CPT (M?) symmetries in Kerr-Newman metric

[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

 

[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.

 

[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.

 

[michael879]

Anyone have an answer?

 

[PeterDonis]

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?

 

[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?

 

[PeterDonis]

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.

 

[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

 

[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...

 

[PeterDonis]

Ok well I think I figured out what I was doing wrong. Basically, the metric was derived assuming r is non-negative.

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.

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...

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.)

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...

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.

http://jila.colorado.edu/~ajsh/insidebh/penrose.html

 

[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&#039;\equiv -r. However this messes up the transformation of f, so you must make M&#039;\equiv -M. Similarly, A_\mu no longer transforms correctly so you must make Q&#039;\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&#039; \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?

 

[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.

 

[PeterDonis]

I read up on Kruskal coordinates (which I really should know since Martin Kruskal was my uncle...)

Cool!

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.

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.

 

[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??

 

[michael879]

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

 

[PeterDonis]

I see what you're saying, but aren't tensors defined by how they transform under unitary transformations?

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.

 

[PeterDonis]

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

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.

 

[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.

 

[PeterDonis]

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

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.

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

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).

 

[michael879]

ok let's 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. Let's 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.

 

[PeterDonis]

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).

Yes, I see what you're saying. On thinking it over, it's quite possible that we're both actually looking at the same thing from different viewpoints. See below.

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.

And the same would be true for time reversal, by this reasoning: the time-time component would not change sign, but the time-space components would.

This is consistent with the way the Kerr metric behaves. It is also consistent with the way the Painleve form of the metric for Schwarzschild spacetime behaves, which is the other quick example I can come up with of a metric with a time-space component (dtdr in this case); under time reversal the dtdr term switches sign, but all the others remain the same.

So perhaps my statement about the functional form of the components is really a way of stating how your constraint is *enforced*; any time-space component of a metric must have a functional form that makes it switch sign under parity or time reversal. Let's see how this applies to the Painleve case: in this case the coefficient of the dtdr term is the "escape velocity" as a function of r. The reason it switches sign under time reversal is that the time-reversed Painleve coordinates are the natural ones for an "outgoing" Painleve observer--one that moves *outward* at escape velocity, out "to infinity". The "normal" (non-time-reversed) Painleve coordinates are the natural ones for an "ingoing" Painleve observer--one that is falling *inward* at escape velocity "from infinity". So one would expect the sign of the "velocity" of the observer to switch under time reversal.

Edit: The example of the Painleve metric seems to lead to a slight modification of the above. The above reasoning, as given (*all* time-space components switch sign under parity), would also imply that a parity transformation on the Painleve metric should switch the sign of the dtdr term. But that doesn't make sense; switching the handedness of the spatial coordinates doesn't change an ingoing observer to an outgoing one. But I think I commented in an earlier post that the correct parity transformation in spherical coordinates does *not* switch the sign of r, only of the angular coordinates. So the correct specification of which time-space terms should switch sign under parity has to at least take that into account--only spatial coordinates which switch sign under parity would have their corresponding time-space tensor components switch sign. So the Kerr metric's term in J would still switch sign (since it's a dt dphi term), but the Painleve metric's dtdr term wouldn't.

 

[michael879]

I can see we're on the same page now, and that's great, but I still have some comments/questions.

1) I'm not familiar with Painleve coordinates, but I looked them up and the functional form of the dtdr element of the metric doesn't appear to have any dependence on t or the sign of r. According to wikipedia its just -2\sqrt{2M/r}. How does this change under time reversal?

2) that's a great point you made about sign changes in spherical coordinates, and I think I hadn't taken that into account before. The components of the metric in spherical coordinates would not simply change sign under a parity transformation. However I think the transformation is much more complicated than just a sign transformation, since in spherical coordinates (r,\theta,\phi)→(r,\pi-\theta,\phi+pi) and no element simply changes sign.

3) Although the above is true, I was dealing with the Kerr-Newman metric in its Kerr–Schild form, which uses cartesian coordinates. What bothered me is that the *functional* form of the metric does not lead to the correct sign changes under parity OR time reversal transformations. Under parity transformations it simply doesn't change at all, which seems wrong but might be fixable. However under time reversal J changes sign, which completely changes the form of the metric, instead of just an overall sign!

