Is General Relativity Time Symmetric?

[JustinLevy]

bcrowell,
I think your last comment is getting close to the mark, but I'm worried we're both still misunderstanding what is meant by symmetry in these contexts. I don't think coordinate independence itself has the physical meaning you are imparting to it. And even if I'm wrong in that, I'd like to dig into the meaning of symmetry here to learn more from it.

Would you claim the standard model has Galilean symmetry?

I don't think the standard model can be written completely in tensor notation.

Let's look at the lagrangian. There are no uncontracted spacetime or spinor indices. This means the action is, at a minimum, a psuedo-scalar. It cannot change under a coordinate transformation like a Galilean transformation ... the Lagrangian stays the same.

There are all kinds of issues there. For instance, time is treated differently than position in quantum mechanics (time is not an operator), and that distinction can't be expressed in tensor language. Raising and lowering indices is done in the SM by using a fixed background metric, which doesn't transform like a tensor.

First responding to your comments on the standard model and time. In non-relativistic quantum "particle" theory, position is an operator, and time is merely a parameterizing label. But the standard model is a field theory. In relativistic quantum field theory, time and space must be treated equally, so either time must be promoted to an operator or position demoted to a label. The standard approach is coordinates are just labels in quantum field theory (and the field itself promoted to an operator). The Lagrangian / Path integral formulation helps really make this lorentz symmetry clear.

Regarding the "fixed background metric" comment, I'll respond to this in two ways:
1] I may be misunderstanding you here. What do you mean by a background metric not transforming correctly? For a simple example, let's look at the free-field term in QED
F_{ab}F_{cd}g^{ac}g^{bd}
are you claiming that if I choose a coordinate system and calculate the value of this, then do a Galilean coordinate transformation into a new coordinate system, that the value of this term changes? It shouldn't since it is a coordinate independent scalar.

2] Maybe you meant something along "prior geometry". Change coordinates all you want, and we still have the geometry we started with. Well fine. Neither the metric, nor any prior geomtery is specified by the standard model. Heck we can look at quantum field theory in curved spacetimes if you wish. (my understanding from others responding to this on this forum is that we can do this, but insurmountable issues arise if we try to take the next step and allow spacetime itself to be dynamicly interacting using the current quantum field theory framework)

Well, basically I think electrodynamics does have symmetry under Galilean transformations, but I don't necessarily mean what you think I mean by that. First off, there is the kind of difficulty that atyy was talking about above. Atyy pointed out that the meaning of a global time reversal is not even necessarily well defined in GR, on an arbitrary spacetime. This is even more of a problem when it comes to trying to define something like Galilean symmetry. You don't have global coordinate systems, so you don't necessarily have any clearcut way to define what you even mean by a global Galilean transformation.

You're turning this into a "global" vs "local" issue, which I feel is missing the point.

Would you agree with the following:
GR doesn't have global Lorentz symmetry.
GR has local Lorentz symmetry.

Can we show this by merely looking at how the action changes with a change in coordinate system? I don't believe so. Therefore I feel the ability to write a theory in tensor notation can't possibly hold as much physical significance as you feel. I've even seen people argue in this very subforum that Newton-Cartan is a great counter example for this.

So let's pause and discuss what we even mean by symmetry here.

A possible definition of symmetry is: If the operator generating some symmetry commutes with the Hamiltonian, then the theory is said to have that symmetry.

Does this solve the issue? I don't know. I still need to think about it.
In particular, does this definition give different answers to whether a theory has a particular symmetry or not? I'm not sure.

A secondary issue is that when you refer to Galilean symmetry, you're referring to something that carries a lot more baggage than a simple coordinate transformation. It's a coordinate transformation that is part of a continuous group, with each member of the group labeled by a velocity vector v. Transformations corresponding to v and -v are inverses. There is also the notion that v is related in some specific way to tangent vectors of observers' world-lines. All of these additional notions are separate things that do not follow from diff-invariance.

I think this is starting to hit on why coordinate independence itself is not sufficient to tell us whether theories have Lorentz symmetry, etc. Because they are coordinate independent, the coordinates lose physical meaning. They are merely labels. We can choose all kinds of bizarre coordinate systems. So merely doing the coordinate transformation which inverts the sign on the 0th component of the coordinate labels everywhere, may not even mean time inversion locally.

Only in specific coordinate systems can we associate the coordinate labels directly to physical meaning such as relating to lengths measured by local rulers or relating to time measured by local clocks.

If I did the coordinate transformation:
t' = t + x
x' = t - x
y' = y
z' = z
which one is the local time coordinate?

This lack of physical meaning is why coordinate velocity can be arbitrarily larger than c, and one can do all kinds of bizarre things like make your "time coordinate" go in loops ... all in flat spacetime.

