Is General Relativity Time Symmetric?

[Phrak]

Is General Relativity Time Symmetric?

 

[Mentz114]

I'm not sure what you mean. Presumably invariance of something under the substitution $t\rightarrow -t$.

I believe some solutions will work if time is reversed, an expanding dust can become a collapsing dust. The theory of the Riemannian manifold specifies that the time component of tangent space basis vector must point in the t-direction. I have no idea if this can be safely reversed.

In short, I don't know, but maybe someone else does ?

 

[Demystifier]

Is General Relativity Time Symmetric?

Yes it is.

For example, black holes and their time-inverted configurations - white holes - are both solutions of GR.

 

[bcrowell]

Yes, relativity is symmetric under time-reversal. The laws of physics in general relativity have the same form regardless of any smooth coordinate transformation whatsoever. Time-reversal is a smooth coordinate transformation.

 

[Phrak]

It occurs to me that I don't know how to show that it is true. Any hints or speculation as to how to proceed?

 

[JustinLevy]

bcrowell: "Time-reversal is a smooth coordinate transformation."

Hmm... I think I see what you mean here by "smooth" : the mapping of old coordinates to new coordinates is a smooth function.

But I have also heard of parity or time-reversal to not be considered smooth transformations because one cannot smoothly transition from the initial to final coordinate maps by a series of infinitesimal steps (like rotations, or stretching/scaling, etc.).

--

Anyway, how do we show this more rigorously?
Merely being able to write the equations in tensor notation doesn't seem sufficient, as we can do that for the standard model, but it isn't time reversal invariant.

Under time reversal, the metric g_uv -> g'_uv will change sign on the dx dt, dy dt, dz dt, pieces. This will effect some terms of the curvature, maybe? This looks like it will get complicated. I'm not sure how it will affect the field equations.

Maybe a better place to look is the Lagrangian. Is the ricci curvature scalar affected by a time reversal? There's probably an easy way to see this, but I'm not sure how without the messy method of looking at how it depends on the metric components.

 

[bcrowell]

bcrowell: "Time-reversal is a smooth coordinate transformation."

Hmm... I think I see what you mean here by "smooth" : the mapping of old coordinates to new coordinates is a smooth function.

But I have also heard of parity or time-reversal to not be considered smooth transformations because one cannot smoothly transition from the initial to final coordinate maps by a series of infinitesimal steps (like rotations, or stretching/scaling, etc.).

To put it in more formal terminology, the field equations of GR are form-invariant under diffeomorphisms, and time-reversal is a diffeomorphism. Therefore any solution is also a solution under time-reversal.

Anyway, how do we show this more rigorously?

I claim that it is rigorous, as expressed above.

Merely being able to write the equations in tensor notation doesn't seem sufficient, as we can do that for the standard model, but it isn't time reversal invariant.

I think what you've proved is that the standard model can't be written in terms of tensors. That is, I don't think the standard model is diffeomorphism invariant. (It doesn't have general covariance.) If it were diffeomorphism invariant, it would be expressed without a background spacetime; but the standard model is expressed with a certain assumed background spacetime (flat spacetime).

[EDIT] Sorry, this was not quite right. The existence of time-reversal symmetry in the SM doesn't necessarily show that the SM lacks diffeomorphism invariance, because it could be a case of spontaneous symmetry breaking. However, I don't think it's true that the SM has diffeomorphism invariance. For example, whenever you write a derivative with respect to time, that's breaking diffeomorphism invariance; time is treated asymmetrically in quantum mechanics, not as an operator like position.

Under time reversal, the metric g_uv -> g'_uv will change sign on the dx dt, dy dt, dz dt, pieces. This will effect some terms of the curvature, maybe? This looks like it will get complicated. I'm not sure how it will affect the field equations.

You don't have to crank through the complete rederivation of the curvature tensors. The curvature tensors change under a change of coordinates according to the tensor transformation laws. Since both sides of the Einstein field equations are tensors, they transform identically, and a time-reversed solution is still a solution.

Is the ricci curvature scalar affected by a time reversal?

It's a scalar, so it's invariant under diffomorphisms, including time reversal.

 

[Dale]

But I have also heard of parity or time-reversal to not be considered smooth transformations because one cannot smoothly transition from the initial to final coordinate maps by a series of infinitesimal steps (like rotations, or stretching/scaling, etc.).

You are probably thinking in terms of the differentiable symmetries of Noether's theorem. Parity is not an infinitesimal transform of the sort that would indicate a differential symmetry, but it is a smooth transform that indicates another kind of symmetry.