*edit* I think I figured out 1) and part of 3). In the line element equation, under time reversal dt→-dt, and under parity dx_i→-dx_i. This can be rephrased as certain metric elements changing sign, and therefore the Painleve metric transforms correctly under time and parity transformations, and the Kerr-Newman metric transforms correctly under parity transformations. However there is still the issue of time reversal in the Kerr-Newman metric, and I'm not sure how to explain that.

*edit again* Now I'm just thoroughly confused. In Boyer-Linquist, time reversal trivially leaves the metric unchanged. The sign change in dt cancels out the sign change in J, and you're left an identical line element (dtd\phi term doesn't change sign like it should). This is wrong, but not nearly as wrong as in the cartesian case where the x and y components of the metric are drastically altered..

 

[PeterDonis]

The components of the metric in spherical coordinates would not simply change sign under a parity transformation. However I think the transformation is much more complicated than just a sign transformation, since in spherical coordinates (r,\theta,\phi)→(r,\pi-\theta,\phi+pi) and no element simply changes sign.

Yes, good point, "sign change" is also too simplistic a term for the actual transformations.

*edit* I think I figured out 1) and part of 3). In the line element equation, under time reversal dt→-dt, and under parity dx_i→-dx_i. This can be rephrased as certain metric elements changing sign, and therefore the Painleve metric transforms correctly under time and parity transformations, and the Kerr-Newman metric transforms correctly under parity transformations. However there is still the issue of time reversal in the Kerr-Newman metric, and I'm not sure how to explain that.

*edit again* Now I'm just thoroughly confused. In Boyer-Linquist, time reversal trivially leaves the metric unchanged. The sign change in dt cancels out the sign change in J, and you're left an identical line element (dtd\phi term doesn't change sign like it should). This is wrong, but not nearly as wrong as in the cartesian case where the x and y components of the metric are drastically altered..

I think you've got the right idea in looking at how the coordinate *differentials* change under the transformations, as well as the coefficients. However, I come up with what might be a slightly different set of transformation laws for them. Here's what I come up with for the coordinate differentials:

Parity

dt -> dt
dr -> dr
d\theta -> - d\theta
d\phi -> - d\phi

Time Reversal

dt -> - dt
dr -> dr
d\theta -> d\theta
d\phi -> d\phi

For the transformation of d\phi under parity, changing the handedness of the coordinates ought to change the "direction" of \phi as well as \theta. I think the correct parity transformation for \phi ought to read

\phi -> 2\pi - \phi

which would make the differential change sign. (It also keeps the range of \phi between 0 and 2 pi, which your transformation law does not.)

The question, of course, then becomes what happens to the angular momentum J under parity and time reversal? And does that create an issue with the behavior of the Kerr-Newman metric? Different references appear to say different things. For example, the Wikipedia pages on parity and time reversal say that angular momentum (they call it L) does *not* change sign under parity, but does under time reversal:

http://en.wikipedia.org/wiki/Parity_(physics )

http://en.wikipedia.org/wiki/T-symmetry

If this is correct, it would mean the Kerr-Newman metric is not invariant under parity (the dt dphi term changes sign) but *is* under time reversal (the dt dphi term does *not* change sign). Stephani's introductory GR text appears to agree with this, sort of; it says (p. 243) that "the Kerr metric is invariant under the transformation t -> -t, a -> -a (time reversal and simultaneous reversal of the sense of rotation)". However, I'm pretty sure other sources say that Kerr is *not* "time reversal invariant". I'll have to dig out my copy of MTW to refresh my memory on what it says.

 

[PeterDonis]

I think the correct parity transformation for \phi ought to read

\phi -> 2\pi - \phi

which would make the differential change sign.

Hm, this doesn't work either because it doesn't move all points on a circle to their opposite points. Need to dig up some references.

 

[michael879]

all of the above is why i switched back to cartesian coordinates lol. In boyer-linquist (or spherical) coordinates, the parity transformation is:

r → r
θ → π - θ
ϕ → ϕ + π

this transformation is "unique" as it is the only one that keeps θ and r within their respective domains. The differential transformations corresponding to the above are:

dr → dr
dθ → -dθ
dϕ → dϕ

The reason you found that the metric is not invariant under parity, is because u got the transformation of ϕ wrong. In both Kerr-Schild and Boyer-Linquist coordinates, the line element is easily seen to be invariant under parity, and the metric transforms as a true metric (at least in the Kerr-Schild form it clearly transforms as a tensor... However in the Boyer-Linquist form the metric doesn't appear to change at all, even in the spatial-time components... But with the weirdness of the transformation I wouldn't worry about it).