 

[Phrak]

Charge conservation via Noether's theorem comes from U(1) as a global symmetry, not a gauge symmetry.

Classically, without using Noether's theorem, Maxwell's equations predict that charge is conserved. Because the charge continuity equation works for any E and B, it can be regauged to any E and B. As gauge theories go, it's anticlimatic. A little more deeply it says that the values of charge and current density depend upon E and B up to a constant of integration. Physically it says that many different arrangements of electric and magnetic fields will have the same distribution of charges and currents.

 

[bcrowell]

Let's look at the lagrangian. There are no uncontracted spacetime or spinor indices. This means the action is, at a minimum, a psuedo-scalar. It cannot change under a coordinate transformation like a Galilean transformation ... the Lagrangian stays the same.

I think the issue here is what the contractions mean. The contractions involve raising and lowering indices. In QFT the raising and lowering of the indices is done using a fixed metric, which doesn't transform like a tensor. Therefore the Lagrangian is written to look like a tensor, but it isn't actually a tensor.

Heck we can look at quantum field theory in curved spacetimes if you wish. (my understanding from others responding to this on this forum is that we can do this, but insurmountable issues arise if we try to take the next step and allow spacetime itself to be dynamicly interacting using the current quantum field theory framework)

If the spacetime is a fixed background that doesn't obey the Einstein field equations, then we're no longer discussing the symmetry properties of GR; we're discussing the symmetry properties of some theory that violates the Einstein field equations.

You're turning this into a "global" vs "local" issue, which I feel is missing the point.

Would you agree with the following:
GR doesn't have global Lorentz symmetry.
GR has local Lorentz symmetry.

To answer this question, you would have to define what was meant by a global Lorentz transformation. Since there is in general no way to define such a thing, there is no way to define whether GR has global Lorentz symmetry.

A possible definition of symmetry is: If the operator generating some symmetry commutes with the Hamiltonian, then the theory is said to have that symmetry.

The theory has a certain symmetry if the laws of physics retain the same form under that transformation. It doesn't matter if you write the laws of physics as the Einstein field equations, a Lagrangian, or a Hamiltonian; they all have the same properties under the same transformations.

I think the long and the short of it is this. Almost all spacetimes that are of any physical interest are time-orientable. Time-orientability is an extremely weak condition. Even a spacetime that has closed, timelike curves is typically still time-orientable. For any time-orientable spacetime, there is a unique and physically reasonable way to define a time-reversal operator. In particular, this operator acts in the same way locally (in the tangent space) as the time-reversal from the Poincare group. This operator is a diffeomorphism, so by diff-invariance, a spacetime that has been subjected to this operation remains a valid solution to the Einstein field equations. In this sense, GR has time-reversal symmetry.

Few spacetimes of physical interest in GR allow any natural way of defining anything like a global Galilean transformation, global Lorentz transformation, etc.; if you insist on defining such a transformation, it has no natural or compelling physical interpretation. Therefore although such a transformation does leave the laws of physics in GR form-invariant (assuming it's smooth enough to be a diffeomorphism), that fact can't be interpreted as a confirmation that GR has a specific global symmetry with any interesting physical or geometrical interpretation.

 

[JustinLevy]

I think the issue here is what the contractions mean. The contractions involve raising and lowering indices. In QFT the raising and lowering of the indices is done using a fixed metric, which doesn't transform like a tensor. Therefore the Lagrangian is written to look like a tensor, but it isn't actually a tensor.

Wait, you're claiming terms like the free field in QED:
F_{ab}F_{cd}g^{ac}g^{bd}
are not actually a scalar? That if we start in a coordinate system in which:
g^{ac} = <br /> \left( \begin{array}{cccc}<br /> -1 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 1<br /> \end{array} \right)\)<br />
then transform into another coordinate system, that F_ab will transform like a tensor, but g^ac will not transform appropriately and therefore that term is actually coordinate system dependent and not a scalar?

I'm really hoping I'm misunderstanding you here, because that claim does not make any sense. The geometry is fixed, not the components of the metric. That term is indeed a scalar.

If the spacetime is a fixed background that doesn't obey the Einstein field equations, then we're no longer discussing the symmetry properties of GR; we're discussing the symmetry properties of some theory that violates the Einstein field equations.

You were using the ability to write the lagrangian in coordinate free notation (and thus be valid under any coordinate system, or coordinate transformation), to make claims about physical symmetries. I brought up non GR examples to simplify the situation and point out what I hoped would be obvious counter-examples.

My point is, and remains:
I feel the ability to write a theory in tensor notation can't possibly hold as much physical significance as you feel it does.

