Classical black holes?

• How big should the black hole be so we can consider it to be classical?

• When they claim that we can not probe shorter distances than the Planck length, can it be true?

The argument says that, doing experiments with higher energies we would create even bigger black hole. But if Quantum Mechanics is valid up to the shortest distances (don't see the reason why it shouldn't be), then, by creating the black holes the informations about interactions would be conserved on the horizon.

• Would this return us to observe larger distances again, or could there be some new things going on even at trans-planckian lengths?
-

Consider the neutral, Schwarzschild black holes in $d=4$. If they're large, they're solutions to Einstein's equations. However, Einstein's equations have higher-derivative correction terms that go like $(L^2 R)^k R$ where $L$ is a characteristic length and $R$ is the Riemann curvature, schematically: many possible contractions are possible.

These terms may only be neglected if $L^2 R\ll 1$ i.e. if the characteristic size (curvature radius) of the black hole is much greater than the coefficient $L$. Because the coefficient $L$ in the effective action (governing the higher-derivative terms) is of order the Planck length, the higher-derivative terms may only be neglected for black holes much larger than the Planck length.

At least parameterically (up to numerical constants of order one), it's the same condition as having the mass much greater than the Planck mass, too. Also, black holes obeying such an inequality have the lifetime (of Hawking evaporation) that is much longer than the Planck time, too.

It makes no sense to consider black holes that are smaller than the Planck length because the solutions to Einstein's equations aren't relevant: one should consider the solutions to the equations including all the higher-derivative corrections. The nature of geometry changes dramatically below the Planck scale. Also, black holes that are smaller than the Planck length have the evaporation lifetime shorter than the time needed for light to get from one side of the hole to the other. It makes no sense. Such an object is so unstable that it makes no sense to consider it.

In reality, the black holes are described by microstates that are discrete: there are only $\exp(A/4G)$ microstates for a given black hole. This quantization becomes important for very light black holes. The number of "black hole microstates" that are lighter than the Planck mass is actually finite: the microstates are the elementary particles we know. The approximation of Einstein's equations totally breaks down. Instead, these particles should be described as point-like perturbations on a pre-existing flat spacetime background.

When one tries to collide particles (e.g. protons) of ever greater energies, he may probe ever shorter distances. However, this process breaks down once the energy of the center-of-mass energy of the protons approaches the Planck mass (a universe-sized collider is needed). In the critical transition regime, one expects to probe distances of order the Planck length. However, one actually creates a black hole of the same radius. When the energy of the protons is increased beyond that, one no longer studies "ever shorter distances"; instead, one creates ever more massive i.e. ever larger black hole. You can see that the minimum distance one may really probe is comparable to the Planck length.

In various mathematical contexts, one may "consider" distances that are shorter than that. However, it's important that physics doesn't obey the naive laws of physics in an ordinary classical geometry, as we know it from long distances. The most general generalization has to be assumed to replace our usual framework of "objects on a pre-existing spacetime geometry".

On the other hand, one must realize that the new Planckian effects and restrictions only occur if some Lorentz-invariant distances become formally shorter than the Planck length. For example, the wavelength of a photon isn't Lorentz-invariant. That's why the wavelength may be arbitrarily short, arbitrarily shorter than the Planck length, too. That's guaranteed by the Lorentz contraction in special relativity. Only distances describing the internal structure of something as seen from the "rest frame" start to behave wildly once we reach the Planck length.

The information preservation is an entirely different and huge topic. Yes, in some sense, the information is preserved at the horizon. It returns when the black hole evaporates. For very small black holes, this description isn't too useful because we can't take the "location of the horizon" too seriously as the quantum corrections and fluctuations are very strong, as argued above.

-
Thanks. Excited when saw your answer, than carefully read it. – Newman Apr 26 '12 at 17:19
@luboš-motl It's not clear from your post which theory you base your calculations on, in particular the (L^2R)^kR part is very presice, but unclear where you get it from. And your part about needing a particle collider the size of the universe to get a plank energy of center of mass collision for protons, the size of the universe is not constant, nor it is a known value, and if you used the current size of the observable universe that sounds like an accident. They both make it unclear where your numbers are coming from (though most of the rest I know where it is coming from). – Timaeus Jan 4 '15 at 3:32
Dear Timaeus, every quantum mechanical theory will imply that the long-distance effective theory contains not just the Einstein-Hilbert term $R$ but also the higher-derivative terms of the form I mentioned. One may see the explicit form of these terms in string theory which is really the only consistent theory of quantum gravity we know but the arguments that these terms arise may be presented independently of string theory, just with the knowledge of the renormalization group etc. So what you want to learn is either "renormalization group" or "string theory" or both. – Luboš Motl Jan 4 '15 at 18:48
The size of the Universe was not constant in the past but the size of the visible Universe actually is converging to a finite constant, very close to the current size, because in the asymptotically far future, and it's almost the case now as well, the Universe will be (close to) a de Sitter space with a fixed spacetime curvature radius (given by dark energy, i.e. the cosmological constant), and the size of the visible universe up to the event horizon is a fixed multiple of the curvature radius. All these radii are close to "dozens of billions of light years". – Luboš Motl Jan 4 '15 at 18:51