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CPT (M?) symmetries in KerrNewman metric 
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#37
Mar2812, 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 4vector. 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 BoyerLinquist 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. 


#38
Mar2812, 11:41 AM

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#39
Mar2812, 12:18 PM

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#40
Mar2812, 01:58 PM

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Let me expand on that some more. An excellent reference for KerrNewman spacetime is Matt Visser's article here: http://arxiv.org/abs/0706.0622 He describes the KerrSchild 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 EddingtonFinkelstein 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 nondiagonal 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 coordinatesit 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 KerrSchild coordinates on KerrNewman spacetime, as compared to BoyerLindquist coordinates; time reversal leaves the surfaces of constant BoyerLindquist time invariant, but it does *not* leave the surfaces of constant "ingoing" KerrSchild time invariant. So all the sign changes must cancel in the BoyerLindquist form to leave the ds^2 of an actual, physical line element invariant (and they do), but they should *not* cancel in the KerrSchild form (and they don't). 


#41
Mar2812, 03:05 PM

P: 630

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 4potential issue? The 4potential is a "physical" object with well defined transformation properties. Why would time reversal completely change the 4potential?



#42
Mar2812, 03:57 PM

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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 2spheres, *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. 


#43
Mar2812, 04:03 PM

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#44
Mar2812, 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^22Mr+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\rightarrowdt[/itex], you must flip M and r. 


#45
Mar2812, 10:20 PM

P: 630

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)^22(M)(r)+a^2}(dr)\rightarrowd\bar{t}_{M\rightarrowM,r\rightarrowr}[/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)yax}{(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\rightarrowM,Q\rightarrowQ,r\rightarrowr, J\rightarrow J}[/itex] and [itex]ds^2 = g_{\mu\nu}dx^\mu dx^\nu \rightarrow g_{\mu\nu}dx^\mu dx^\nu_{M\rightarrowM,Q\rightarrowQ, J\rightarrow J,r\rightarrowr}[/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 KerrNewman spacetime is time reversal invariant! This is somewhat disappointing considering my original goal of showing a CPT symmetry in the KerrNewman geometry (which should be the geometry of fundamental particles). It seems to be trivial now though, and there is no CP violation. 


#46
Mar2912, 02:56 AM

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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 freefalling 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). 


#47
Mar2912, 11:43 AM

P: 630

[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. 


#48
Mar2912, 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:
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. 


#49
Mar2912, 01:08 PM

P: 630

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 antiparticle. I'm actually kindof 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. 


#50
Mar2912, 01:40 PM

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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 entropyi.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. 


#51
Mar2912, 02:19 PM

P: 630

*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) 


#52
Mar2912, 02:33 PM

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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.) 


#53
Mar2912, 03:45 PM

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#54
Mar2912, 07:02 PM

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