To answer this question, you would have to define what was meant by a global Lorentz transformation. Since there is in general no way to define such a thing, there is no way to define whether GR has global Lorentz symmetry.

By "meant by" you seem to want to inject some kind of physical / coordinate independent meaning, to what the coordinate transformation is doing. So you seem to have some sense of the lorentz symmetry, not being just a coordinate symmetry of the equations ... which is exactly what I'm trying to get at. You can't just take the Lagrangian, which is a scalar and coordinate indepedent, and claim because it is coordinate independent that automatically means it has any coordinate symmetry.

To make this more clear, let's look at a local instead of global symmetry.
Would you claim GR has local Galilean symmetry, just because the lagrangian is a scalar?
Would you claim classical electrodynamics has local Galilean symmetry, just because the lagrangian is a scalar?

The theory has a certain symmetry if the laws of physics retain the same form under that transformation. It doesn't matter if you write the laws of physics as the Einstein field equations, a Lagrangian, or a Hamiltonian; they all have the same properties under the same transformations.

No. I would maintain that classical electrodynamics, written in tensor notation, retains that "form" regardless of the coordinate system you use. But if you look at how the evolution equations look in terms of E and B fields, they will NOT look the same after applying a galilean transformation. So if you use "same form" to define symmetry, then the resulting symmetries do depend on how you write it. I don't think this "same form" definition is precise enough, or useful, when we have tensor equations which are coordinate system independent.

Writing in tensor notation allows the equations to be true in all coordinate systems. So the idea of symmetry being "The theory has a certain symmetry if the laws of physics retain the same form under that transformation." cannot be maintained in such a coordinate free form.

I think the long and the short of it is this. Almost all spacetimes that are of any physical interest are time-orientable. Time-orientability is an extremely weak condition. Even a spacetime that has closed, timelike curves is typically still time-orientable. For any time-orientable spacetime, there is a unique and physically reasonable way to define a time-reversal operator. In particular, this operator acts in the same way locally (in the tangent space) as the time-reversal from the Poincare group. This operator is a diffeomorphism, so by diff-invariance, a spacetime that has been subjected to this operation remains a valid solution to the Einstein field equations. In this sense, GR has time-reversal symmetry.

Again, by this argument, GR has local Galilean symmetry.

I'm not disagreeing that GR probably has time reversal symmetry. I'm saying that I disagree with the claims that the ability to write a theory in tensor notation holds as much physical significance as you feel it does.

If we write the physics in coordinate free form, there must be a way to discuss the symmetries in a "physical" / "coordinate free" method. In your last paragraph, you too start to refer to trying to get a physical interpretation of a transformation. So you seem to be groping towards this as well.

I'm hoping someone else can come along and point us in a more "correct" direction of how to talk about symmetries, when dealing with coordinate free expressions.

---
EDIT: Are Lie groups a way to get at the symmetries in a coordinate independent way? I am not very familiar with the the ability / power of that method of approaching symmetries.

 

[bcrowell]

The geometry is fixed, not the components of the metric. That term is indeed a scalar.

You were right about this point and I was wrong. I wasn't appropriately distinguishing between a fixed geometry and a fixed form for the metric.

Would you claim GR has local Galilean symmetry, just because the lagrangian is a scalar?

Yes, if you don't take that to imply any of the "baggage" referred to in #29. Otherwise no.

No. I would maintain that classical electrodynamics, written in tensor notation, retains that "form" regardless of the coordinate system you use. But if you look at how the evolution equations look in terms of E and B fields, they will NOT look the same after applying a galilean transformation. So if you use "same form" to define symmetry, then the resulting symmetries do depend on how you write it. I don't think this "same form" definition is precise enough, or useful, when we have tensor equations which are coordinate system independent.

The field equations of classical electrodynamics as expressed in GR, on a dynamical background, are certainly form-invariant under a local Galilean transformation. But again, that doesn't imply any of the "baggage," and there is also no sensible way to define a global Galilean transformation, because we don't have global coordinates that have the right interpretation.

 

[JustinLevy]

... because we don't have global coordinates that have the right interpretation.

Hmm... we can both feel it; there is some kind of physical "meaning"/"interpretation" we want to associate with the transformations ... and for the Galilean transformation, we can "feel" that it doesn't work.

But there must be some mathematical way to be explicit about this physical "baggage", so we can ask about symmetries in a methodical and clear manner.

I've asked a couple friends in physics and no one has come up with an answer. Maybe our starting point of associating symmetries with coordinate transformations is a crutch from when we started physics where it was just assumed the coordinates themselves had direct physical meaning. Maybe there is a deeper / coordinate independent method that demonstrates symmetries, and which reduces to merely playing with coordinate transformations in our "special case inertial frames" or something.

