CPT (M?) symmetries in Kerr-Newman metric

PeterDonis

My terminology here is based on the connection between thermodynamic entropy and information-theoretic (or Shannon) entropy. See here:

http://en.wikipedia.org/wiki/Entropy...rmation_theory

Basically, the number of bits of entropy is the number of bits of information that would be required to completely specify the system's state, given the macroscopic information you have. For a BH, the "macroscopic information" would be its mass, charge, and spin.

Strictly speaking, a Planck mass BH would have *zero* bits of entropy since it has only 1 possible internal state (and therefore no information is required to specify what state it is in); so I was off by 1 bit in my previous post. (Also, strictly speaking, by "Planck mass BH" I really meant "Planck mass BH with zero charge or spin"--I don't know offhand how the presence of charge or spin changes things at this scale.)

True, the reasoning I gave only applies to BH's, not to naked singularities. So there could still be room for a theory that said that, for example, "fundamental particles" are really Kerr-Newman geometries with charge/spin greater than the "critical value" that produces a naked singularity.

michael879

Why would a planck mass BH have 1 internal state while a black hole with mass M > M_P would have more? Sorry to be nitpicking, but your logic just seems kind of circular. From the papers I've read I had the impression "1 bit" was defined as the entropy contained in a planck mass black hole. This number happens to be 4π, so I don't really see why 1 bit would necessary correspond to an entropy of 4π.

Well every fundamental particle found does satisfy M² < a² + Q², so general relativity would describe all of them as naked Kerr-Newman ring singularities. And since we're talking about fundamental particles and black holes, entropy really doesn't come into the equation at all (I've been assuming a naked singularity this whole time).

PeterDonis

That's one of the $64M questions of quantum gravity. Nobody knows the answer because nobody has a "kinetic" model of BH's that derives their macroscopic properties by statistical averaging over microstates. But we can infer just from the macroscopic facts about BH entropy that the number of internal states must go up with horizon area (and hence with the mass), since the entropy itself does. The fact that entropy = number of bits needed to completely specify the internal state (or, equivalently, the logarithm of the total number of possible internal states) is not specific to BH's; it's a general fact about entropy in any physical system. We just don't know precisely *what* the internal states are for a BH that the macroscopic facts about BH's are derived from.

PeterDonis

Only if one accepts that GR continues to be valid in this regime. Most physicists believe that the singularities are signs that GR, as a theory, cannot cover these regimes; or, equivalently, that GR is only valid up to a certain very high but finite spacetime curvature, above which some other theory has to take over. This is connected to the view of GR as an "effective field theory" that I mentioned before.

PeterDonis

Also, for fundamental particles to actually be Kerr-Newman "super-extremal" BH's, they would have to be confined within a small enough radius--which since their masses are all much, much smaller than the Planck mass, would have to be much, much smaller than the Planck length. But the smallest lengths that we have observed fundamental particles to be confined to are much, much larger than the Planck length. So at best, we could possibly model them as rotating massive objects with r >> M. But it's actually not clear that the exterior vacuum geometry of rotating massive objects is still Kerr-Newman, except asymptotically; there is no analogue to Birkhoff's theorem for rotating spacetimes.

None of this is intended to say that this set of ideas is not very interesting; it is. It's just not as straightforward as it may look at first.

michael879

Thats my point, setting the planck mass as the "1 bit" scale is complete speculation. By saying you can't have a black hole with a mass less than the planck mass you're making an assumption with no experimental or theoretical evidence. Don't get me wrong, its a very sensible limit, since the planck scale is where quantum gravity effects should starts to dominate. But to say the minimum mass black hole is exactly 1 planck mass seems overly restrictive given the limit information we have.

I know all that, all I was saying was that GR would describe fundamental particles as kerr-newman naked singularities. Bringing GR to that scale is highly speculative though, as you mentioned.

Actually you'd be suprised at how big fundamental particles would actually be if described by the Kerr-Newman geometry. The electron, for example, would have a "radius" of a=3x10^-13. Because of the warped spacetime the a parameter isn't really the radius, but 2πa is the circumference, so from far away the radius would appear to be a. The more massive particles would have smaller radii, but not significantly.

Also, I was treating the particles as real ring singularities, not extended objects so the Kerr-Newman geometry would apply.