The problem I'm having is with time inversion (t→-t). There are two situations we could be in:

a→-a: In the Boyer-Linquist coordinates the dt sign changes cancel out all the J sign changes. This leaves the line element unchanged, and shows that the metric correctly transforms as a tensor (all of the spatial-time components flip sign). However, in the Kerr-Schild form there are huge apparent problems. Setting a → -a completely changes the structure of the metric, and it is very clear the line element does NOT stay the same, and the metric does NOT transform like any kind of tensor. Aren't these two forms of the metric supposed to be identical??

a→a: I don't have any way to prove this, but based on what I've been reading up on I would suspect M→-M might occur under a time reversal. This means the sign of a would NOT change. Now in both coordinate systems the metric does not change at all (which is wrong), and the line element is changed in a very non-trivial way.

*edit*
However, if you look at the second case a little further, you can see that there are possible ways to rectify the situation. By changing the sign of J a second time (i.e. J→-J before time reversal) and set r→-r, you will find that the new metric is what the time-reversed metric SHOULD look like, and the new line element is identical to the old one, as it SHOULD be. This works in both coordinate systems, and I think this should warrant further investigation. Why would time reversal change the sign of M but not J? And why would time reversal only work if you move into the alternate universe??

 

[michael879]

I've been thinking about it a little more, and while the classical parity operation trivially works, the time reversal operation is very interesting.

1) Let's separate local and global transformations. CPT symmetries are generally used on single particles, rather than the entire universe (i.e. a positron is identical to an electron of opposite parity going backwards in time, regardless of what happens in the rest of the universe). Global symmetries would be applied to the entire universe, and are more of coordinate transformations. For example the charge reversal symmetry in E&M, parity inversion in all classical theories, and time reversal theories.

2) I mentioned mass should change sign with time, but I have a better argument now. Imagine two chargeless masses attracting via gravity. Under time reversal, these masses should repel, but follow the same laws of gravity. Therefore their masses, which is the only real free parameter) must go negative.

3) The two parity operations I mentioned before are actually very different. The classical global parity transformation (x,y,z)→(-x,-y,-z) transforms the metric correctly, and is a symmetry of general relativity as well as the rest of classical physics. The transformation r→-r can be viewed of as a local transformation, where the black hole "flips". This transformation will make the metric look (in our universe) as it normally does in the 2nd analytically extended universe. However r→-r flips the signs of x,y, and z, and therefore also does a classical parity transformation. So (x,y,z,r)→(x,y,z,-r) is a mixture of two global transformations (classical parity + r→-r) that really only changes the structure of the black hole, and not the rest of the universe. This transformation seems very similar to the P transformation from QFT.

Given all of the above, let's look at the time reversal of a Kerr-Newman black hole, analytically extended for -∞<r<∞. If you take the above to be true, M changes sign. Therefore J changes sign twice, leaving it unchanged. This means a→-a, canceling all the t and dt sign changes (in the Boyer-Linquist form at least). However, the structure of the Δ term changes in the Boyer-Linquist formulation, and the Kerr-Schild form completely changes. The only way I can see to recover the same line element, and a metric that is transformed correctly, is to do the local "P transformation" mentioned above. This sets r to -r, but leaves x, y, and z unchanged.

Next, you have to make sure the 4-potential has transformed correctly. The E&M 4-potential is an axial 4-vector (i.e. under time and parity transformations it remains the same). Under both time reversal, and parity transformations the spatial component changes sign, and the time component remains the same (its a scalar). Looking at the Kerr-Schild form of the 4-potential, it is easy to see that under the above transformations, the scalar component has changed signed while the vector component has remained the same (which is backwards). Therefore, you also need a local charge transformation to make the 4-potential transform correctly.

In conclusion, in order to the time reversal to work correctly, it must be coupled with local parity and charge transformations. If you don't buy my argument for the mass automatically changing sign with time, then you could say that what I've done is:
global time reversal + local mass, parity, and charge reversal.

 

[PeterDonis]

all of the above is why i switched back to cartesian coordinates lol. In boyer-linquist (or spherical) coordinates, the parity transformation is:

r → r
θ → π - θ
ϕ → ϕ + π

this transformation is "unique" as it is the only one that keeps θ and r within their respective domains.

But does it? It looks to me like it changes the domain of ϕ from (0, 2π) to (π, 3π). At the very least there needs to be a mod (2π) operation in there somewhere.