Thanks for your help and discussion. I clearly need to ruminate on this more. But if you run into any further insights, please do let me know.

 

[Phrak]

I think I can supply an operational meaning behind temporal inversion that should be seen to be coordinate free despite reference to indices.

Each upper or lower index that appears in a tensor, should the tensor be indexed in any coordinate system, is multiplied by a tensor with signature (-1,1,1,1), having all other entries zero.

With this as a definition, every tensor equation is, as bcrowell has called it, form invariant.

(Is the Jacobian of the metric unchanged?)

bcrowell, I don't understand your deference to diffeomorphic invarance. In my understanding, it doesn't seem to apply. Though please correct me if I'm wrong. Diffeomorphic invariance seems to be simply a generalization of what was once called a continuous transformation, which temporal inversion is not.

A general coordinate transformation involves dilations, shears and rotations. The Lorentz transformation that involves a single spatial coordinate and the time coordinate has a parameter theta, where elements of the Lorentz tensor are cosh(theta) and sinch(theta). For any finite theta, time never changes sign. I think that if you do find a parameter that inverts t, then at some value of the parameter t-->infinity.

bcrowell and Dale, I don't know what a smooth transformation is unless you meant a smooth transformation over a parameter(s) as the above.

 

[Dale]

I have been thinking about this thread but not posting, but the more I think about it the less I am sure that it makes sense to talk about GR having or not having time reversal symmetry. A tensor doesn't have time, and you can even use coordinate systems such as light-cone coordinates where there is not even a timelike coordinate. If a theory is not expressed in terms of some variable it seems to not make sense to ask if it is symmetric wrt changes in that variable. E.g. would you ask if Newton's laws were symmetric wrt charge reversal? I don't think so, since Newton's laws don't describe charge.

The only way that you get time in GR is when you are expressing it in terms of some specific coordinate system. So, while I don't think that you can ask if GR is time reversal symmetric, you could ask if some specific metric in some specific coordinate system is time reversal symmetric. That should be pretty easy to determine based on the power of any t and dt terms in the given coordinate system and metric.

Anyway, I am not terribly confident about this position, but I thought I would go ahead and post it.

 

[bcrowell]

I have been thinking about this thread but not posting, but the more I think about it the less I am sure that it makes sense to talk about GR having or not having time reversal symmetry. A tensor doesn't have time, and you can even use coordinate systems such as light-cone coordinates where there is not even a timelike coordinate. If a theory is not expressed in terms of some variable it seems to not make sense to ask if it is symmetric wrt changes in that variable. E.g. would you ask if Newton's laws were symmetric wrt charge reversal? I don't think so, since Newton's laws don't describe charge.

The only way that you get time in GR is when you are expressing it in terms of some specific coordinate system. So, while I don't think that you can ask if GR is time reversal symmetric, you could ask if some specific metric in some specific coordinate system is time reversal symmetric. That should be pretty easy to determine based on the power of any t and dt terms in the given coordinate system and metric.

I would pretty much agree with this analysis, but maybe with a different spin on it. The role of time in GR is not quite analogous to the role of charge in Newton's laws. At least locally, GR clearly does have something that plays the role of time. It's true that the field equations don't explicitly single out a different role for time, and in fact the field equations function very happily if you feed them a metric with a ++++ signature. But in practice we always look for solutions that have a +--- metric, and then we do have light cones and a notion of time defined, at least locally. What I'm claiming is that for the vast majority of spacetimes of physical interest, this makes it possible to define a unique and sensible time-reversal operation. This is because nearly all spacetimes of physical interest are time-orientable.

The part that I still don't feel satisfied with is my understanding of is the role of background-independence. For example, consider Lorentz invariance versus Galilean invariance. Let theory T1 be Maxwell's equations in vacuum as expressed in the language of x, y, z, t, div, and curl. I'm pretty sure that T1 is form-invariant under a Lorentz transformation, but not form-invariant under a Gaillean transformation. On the other hand, let theory T2 be the full electrovac field equations of GR ( http://en.wikipedia.org/wiki/Electrovacuum_solution ), I'm pretty sure T2 is form-invariant under any diffeomorphism. In particular, say you pick some point P in your spacetime, and you write down some coordinate transformation G that, in the neighborhood of P, is a Galilean transformation. As long as G is defined globally (i.e., on every coordinate patch in your manifold), and is smooth enough to be a diffeomorphism, then I'm pretty sure T2 is form-invariant under G. That doesn't mean that G deserves to be interpreted as a global Galilean transformation -- there is no such thing in a general spacetime.

