CPT (M?) symmetries in Kerr-Newman metric


by michael879
Tags: kerrnewman, metric, symmetries
michael879
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#37
Mar28-12, 11:09 AM
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Also, I just noticed you're over complicating the situation. The line element is given simply by
[itex]ds^2 = g_{\mu\nu}dx^\mu dx^\nu[/itex]
The "metric terms" are contained in g, and dx is a 4-vector. Both g and dx will change under Lorentz transformations, but the changes will cancel out to leave [itex]ds^2[/itex] unchanged. It is trivial to show that the Boyer-Linquist line element and metric transform correctly under time reversal.

*edit* Also, none of the terms in the line element equation will change sign. You had the right idea but you made a few sign mistakes.
PeterDonis
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Mar28-12, 11:41 AM
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Quote Quote by michael879 View Post
I totally agree with what you just said, but I'm talking about the Kerr-Schild form of the metric not the Boyer-Linquist one.
But the property of time reversal invariance is independent of the form of the metric. Changing the form of the metric just corresponds to changing coordinate charts, which does not affect any physical quantities. So if two different forms of the metric appear to be giving two different answers, we must be doing something wrong. I'll take another look at the Kerr-Schild form and work through the computation in more detail.

Quote Quote by michael879 View Post
Also, I just noticed you're over complicating the situation. The line element is given simply by
[itex]ds^2 = g_{\mu\nu}dx^\mu dx^\nu[/itex]
The "metric terms" are contained in g, and dx is a 4-vector. Both g and dx will change under Lorentz transformations, but the changes will cancel out to leave [itex]ds^2[/itex] unchanged.
Yes, I know; that was my point, that you have to look at sign changes in both g *and* dx to show time reversal invariance. Just looking at the formula for g is not enough. See below.

Quote Quote by michael879 View Post
*edit* Also, none of the terms in the line element equation will change sign. You had the right idea but you made a few sign mistakes.
I may have; but again, my general point is that "none of the terms in the line element equation will change sign" is true *only* if you correctly analyze the correspondence between specific line element expressions (i.e., with specific values inserted for the coordinate differentials) and specific physical line elements. If you just look at the general *formula* for the line element, with coordinate differentials not assigned any specific values, formally the sign of the J dt dphi term *does* change under time reversal, because you have changed coordinate charts; the line element equation with the sign of J flipped refers to a different chart than the original equation did, one in which the direction of the time axis is reversed. But a given *physical* line element will have the same actual number for ds^2 as it had before, because the coordinate differentials describing it will change--specifically, the sign of dt that describes that physical line element will flip.
michael879
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Mar28-12, 12:18 PM
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Quote Quote by PeterDonis View Post
But the property of time reversal invariance is independent of the form of the metric. Changing the form of the metric just corresponds to changing coordinate charts, which does not affect any physical quantities.
Well that was kindof my point. Lets just assume time reversal causes R, M, and Q sign changes. This would make the Kerr-Schild form of the metric transform correctly, and keep the line element invariant. It would also work for the Boyer-Lindquist form. The Boyer-Lindquist form doesn't disagree with the Kerr-Schild form if you make the above assumption, it just doesn't enforce that assumption (i.e. it doesn't prove or disprove it). And if the Boyer-Lindquist form is so clearly time reversal invariant, the Kerr-Schild form needs to be as well (which forces the RMQ changes).
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Mar28-12, 01:58 PM
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Quote Quote by michael879 View Post
Lets just assume time reversal causes R, M, and Q sign changes...And if the Boyer-Lindquist form is so clearly time reversal invariant, the Kerr-Schild form needs to be as well (which forces the RMQ changes).
No, I don't agree. As I've said before, I don't think these additional sign changes have physical meaning. I think what's actually going on is that the Kerr-Schild form of the metric has two forms, "ingoing" and "outgoing", which are not identical; instead, one is the time reverse of the other. This is different from the Boyer-Lindquist form, which is its own time reverse.

Let me expand on that some more. An excellent reference for Kerr-Newman spacetime is Matt Visser's article here:

http://arxiv.org/abs/0706.0622

He describes the Kerr-Schild form in "spherical" coordinates first, which makes it easier to understand what's going on (at least for me). As you can see from his article, there are indeed two forms of the metric in these coordinates: an "ingoing" form and an "outgoing" form. This is similar to Eddington-Finkelstein or Painleve coordinates for Schwarzschild spacetime. And just as with those coordinates, the "ingoing" form, when time reversed, gives the "outgoing" form, and vice versa. So the sign changes you are seeing on time reversal are *supposed* to be there.