 

[nicksauce]

So I'm not so sure about this, but maybe I'm missing something. I figure it would be easy to imagine a metric, which is not invariant under t-> -t, (just put some term in the metric proportional to t, for example). As long as the stress tensor is not completely ridiculous, you'd end up with a solution that is not time symmetric.

 

[bcrowell]

So I'm not so sure about this, but maybe I'm missing something. I figure it would be easy to imagine a metric, which is not invariant under t-> -t, (just put some term in the metric proportional to t, for example). As long as the stress tensor is not completely ridiculous, you'd end up with a solution that is not time symmetric.

Equations that are invariant under a certain transformation can have solutions that are not invariant. For example, the equation $x^2=1$ is invariant under $x\rightarrow -x$, but the solution x=1 is not invariant under that transformation. Or consider that Newton's laws are invariant under rotations, but a particle moving along the x-axis in the positive x direction is a solution that is not invariant under rotations.

What you can say is that if a set of differential equations has a certain symmetry, and you find a certain solution under boundary conditions that respect that symmetry, then the solution will have that symmetry as well. For example, this is how we know that the Schwarzschild spacetime is time-reversal symmetric.

[EDIT] Actually this is not quite right, as I realized by asking some questions over in Beyond the Standard Model and browsing WP. You can have spontaneous symmetry breaking. In cosmological solutions, Lorentz invariance is spontaneously broken by the occurrence of a preferred frame (the CMB's frame, or the frame of the Hubble flow). I guess T symmetry is also broken in perpetually expanding cosmologies.

 

[Parlyne]

[EDIT] Actually this is not quite right, as I realized by asking some questions over in Beyond the Standard Model and browsing WP. You can have spontaneous symmetry breaking. In cosmological solutions, Lorentz invariance is spontaneously broken by the occurrence of a preferred frame (the CMB's frame, or the frame of the Hubble flow). I guess T symmetry is also broken in perpetually expanding cosmologies.

Of course, in such cases the distinct solution you get by applying the symmetry operation to the symmetry-broken state will also be a valid solution if the underlying dynamics respect the symmetry.

 

[atyy]

But diff invariance is a gauge symmetry, which isn't a real symmetry, is it?

I tend to think of gauge symmetries as redundancies of description.

Certainly at the level of the equations of motion every theory can be made diff invariant, and apparently even at the level of the action http://arxiv.org/abs/1010.2535: "The point here is that any local theory (e.g., a single free scalar field) can be written in diffeomorphism-invariant form through a process known as parametrization"

In Newtonian mechanics, time reversal symmetry means if you reverse time and momenta, the different physical situation obtained is also a solution of the equations of motion. I think Demystifier was giving a comment along these lines, which seems to be different from the diff invariance perspective.

 

[bcrowell]

Certainly at the level of the equations of motion every theory can be made diff invariant, and apparently even at the level of the action http://arxiv.org/abs/1010.2535: "The point here is that any local theory (e.g., a single free scalar field) can be written in diffeomorphism-invariant form through a process known as parametrization"

This paper might be relevant: http://arxiv.org/abs/gr-qc/0603087

Take a look at p. 7. They show an example of a heat diffusion equation, which isn't diff-invariant, and definitely not time-reversal invariant. They turn it into a diff-invariant equation, but in order to do that they have to introduce a background field that points in the preferred direction of time. I think the key is that this background field is written as if it were a tensor, but it doesn't transform like a tensor; it's postulated to be constant in its upper-index form.

The field equations of GR aren't like this. They're written in terms of things that really transform like tensors -- no fixed background.

In Newtonian mechanics, time reversal symmetry means if you reverse time and momenta, the different physical situation obtained is also a solution of the equations of motion. I think Demystifier was giving a comment along these lines, which seems to be different from the diff invariance perspective.

I think diff invariance implies exactly the same thing as what you're saying about Newtonian mechanics.

In the standard formulation of Newtonian mechanics, the laws of physics are form-invariant under time reversal. Therefore any solution can be time-reversed, and it's still a solution.

In the standard formulation of GR, the laws of physics are form-invariant under diffeomorphisms, which include time reversal. Therefore any solution can be time-reversed, and it's still a solution.

I don't think there's anything mysterious or deep going on here. If you write down the tensor transformation law as applied to the rank-2 tensors in the Einstein field equations, they operate in the same way on both sides of the equation, and the only requirement is that the transformation be a diffeomorphism (so that the derivatives appearing in the transformation law don't blow up or fail to exist). The transformation can be any diffeomorphism you like. Time reversal is simply one special case.