What I would like to understand better is how and where this difference in behavior between T1 and T2 creeps in. For example, we could imagine intermediate levels between T1 and T2. Maybe T1.5 could be Maxwell's equations written on a background of fixed curvature. Is T1.5 form-invariant under G, or not? How can we tell? We would have to consider things like whether the derivativatives are covariant derivatives or plain old partial derivatives, etc. It also gets confusing because in the language of T1, we don't distinguish between the covariant and contravariant forms of something like the electric field vector. Another thing I have to understand better is the implications of the fact that a theory like T2 needs a supplementary field equation on the Maxwell tensor F, rather than just the Einstein field equations.

Probably this discussion can be simplified quite a bit by taking something simpler than G. For example, we could take a point P, define local Minkowski coordinates t,x,y,z, and then define a dilation D that only acts on one coordinate, so x'=kx. If we like, we can extend this in some stupid way to a global coordinate chart, with the understanding that at points far from P it loses its geometrical interpretation as a one-dimensional dilation. I claim that T1 is not form-invariant under D, T2 is form-invariant under D, and that if we could understand exactly why this is, that understanding would also apply to G.

 

[JustinLevy]

Beautiful summary, both of you. Much better than I was able to say it.

Dalespam's view on discussing these symmetries in GR is a bit more pessimistic than the outcome I was hoping for, but it is hard to find a good way to argue with mathematically. My gut feeling still currently falls more towards bcrowell's view that we should still be able to discuss local symmetries in a meaningful way ... I just don't know how.

It also gets confusing because in the language of T1, we don't distinguish between the covariant and contravariant forms of something like the electric field vector. Another thing I have to understand better is the implications of the fact that a theory like T2 needs a supplementary field equation on the Maxwell tensor F, rather than just the Einstein field equations.

For the first one, when forced to choose I've seen people preferably choose to identify the E and B fields as components of the covariant field tensor (or the contravariant dual to the field tensor), so as to leave the "source free" maxwell's equations least "changed" as opposed to the source equation. However I think it is just purely convention choice, with no real meaning (like the sign choice of the inertial frame metric as -1,1,1,1 or 1,-1,-1,-1). So our choice shouldn't matter.

I'm not sure what exactly you are pointing out in the second one. Maybe I'm missing some implications here entirely.

For example, we could imagine intermediate levels between T1 and T2. Maybe T1.5 could be Maxwell's equations written on a background of fixed curvature. Is T1.5 form-invariant under G, or not? How can we tell? We would have to consider things like whether the derivativatives are covariant derivatives or plain old partial derivatives, etc.

Regarding partial vs covariant derivatives, I think we could just start by looking at Maxwell's equations in flat spacetime. I may be making a mistake here, but I think if we want to write Maxwell's equations in tensor notation for a fixed geometry, we'd already have to write it as covariant derivatives.

Consider a simple case of coordinate system in which the components of the metric in flat spacetime depends on the position: Rindler coordinates for an "accelerating frame". We're still in flat spacetime, but for the equations to work, we need to use covariant derivatives. So maxwell's equations as tensor equations seem to already require:
F^{\alpha\beta}{}_{;\beta} = \mu_0 J^{\alpha}
F_{ \alpha \beta ; \gamma } + F_{ \beta \gamma ; \alpha } + F_{ \gamma \alpha ; \beta } = 0
or in terms of the dual electromagnetic field tensor G^{\gamma\delta} = \frac{1}{2}\epsilon^{\alpha\beta\gamma\delta}F_{\alpha\beta}
G^{\alpha\beta}{}_{;\beta} = 0
instead of partial derivatives (which it reduces to when the components of the metric don't depend on position).

Which would appear to tell us yes, T1.5 is invariant to G. So it appears merely writing it in full tensor notation, removes our ability to discuss symmetries as related to coordinate transformations ... which makes sense along DaleSpam's reasoning.

It really seems to me, we are missing a mathematical tool for asking symmetry questions for tensor equations. This has most likely been considered and solved long ago. So what is this missing tool?

 

[bcrowell]

Which would appear to tell us yes, T1.5 is invariant to G. So it appears merely writing it in full tensor notation, removes our ability to discuss symmetries as related to coordinate transformations

I disagree. I think generalizing to curved spacetime is what removes our ability to discuss a global symmetry like G. Say I ask question Q: "Does theory T have invariance under a global Galilean transformation?" Question Q is a sensible question if T is QED, but if T is GR then question Q is a nonsensical question, like asking whether love smells good.

It really seems to me, we are missing a mathematical tool for asking symmetry questions for tensor equations. This has most likely been considered and solved long ago. So what is this missing tool?