The reason you found that the metric is not invariant under parity, is because u got the transformation of ϕ wrong.

Agreed, that's why I added a quick post later on that.

Deferring comment on the rest of your latest posts until I've had some more time to digest them.

 

[PeterDonis]

The differential transformations corresponding to the above are:

dr → dr
dθ → -dθ
dϕ → dϕ

I'm still not sure I agree that the sign of dϕ remains unchanged under parity. I need to think about this some more. (But see below.)

a→-a: In the Boyer-Linquist coordinates the dt sign changes cancel out all the J sign changes. This leaves the line element unchanged, and shows that the metric correctly transforms as a tensor (all of the spatial-time components flip sign). However, in the Kerr-Schild form there are huge apparent problems. Setting a → -a completely changes the structure of the metric, and it is very clear the line element does NOT stay the same, and the metric does NOT transform like any kind of tensor. Aren't these two forms of the metric supposed to be identical??

Can you give references to where you're getting these forms of the metric? From what I can see, in Boyer-Lindquist coordinates there's an a dt dphi term, and in Kerr-Schild form there's an a dt (y dx - x dy) term. If you are right and the sign of dphi doesn't change (and that looks more reasonable now that I observe that dphi = y dx - x dy makes the two terms the same, and also requires that the sign of dphi not change, since x, y, dx, dy all flip sign under parity and are multiplied in pairs so the sign flips cancel), then the a and dt sign changes cancel in both cases and both metrics are invariant under parity.

2) I mentioned mass should change sign with time, but I have a better argument now. Imagine two chargeless masses attracting via gravity. Under time reversal, these masses should repel, but follow the same laws of gravity. Therefore their masses, which is the only real free parameter) must go negative.

No, this is wrong. The time reverse of attractive gravity is still attractive gravity. Think of a ball going up and coming back down. The time reverse of that is...a ball going up and coming back down. Gravity is attractive in both cases.

Or, if you don't like that case because it's a manifestly time-symmetric solution, consider an expanding matter-dominated universe, whose expansion is decelerating due to gravity, because gravity is attractive. The time-reverse of that is a contracting matter-dominated universe, whose contraction is accelerating due to gravity, because gravity is attractive.

The transformation r→-r can be viewed of as a local transformation, where the black hole "flips". This transformation will make the metric look (in our universe) as it normally does in the 2nd analytically extended universe.

Hmm; I see what you're saying but I'm not sure it's right. I'm sure it's not in the simpler Schwarzschild case--the maximum analytically extended spacetime there has no r < 0 region, the entire spacetime has r > 0 (and the two "exterior" regions both have r > 2M). I'm pretty sure the same applies to the exterior regions in the Kerr-Newman case. (And of course there already is an r < 0 region in the Kerr-Newman case, but it's "inside" the ring singularity, not in the exterior.)

However r→-r flips the signs of x,y, and z, and therefore also does a classical parity transformation.

No, it doesn't. This is what I objected to before. There is *no* "standard" transformation from spherical polar to Cartesian coordinates for which this is true. If you want to make it true, you need to figure out a completely *new* coordinate transformation that somehow maps the r > 0 coordinate patch to only "half" of Euclidean 3-space, instead of all of it.

In conclusion, in order to the time reversal to work correctly, it must be coupled with local parity and charge transformations. If you don't buy my argument for the mass automatically changing sign with time, then you could say that what I've done is:
global time reversal + local mass, parity, and charge reversal.

This may be along the correct lines (although as I said above I don't think the mass changes sign under any of these transformations); the language in the Stephani book that I quoted earlier makes it seem as though "time reversal" is supposed to include other "local" transformations as well.

 

[michael879]

You're right, I was a little jet-lagged when I wrote that and I realized a lot of the mistakes later. Give me a day to organize my thoughts and I'll try to make it clearer (it makes perfect sense in my head)

 

[michael879]

Ok I'm thoroughly confused now. What exactly does time reversal entail? And how do you represent it mathematically?? From everything I've read time reversal is just an improper lorentz transformation with the time-time component negative (diagonal, and the space-space components are positive). This would imply that to apply time reversal, you just apply that lorentz transformation to the 4-tensor you wish to transform. The obvious problem with this is that it doesn't work. Take 4-momentum for example: the time reversal of 4-momentum is negative of what it should be. Does this mean that 4-momentum isn't a true 4-vector? Also, look at 4-velocity: no component depends on the sign of time, so it is unchanged by time reversal.