Let's go back to Painleve coordinates on Schwarzschild spacetime as an easier example to see what's actually going on. I think we established in earlier posts that the time reverse of the standard "ingoing" Painleve coordinates, which have a - dt dr term as the only non-diagonal term, is a line element with a + dt dr term (because the sign of dt flips), which is what you see in "outgoing" Painleve coordinates. So this line element expression *does* change under time reversal, *even* when we take into account the sign flips in both the form of g and in the coordinate differentials themselves (and I agree that we have to do that to see what happens to an actual physical line element).

Now suppose we thought this sign flip in the dt dr term was "wrong", and tried to "fix" it by changing the signs of M and/or r. Changing the sign of M alone, which would "fix" the sign of the dt dr term, won't work, because M also appears in the dt^2 term, and switching M's sign does *not* leave that term invariant. The same goes for changing the sign of r alone. And switching the sign of *both* M and r, which would leave the dt^2 term invariant, doesn't work either, because that would make the sign of the dt dr term flip again (the sign flips in M and r cancel and we're left with the sign flip in dt)! So trying to "fix" things by flipping the signs of M and/or r won't work (and that's OK, since as I've said, those sign flips don't make sense physically anyway).

Instead, we have to recognize that in Painleve coordinates, unlike in Schwarzschild coordinates, the surfaces of "constant time" are *not* left invariant by time reversal! Put another way, in Schwarzschild spacetime, the surfaces of constant Schwarzschild time t are taken into themselves by the time reversal transformation. But the surfaces of constant Painleve time are *not*; rather, time reversal takes the surfaces of constant "ingoing" Painleve time into surfaces of constant "outgoing" Painleve time (which are *different* surfaces), and vice versa. And *that* means that a line element which is, say, "pure" dr in "ingoing" Painleve coordinates (i.e., it lies completely within a single surface of constant ingoing time) will *not* be "pure" dr in "outgoing" Painleve coordinates--it will have a dt component as well. And vice versa.

So with Painleve coordinates, when figuring out the impact of time reversal, it's not enough just to look at the sign flips in dt and other things; you also have to take into account that coordinate differentials which were zero before may be nonzero after (or vice versa). The same thing applies to Kerr-Schild coordinates on Kerr-Newman spacetime, as compared to Boyer-Lindquist coordinates; time reversal leaves the surfaces of constant Boyer-Lindquist time invariant, but it does *not* leave the surfaces of constant "ingoing" Kerr-Schild time invariant. So all the sign changes must cancel in the Boyer-Lindquist form to leave the ds^2 of an actual, physical line element invariant (and they do), but they should *not* cancel in the Kerr-Schild form (and they don't).
michael879
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#41
Mar28-12, 03:05 PM
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ok I'm still not really satisfied with just accepting that the line element, which should be a Lorentz scalar, just changes for these coordinate systems... However, ignoring that for a second, how do you explain the 4-potential issue? The 4-potential is a "physical" object with well defined transformation properties. Why would time reversal completely change the 4-potential?
PeterDonis
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Mar28-12, 03:57 PM
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Quote Quote by michael879 View Post
ok I'm still not really satisfied with just accepting that the line element, which should be a Lorentz scalar, just changes for these coordinate systems...
Once again, the actual, physical length of a given actual, physical line element does *not* change. That's what "Lorentz scalar" means. But the exact *description* of a given actual, physical line element in terms of coordinate differentials *does* change, because time reversal means changing coordinate charts, and any change of coordinate charts changes the description of a given actual, physical line element in terms of coordinate differentials. What was confusing us was the assumption that the *only* change "time reversal" can make to the coordinate differentials is to flip the sign of dt; but in fact, for coordinates like Painleve or Kerr-Schild, that is *not* the case.

So to obtain the same actual number for ds^2 for the same actual, physical line element, the functional form of the line element in terms of the coordinate differentials has to change in other ways besides just what is needed to balance the sign change in dt. We simply haven't computed exactly what those other changes are. For example, I didn't compute what *other* coordinate differentials would change upon "time reversal" of Painleve coordinates; I simply showed that flipping the sign of dt alone does not leave ds^2 invariant, and no combination of sign flips of M and r can fix that. So obviously other differentials must change as well to keep ds^2 invariant.

The simplest case would be the one I described in my last post: a line element that is purely radial (only dr nonzero) in "ingoing" Painleve coordinates. Time reversal would leave dr the same (since the surfaces of constant r, the 2-spheres, *are* left invariant), but would add a nonzero dt; so there would now be nonzero dt^2 and dt dr terms in the formula for ds^2 in "outgoing" Painleve coordinates, in addition to the nonzero dr^2 term. The signs of these two new terms are opposite, and they should cancel each other, leaving the (unchanged) dr^2 term as the value of ds^2.