PeterDonis

I certainly agree that there is no experimental evidence, but I wouldn't say there is no theoretical evidence. The formula for BH entropy in terms of area in Planck units is not an approximate formula, and it requires a BH with area of 4 in Planck units to have an entropy of 1. I would say that counts as theoretical evidence. (I realize that area of 4 in Planck units does not translate to mass of exactly 1 in Planck units, because there is a factor of pi in there; I didn't actually mean to imply that the "minimum" BH mass on this view was *exactly* the Planck mass, just that that's the right scale.)

How are you deriving this number? It looks like the electron's Compton wavelength.

michael879

oooo, ok I see what you're saying now. Thats actually an interesting point, since an entropy of O(1) clearly implies O(1) microstates, which will decrease logarithmically. In order to have a microstate of exactly 1, you would need 0 entropy, which doesn't appear to be possible with the entropy formula. Of course non-integer numbers of microstates makes no sense (although it might for a black hole where the microstates are hidden...), so by the time you're in O(1) something must change.

However if non-integer "number of microstates" is possible, the minimum black hole mass is 0 which would correspond to 0 entropy. Whether or not these are possible I don't know. Part of me is saying no, because black holes lose entropy to the outside world through radiation, and in the real world non-integer microstates is not possible. The other part of me is thinking that if you take quantum mechanics into account non-integer microstates may be possible (i.e. continuous rather than discretized entropy). I only know that entropy is discretized classically. I don't really remember quantum statistical mechanics very well.


a = J/(Mc), J = h/2π, M = m_e
And yes it is exactly the Compton radius. Just another of the striking "coincidences" I was mentioning earlier

PeterDonis

Ah, OK, so basically the geometric "radius" associated with the electron's spin turns out to be its Compton radius. Small point: J should be h/4 pi since the electron has spin 1/2.

michael879

yea but a = R to an observer at infinity, where R is the radius of the ring singularity. And yea, I missed a factor of 1/2 .

But anyway, the point I was trying to make way back was that most physicists start with the assumption that QFT is correct, and we are just lacking the correct model to describe the universe (i.e. the quantum gravity terms, and whatever extra terms are needed to solve the various problems). This boils down to using QFT to recreate gravity as an "emergent" macroscopic approximation. There is, however, the other route of assuming GR is correct and trying to recreate QFT with it. I'd say the experimental confirmation of GR is much more convincing than the experimental confirmations of QFT (which isn't to say they aren't convincing, but all of the experiments are riddled with unknowns and tiny invisible "objects").

If you remove the cosmic censorship hypothesis (which is complete speculation IMO), you can include naked singularities in GR. These objects possess many of the qualities present in quantum particles (non-determinism being the main one). Additionally, I don't know how true this is but I read in a paper on the subject that an infinitesimal number of initial conditions will lead to a direct interaction with the naked singularity, making it virtually invisible. Clearly a many-body naked singularity solution is intractable, so proving or disproving that GR can recreate quantum effects is not an easy task. I was trying to make some progress with CPT symmetry, but it turns out I've shown nothing. Black holes obey C, P, and T symmetries independently, and so do fundamental particles if you ignore the strong and weak forces. It would be interesting to try to give a black hole charge in an SU(3) gauge field though...

PeterDonis

This is one thing I've always wondered about: shouldn't conserved charges for other fields besides the electromagnetic field count as additional "hair" that a black hole could have? And therefore shouldn't there be a generalization of the Kerr-Newman metric that includes the additional charges as well as Q (and J, or a, for angular momentum)?

michael879

Yea I wondered this myself a while back. I derived the classical "Maxwell's" equations for an SU(N) gauge field, but they were so much more complicated than the U(1) case I gave up trying to apply them to a black hole. Hypothetically a black hole should be capable of holding any charge. However I read a paper that claimed to prove that a black hole couldnt be charged with respect to a massive vector field (weak force), or a massive scalar field (nuclear strong force), although I couldn't really follow it. Their conclusion was that a black hole couldnt interact weakly or have baryonic charge. However a black hole could probably have charge in a massless SU(2) field (which is what the weak force is before the Higgs field is applied), and a color charge (which would just be hadronized away in the macroscopic case).

Now that I was reminded of it I'm actually going to try to work out what the metric of a black hole charged under an SU(N) gauge field would be.