A stationary observer at infinity sees a particle of mass m falling in a superma

Question

A stationary observer at infinity sees a particle of mass m falling in a supermassive Schwarzschild black hole. He observes an increasing redshift and sees the particle ceasing to progress when it approaches the black hole's horizon. What happens to the positional uncertainty of this particle in the reference frame of the distant observer?

A straightforward scaling argument (inserting the Hawking temperature into the equation for the thermal de Broglie wavelength for a particle of mass m) yields a thermal areal uncertainty scaling as the black hole circumference times the particle's Compton wavelength.

Is this the correct limiting behavior?

Explanation / Answer

Indeed, if you solve the Klein-Gordon equation in the vicinity of a black hole you find that for a wave-packet falling radially inwards the angular spread starts to grow exponentially with Schwarzschild time and quickly fills out an angular size of 4 Pi. This may also be seen by applying the "UV/IR" relationship in conjunction with the Heisenberg uncertainty principle. Basically, the closer a particle is to the horizon the hotter it is, so the more it gets kicked around. From the UV/IR relation one knows that the radius of spreading (on the horizon) goes like the temperature. We also know that the temperature goes like the inverse uncertainty in the proper time. Knowing the relationship between the proper and Schwarzschild times will show you that spread is exponential with time for the asymptotic observer, at least initially.

However, one should not make any naive extrapolations once the size of the wave packet is comparable to the Schwarzschild radius. This applies to the argument you've made above as well. The area uncertainty will tend to grow until it covers the black hole and then stop. As for the uncertainty in the radial position, this ap

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