Why does an infalling observer not see Hawking radiation?

[beno23]

here is my question i hope someone can explain

A black hole complemenarity says that when someone jump into a a large enough black hole he would pass the event horizon and nothin would happen until he hits the singularity. while for the people observing this from far distance they would see the infalling person scrambled and destroyed by hawking radiation just before he hit the event horizont.
according to Susskind and many other this is no paradox at all since the infalling observer cannot compare notes with people who saw him get scrambled...so there is no apparent contradiction

...well my question is...why doesn't an infalling observer see hawking radiation? how is it possible for him to no get affected by it? if hawking radiation exist just above the black hole horizon why would he be imune to it?

 

[George Jones]

A bit technical, bu according to the book Quantum Fields in Curved Space by Birrell and Davies, pages 268-269,

These consideration resolve an apparent paradox concerning the Hawking effect. The proper time for a freely-falling observer to reach the event horizon is finite, yet the free-fall time as measured at infinity is infinite. Ignoring back-reaction, the black hole will emit an infinite amount of radiation during the time that the falling observer is seen, from a distance to reach the event horizon. Hence it would appear that, in the falling frame, the observer should encounter an infinite amount of radiation in a finite time, and so be destroyed. On the other hand, the event horizon is a global construct, and has no local significance, so it is absurd too conclude that it acts as physical barrier to the falling observer.

The paradox is resolved when a careful distinction is made between particle number and energy density. When the observer approaches the horizon, the notion of a well-defined particle number loses its meaning at the wavelengths of interest in the Hawking radiation; the observer is 'inside' the particles. We need not, therefore, worry about the observer encountering an infinite number of particles. On the other hand, energy does have a local significance. In this case, however, although the Hawking flux does diverge as the horizon is approached, so does the static vacuum polarization, and the latter is negative. The falling observer cannot distinguish operationally between the energy flux due to oncoming Hawking radiation and that due to the fact that he is sweeping through the cloud of vacuum polarization. The net result is to cancel the divergence on the event horizon, and yield a finite result, ...

This finite amount of radiation is negligible for observers freely falling into an astrophysical black hole.

 

[beno23]

thx alot

 

[Q-reeus]

There seems to be a consistency problem here. My position is that 'time freezing' of an in-falling object at the event horizon (as determined by a distant observer) is not a Doppler shift illusion, but real in the following sense. Lower a clock close to the event horizon and raise it again after a given time interval as determined by a 'background metric' observer. Repeat but with a different time interval. We are now in a position to correct for the lowering-raising portion and determine the 'absolute' time retardation of the clock as a function of radius only. We can repeat the above procedure using a different inner radius location wrt the event horizon. Plot the graph of time retardation vs radius and the result is the 'real' time retardation goes to infinity at the event horizon. In other words, the in-falling object 'really' freezes at the horizon (as seen from 'infinity'). By equivalent procedure we find that the object 'really' shrinks spatially to zero size at the horizon. Conversely from the perspective of the in-falling object, the horizon radius blows up to infinity, and the exterior universe has blue-shifted/aged infinitely.

But the latter has to take account of a finite BH lifetime (assuming Hawking radiation is real - not all experts agree). The scenario as I see it is of an in-falling observer that experiences not a fixed BH but one that shrinks to zero size via 'instantaneous' Hawking radiation evaporation. This is a heuristic argument but is it wrong in principle?
Edit:
Looks like I'll have to back away from the above arguments , which it seems are an artifact of mixing different coordinate systems. From http://cosmology.berkeley.edu/Education/BHfaq.html#q9", under
"My friend Penelope is sitting still at a safe distance, watching me fall into the black hole. What does she see?"
"So which of these two explanation (the optical-illusion one or the time-slowing-down one) is really right? The answer depends on what system of coordinates you use to describe the black hole. According to the usual system of coordinates, called "Schwarzschild coordinates," you cross the horizon when the time coordinate t is infinity. So in these coordinates it really does take you infinite time to cross the horizon. But the reason for that is that Schwarzschild coordinates provide a highly distorted view of what's going on near the horizon. In fact, right at the horizon the coordinates are infinitely distorted (or, to use the standard terminology, "singular"). If you choose to use coordinates that are not singular near the horizon, then you find that the time when you cross the horizon is indeed finite, but the time when Penelope sees you cross the horizon is infinite. It took the radiation an infinite amount of time to reach her. In fact, though, you're allowed to use either coordinate system, and so both explanations are valid. They're just different ways of saying the same thing."
Further down, under "Won't the black hole have evaporated out from under me before I reach it?", the author argues NO based purely on the optical illusion (infinite Doppler shift) viewpoint. Will have to think some more.