# CPT (M?) symmetries in Kerr-Newman metric

1. Feb 29, 2012

### michael879

So the confusion I'm having here really has to do with parity inversion in spherical (or boyer-linquist) coordinates. I've been looking at the discrete symmetries of the Kerr-Newman metric, and I've noticed that depending on how you define parity-inversion, you can get very different results.

Method #1: r->r, theta->pi-theta, phi->pi+phi => x->-x, y->-y, z->-z
When you use this parity inversion, you find that C, P, and T symmetries are all independent symmetries of the metric.

Method #2: r->-r, theta->theta, phi->pi+phi => x->-x, y->-y, z->-z
Using this parity inversion, you find that the resulting metric has not transformed correctly. However if you also do a time inversion (t->-t), you find that the resulting metric describes a Kerr-Newman black hole with parameters (-M,-Q,-J). This suggests that the Kerr-Newman metric only has an MCPT symmetry, rather than independent C, P, T symmetries.

I know I must be missing something, but these two results are astoundingly different. In the first case, you just get the trivial classical result that C, P, and T symmetries all hold independently. However in the second case you find that they are not independent, that there is a new M symmetry, and that the only true symmetry is MCPT.

What am I doing wrong?

*note* M is the active gravitational mass, not the passive gravitational or inertial mass. Method #2, if true, would suggest that anti-matter has negative gravitational mass. However, this is not in conflict with either the equivalence principle or the experimental result that antimatter has positive inertial mass. Hypothetically, it would be possible, within the general relativistic framework, to have an object with gravitational mass of an opposite sign than its inertial and passive gravitational mass. Since the only anti-matter objects we observe are microscopic, there would be absolutely no way to measure its active gravitational mass.

thanks,
Mike

2. Feb 29, 2012

### michael879

Sorry, just a small correction:

Method #2
After the parity transformation the new metric would represent a Kerr-Newman black hole with the parameters (-M, -Q, J)

Also,

Method #3: r->-r, phi->phi, theta->theta, t->-t => x->x, y->y, z->-z
Would result also in the metric of a Kerr-Newman black hole with parameters (-M, -Q, J). This is basically the same as Method #2, except it is simply just a reflection in the x-y plane. Doing the analog for Method #1, where phi->phi, theta->pi-theta, would result in C, P, and T inversions all leaving the metric unchanged, and M symmetry would not hold at all.

3. Feb 29, 2012

### michael879

I've been thinking about it some more, and I might have a possible explanation.

If you think about how the two different parity transformations "move" points through space, Method #1 never sends any points through the center of the ring singularity. It merely rotates the theta coordinate of each point and then the phi coordinate, always staying outside the ring area.
Method #2, on the other hand, first rotates the point in phi, and then send it directly through the ring singularity to -r. I have read that Kerr ring singularities represent "wormholes" to another universe, but it never made much sense to me. However this ambiguity in parity transformations that normally produces no inconsistencies, produces one here and I believe this is why.

If anyone could explain this further it would be great though (possibly by explaining mathematically why the Kerr-Newman metric connects two virtually independent space-times), because I am still very confused.

4. Mar 1, 2012

5. Mar 1, 2012

### Staff: Mentor

Hi michael879, I would suggest trying the same parity transformations on a rotating, charged spherical object (i.e., a rotating charged "star" or planet, instead of a rotating charged BH). Probably even the simpler rotating case would do (no charge). What do you get, and does it depend on which transformation you do?

6. Mar 1, 2012

### michael879

I don't think the metric solution for a rotating charged star is known exactly is it? Anyway, "classically", these transformations are indistinguishable (newton's laws, E&M, special relativity). The thing that I believe makes them different is the presence of the ring singularity. Also, I am treating this singularity as naked, since if it were bounded by 2 event horizons the transformation r->-r would not be physically possible.

*edit* anyway, black holes are the simplest objects to deal with in GR, so why would I go to something more complicated?

7. Mar 1, 2012

### Staff: Mentor

First of all, to make sure the "purely classical" effects of the parity transformations (i.e., in a Euclidean space, with no funky topology like ring singularity) are clear. The transformations you are describing are not all "classically" indistinguishable, at least not if I'm understanding your descriptions right.