This argument only depends on the assumption that everything in the field equations is a tensor that really transforms like a tensor -- not something that's written so as to look like a tensor, but doesn't transform like one.

 

[atyy]

This paper might be relevant: http://arxiv.org/abs/gr-qc/0603087

Take a look at p. 7. They show an example of a heat diffusion equation, which isn't diff-invariant, and definitely not time-reversal invariant. They turn it into a diff-invariant equation, but in order to do that they have to introduce a background field that points in the preferred direction of time. I think the key is that this background field is written as if it were a tensor, but it doesn't transform like a tensor; it's postulated to be constant in its upper-index form.

The field equations of GR aren't like this. They're written in terms of things that really transform like tensors -- no fixed background.

I agree GR has no fixed background, which is different from diff invariance (or in the language of that paper, diff covariance). But what has no fixed background got to do with time reversal?

I think diff invariance implies exactly the same thing as what you're saying about Newtonian mechanics.

In the standard formulation of Newtonian mechanics, the laws of physics are form-invariant under time reversal. Therefore any solution can be time-reversed, and it's still a solution.

In the standard formulation of GR, the laws of physics are form-invariant under diffeomorphisms, which include time reversal. Therefore any solution can be time-reversed, and it's still a solution.

I don't think there's anything mysterious or deep going on here. If you write down the tensor transformation law as applied to the rank-2 tensors in the Einstein field equations, they operate in the same way on both sides of the equation, and the only requirement is that the transformation be a diffeomorphism (so that the derivatives appearing in the transformation law don't blow up or fail to exist). The transformation can be any diffeomorphism you like. Time reversal is simply one special case.

This argument only depends on the assumption that everything in the field equations is a tensor that really transforms like a tensor -- not something that's written so as to look like a tensor, but doesn't transform like one.

Time is absolute in Newtonian mechanics. The time you are referring to seems to be coordinate time, which is not gauge invariant. Is there any formulation using the absolute notion of timelike?

 

[JustinLevy]

I think you are reading way too much into the ability to write an equation in tensor notation.

I'm not arguing that GR doesn't have T symmetry, I'm merely pointing out that I feel you are overstating what a tensor equation gives us.

I don't think there's anything mysterious or deep going on here. If you write down the tensor transformation law as applied to the rank-2 tensors in the Einstein field equations, they operate in the same way on both sides of the equation

This ISN'T enough.

For example, I can write the equations of electrodynamics in tensor notation. I can apply a Galilean transformation to both sides of an evolution equation. And of course both sides still equal each other.

Does this mean electrodynamics has Galilean symmetry? No.
Electrodynamics has Lorentz symmetry, not Galilean symmetry.

I think what you've proved is that the standard model can't be written in terms of tensors.

I believe the standard model can be written in terms of tensors. Using your limited requirement for "time reversal symmetry", we can apply any coordinate transformation to both sides of a tensor equation, so yes, in that sense (which I feel lacks meaning) it has time reversal symmetry. But the "form" of the equations changed. The weak force instead of dealing only with left handed neutrinos, now only deals with right handed neutrinos.

So I think we are disagreeing on what is even meant by Lorentz symmetry, Galilean symmetry, time translation symmetry, spatial translation symmetry, parity symmetry, time reversal symmetry, etc. which can be worded as coordinate transformations. To me, these mean much more than just requiring equations can be written in tensor notation.

I'm not sure how to proceed. So maybe someone else can suggest a precise definition of "symmetry" in this context.

 

[Dale]

But diff invariance is a gauge symmetry, which isn't a real symmetry, is it?

Gauge symmetries certainly are real symmetries. They are the most fundamental symmetries of the standard model or any field theory. Also, per Noether's theorem they lead to conservation laws, e.g. the U(1) gauge symmetry leads to the conservation of charge.

EDIT: I just realize that I have violated my usual policy of not getting involved in discussions about whether or not something is "real". So I would like to revise my above statement and simply say that a gauge symmetry is a differential symmetry that leads to a conservation law as per Noether's theorem. As to whether or not that qualifies it for status of a "real" symmetry depends on the definition of "real".

 

[Phrak]

Does it make any sense to ask if General Relativity is time symmetrical, or is this an ill posed question?

I could ask the same of the spatial manifold+time of Newtonian mechanics. It makes no sense. Its the stuff that happens on the manifold that may or may not be time symmetrical. (Who asked this ridiculous question in the first place? Oh, that was me...)