Quote Quote by michael879 View Post
The 4-potential is a "physical" object with well defined transformation properties. Why would time reversal completely change the 4-potential?
It would change the *components* of the 4-potential in the new coordinate chart, as compared to the old. It would not change the actual, physical 4-vector that those components describe.
PeterDonis
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#43
Mar28-12, 04:03 PM
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Quote Quote by PeterDonis View Post
I think we established in earlier posts that the time reverse of the standard "ingoing" Painleve coordinates, which have a - dt dr term as the only non-diagonal term, is a line element with a + dt dr term (because the sign of dt flips), which is what you see in "outgoing" Painleve coordinates.
I should clarify here that this is using the (+---) sign convention, which is the one used in the Wikipedia article on Painleve coordinates.
michael879
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#44
Mar28-12, 07:10 PM
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*sigh* I finally get it, what a let down.

The "dt" coordinate I've been using is actually [itex]\bar{dt} = dt - \dfrac{2Mr}{r^2-2Mr+a^2}dr[/itex]. waaay too much algebra for me to work out, but I'm willing to bet that under a real time reversal, everything will transform correctly.

*edit* that equation alone explains why I was forced to flip the sign of M and r. If [itex]\bar{dt}\rightarrow-\bar{dt}[/itex] and you only want to get [itex]dt\rightarrow-dt[/itex], you must flip M and r.
michael879
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#45
Mar28-12, 10:20 PM
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Ok, so given this new information, lets do a real time reversal: dt→-dt

[itex]d\bar{t} \rightarrow -dt - \dfrac{2(-M)(-r)}{(-r)^2-2(-M)(-r)+a^2}(-dr)\rightarrow-d\bar{t}|_{M\rightarrow-M,r\rightarrow-r}[/itex]

[itex]f \rightarrow \dfrac{G(-r)^2}{(-r)^4+a^2z^2}\left(2(-M)(-r)-(-Q)^2\right)[/itex]
[itex]k_\bar{t} \rightarrow 1[/itex]
[itex]k_x \rightarrow -\dfrac{(-r)x+ay}{(-r)^2+a^2}[/itex]
[itex]k_y \rightarrow -\dfrac{(-r)y-ax}{(-r)^2+a^2}[/itex]
[itex]k_z \rightarrow -\dfrac{z}{(-r)}[/itex]
[itex]A_\mu \rightarrow \dfrac{(-Q)(-r)^3}{(-r)^4+a^2z^2}k_\mu[/itex]

so therefore,
[itex]g_{\mu\nu} \rightarrow \bar{T}[g_{\mu\nu}]|_{M\rightarrow-M,Q\rightarrow-Q,r\rightarrow-r, J\rightarrow J}[/itex]
and
[itex]ds^2 = g_{\mu\nu}dx^\mu dx^\nu \rightarrow g_{\mu\nu}dx^\mu dx^\nu|_{M\rightarrow-M,Q\rightarrow-Q, J\rightarrow J,r\rightarrow-r}[/itex]

Now assuming that the line element for [M,J,Q,r] is the same as that for [itex]\bar{T}[/itex][-M, J, -Q, -r], the line element is invariant under time reversal. And it turns out that it is, so although its far from obvious, the Kerr-Newman space-time is time reversal invariant! This is somewhat disappointing considering my original goal of showing a CPT symmetry in the Kerr-Newman geometry (which should be the geometry of fundamental particles). It seems to be trivial now though, and there is no CP violation.
PeterDonis
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Mar29-12, 02:56 AM
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Quote Quote by michael879 View Post
The "dt" coordinate I've been using is actually [itex]\bar{dt} = dt - \dfrac{2Mr}{r^2-2Mr+a^2}dr[/itex]
For the Kerr-Schild form of the metric, yes, the "time" coordinate is not the same as the "t" coordinate in Boyer-Lindquist coordinates. See below for more discussion of this.

Quote Quote by michael879 View Post
Ok, so given this new information, lets do a real time reversal:
I still don't understand why you're flipping the signs of M and r (or Q either, for that matter). Let me work through this in Painleve coordinates on Schwarzschild spacetime to illustrate how I'm looking at this.