Or perhaps you intended them to be classically indistinguishable? In that case, I think you need to reconsider your descriptions:

Method #1 looks fine, and corresponds to a complete "parity inversion" in all the spatial coordinates (P); in fact it is the standard "classical" description of the P transformation.

Method #2, as you describe it, can't be done classically at all: classically, r < 0 is not even allowed, so there is no transformation at all (including a parity one) that swaps the sign of r. This should be obvious for parity transformations since any such transformation takes a sphere centered on the origin into itself, so it can't change r at all.

Also, in method #2 you are transforming phi but not theta, which corresponds to inverting x and y but not z; you have indicated z -> -z, so unless that's a typo you are "mis-translating" Method #2 from spherical into rectilinear coordinates.

Method #3: Same issue as method #2 with inverting the sign of r--as you state it this is not a "classical" transformation at all.

Also, in method #3 you are not inverting theta or phi, so none of x, y, z are inverted; only t is inverted, but that doesn't affect x, y, z at all. In other words, if we correct the "r" issue, classically #3 is a time reversal (T), not a parity transformation at all.

8. Mar 2, 2012

### michael879

Peter, I get what you're saying, but I would argue it's just semantics. Yes, technically spherical (or boyer-linquist) coordinates are only defined for 0 <= r <= infinity, 0 <= theta <= pi, 0 <= phi < 2pi. However, these domains are extended all the time in physics, and in general as long as you are using the correct (x,y,z) vector, the physics all works out.

Take for example a particle rotating around the z axis at a constant angular velocity v. You would describe this particles motion by d(phi)/dt = v. However this is clearly "technically" wrong, since phi would go outside its domain after a single loop.

Instead of further arguing the above (and straying from my OP), let me just make a "new" coordinate system(s) defined by the parameters (r,theta,phi) in terms of (x,y,z) such that:

$x = rsin\theta cos\phi \textit{ } \left(\sqrt{r^2+a^2}sin\theta cos\phi\right)$
$y = rsin\theta sin\phi \textit{ } \left(\sqrt{r^2+a^2}sin\theta sin\phi\right)$
$z = rcos\theta$

where the boyer-linquist definition is in parenthesis.

Assuming no bounds on r, theta, or phi, there are two ways to do the transformation (x,y,z)->(-x,-y,-z). The first one is the usual one I called method #1. The second is to just r->-r (and phi->pi+phi in boyer-liquist). I would argue that in any classical situation, these two transformations are indistinguishable, since in both cases (x,y,z)->(-x,-y,-z) and therefore the same physics must apply to both transformations.

Method #3 was just a simplification where instead of doing the usual parity inversion, I did the inversion z->-z by just doing r->-r

If you disagree with the above, I would ask that you show me some classical system that is invariant under #1 but not #2

Last edited: Mar 2, 2012
9. Mar 5, 2012

### michael879

Ok well I think I figured out what I was doing wrong. Basically, the metric was derived assuming r is non-negative. I noticed that the Schwarzschild had a similar issue, and is much simpler to deal with (since in that geometry r actually represents $|\vec{x}-\vec{0}|$). I rederived the metric assuming r could go negative, and wound up with a similar metric with sign(r) showing up in a lot of places. This new metric was completely invariant under parity transformations.

What bugs me though, is that this very slight mistake led to a prediction that a non-quantum theory is not invariant C, P, and T symmetries, and only invariant under all 3 (and M). Additionally, I have seen MANY references to the ring singularity being a "wormhole" between two asymptotically flat universes. I don't see how this can come out of the metric without making the mistake I made...

Even in the non-naked case, imagine a particle falling into a kerr black hole. When it crosses the outer event horizon, r becomes time-like, and it is forced through the space-like region, past the inner event horizon, into the inner time-like region. Once there, what is stopping it from once again, passing through the inner event horizon? Once passed r would become time-like again and the particle would be forced out of the black hole. I have read that it would emerge in a new universe, where the black hole is naked, rather than exit the black hole in the same universe. However, I just can't see how the metric predicts this...