*edit*
ok so apparently the clear distinction between pseudo-tensors and tensors doesn't apply when you have time coordinates. For example, position 4-vectors are true 4-vectors under both parity inversion and time reversal. However 4-velocity and 4-momentum are pseudo-vectors under time reversal (they pick up a negative sign), and normal vectors under parity inversion. However the line element is necessarily a 4-scalar, and it is the tensor product of the metric with true 4-vectors. Therefore by necessity the metric must be a true 4-tensor, so that it transforms by lorentz transforming each index with no additional sign changes. Give me a few hours and Ill work out what I was trying to explain above in a more formal (and correct) way.

 

[PeterDonis]

What exactly does time reversal entail? And how do you represent it mathematically??

I think at this point we need to go back to basics. First, let's restrict attention to what you called "local" transformations; i.e., we're looking at a specific event in spacetime and what happens when we do various transformations on the local coordinates. But we can always write local coordinates in Minkowski form, so what we are really looking at with the local transformations is transformations on Minkowski spacetime--more precisely, we are looking at *isometries* of Minkowski spacetime, transformations that leave the Minkowski metric invariant.

The full set of such transformations is the 10-parameter Poincare group:

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

The 10 parameters break up into 4 translations (one in each of the four spacetime dimensions) and 6 "rotations", and the latter, if you choose a particular local inertial frame, can be further broken up into 3 spatial rotations and 3 boosts.

However, there's also another way of looking at this group, as the Wikipedia article states:

"As a topological space, the group has four connected components: the component of the identity; the time reversed component; the spatial inversion component; and the component which is both time reversed and spatially inverted."

In other words, there are four transformations we can pick out of the whole Poincare group that have special significance: 1 (the identity), P (parity), T (time reversal), and PT (combined parity and time reversal). These four are special because they form a sort of "basis" of the group: any arbitrary element of the Poincare group can be expressed as the composition of one of these four transformations with a transformation from the restricted Lorentz group, the group of "Lorentz transformations" that is usually talked about in basic special relativity courses (this restricted group is really just the component of the Poincare group that is connected to the identity).

So to find a suitable mathematical expression for time reversal, we need to first pick a representation of the Poincare group to work with. But since the full Poincare group is basically just four "copies" of the Lorentz group, each multiplied by one of (1, P, T, PT), we really just need to pick a representation of the Lorentz group:

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

Then we need to make sure that this representation still works when we extend it to cover the full Poincare group by multiplying by (1, P, T, PT), with suitable representations for those four transformations. Doing that last step should give us a mathematical representation of time reversal in a local inertial frame. We could then look at how to transform that to a global set of coordinates.

Of course the problem with the above is that there are lots of representations! However, as you can see from the Wiki article I just linked to, which representation you want to use is really dictated by what you are trying to represent, e.g., if you want to represent the electromagnetic 4-potential you use the (1/2, 1/2) or 4-vector representation. So when we get to the point of looking at how time reversal acts on a complicated expression like that for the Kerr metric, we may actually be dealing with compositions of objects from different representations.

Having just thrown a whole bunch of additional complexity into the mix , I think it's best to stop for now and take some time to digest.

Edit: Saw your edit after posting, the above may be moot but I think it's still useful background. Will wait to see your further post.

 

[michael879]

Ok I've stepped through this pretty rigourously, and I'm confident of, although confused by, my results. First off, the source I've been using to get the metric is http://en.wikipedia.org/wiki/Kerr%E2%80%93Newman_metric. I've also been using a general relativity book I have to brush up on some things.

Now for the analysis: the line element is given by d\tau^2 = g_{\mu\nu}dx^\mu dx^\nu. d\tau^2 is a Lorentz scalar, and so it should remain unchanged under any Lorentz transformations. Since dx^\mu is a "true" 4-vector ("true" in the sense that it transforms as a 4-vector for any Lorentz transformation without any extra sign changes), g_{\mu\nu} must be a "true" rank 2 4-tensor. The Kerr-Newman metric in Kerr-Schild form is given by:

g_{\mu\nu} = \eta_{\mu\nu} + fk_\mu k_\nu

given the transformation properties of g and η, f must be a Lorentz scalar, and k must be a 4-vector or a 4-pseudovector (or any combination of the two). These two quantities are defined as
f = \dfrac{Gr^2}{r^4+a^2z^2}\left(2Mr-Q^2\right)
k_t = 1
k_x = \dfrac{rx+ay}{r^2+a^2}
k_y = \dfrac{ry-ax}{r^2+a^2}
k_z = \dfrac{z}{r}
and a is the usual J/M.