Consider a line element where only dT and dR (I'll capitalize the T and R to make it clear that I mean Painleve T and R, *not* Schwarzschild T and R) are nonzero. I will also write "v" for the "escape velocity", [itex]\sqrt{2M / r}[/tex] to make the formulas easier to read. So in the "ingoing" Painleve coordinates we have (using the +--- sign convention I used earlier):

[tex]ds^{2} = \left( 1 - v^{2} \right) dT^{2} - 2 v dT dR - dR^{2}[/tex]

Now, to make the signs of each term absolutely clear, let's put in absolute value signs and make the sign of dT and dR explicit. Suppose we are talking about the line element along the worldline of a "Painleve observer" who is free-falling into the black hole. Then we have dT = |dT| and dR = - |dR| = - |v| |dT|, where I have also put absolute value signs around v to make it clear that we are talking about the *magnitude* of v only. So we have the actual *number* ds^2 for this physical line element given by:

[tex]ds^{2} = \left( 1 - v^{2} \right) dT^{2} - 2 |v| |dT| (- |v| |dT|) - v^{2} dT^{2} = dT^{2} \left(1 - v^{2} + 2 v^{2} - v^{2} \right) = dT^{2}[/tex]

Now this actual number has to be invariant under time reversal. What does the line element look like in the time reversed coordinate chart T', R'? It looks like this: dT' = - dT, dR' = dR. In other words: dT' = - |dT|, dR' = - |v| |dT|. But that means that, to keep the actual number ds^2 invariant for the same physical line element, the line element equation in T', R' must look like this:

[tex]ds^{2} = \left( 1 - v^{2} \right) dT'^{2} + 2 v dT' dR' - dR'^{2}[/tex]

The sign of the second term must flip to compensate for the fact that dT' = - dT. Substituting dT' = - |dT| and dR' = - |v| |dT| into the above line element will give the same result, ds^{2} = dT^{2}, but *only* with the sign flip in the second term of the above equation. In other words, the time reverse of ingoing Painleve coordinates is outgoing Painleve coordinates. And as I said before, flipping the signs of M and/or r will *not* keep ds^2 invariant under time reversal (because flipping the sign of either one individually flips the sign of *both* v and v^2, which does not leave ds^2 invariant, and flipping the sign of both leaves both v and v^2 invariant, so it does nothing). You *have* to change the sign of the second term in the line element equation.

Now, what about the fact that the Painleve time coordinate T is not the same as the Schwarzschild time coordinate t? The Wikipedia page gives this (using my notation):

[tex]dT = dt - \frac{v}{1 - v^{2}} dr[/tex]

Time reversing this, which is expressed as dt' = - dt and dr' = dr, gives:

[tex]dT' = dt' + \frac{v}{1 - v^{2}} dr'[/tex]

Notice that the sign of dT' is still the same as the sign of dt'; the only change is the sign of the dr term. And since dr = dR (the radial coordinate is unchanged in going from the Schwarzschild chart to the Painleve chart), this shows *why* the sign of the second term in the Painleve line element formula above changes sign under time reversal (whereas the Schwarzschild line element formula is unchanged).

Quote Quote by michael879 View Post
And it turns out that it is, so although its far from obvious, the Kerr-Newman space-time is time reversal invariant!
Still not sure about this because of the sign change in J; as I said before, the time reverse of an object rotating counterclockwise should be an object rotating clockwise (speaking somewhat loosely), which is also a valid solution of the applicable equations, but not the *same* solution.
michael879
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#47
Mar29-12, 11:43 AM
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Quote Quote by PeterDonis View Post
I still don't understand why you're flipping the signs of M and r (or Q either, for that matter).
This time I'm not, I was merely using an algebraic trick to show that
[itex]T[M,J,Q,r] \equiv \bar{T}[-M,-J,-Q,-r][/itex]
which can easily be shown to leave the line element unchanged. I made some minor mistakes regarding J, but I think my conclusion was correct.
Quote Quote by PeterDonis View Post
[itex]dT' = dt' + \frac{v}{1 - v^{2}} dr'[/itex]
this is the main difference between the Painleve metric and the Kerr-Schild metric. The "time" coordinate in the Painleve metric changes sign under time reversal. Time reversal on the Kerr-Schild metric does not change the sign of the "time" coordinate, because the second term is just a function of M and r. If you want to make time reversal change the sign of the time coordinate, you must make M and r negative. I now understand that this is merely an algebraic trick to relate the two time coordinates, and not a physical phenomena

