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?

  • 4
    $\begingroup$ I'd be interested to know how to think about this problem too. If I drop a localized Klein Gordon wavepacket into a BH, since wavepackets spread with time and here we have runaway time dilation, won't the wavepacket spread uncontrollably? $\endgroup$
    – twistor59
    Dec 1, 2012 at 9:08
  • $\begingroup$ How does an observer look at a wave packet at infinity without disturbing it? $\endgroup$
    – Prathyush
    Dec 4, 2012 at 20:16

1 Answer 1


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

  • $\begingroup$ Won't an object failing in the BH lose its structure due to such high temperature? The chemical forces evidently could not bind the particles at distances comparable to the BH circumference. $\endgroup$
    – Anixx
    Dec 4, 2012 at 20:27
  • $\begingroup$ I can think of several arguments why the uncertainty would grow without apparent limit. But my question is "what is the limit?" You suggest the growth stops when an uncertainty is reached roughly corresponding to the horizon's radius of curvature. But why is this the relevant length scale? The thermal wavelength argument suggests the growth stops when an uncertainty is reached corresponding to the geometrical average of the particle's Compton wavelength and the black hole radius. Somehow it seems one has to apply the right stretched horizon concept? Or is my thermal wavelength argument flawed? $\endgroup$
    – Johannes
    Dec 5, 2012 at 16:57
  • $\begingroup$ @user8260 I think your last sentence got chopped off. (and it was an interesting one!) $\endgroup$
    – twistor59
    Dec 6, 2012 at 10:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.