to the observer-dependent / horizon.

"My point is exactly this--nothing special happens at a future horizon. If you were there you could verify this claim." - Nick

Nick, your statement is meaningless. You are not asking the right question. The right question is does anything special happen for WHO? In the case of observer-dependent cosmological / horizons nothing special happens for us if we are co-moving with zero covariant acceleration, but something special does happen for a photon from the source whose future event horizon we are crossing at same place and time as the photon. The virtual electron-positron pairs see that photon with a huge blue shift > 2mc^2 that pulls them out of the vacuum.

"If Sarfatti and Ibison claim that all sorts of weird stuff happens at future horizons, then our descendants will certainly be in a perfect position to witness this weird stuff."

No Nick because you don't understand the new physics here of horizon complementarity.

[hep-th/9306069] The Stretched Horizon and Black Hole Complementarity

L Susskind - 1993 "Leonard Susskind[3] proposed a radical resolution to this problem by claiming that the information is both reflected at the event horizon and passes through the event horizon and can't escape, with the catch being no observer can confirm both stories simultaneously. According to an external observer, the infinite time dilation at the horizon itself makes it appear as if it takes an infinite amount of time to reach the horizon. He also postulated a streched horizon, which is a membrane hovering about a Planck length outside the horizon which is both physical and hot. According to the external observer, infalling information heats up the stretched horizon, which then reradiates it as Hawking radiation, with the entire evolution being unitary. However, according to an infalling observer, nothing special happens at the event horizon itself, and both the observer and the information will hit the singularity. This isn't to say there are two copies of the information lying about — one at or just outside the horizon, and the other inside the black hole — as that would violate the no cloning theorem. Instead, an observer can only detect the information at the horizon itself, or inside, but never both simultaneously. Complementarity is a feature of the quantum mechanics of noncommuting observables, and Susskind proposed that both stories are complementarity in the quantum sense.

Interestingly enough, an infalling observer will see the point of entry of the information as being localized on the event horizon, while an external observer will notice the information being spread out uniformly over the entire stretched horizon before being re-radiated."

Black hole complementarity - Wikipedia, the free encyclopedia

However, there are some qualitative differences between the observer-independent black hole horizon and the observer-dependent dark energy / cosmological horizon.

#1 We are outside black hole horizons and can receive retarded light from sources close to but outside them.

#2 We are inside our past and future cosmological horizon.

#3 Only our future horizon is a / horizon - not our past particle horizon.

#3 The emission of a photon and its interaction with the future cosmological horizon is that of Susskind's stretched membrane horizon in an undivided Cramer transaction.

"a streched horizon, which is a membrane hovering about a Planck length outside the horizon which is both physical and hot. According to the external observer, infalling information heats up the stretched horizon, which then reradiates it as Hawking radiation, with the entire evolution being unitary."

The distant external observer and the local static LNIF virtual particles STUCK at the horizon both see a physical hot plasma from the Unruh effect caused by the huge blue shift of the photon from the emitter. However, a locally coincident unaccelerated geodesic observer with no connection to the emitter of the photon sees nothing odd - no hot plasma.

On Jan 27, 2011, at 10:11 PM, nick herbert wrote:

"Since the Unruh effect depends only on acceleration and not on velocity and assuming that today's universal acceleration remains constant your homework problem for today, Jack, is to calculate the Unruh temperature corresponding to an observer (say the photon travelling from Galaxy X experiencing the effect of this cosmic acceleration. Number please, Herr Professor."

Wake up and smell the coffee Nick. I gave that calculation in detail days ago. Here it is again.

The static LNIF virtual electrons stuck to the r = /^-1/2 future horizon of Galaxy X located at r = 0 accelerate at distance H from the / horizon

H << /^-1/2

g(r) ~ c^2/^1/2(1 - /(/^-1/2 - H))^2^-1/2 ~ c^2/^1/2(2H/^1/2)^-1/2

= c^2/^1/2/(H/^1/2)^1/2 = c^2/^1/4/H^1/2 ---> infinity as H ---> 0

The Unruh temperature is

TUnruh(H) = hg(H)/ckB

Jan
26

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Note to Ray Chiao

Because the light cone tips over so much I don't think a stable circular orbit inside the event horizon is possible - at least classically for a non-rotating black hole.

I do agree that there will be a real electron-positron plasma stuck just outside at the event horizon where the tensor acceleration of virtual quanta off the timelike geodesic is

g(r) = (c^2rs/2r^2)(1 - rs/r)^-1/2

where r = rs + H

H << rs

1 - rs/(rs + H) = 1 - 1/(1 + H/rs) ~ 1 - 1 + H/rs = H/rs

g(r) ~ (c^2rs/2rs^2)(rs/H)^1/2 = (c^2/rs)(rs/H)^1/2

The Unruh energy is

kBT = hg(r)/c ~ (hc/rs)(rs/H)^2

to get a real electron-positron plasma requires

2mc^2 ~ (hc/rs)(rs/H)^2 = hc/(rsH)^1/2

note geometric mean of rs with H.

2(rsH)^1/2 ~ h/mc