CPT (M?) symmetries in Kerr-Newman metric


by michael879
Tags: kerrnewman, metric, symmetries
PeterDonis
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Mar10-12, 11:28 AM
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Quote Quote by michael879 View Post
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.

Quote Quote by michael879 View Post
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
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Mar11-12, 04:05 AM
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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 [itex]a \rightarrow a[/itex]
arbitrary vector [itex]b_i \rightarrow M_{ij}b_j[/itex]
arbitrary rank 2 tensor [itex]c_{ij} \rightarrow M_{ik}M_{jl}c_{kl}[/itex]
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
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Quote Quote by michael879 View Post
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.

Quote Quote by michael879 View Post
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
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Mar14-12, 04:34 AM
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I can see we're on the same page now, and thats 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 [itex]-2\sqrt{2M/r}[/itex]. How does this change under time reversal?

2) thats 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,[itex]\theta[/itex],[itex]\phi[/itex])→(r,[itex]\pi-\theta[/itex],[itex]\phi+pi[/itex]) 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 dxi→-dxi. 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[itex]\phi[/itex] 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
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Mar14-12, 01:25 PM
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Quote Quote by michael879 View Post
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,[itex]\theta[/itex],[itex]\phi[/itex])→(r,[itex]\pi-\theta[/itex],[itex]\phi+pi[/itex]) and no element simply changes sign.
Yes, good point, "sign change" is also too simplistic a term for the actual transformations.

Quote Quote by michael879 View Post
*edit* I think I figured out 1) and part of 3). In the line element equation, under time reversal dt→-dt, and under parity dxi→-dxi. 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[itex]\phi[/itex] 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

[itex]dt[/itex] -> [itex]dt[/itex]
[itex]dr[/itex] -> [itex]dr[/itex]
[itex]d\theta[/itex] -> [itex]- d\theta[/itex]
[itex]d\phi[/itex] -> [itex]- d\phi[/itex]

Time Reversal

[itex]dt[/itex] -> [itex]- dt[/itex]
[itex]dr[/itex] -> [itex]dr[/itex]
[itex]d\theta[/itex] -> [itex]d\theta[/itex]
[itex]d\phi[/itex] -> [itex]d\phi[/itex]

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

[itex]\phi[/itex] -> [itex]2\pi - \phi[/itex]

which would make the differential change sign. (It also keeps the range of [itex]\phi[/itex] 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
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Mar14-12, 01:51 PM
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Quote Quote by PeterDonis View Post
I think the correct parity transformation for [itex]\phi[/itex] ought to read

[itex]\phi[/itex] -> [itex]2\pi - \phi[/itex]

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
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Mar18-12, 10:41 PM
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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 wouldnt 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
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Mar19-12, 09:47 AM
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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) Lets 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, lets 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 Ive done is:
global time reversal + local mass, parity, and charge reversal.
PeterDonis
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Quote Quote by michael879 View Post
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.

Quote Quote by michael879 View Post
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.
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Quote Quote by michael879 View Post
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.)

Quote Quote by michael879 View Post
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.

Quote Quote by michael879 View Post
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.

Quote Quote by michael879 View Post
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.)

Quote Quote by michael879 View Post
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.

Quote Quote by michael879 View Post
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 Ive 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
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#29
Mar20-12, 07:30 AM
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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
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Mar22-12, 04:04 PM
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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 doesnt 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 isnt 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
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Mar22-12, 04:40 PM
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Quote Quote by michael879 View Post
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/Poincar%C3%A9_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/Represe..._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
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Mar23-12, 12:01 PM
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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 [itex]d\tau^2 = g_{\mu\nu}dx^\mu dx^\nu[/itex]. [itex]d\tau^2[/itex] is a Lorentz scalar, and so it should remain unchanged under any Lorentz transformations. Since [itex]dx^\mu[/itex] 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), [itex]g_{\mu\nu}[/itex] must be a "true" rank 2 4-tensor. The Kerr-Newman metric in Kerr-Schild form is given by:

[itex]g_{\mu\nu} = \eta_{\mu\nu} + fk_\mu k_\nu[/itex]

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
[itex]f = \dfrac{Gr^2}{r^4+a^2z^2}\left(2Mr-Q^2\right)[/itex]
[itex]k_t = 1[/itex]
[itex]k_x = \dfrac{rx+ay}{r^2+a^2}[/itex]
[itex]k_y = \dfrac{ry-ax}{r^2+a^2}[/itex]
[itex]k_z = \dfrac{z}{r}[/itex]
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
[itex]A_\mu = \dfrac{Qr^3}{r^4+a^2z^2}k_\mu[/itex]

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:
[itex]f = \dfrac{Gr^2}{r^4+(-a)^2(-z)^2}\left(2(-M)r-(-Q)^2\right)[/itex]
[itex]k_t = 1[/itex]
[itex]k_x = \dfrac{r(-x)+(-a)(-y)}{r^2+(-a)^2}[/itex]
[itex]k_y = \dfrac{r(-y)-(-a)(-x)}{r^2+(-a)^2}[/itex]
[itex]k_z = \dfrac{(-z)}{r}[/itex]
[itex]A_\mu = \dfrac{(-Q)r^3}{r^4+(-a)^2(-z)^2}k_\mu[/itex]
It now becomes clear what the black hole would "look" like after an R transformation. R[M,J,Q] [itex]\equiv[/itex] 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 [itex]r^3[/itex] 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 [itex]d\tau^2[/itex] 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] [itex]\equiv[/itex] 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
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Mar26-12, 12:30 AM
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Quote Quote by michael879 View Post
R: r→-r
I understand what this transformation is doing mathematically, but I'm not sure it has any actual physical meaning.

Quote Quote by michael879 View Post
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.

Quote Quote by michael879 View Post
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
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Mar26-12, 10:27 AM
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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 [itex]\phi[/itex]→[itex]\phi[/itex]. 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
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Mar26-12, 05:35 PM
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Quote Quote by michael879 View Post
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
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Mar28-12, 11:05 AM
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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).


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