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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.


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