The 4-potential (which is not a "true" 4-vector because it is a pseudo vector under time-reversal) is defined as
A_\mu = \dfrac{Qr^3}{r^4+a^2z^2}k_\mu

Now that I've setup the metric, I want to talk about 3 transformations that I will call P, R, and T.

P: x→-x, y→-y, z→-z
=================
Looking at the above equations, it is clear that k (and therefore A) transforms as a 4-vector, and f remains constant. Therefore the symmetry requires are satisfied and parity is a trivial symmetry of the Kerr-Newman black hole.

R: r→-r
=================
Although this is not a Lorentz transformations (so no symmetry arguments can be made), it is particularly interesting. After performing the transformation, you can reorganize the minus signs to get:
f = \dfrac{Gr^2}{r^4+(-a)^2(-z)^2}\left(2(-M)r-(-Q)^2\right)
k_t = 1
k_x = \dfrac{r(-x)+(-a)(-y)}{r^2+(-a)^2}
k_y = \dfrac{r(-y)-(-a)(-x)}{r^2+(-a)^2}
k_z = \dfrac{(-z)}{r}
A_\mu = \dfrac{(-Q)r^3}{r^4+(-a)^2(-z)^2}k_\mu
It now becomes clear what the black hole would "look" like after an R transformation. R[M,J,Q] \equiv P[-M,J,-Q] (not equal, but identical fields), i.e. an R transformation produces a metric which is identical to the original under a parity transformation with opposite signed mass and charge. However, this was just some manipulation of the equations, and the actual black hole still has inertial mass M, and electric charge Q.

What's interesting about this though, is that it connects R and P transformations, making R similar to some kind of parity transformation. Additionally, since the sign of r only affects the black hole, the R transformation is a transformation of some kind of intrinsic property of the black hole. It basically "flips" the black hole between the two alternate universes. I never understood the concept of chirality in QFT intuitively, but the strangeness of this reminds me of it a lot..

T: t→-t
=================
It is this transformation that has me completely confused. The only part of the metric that depends explicitly on time is the a parameter (because J→-J). This alone does not transform the metric even close to correctly. So there are two options:
1) time reversal is not a symmetry of general relativity. However, time reversal is an improper Lorentz transformation and therefore should be a symmetry.
2) there are other values that transform under time reversal that aren't immediately obvious.

After looking through the metric, I found that there is only one way to make it transform correctly. If t→-t implies r→-r and M→-M, f transforms correctly as a scalar, and k transforms as a pseudo 4-vector. Note, that in this case the actual mass must change sign, and not just the active gravitational mass (because changing the sign of inertial mass changes the sign of J, leaving a with an overall minus sign still). In addition to enforcing the metric transformation, the 4-potential also needs to transform as a pseudo-vector. The dependence on r^3 though, makes it transform as a 4-vector (the scalar potential changes sign, and the vector potential doesnt). Therefore t→-t must also make Q→-Q for the electromagnetic fields to transform properly.

So the result is that there are two possibilities. Either d\tau^2 is not a true Lorentz scalar (which I'm pretty sure would invalidate General Relativity), or T[M,J,Q] = R[-M, J, -Q] (i.e. t→-t makes r→-r, M→-M, Q→-Q, and J→-(-J)=J). Another interesting feature is that using the above analysis for R transformations, T[M,J,Q] \equiv P[M,J,Q] (again, not equal because T[M,J,Q] has negative inertial mass and negative electric charge). Although this should be obvious from the transformation properties.

 

[PeterDonis]

R: r→-r

I understand what this transformation is doing mathematically, but I'm not sure it has any actual physical meaning.

T: t→-t
=================
It is this transformation that has me completely confused. The only part of the metric that depends explicitly on time is the a parameter (because J→-J). This alone does not transform the metric even close to correctly.

What would you consider "correctly"? As I read it, J -> -J (or equivalently a -> -a) converts (loosely speaking) a black hole rotating clockwise into a black hole rotating counterclockwise, with no other parameters changed. This seems OK to me after thinking it over, since the same would be true for an ordinary object. See further comments below.

So there are two options:
1) time reversal is not a symmetry of general relativity. However, time reversal is an improper Lorentz transformation and therefore should be a symmetry.