I don't think we're missing a mathematical tool. The tool of tensors just makes it easier to detect that Q is a nonsense question when T is GR. Without tensors, it would just be hard to tell whether (1) Q is an interesting question, and we need to compute the answer by laborious methods, or (2) Q is a nonsense question. Tensors make the computations trivial, so we can focus on whether Q is sensical or nonsensical.

[EDIT] Actually, maybe I would argue for the usefulness of one mathematical tool, which is the notion of orientability. Time-orientability is what makes it sensible to ask whether GR is time-reversal symmetric, because it let's us understand the conditions under which we can meaningfully define a time-reversal operator. It's possible that there are other kinds of orientability that are useful as well. E.g., it might be meaningful to ask whether GR has symmetry under parity inversion.

 

[bcrowell]

Here's a nice easy way to get at some of the issues we've been discussing. Anything that introduces a preferred scale violates diff-invariance. As a really simple example, suppose that I couple GR to classical, pointlike, identical particles of mass m. Then a Schwarzschild spacetime with mass m is a solution of the Einstein field equations. But if I subject this spacetime to a coordinate transformation where all the space and time coordinates are contracted by a factor of 2, this becomes a Schwarzschild spacetime with mass parameter m/2, which isn't a valid spacetime of this theory. IMO this is an example of how GR has a much higher degree of symmetry than other physical theories, the symmetry is expressed by GR's diff-invariance, and it's broken by coupling to specific types of matter.

 

[atyy]

Here's a nice easy way to get at some of the issues we've been discussing. Anything that introduces a preferred scale violates diff-invariance. As a really simple example, suppose that I couple GR to classical, pointlike, identical particles of mass m. Then a Schwarzschild spacetime with mass m is a solution of the Einstein field equations. But if I subject this spacetime to a coordinate transformation where all the space and time coordinates are contracted by a factor of 2, this becomes a Schwarzschild spacetime with mass parameter m/2, which isn't a valid spacetime of this theory. IMO this is an example of how GR has a much higher degree of symmetry than other physical theories, the symmetry is expressed by GR's diff-invariance, and it's broken by coupling to specific types of matter.

By diff invariance, do you mean general covariance?

As I understand it, general covariance is just gauge, and not meaningful.

 

[bcrowell]

By diff invariance, do you mean general covariance?

Yes. (Diffeomorphism invariance=general covariance.)

As I understand it, general covariance is just gauge,

Some people like to refer to it a gauge symmetry, others don't. It's a matter of taste. In any case, it's certainly at least analogous to a gauge symmetry.

and not meaningful.

...which is a whole different issue, and would depend on what you meant by "meaningful." That's pretty much what we've been debating in this thread.

 

[atyy]

...which is a whole different issue, and would depend on what you meant by "meaningful." That's pretty much what we've been debating in this thread.

I thought you had decided that it should be defined in terms of timelike worldlines, time orientability etc, just like Demystifier's, Wald and Strominger's discussions of black holes, white holes etc, which are gauge invariant?

Wrt to Maxwell's equations in Minkowski spacetime, the solutions may break the symmetries, but in contrast, because diff invariance is a gauge symmetry, it cannot be broken by any solution of GR.

 

[atyy]

I don't think we're missing a mathematical tool. The tool of tensors just makes it easier to detect that Q is a nonsense question when T is GR. Without tensors, it would just be hard to tell whether (1) Q is an interesting question, and we need to compute the answer by laborious methods, or (2) Q is a nonsense question. Tensors make the computations trivial, so we can focus on whether Q is sensical or nonsensical.

I think it's related to the question - how do you know your diff eq is really nonlinear, and not just a linear equation to which you have applied a nonliear change of coordinates?

Apparently, there is a method to find this out using Lie groups.
http://books.google.com/books?id=nFSJn7dIYysC&dq=stephani+lie+hans&source=gbs_navlinks_s
http://books.google.com/books?id=Yu...etries+invariants+olver&source=gbs_navlinks_s

I don't know how far it extends to other symmetries. For example, http://arxiv.org/abs/1007.5472 talks about "known or unknown 'hidden symmetries'" of a well-studied theory, so I'd guess there's no simple way to uncover all symmetries.

 

[Haelfix]

I must admit I am a little stumped by this question. Locally, surely everyone agrees that there is a good notion of time reversal symmetry.

Globally, I am not so sure and I think this is subtle.

The problem as I see it is topological. Certain spacetimes do not admit time orientability, b/c there is an obstruction to this identification (think pure De Sitter space).