10. Mar 5, 2012

### Staff: Mentor

Not quite; the *coordinate transformation* you defined assumed that r was non-negative. More precisely, if you let r be negative, then you have *two* sets of (r, theta, phi) coordinate values corresponding to a single set of (x, y, z) values, instead of one, so the coordinate mapping is no longer one-to-one. For the "classical" case, where the space you are dealing with is Euclidean 3-space, that's what invalidates your "method #2" transformation.

Because that arises from something different: recognizing that the manifold in question can be analytically extended. In other words, the "r < 0" portion of Kerr-Newman spacetime does not correspond to *any* set of (x, y, z) values from the Cartesian coordinates on the original "r >= 0" portion. The r < 0 coordinate values refer to a new "piece" of spacetime that is not covered by the original r >= 0 coordinates at all.

The reason this works is that there are worldlines that reach r = 0 without hitting a singularity (because the singularity at r = 0 is a "ring singularity", so not all particles reaching r = 0 will hit it). That is, they come to a finite proper time along the worldline and then, if r can't be less than 0, just "stop". But that doesn't make sense; the particle following the worldline has to go *somewhere*. (I'm giving this in English, which is not ideal; there's a lot of math behind this that I don't have time to put into this post, but can elaborate on if needed later. However, it's worth noting that similar reasoning is what tells us that there has to be a region of spacetime inside the event horizon of a Schwarzschild black hole.)

This is "analytic extension" again; since the "ring singularity" at r = 0 is timelike, the region the particle comes out in *can't* be the same as the one it came in from. See the Penrose diagrams at the link below for a better visualization of what's going on.

11. Mar 6, 2012

### michael879

Ok thanks, that clarified a lot for me. I read up on Kruskal coordinates (which I really should know since Martin Kruskal was my uncle...), and think I understand the issue better. I do have one more question that I can't seem to figure out though:

Under the time inversion transformation t->-t, the metric tensor, the electromagnetic 4-potential, and the angular momentum should transform as follows:
$g_{i0}\rightarrow -g_{i0}$
$g_{ij}\rightarrow g_{ij}$
$g_{00}\rightarrow g_{00}$
$A_{i}\rightarrow -A_{i}$
$A_{0}\rightarrow A_{0}$
$a \equiv \dfrac{J}{M}\rightarrow -\dfrac{J}{M}$

The metric for the Kerr-Newman geometry and the 4-potential are given by:
$g_{\mu\nu}=\eta_{\mu\nu} + fk_\mu k_\nu$
$f\equiv \dfrac{Gr^2}{r^4+a^2z^2}(2Mr-Q^2)$
$k_x\equiv\dfrac{rx+ay}{r^2+a^2}$
$k_y\equiv\dfrac{ry-ay}{r^2+a^2}$
$k_z\equiv\dfrac{z}{r}$
$\dfrac{x^2+y^2}{r^2+a^2}+\dfrac{z^2}{r^2}\equiv 1$
$A_\mu=\dfrac{Qr^3}{r^4+a^2z^2}k_\mu$

therefore, in order to get everything to transform correctly, we need $k_i\rightarrow -k_i$, $k_0\rightarrow k_0$, and $f\rightarrow f$. As you can see by simply plugging in $a\rightarrow -a$, $k_i$ does not transform as expected. Now, I may be wrong in assuming the metric must transform the way I suggested above, so please correct me if I'm wrong. However, since it is a tensor, I'm pretty sure it must.

In order to make the $k_i$ to transform correctly, you must first make r negative such that $r'\equiv -r$. However this messes up the transformation of f, so you must make $M'\equiv -M$. Similarly, $A_\mu$ no longer transforms correctly so you must make $Q'\equiv -Q$. The final step is in noticing that by changing the sign M, you've changed the sign of a (again). Therefore you also need to make the change $J' \equiv J$, so that J has actually flipped sign twice.

The result of all this suggests that in order to make the metric transform correctly under time reversal you must enter the -r "universe", which necessarily changes the effective sign of M, Q, and J (regardless of what actually happens, the -r metric is identical to the metric of a +r black hole with parameters -M,-Q,-J).

Can someone either explain what I'm doing wrong or help me interpret what exactly this means?