After thinking about this quite a bit, I think we've been confusing symmetry of a *theory* with symmetry of particular *solutions*.

For example: one could say that special relativity "has time reversal symmetry", but one can interpret that statement in two ways, which both happen to be true in that particular case. One could say that the general *equations* of special relativity are symmetric under time reversal; but one could also say that *Minkowski spacetime*, which happens to be the unique spacetime that satisfies the general equations of SR, is symmetric under time reversal. More precisely, your transformation T is an isometry of Minkowski spacetime; it leaves the metric invariant.

I think what we're seeing with Kerr-Newman spacetime is that these two notions become distinct. The general equation of GR, the Einstein Field Equation, of which the Kerr-Newman spacetimes constitute just one family of solutions, is "time reversal symmetric"; but what that means is not that nothing can change under time reversal, but that the time reverse of any solution of the EFE is also a solution of the EFE.

If a solution's time reverse happens to be that same identical solution (as in the case of Minkowski spacetime, or if that's too simple, take the case of Schwarzschild spacetime, which goes into itself under time reversal), then the individual solution is "time reversal symmetric"--again, your transformation T is an isometry of that spacetime.

But in the case of Kerr-Newman, the time reverse of a given solution is not the same solution, but a solution which looks the same except for the reversal of sign of the angular momentum. So the time reversal symmetry of the underlying theory is manifested by solutions occurring in pairs which are the time reverses of each other (except for the case a = 0, which reduces to Schwarzschild and therefore time reverses into itself--I am leaving out the case of nonzero electric charge here).

 

[michael879]

I get what your saying, but isn't one of the postulates of GR basically that the line element is a Lorentz scalar??

This means under any Lorentz transformation it stays the same (i.e. the infinitesimal proper length is invariant of reference frame).

The only way for this to be true is for the spacetime metric to transform as a tensor under any Lorentz transformation.

If the metric doesn't transform as a tensor under some transformation, then the line element will not transform as a scalar.

Time reversal T is an inproper Lorentz transformation (detT = -1).

The Kerr-Newman metric does not transform as a tensor under T if you only take into account the transformation of a. This means that under time reversal the line element changes, violating the initial postulate.

What I'm saying is this deeper symmetry constraint on the metric is what causes the EF equations to be symmetric under Lorentz transformations, not the other way around. Where is the flaw in the above logic?

Also, how would you explain why the 4-potential of a KN black hole does not transform correctly under time reversal? It is clear from Maxwells equations that under time reversal A→-A and \phi→\phi. However the metric and 4-potential of the KN black hole are related by the 4-vector k. k does not transform as a vector by simply setting a→-a under time reversal, and therefore the 4-potential doesn't transform as a vector...

 

[PeterDonis]

I get what your saying, but isn't one of the postulates of GR basically that the line element is a Lorentz scalar??

Yes, but you have to be careful in specifying exactly what that means.

To lead into what I'm getting at, in going back over the thread I realized that I missed something in my last post. I posted earlier that time reversal changes the sign of J (or a), but also changes the sign of dt (but not dphi). If that's correct, then the dt dphi term of the metric would *not* change sign under time reversal; J changes sign but so does dt, so the two sign changes cancel.

But what does this actually mean, physically? Since dt and dphi are the critical differentials here, consider a small line element where only those two differentials are nonzero. In the "forward time" case, we have a scalar composed of three terms: a dt^2 term, a dphi^2 term, and a dt dphi term (which has a factor of J in it).

What happens when we do a time reversal? Think of the line element as a small line segment, which in the "forward time" case went from (t, phi) to (t + dt, phi + dphi). Under time reversal, it now goes from (t', phi) to (t' - dt', phi + dphi), because the "direction" of the time coordinate has flipped. So the length of the actual physical line element does not change--what happens is that the same physical line element is now described by a slighly different set of coordinates due to the time reversal.

What makes this look a bit confusing is that if you just look at the formula for the line element, without considering exactly what the coordinate differentials mean physically, it *looks* as though the line element has changed, because the sign of the dt dphi term has indeed flipped in the formula. But when you actually look at how the formula *describes* the same actual, physical line element, you see that there's a sign change in dt as well, which results in the same final scalar describing the same physical line element.

 

[michael879]

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. I have no idea what the 4-potential is in the Boyer-Linquist formulism, which is why I've been sticking to the Kerr-Schild form. In the Boyer-Linquist form, time reversal changes the sign of a and dt, which cancel out exactly, preserving the line element. The metric has 1 off diagonal term in phi-time. This component has an overall factor of a, so that it changes sign under time reversal.