Consequently, it strikes me that diffeomorphism invariance alone does not guarantee that solutions of GR are time symmetric. I suppose you could say that the diffeomorphism invariance is spontaneously broken by the choice of topology. Otoh that's a little bit backwards, hence my confusion

 

[Phrak]

I have been thinking about this thread but not posting, but the more I think about it the less I am sure that it makes sense to talk about GR having or not having time reversal symmetry. A tensor doesn't have time, and you can even use coordinate systems such as light-cone coordinates where there is not even a timelike coordinate. If a theory is not expressed in terms of some variable it seems to not make sense to ask if it is symmetric wrt changes in that variable. E.g. would you ask if Newton's laws were symmetric wrt charge reversal? I don't think so, since Newton's laws don't describe charge.

The only way that you get time in GR is when you are expressing it in terms of some specific coordinate system. So, while I don't think that you can ask if GR is time reversal symmetric, you could ask if some specific metric in some specific coordinate system is time reversal symmetric. That should be pretty easy to determine based on the power of any t and dt terms in the given coordinate system and metric.

Anyway, I am not terribly confident about this position, but I thought I would go ahead and post it.

I've vacillated over this many times in 24 hrs. However, the light cone at each point is independent of coordinates. No matter what coordinates we choose, it's invariant.

Assume at first, for simplicity, that spacetime is orientable. In addition, the light cone at any arbitrary point is equipped with labels + and - for each branch. This can be translated over the entire manifold as long as we ignore points that are singular. Doing it this way we can ask about interchanging the labels + and -. Interchanging labels has no effect on any structure that inherits its characteristics from the metric, and as long as we confine our criteria to metric structures and it's derivatives, this is good to know.

The labeled light cone pair is invariant over proper Lorentz transforms. It is not invariant over improper Lorentz transforms that include spacetime inversion, which is why I objected to arguments over diffeomorphic invariance.

The map is not the territory. Anytime we talk about an abstract manifold, implicit or not, an additional element is the direction of +time. This established the light cone labels. Implicitly, when we talk about the Schwarzschild metric, we all think it's a black hole, rather than a white hole, because the labels are implied.

 

[Phrak]

As long as we claim that the answer to the question "Is General Relativity Time Symmetric?" is confined only to the metric and it's derivatives", then the answer is yes, because equalities involving the metric and it's derivatives are invariant with respect to this labeling.

If electric charge is introduced, for instance, the question must be reconsidered.

In the end, Dale, we are seem to be in agreement, in as much as t or dt is invariably and implicitly introduced whenever a model of spacetime is argued to correspond with elements of physical reality, in so doing placing a label on each leg of the cone.

 

[Dale]

In addition, the light cone at any arbitrary point is equipped with labels + and - for each branch.

That is true. But I am not sure that the light cone adequately defines "time". The light cone itself is null, so that is not sufficient, and then any timelike vector can be chosen as your time coordinate. So I agree that you can show light cone reversal symmetry in a coordinate independent tensor form for the usual spacetimes of interest, but my hesitation is in identifying that with time reversal symmetry. I'm certainly not saying your argument is wrong, I am just still not comfortable about how to talk about time reversal symmetry without a coordinate system and a metric expressed in that coordinate system.

 

[bcrowell]

That is true. But I am not sure that the light cone adequately defines "time". The light cone itself is null, so that is not sufficient, and then any timelike vector can be chosen as your time coordinate. So I agree that you can show light cone reversal symmetry in a coordinate independent tensor form for the usual spacetimes of interest, but my hesitation is in identifying that with time reversal symmetry. I'm certainly not saying your argument is wrong, I am just still not comfortable about how to talk about time reversal symmetry without a coordinate system and a metric expressed in that coordinate system.

In GR, we don't even expect to have global coordinates, just an atlas of coordinate patches. It's handy to talk about t->-t when we want to talk about a global time reversal operation, but really that's just a shorthand, because in general it's not possible to cover the whole manifold with a single coordinate patch on which a single t variable is defined. Reversing the light cones is exactly the right notion if you want to talk about time-reversal.

Another way of looking at it is that once you know the null cones, you can determine the metric everywhere, up to a conformal factor. (See Hawking and Ellis, pp. 60-61.) There is no need to fix the conformal factor in order to talk about time reversal. In other words, you can define time-reversal simply in terms of flipping the Penrose diagram upside down.

 

[Phrak]

That is true. But I am not sure that the light cone adequately defines "time". The light cone itself is null, so that is not sufficient, and then any timelike vector can be chosen as your time coordinate.

It was a poor choice of language. I am not well equipped to turn concepts into words.

