4 ms·
I thought perhaps the confusion comes from seeing things in geometrized units c=G=hbar=1 and losing track in dimensional analysis, but now I think that's too ch
by raattgift 7y ago
I thought perhaps the confusion comes from seeing things in geometrized units c=G=hbar=1 and losing track in dimensional analysis, but now I think that's too charitable and you're right. Cool whiteboard Penrose diagram though.
How one gets to an entropy is ultimately via no-hair. If gravitational multipoles "bald" away in about a horizon light-crossing time, then we have no chance of distinguishing between having formed the black hole from (classically) one collapsing homogeneous spherical shell of neutral dust of mass M (and no angular momentum), or two such shells each of mass M/2 and each with identical-but-differently-signed angular momentum, or three such shells whose mass sums to M and whose angular momenta sum to zero, and so on. The shell count and the details of their individual angular momenta are "lost" from the black hole.
The spacetime-filling metric tensor encodes all of this into something that far from the infalls of these shells everything looks like Schwarzschild far from the central mass. This is a bit frustrating if the shells of dust are eternally present, because it doesn't give a strong hint about an observer -- who is far from the central mass at all times -- should slice up spacetime so that the metric is time-dependent. (Breaking the maximal symmetries of the black hole at some point can, though, but we are avoiding that by using these classical shells: if we choose an observer who is sufficiently far away from the shells then at all times they look pointlike (or more precisely always like a central spherically symmetric uncharged mass that is the sum of the black hole and all the shells we throw in, and the observer struggles to determine whether all the mass is within the Schwarszchild radius)).
The frustration arises because we can construct the situation so that for a far future observer there is no way to observe how many of these shells were in that observer's distant past, nor what their individual masses, momenta, and charges were: only that all that remains is a nonzero mass, zero angular momentum, and zero charge occupying such a small solid angle of the observer's sky that the observer concludes it is a black hole.
The Boltzmannian macrostate is the black hole and its mass, angular momentum and spin; the microstates are the details of the shells, starting with their individual macrostates (each shell has some mass, angular momentum, and charge), and then considering their internal degrees of freedom (dust particles can be of arbitrarily small mass without breaking our spherical-and-homogeneous condition). Thus the "balded" state can have arbitrarily high entropy, because we could have thrown in any number of shells each containing any number of dust particles, because we are in a classical setting where infinitesimals are OK. If we massively increase the particle-mass and reduce the particle number of a shell (constant M>0, J=0, Q=0), we still get the same macrostate compared to arbitrarily lowering particle mass and increasing the particle number to keep the same M>0, J=0, Q=0.
If we were to be like Merlin or similar fictional characters and travel with a different arrow-of-time, starting with closer-in view of the black hole as the only thing in the test universe of pure Schwarzschild, we would not be able to predict whether in our personal "future" the black hole will spontaneously spit out one shell of mass M, or two shells each of mass M/2 with an angular momentum summing to 0, or three shells, etc etc. Not being able to predict successor time-indexed spacelike slices of spacetime given the entirety of data on one slice frustrates the initial values approach which works remarkably well in much of physics. (Singularities appear in other physical theories that are far from fundamental, such as Euler/Navier-Stokes fluid dynamics, and they too frustrate initial values studies of systems in which they can arise).
("Initial" is a free choice, and generally means where the physicist starts from, rather than initial as in the sense of in the ultra-distant past. It's fairly common to choose as an initial values surface some final state and work backwards from there.)
The quantum picture is a bit more complicated because of the vacuum state problem (I do not have time to get much into that right now[1]) but one can think of throwing in a bunch of neutral hydrogen atoms of mass M versus a bunch of helium atoms of mass M: after "balding", we are again left with only having mass M, angular momentum summing to zero, and no charge: all the quantum information is in the balded black hole. (There are issues about keeping the metric tensor indistinguishable from Schwarzschild in the quantum case; and about how one defines a quantum vacuum).
The relevant difference here is that we expect that "Merlin" in the quantum case must only see QFT observables in the stress-energy tensor as he watches the stable black hole spontaneously emit matter, and this is where \hbar comes in in the equation you found.
We still think about Boltzmann entropy in the same way though: all the quantum numbers and degrees of freedom in the atoms we throw in result in a no-hair black hole with some mass M, some angular momentum (here we carefully keep it at zero), and some other quantum numbers. The entropy of a black hole is thus extremely high.
Sorry that this comment is a bit messy due to lack of time. Also, sorry, I'm not sure when I'll be able to clarify or comment further if needed.
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[1] (Briefly, some people are unhappier in the quantum case where Merlin cannot predict what QFT observables would appear and when, than in the classical case when he predicts Schwarzschild black hole forever in one arrow-of-time direction, but can't predict shells and shell-components in the other. In particular, the quantum case introduces black hole evaporation, and what gets spit out "spontaneously" in the future does not seem to be what fell in. In the sketch quantum BH above Merlin sees a spitting out of hydrogen or helium atoms, whereas ordinary observer sees an evaporation into mostly just photons (with possibly neutrinos and other massive things in the very late stages of evaporation, but probably never neutral hydrogen or helium atoms (too heavy and too bound during early evaporation, too easy to ionize or transmute during late evaporation)).