Therefore the metric, and the line element transform correctly in the Boyer-Linquist form. However I don't know the form of the 4-potential, so I can't say if that transforms correctly as well.

In the Kerr-Schild form, the line element changes under time reversal (if you just take into account the sign changes of a and dt), and the metric and 4-potential do not transform correctly. The only way to rectify this, that I can see, is to either say the Kerr-Schild form is wrong, or to make the additional transformations I mentioned above. Note, that in the Boyer-Linquist form the additional transformations do not change the line element either, so that everything still transforms correctly. I would be interested to see what the 4-potential is though, because I would suspect it could only transform correctly given the additional transformations (and for whatever reason the Boyer-Linquist form hides them).

 

[michael879]

Also, I just noticed you're over complicating the situation. The line element is given simply by
ds^2 = g_{\mu\nu}dx^\mu dx^\nu
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 ds^2 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]

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.

Also, I just noticed you're over complicating the situation. The line element is given simply by
ds^2 = g_{\mu\nu}dx^\mu dx^\nu
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 ds^2 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.

*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]

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. Let's 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).

 

[PeterDonis]

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]

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]

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.

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]

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]

*sigh* I finally get it, what a let down.

The "dt" coordinate I've been using is actually \bar{dt} = dt - \dfrac{2Mr}{r^2-2Mr+a^2}dr. 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 \bar{dt}\rightarrow-\bar{dt} and you only want to get dt\rightarrow-dt, you must flip M and r.

 

[michael879]

Ok, so given this new information, let's do a real time reversal: dt→-dt

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}

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

so therefore,
g_{\mu\nu} \rightarrow \bar{T}[g_{\mu\nu}]|_{M\rightarrow-M,Q\rightarrow-Q,r\rightarrow-r, J\rightarrow J}
and
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}

Now assuming that the line element for [M,J,Q,r] is the same as that for \bar{T}[-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]

The "dt" coordinate I've been using is actually \bar{dt} = dt - \dfrac{2Mr}{r^2-2Mr+a^2}dr

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.

Ok, so given this new information, let's 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", \sqrt{2M / r}[/tex] to make the formulas easier to read. So in the &quot;ingoing&quot; Painleve coordinates we have (using the +--- sign convention I used earlier):<br /> <br /> ds^{2} = \left( 1 - v^{2} \right) dT^{2} - 2 v dT dR - dR^{2}<br /> <br /> Now, to make the signs of each term absolutely clear, let&#039;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 &quot;Painleve observer&quot; 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:<br /> <br /> 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}<br /> <br /> 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&#039;, R&#039;? It looks like this: dT&#039; = - dT, dR&#039; = dR. In other words: dT&#039; = - |dT|, dR&#039; = - |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&#039;, R&#039; must look like this:<br /> <br /> ds^{2} = \left( 1 - v^{2} \right) dT&amp;#039;^{2} + 2 v dT&amp;#039; dR&amp;#039; - dR&amp;#039;^{2}<br /> <br /> The sign of the second term must flip to compensate for the fact that dT&#039; = - dT. Substituting dT&#039; = - |dT| and dR&#039; = - |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.<br /> <br /> 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):<br /> <br /> dT = dt - \frac{v}{1 - v^{2}} dr<br /> <br /> Time reversing this, which is expressed as dt&#039; = - dt and dr&#039; = dr, gives:<br /> <br /> dT&amp;#039; = dt&amp;#039; + \frac{v}{1 - v^{2}} dr&amp;#039;<br /> <br /> Notice that the sign of dT&#039; is still the same as the sign of dt&#039;; 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).<br /> <br /> <blockquote data-attributes="" data-quote="michael879" data-source="post: 3839366" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> michael879 said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> And it turns out that it is, so although its far from obvious, the Kerr-Newman space-time is time reversal invariant! </div> </div> </blockquote><br /> 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]

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
T[M,J,Q,r] \equiv \bar{T}[-M,-J,-Q,-r]
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.

dT&#039; = dt&#039; + \frac{v}{1 - v^{2}} dr&#039;

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

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, \vec{J}^2, 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]

Looks like we've finally got this one nailed. The only thing I still wanted to comment on is this:

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]

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]

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.

However there are some striking similarities between fundamental particles and black holes.

Yes, this whole area is a fascinating field for speculation.