I could combine thermodynamics with a curved manifold and come to the conclusion that, no, General Relativity is Not time symmetric. All I have to do is smash two planets together. So this combination doesn't work for temporal symmetry.

But, never fear. We can still ask if the vacuum metric and all it's derivative quantities are time symmetric in Einstein's spacetime manifold.

Secondly, thermodynamics sucks. The world is not a roulette wheel where the value of a function is some obscure value determined by an unknown gambler wasting his life in Monte Carlo. God does not have to play dice to utterly confuse us. We can still leave out thermodynamics and substitute the minutia of particle interaction.

 

[atyy]

Gauge symmetries certainly are real symmetries. They are the most fundamental symmetries of the standard model or any field theory. Also, per Noether's theorem they lead to conservation laws, e.g. the U(1) gauge symmetry leads to the conservation of charge.

EDIT: I just realize that I have violated my usual policy of not getting involved in discussions about whether or not something is "real". So I would like to revise my above statement and simply say that a gauge symmetry is a differential symmetry that leads to a conservation law as per Noether's theorem. As to whether or not that qualifies it for status of a "real" symmetry depends on the definition of "real".

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

 

[Dale]

I don't know the distinction. I thought U(1) was the gauge symmetry.

 

[bcrowell]

I agree GR has no fixed background, which is different from diff invariance (or in the language of that paper, diff covariance). But what has no fixed background got to do with time reversal?

If you look at that example in the paper, the only way they were able to recast the heat diffusion equation into a diff-invariant form was by writing the field equations with a background "tensor" that doesn't really transform like a tensor. This background points in the forward time direction, and it's the only thing in the field equations that defines a direction in time.

 

[bcrowell]

Time is absolute in Newtonian mechanics. The time you are referring to seems to be coordinate time, which is not gauge invariant. Is there any formulation using the absolute notion of timelike?

Hmm...you lost me here. It seems to me that the gauge symmetry of GR is symmetry under diffeomorphisms. We don't want time to be invariant under diffeomorphisms, because then it would be absolute...?

The coordinate-independent way of singling out the role of time, as opposed to some other coordinate, is that the signature of the metric is +---, as opposed to, say, ++++ or something. The signature is diff-invariant, by Sylvester's law of inertia. There is nothing in the field equations of GR that prevents us from constructing solutions with signatures like ++++ or ++--. When we look for solutions that have the +--- signature, we are externally imposing a requirement that is not present in the field equations, and that in some sense artificially breaks the symmetry between time and space.

 

[atyy]

I don't know the distinction. I thought U(1) was the gauge symmetry.

I think there are 2 different things that go by the same name ("U(1) gauge symmetry"). The first thing is a global symmetry gives charge conservation. The second thing is more like a principle of minimal coupling between electric charge and field, so it's analogous to the principle of equivalence which says that matter and spacetime geometry are minimally coupled.

There's a third use of "gauge symmetry", as in Higgs boson and "spontaneous gauge symmetry breaking", which apparently actually means "explicit symmetry breaking", since the gauge (global) symmetry isn't actually "spontaneously" broken. http://www.scholarpedia.org/article/Englert-Brout-Higgs-Guralnik-Hagen-Kibble_mechanism

 

[Dale]

OK, then I guess my comments refer to the first meaning and I don't know about the other meaning.

 

[atyy]

If you look at that example in the paper, the only way they were able to recast the heat diffusion equation into a diff-invariant form was by writing the field equations with a background "tensor" that doesn't really transform like a tensor. This background points in the forward time direction, and it's the only thing in the field equations that defines a direction in time.

But how do we know that there is no theory without prior geometry that isn't time reversal invariant?

Hmm...you lost me here. It seems to me that the gauge symmetry of GR is symmetry under diffeomorphisms. We don't want time to be invariant under diffeomorphisms, because then it would be absolute...?

The coordinate-independent way of singling out the role of time, as opposed to some other coordinate, is that the signature of the metric is +---, as opposed to, say, ++++ or something. The signature is diff-invariant, by Sylvester's law of inertia. There is nothing in the field equations of GR that prevents us from constructing solutions with signatures like ++++ or ++--. When we look for solutions that have the +--- signature, we are externally imposing a requirement that is not present in the field equations, and that in some sense artificially breaks the symmetry between time and space.

But what has the signature got to do with diffeomorphisms or no prior geometry? Special relativity also has Lorentzian signature, but the standard model of particle physics is a specail relativistic theory which isn't time reversal invariant.

