- #36
lukesfn
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dauto said:Let me add another voice to the discussion. The free falling observers never see any Hawking radiation. Not before they cross the Horizon, not while they cross the Horizon, not after they cross the horizon. That's a simple consequence of the equivalence principle valid for any B-hole no matter its mass.
Thanks, but this appears to be a common miss-perception but as far as I can tell is only true in one specific situation rather then being the general case. (After all, why would Hawking radiation disappear if you started free falling) This has already been discussed in this thread in detain referencing several papers with the math, but see below for the most relevant bit.
lukesfn said:
According to [2], the only a free falling observer with zero instantaneous radial velocity at the horizon will see no hawking radiation.
PeterDonis said:Do you have any references where you've seen this discussed? I haven't seen any calculations of the behavior of quantum fields close to the singularity, so I'm not sure what the basis is for the temperature behavior you're referring to.
I'm referring to the discussion we had earlier in this thread some time ago now. See below where I extrapolated a formula for Hawking radiation inside the BH. I guess the accuracy of the formula would come under question at some point near the singularity, but it is still a curious question.
This isn't really a meaningful question; to an infaller that's falling through the horizon, the BH isn't an "object" that the concept of "relativistic mass" or energy can be applied to; that sort of concept can only be meaningfully applied to the BH by an observer very far away, to whom the BH just looks like an ordinary gravitating object. There is an ##M## parameter in the metric, but that parameter doesn't depend on the observer's state of motion.[/QUOTE]lukesfn said:I will return to the formula I gave earlier. Under certain assumptions, this gives observed hawking radiation for a free faller where m is mass of the BH, r is the observers radius, and /r0 is the radius the object was dropped from. (Before being dropped, the object was hovering at a constant radius)
1/(4m) (1 + 2m/r) (1 - 2m/r)1/2
×{(1 - 2m/r0)1/2 + (2m/r×(1-r/r0))1/2 }1/2
×{(1 - 2m/r0)1/2 - (2m/r×(1-r/r0))1/2 }-1/2
Now if we simplify things by taking the limit /r0→∞
1/(4m) (1 + 2m/r) (1 - 2m/r)1/2
×{1 + (2m/r)1/2 }1/2
×{1 - (2m/r)1/2 }-1/2
In the approximation in the paper it came from, this formula now gives a temperature of hawking radiation for an observer free falling from ∞. This formula can be continued all the way to the singularity. (Note that I don't think that the paper ever tried to extend this formula inside the horizon, I am doing that on my on initiative) Anyway, towards the singularity, the observed temperature tends towards ∞. This implies that at some finite radius, the observer will see a radiation with energy greater then the mass of the BH. Therefore, I've shown that the speculation in my original post has some mathematical basis. I still find it quite an interesting question.
I suppose that would make sense why my head hurts when I try to think about it.
I assume M is the rest mass, which obviously should not depend on motion.
So you are saying the concept of "relativistic mass" breaks down strongly curved space? I must say that I am struggling with the concept of velocity differences in strongly curved space.
Even so, I am not convinced question is completely without meaning. Perhaps it is time for some more reading...