However, the light cones serve to separate spacetime into regions. We can take a pair of cones at each point in a model spacetime, and move it all around, so long as we don't run into a singularity such as a Schwarzschild coordinate singularity, we can populate the entire mainifold with cone-pairs at each point. Fill one of the cone pairs with blue and the other fill with red. At each point on the manifold there is a corresponding red-blue pair of cones. This, with the labeling, serves as the model to examine time reversal symmetry, as I see it, so long as spacetime is not a Klein bottle.

So I agree that you can show light cone reversal symmetry in a coordinate independent tensor form for the usual spacetimes of interest, but my hesitation is in identifying that with time reversal symmetry. I'm certainly not saying your argument is wrong, I am just still not comfortable about how to talk about time reversal symmetry without a coordinate system and a metric expressed in that coordinate system.

 

[bcrowell]

I thought you had decided that it should be defined in terms of timelike worldlines, time orientability etc, just like Demystifier's, Wald and Strominger's discussions of black holes, white holes etc, which are gauge invariant?

There are two different issues: (1) Can you define time-reversal? (2) If so, is GR form-invariant under it? You need orientability to decide on which spacetimes the answer to #1 is yes. But I would still claim that the affirmative answer to #2 (for spacetimes where the answer to #1 is yes) follows from diff-invariance.

Wrt to Maxwell's equations in Minkowski spacetime, the solutions may break the symmetries, but in contrast, because diff invariance is a gauge symmetry, it cannot be broken by any solution of GR.

I'm willing to be convinced, but so far this doesn't work for me. On a time-orientable spacetime, there is a uniquely defined time-reversal operator T, and this operator is an element of the diffeomorphism group D. Minkowski space doesn't break T symmetry, and in fact it's invariant under every element of D. A Schwarzshild metric fails to be invariant under almost every element of D, except for certain special ones like rotations and T. Realistic cosmological models break T.

Maybe we have different ideas of what it would mean to be invariant under a member of D. The way I see it, e.g., a Killing vector would generate a family of diffeomorphisms that are all members of D. The spacetime might or might not have the symmetry represented by that Killing vector.

I suppose you could take the point of view that a diffeomorphism is just a change of coordinates, and coordinates are just names for points, so a diffeomorphism is just a renaming, like calling the t coordinate \tau, in which case obviously a point in a spacetime can't "care" what it's named.

Say I tell you, "The stress-energy tensor is zero at point P." Then you apply a diffeomorphism. You could take the diffeomorphism as simply translating into a new language, in which P isn't even a word, and the new word for that point is Q. In that case obviously the same sentence is still true. On the other hand, you could take it as saying that P has been redefined to refer to some other point in space, say the center of the earth, in which case the sentence is now false.

Possibly both of these ways of looking at it are equally valid, in which case it could be valid to prove time-reversal symmetry by my argument, but one would have to specify which point of view was being adopted.

 

[atyy]

There are two different issues: (1) Can you define time-reversal? (2) If so, is GR form-invariant under it? You need orientability to decide on which spacetimes the answer to #1 is yes. But I would still claim that the affirmative answer to #2 (for spacetimes where the answer to #1 is yes) follows from diff-invariance.

But if you use orientability in #1, that is already using #2, since orientability is invariant under change of coordinates, so #1 and #2 are not distinct.

Possibly both of these ways of looking at it are equally valid, in which case it could be valid to prove time-reversal symmetry by my argument, but one would have to specify which point of view was being adopted.

I was using the second one because you indicated that you wanted to take diff invariance - general covariance.

 

[atyy]

Maybe we have different ideas of what it would mean to be invariant under a member of D. The way I see it, e.g., a Killing vector would generate a family of diffeomorphisms that are all members of D. The spacetime might or might not have the symmetry represented by that Killing vector.

Also, does the qualification "local" diffeomorphism for a Killing vector make any difference to your argument?

 

[bcrowell]

Maybe we have different ideas of what it would mean to be invariant under a member of D. The way I see it, e.g., a Killing vector would generate a family of diffeomorphisms that are all members of D. The spacetime might or might not have the symmetry represented by that Killing vector.

Also, does the qualification "local" diffeomorphism for a Killing vector make any difference to your argument?

That's why I said "e.g." All Killing vectors generate diffeomorphisms, but not all diffeomorphisms can be generated from Killing vectors. No, I don't think this has any effect on my argument. The only reason I mentioned Killing vectors was that I wanted to give some concrete examples of how a spacetime could have some proper subset of the symmetries given by the diffeomorphism group.

 

[atyy]

That's why I said "e.g." All Killing vectors generate diffeomorphisms, but not all diffeomorphisms can be generated from Killing vectors. No, I don't think this has any effect on my argument. The only reason I mentioned Killing vectors was that I wanted to give some concrete examples of how a spacetime could have some proper subset of the symmetries given by the diffeomorphism group.

OK, I agree.