 

[atyy]

A reference which may be useful:
http://prd.aps.org/abstract/PRD/v21/i10/p2742_1
Quantum gravity and time reversibility
Robert M. Wald
The meaning of time-reversal and CPT invariances of a theory is discussed both in the context of theories defined on flat spacetime as well as in general relativity. It is argued that quantum gravity cannot be time-reversal or CPT invariant; that an "arrow of time" must be fundamentally built into the theory. However, a weaker form of CPT invariance could still hold, in which case the fundamental "arrow of time" would not show up in the measurements of observers who perform scattering experiments. Consequences of this weaker hypothesis are explored.

Also interesting, even if not GR-gravity:
http://arxiv.org/abs/hep-th/9307079
White Holes, Black Holes and Cpt in Two Dimensions
Andrew Strominger
It is argued that a unitarity-violating but weakly CPT invariant superscattering matrix exists for leading-order large-N dilaton gravity, if and only if one includes in the Hilbert space Planckian "thunderpop" excitations which create white holes. CPT apparently cannot be realized in a low-energy effective theory in which such states have been integrated out. Rules for computing the leading-large-N superscattering are described in terms of quantum field theory on a single multiply-connected spacetime obtained by sewing the future (past) horizons of the original spacetime with the past (future) horizons of its CPT conjugate. Some difficulties which may arise in going beyond leading order in 1/N are briefly discussed.

 

[bcrowell]

But how do we know that there is no theory without prior geometry that isn't time reversal invariant?

If time reversal means taking some coordinate and subjecting it to a transformation $t\rightarrow -t$, then it's a diffeomorphism, and therefore a diff-invariant theory can't distinguish between a solution and its time-reversal. You don't have to say anything about "without prior geometry." If the equations of the theory are written solely in terms of tensors (meaning things that transform like tensors), then it's diff-invariant.

But what has the signature got to do with diffeomorphisms or no prior geometry? Special relativity also has Lorentzian signature, but the standard model of particle physics is a specail relativistic theory which isn't time reversal invariant.

You seem to be saying that Lorentzian signature doesn't imply time-reversal invariance. That's true, but I never said anything to the contrary.

The reason I brought up Lorentz signature was in reply to your post where you asked this: "Time is absolute in Newtonian mechanics. The time you are referring to seems to be coordinate time, which is not gauge invariant. Is there any formulation using the absolute notion of timelike?"

But maybe we're just talking past each other here...?

I don't seem to be understanding what you're saying, and you don't seem to be understanding what I'm saying.

One thing to keep in mind is that GR itself has an extremely high level of symmetry, but that doesn't mean that the matter fields you plug into the stress-energy tensor have that same level of symmetry.

So if the OP asked the question "Is GR time-reversal-symmetric?" (paraphrased), possibly the reason we're all confusing each other is that some of us may be talking about GR itself, whereas others may be talking about GR coupled to some matter fields.

Suppose you want to couple GR to a scalar field that obeys some kind of differential equation analogous to heat diffusion. What's going to happen is that you're not going to be able to write the theory of this scalar field in terms of tensor equations, because the grammar and vocabulary of tensors doesn't allow one to say anything that distinguishes the forward direction of time (in which heat differences even out) from the backward direction of time.

 

[atyy]

If time reversal means taking some coordinate and subjecting it to a transformation $t\rightarrow -t$, then it's a diffeomorphism, and therefore a diff-invariant theory can't distinguish between a solution and its time-reversal. You don't have to say anything about "without prior geometry." If the equations of the theory are written solely in terms of tensors (meaning things that transform like tensors), then it's diff-invariant.

You seem to be saying that Lorentzian signature doesn't imply time-reversal invariance. That's true, but I never said anything to the contrary.

The reason I brought up Lorentz signature was in reply to your post where you asked this: "Time is absolute in Newtonian mechanics. The time you are referring to seems to be coordinate time, which is not gauge invariant. Is there any formulation using the absolute notion of timelike?"

Well, basically I'm trying to understand if Demystifier's and your answers mean the same thing. It didn't seem obvious to me that "time reversal means taking some coordinate and subjecting it to a transformation $t\rightarrow -t$" , because it's not clear in GR which coordinate is the "time" coordinate. So the definition of time reversal should involve a a diff-invariant quantity in GR, and I thought you were bringing in the signature as the meaning of time in GR.

One thing to keep in mind is that GR itself has an extremely high level of symmetry, but that doesn't mean that the matter fields you plug into the stress-energy tensor have that same level of symmetry.

