You are here:
Home Jack Sarfatti's Blog Blog (Full Text Display) Light we emit to the future blue shifts in the dark energy universe (2nd corrected draft)

Jan
31

Tagged in:

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