Quote Quote by PeterDonis View Post
Still not sure about this because of the sign change in J; as I said before, the time reverse of an object rotating counterclockwise should be an object rotating clockwise (speaking somewhat loosely), which is also a valid solution of the applicable equations, but not the *same* solution.
Yes the time reverse of an object rotating counter clockwise IS an object rotating clockwise. However any Lorentz invariants regarding that object should not change under time reversal. Classical invariants of such an object are M, Q, [itex]\vec{J}^2[/itex], etc. The line element is just another Lorentz invariant and should never change under a Lorentz transformation. The metric, however, does change and becomes the metric of a Kerr-Newman black hole spinning in the opposite direction.
PeterDonis
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Mar29-12, 12:46 PM
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Looks like we've finally got this one nailed. The only thing I still wanted to comment on is this:

Quote Quote by michael879 View Post
the Kerr-Newman geometry (which should be the geometry of fundamental particles)
Only if fundamental particles are actually black holes. The problem with that hypothesis is that all known fundamental particles have masses well below the Planck mass, and although we don't have a complete theory of quantum gravity yet, what we do know about quantum gravity strongly suggests, AFAIK, that the smallest mass a black hole can have is the Planck mass.

I believe at least one physicist (can't remember who) has speculated that what we call "fundamental particles" are really just bound states of the fundamental entities of some underlying theory (perhaps string theory), while what we call "black holes" are just free states of the same fundamental entities. The motivation for this is that "fundamental particles" have a discrete mass spectrum, whereas "black holes" have a continuous mass spectrum. I don't know if any attempt has ever been made to expand on this theoretically.
michael879
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Mar29-12, 01:08 PM
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The planck mass thing is just some order of magnitude thing I believe. Below the planck mass is when quantum effects of gravity should kick in. However there are some striking similarities between fundamental particles and black holes. The most surprising is probably the g=2 factor of both. There's also the no hair theorem, and some others, but what got me interested was the fact that every fundamental particle, if it were a black hole, would be naked. When you introduce naked singularities you get all kinds of strange stuff like nonlocality, nondeterminism, and strange causal behavior. However these three things are fundamental concepts in quantum, and bell's inequality actually proves that no interpretation can escape all three.

Regardless of whether or not particles are black holes, its hard to argue with the similarities they have. I was just trying to investigate the CP symmetry of QFT that takes a particle to its anti-particle. I'm actually kind-of leaning towards calling my R transformation a CP transformation, and my original T' reversal (which is not a time reversal) a T transformation. At distances far away from the black hole, R = CP and T' = T. The black hole would have a CPT symmetry, but applying CP or T would give it negative charge, negative parity, and negative active gravitational mass.

*edit* also, I know that the mass/charge/angular momentum are all discretized for fundamental particles, but charge and angular momentum are related by magnetic monopoles, and I was hoping to find some stability constraint on naked singularities that might give an additional requirement to constrain mass.
PeterDonis
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Mar29-12, 01:40 PM
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Quote Quote by michael879 View Post
The planck mass thing is just some order of magnitude thing I believe. Below the planck mass is when quantum effects of gravity should kick in.
Not necessarily *below* the Planck mass, but around that order of magnitude of spacetime curvature--that condition is more easily expressed as radius of curvature approximately equal to the Planck length, or energy density equal to one Planck mass per Planck volume (Planck length cubed).

However, that wasn't quite what I was referring to. I was referring to the fact that, as we currently understand black holes, a Planck mass BH has 1 bit of entropy--i.e., the minimum possible. Another way of putting this is that there is only one possible quantum state corresponding to a Planck mass BH. BH's of larger mass basically just add more bits, meaning more possible states and more entropy. But there's no way to subtract any more bits to get a BH of smaller mass.

Quote Quote by michael879 View Post
However there are some striking similarities between fundamental particles and black holes.
Yes, this whole area is a fascinating field for speculation.
michael879
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Mar29-12, 02:19 PM
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Quote Quote by PeterDonis View Post
a Planck mass BH has 1 bit of entropy
Define 1 bit? (in the sense you are using it)

*edit* also, naked singularities aren't black bodies so they don't obey the same thermodynamic properties as black holes (i.e. they have no temperature)
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Quote Quote by michael879 View Post
Define 1 bit? (in the sense you are using it)
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.)

Quote Quote by michael879 View Post
*edit* also, naked singularities aren't black bodies so they don't obey the same thermodynamic properties as black holes (i.e. they have no temperature)
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
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Mar29-12, 03:45 PM
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Quote Quote by PeterDonis View Post
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.)
Why would a planck mass BH have 1 internal state while a black hole with mass M > MP 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.

Quote Quote by PeterDonis View Post
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.
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).
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Mar29-12, 07:02 PM
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Quote Quote by michael879 View Post
Why would a planck mass BH have 1 internal state while a black hole with mass M > MP would have more?
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.


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