So if the OP asked the question "Is GR time-reversal-symmetric?" (paraphrased), possibly the reason we're all confusing each other is that some of us may be talking about GR itself, whereas others may be talking about GR coupled to some matter fields.

Yes, I think I confused those two. It's also not clear to me whether I am defining CP violation to be time reversal violation. Do you have access to the Wald article above? Apparently CP violation without CPT violation is not "fundamental" time reversal, whereas CPT violation would be "fundamental" time reversal violation. Apparently Poincare invariance in QFT is closely tied to CPT invariance. Is Demystifier's answer the same as saying that both every [-+++] GR spacetime has a [+---] counterpart which is its time reverse and also a GR spacetime?

 

[JustinLevy]

If time reversal means taking some coordinate and subjecting it to a transformation $t\rightarrow -t$, then it's a diffeomorphism, and therefore a diff-invariant theory can't distinguish between a solution and its time-reversal. You don't have to say anything about "without prior geometry." If the equations of the theory are written solely in terms of tensors (meaning things that transform like tensors), then it's diff-invariant.

Again, I have to object here, because then by that reasoning electrodynamics has Galilean symmetry.
I feel you are reading too much into the ability to write things in tensor notation.

Would you claim the standard model has Galilean symmetry?
Would you claim electrodynamics has Galilean symmetry?

By your argument, anything that can be written in tensor notation automatically has every possible symmetry that we can phrase as a coordinate transformation. I think we need to at least pause and consider: what do we mean when we say a theory has Lorentz symmetry? or time reversal symmetry? etc.

I do not feel coordinate independence automatically yields parity, time, or galilean symmetry. You are equating too many things with mere coordinate independence.

 

[bcrowell]

Well, basically I'm trying to understand if Demystifier's and your answers mean the same thing. It didn't seem obvious to me that "time reversal means taking some coordinate and subjecting it to a transformation $t\rightarrow -t$" , because it's not clear in GR which coordinate is the "time" coordinate. So the definition of time reversal should involve a a diff-invariant quantity in GR, and I thought you were bringing in the signature as the meaning of time in GR.

I see. Yeah, that's a very good point that I hadn't thought about carefully enough before.

The fact that the signature is Lorentzian means that locally, you can always define a distinction between two light cones; locally, time reversal means swapping those two light cones. But to talk about global time-reversal, I think you have to assume that the spacetime is time-orientable, meaning that the distinction between timelike vectors in the two light cones can be made in a way that is continuous. Not all spacetimes are time-orientable.

So if your spacetime is not time-orientable, then I guess you definitely can't define anything like a global time-reversal.

Is Demystifier's answer the same as saying that both every [-+++] GR spacetime has a [+---] counterpart which is its time reverse and also a GR spacetime?

I don't think swapping the signature from -+++ to +--- has anything to do with time reversal.

Would you claim the standard model has Galilean symmetry?

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

Again, I have to object here, because then by that reasoning electrodynamics has Galilean symmetry.

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. But in any case, I think it's certainly true that, for example, any electrovac solution of GR is still a solution under an arbitrary diffeomorphism. The electrovac field equations basically make statements about purely topological facts, e.g., that magnetic field lines never terminate at a point, or that electric field lines don't pass through one another. They also relate the fields to the metric, but again only in ways that are diff-invariant. If you give me an electrovac solution that has the global topology of $\mathbb{R}^4$ and is time-orientable, then I can label it with a single coordinate chart consisting of four coordinates (p,q,r,s), where, say, p is everywhere timelike. Then it is certainly true that this electrovac solution is still a solution after I apply a diffeomorphism that takes q to q'=q+kp, where k is some constant. This diffeomorphism does sort of look like a Galilean transformation, if you think of p as time, q as position along some axis, and k as a velocity -- but there is no guarantee that this is really their physical interpretation in any meaningful way.

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.

[EDIT] Corrected an overly broad, incorrect claim about the purely topological nature of the electrovac field equations.

 

[atyy]

The fact that the signature is Lorentzian means that locally, you can always define a distinction between two light cones; locally, time reversal means swapping those two light cones. But to talk about global time-reversal, I think you have to assume that the spacetime is time-orientable, meaning that the distinction between timelike vectors in the two light cones can be made in a way that is continuous. Not all spacetimes are time-orientable.

So if your spacetime is not time-orientable, then I guess you definitely can't define anything like a global time-reversal.

OK, that seems much closer to what Demystifier was saying about black holes and white holes.

 

[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} = \left( \begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)\)$$
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.