Dangermouse schrieb in Nachricht <7lmc1o$jbl$1_at_pollux.ip-plus.net>...
>E se mi sbaglio di nuovo, perch� la memoria dell'orizzonte degli eventi �
>una frottola?
Mi rispondo da solo e cosi' ti risparmio la fatica:
A pagina
http://casa.colorado.edu/~ajsh/approach.html Andrew Hamilton
scrive:
______________
Actually, the Schwarzschild surface is not completely black. It emits
Hawking radiation, which is thermal blackbody radiation, produced by quantum
mechanical
effects: the strong gravitational acceleration near the black hole modifies
vacuum fluctuations. For a 30 solar mass black hole, the temperature of the
Hawking
radiation is tiny, only 2�10-9 Kelvin. The wavelength at the peak of the
blackbody spectrum is about equal to the Schwarzschild radius. See the page
on Hawking
radiation for more.
Although classically the light from an infalling person would appear to an
outside observer to redshift without limit at the horizon, quantum
mechanically the light
redshifts only until the wavelength becomes comparable to the size of the
black hole. Thereafter the person becomes indiscernible from the thermal
Hawking
radiation. Thus although classically a person would appear to an outside
observer to take an infinite time to fall in, quantum mechanically the
person disappears in a
finite time.
_______________
Quindi non si vede nulla per via della radiazione di Hawking.
E quello che dico nel punto 2 e' pure sbagliato: nella versione classica se
l'orizzonte degli eventi si espande
porta con se anche i fotoni-bit che contengono gli oggetti delle immagini.
L'ipertesto di Andrew Hamilton e' fatto molto bene, soltanto che quando lo
leggo salto di qua e di la'!!
Ciao
Dangermouse
Received on Mon Jul 05 1999 - 00:00:00 CEST