Last edited: Mar 6, 2012
12. Mar 6, 2012

### michael879

Also, something interesting about this is that for very large r, and very small M, the kerr-newman solution approximately symmetric under a CP transformation. However, the P transformation changes the pertubative part of the metric, so it is not an exact symmetry. The exact symmetry requires r->-r and M->-M. The metric has another exact symmetry which is J->-J and t->-t. Let's call these the MCPU (U for universe) and JT symmetries. Under neither of these symmetries does the metric transform as a tensor, as it is unchanged in both. However, under a MJU/CPT transformation, both the metric and the 4-potential are unchanged and both transform correctly.

Assuming all my assumptions are true (which I'm skeptical of and would really appreciate someone else double checking them), and noting the similarities between fundamental particles and the naked kerr-newman solution, the above symmetry arguments make some serious predictions about the nature of matter/anti-matter. First of all, it says that CPT symmetry is very close to exact (since J is usually unknown anyway, M is negligible, and U isn't really measurable), but not exact.

The exact symmetry suggests that anti-matter has negative active gravitational mass (which might explain why anti-matter is spread out "evenly" throughout the universe while matter is condensed into planets/stars/galaxies etc). It also suggests a flip in J, which doesn't really mean much when quantum mechanics is taken into account. Finally, it suggests that in order to get the exact symmetry, you must look at the particle from the universe it connects us to.

13. Mar 6, 2012

### Staff: Mentor

Cool!

The fact that it is a tensor does not tell you enough by itself. You need to actually look at the functional form of the metric components. Any metric component that is a function of t will be affected by the transformation; any component that isn't, will not.

Two examples: first, the Schwarzschild metric:

$$d\tau^{2} = \left( 1 - \frac{2M}{r} \right) dt^{2} - \frac{1}{1 - \frac{2M}{r}} dr^{2} - r^{2} \left( d\theta^{2} + r^{2} sin^{2} \theta d\phi^{2} \right)$$

No metric coefficients depend on t at all, so this metric is invariant under a time reversal transformation.

Second, the FRW metric (for the "flat" universe case, k = 0):

$$d\tau^{2} = dt^{2} - a^{2}(t) \left[ dr^{2} + r^{2} \left( d\theta^{2} + r^{2} sin^{2} \theta d\phi^{2} \right) \right]$$

This metric depends on t, but only in the spatial part; so the spatial components (everything that is multiplied by a(t)) change with a time reversal transformation, but the time component does *not*. (In plain English, the time reverse of an expanding universe is a contracting universe, and vice versa, but this reversal only shows up in the *spatial* components, because they are where the scale factor a(t) shows up--a(t) increasing with time turns into a(t) decreasing with time, and vice versa, without affecting the "rate of time flow" at all).

This general principle doesn't just apply to the metric, btw; it applies to everything (including the electromagnetic 4-potential). The only thing in your starting list that I can see that would change is the angular momentum J; but that's because J is *defined* (roughly speaking) in terms of the time rate of change of an angular coordinate (phi) for an observer that is "rotating with the hole" as compared to an observer that is "static at infinity". So J already has a dependence on time "built into" it. But no other variable in the metric does.

14. Mar 7, 2012

### michael879

I see what you're saying, but aren't tensors defined by how they transform under unitary transformations? Under a parity transform every spatial index of a tensor gets a factor of -1 (pseudotensors are of course different). So under a time reversal transformation shouldn't every time index of a true tensor get a factor of -1??

15. Mar 7, 2012

### michael879

O and also, by the above definition only off-diagonal elements would change sign and neither of your examples contain any

16. Mar 9, 2012

### Staff: Mentor

A decent overview of how tensors are defined is given in the Wikipedia page here:

http://en.wikipedia.org/wiki/Tensor

The article talks about transformation properties, but the "transformations" referred to are changes of basis, not unitary transformations. The transformations you are talking about, like parity and time reversal, are not changes of basis.

17. Mar 9, 2012

### Staff: Mentor

Try the Kerr metric, which does have off-diagonal components. It is also independent of the time coordinate t and so is invariant under time reversal.

18. Mar 10, 2012

### michael879

Parity and time reversal ARE just basis transformations, as long as they are applied globally. Saying all (x,y,z)->(-x,-y,-z) is identical to saying the unit vectors x,y,z->-x,-y,-z

Also, the Kerr metric off diagonal components do depend explicitly on time, since a = J/M and J changes sign under time reversal.

19. Mar 10, 2012

### Staff: Mentor

Part of this may be terminology; I have never seen the term "change of basis" applied to a transformation like parity or time reversal in spacetime. I have only seen it applied to transformations that can be expressed locally as "proper orthochronous" Lorentz transformations--i.e., as expressing a possible "relative velocity" of two timelike worldlines passing through the same event. Parity and time reversal violate that; they either reverse the handedness of the spatial vectors or reverse the direction of time.

But even if we allow a wider definition of "change of basis", the general statement I made about transforming the metric (and tensors in general) still holds; you have to look at the actual functional form of the tensor components to see how they transform; just knowing which component it is (which basis vector indices it has) isn't enough.

For example, look at the FRW metric again: its space-space components (r-r, theta-theta, phi-phi) will change under time reversal (because the function a(t) will change from increasing to decreasing in time, or vice versa). See also my further comment below on the Kerr metric.

Ah, yes, my bad (should have gone back and read my own earlier post where I said that J does change sign under time reversal ). But that is the *only* term that changes sign under time reversal; the other terms all have a^2 so their sign doesn't change if the sign of a changes.

Also, the *reason* that the off diagonal component in the Kerr metric changes sign under time reversal is not that it is an off-diagonal component, per se; it's that it contains a factor (a) that changes sign under time reversal, because of how that factor is defined.

So I think the general prescription should be: look at the actual *functional form* of the metric components to determine which ones change sign under parity/time reversal (or indeed any transformation). That means your specification of a changing sign under time reversal is fine; also there may well be a sign change in the electromagnetic 4-potential (though you haven't really given a specification of *what* 4-potential is present, which you would need to do to really know if it changes sign).

20. Mar 11, 2012

### michael879

ok lets just stick to 3 dimensional, flat space for simplicity until we clear up this confusion. Tensors are defined by how they transform under unitary coordinate transformations. Lets say matrix M is the 3x3 representation of a U(3) coordinate transformation. The following definitions apply for rank 0,1, and 2 tensors:

arbitrary scalar $a \rightarrow a$
arbitrary vector $b_i \rightarrow M_{ij}b_j$
arbitrary rank 2 tensor $c_{ij} \rightarrow M_{ik}M_{jl}c_{kl}$
and any pseudo tensor picks up a factor of det(M)

Now the constant vector (a,b,c) does not depend explicitly on x, y, or z, as it is a constant. However under a parity transformation it still transforms to (-a,-b,-c). Similarly, any tensor would transform as above regardless of how its components are explicitly defined. If it did not transform like that, then it would no longer be a tensor since it wouldn't transform as one.

So yes, you can give me 3x3 matrices that do not transform like a tensor, but they would therefore not be tensors! When looking explicitly at the components of a tensor, they must transform individually in such a way as to make the entire tensor transform correctly.

So, for example, any spatial-spatial component of a metric must be composed in such a way that it does not change sign under parity transformations. Any time-spatial component must be composed such that it does change sign.

Am I doing something wrong here?? Because what I've stated above just seem trivially true, and yet it conflicts with what your saying (which also seems trivially true).

*edit*
and just to give an example outside of general relativity, look at the EM field tensor. The time-space components are the components of the electric field (a vector), and the space-space components are those of the magnetic field (a pseudo-vector). Under a parity transformations, all of the time-space components (the electric field) change sign, and the space-space components (the magnetic field) remain unchanged. Under a time inversion transformation, all of the time-space components remain unchanged while the space-space components change sign. This means that under spatial transformations this is a true tensor. It appears to be a pseudo-tensor under time transformations, but considering the fact that time is fundamentally different from the spatial coordinates, it may be that a rank (3+1) tensor is actually defined to transform like that (in which case my analysis of the metric may need to be modified slightly, but not significantly).

*edit again*
Another example is the EM stress-energy tensor. Under both parity and time transformations the space-space and diagonal components remain unchanged, while the time-space components change sign. Considering this is a much more fundamental quantity, I'd be tempted to say that THESE are the true transformation properties of a rank (3+1) tensor, and the last comment above can be disregarded.

Last edited: Mar 11, 2012