Diffeomorphic Invariance implies Poincare Invariance?

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I have been quite puzzled for some time with the concept of Diffeomorphic Invariance.

Here is what I think about it,

1) Diffeomorphic Invariance is the invariance of the theory under general coordinate transformations. For instance the Einstein-Hilbert action is diffeomorphic invariant.

2) Poincare Invariance : The invariance of the equations of physics under rotations and translations.

Should Poincare be thought of as an Active Diffeomorphism? And if it really can be, would it mean that poincare invariance is always implied when a system has diffeomorphsim invariance?

Thanks!
 
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This is very handwaving, but here is how I see it: classically the difference between active and passive coordinate transformations is that for the passive ones you transform the background, and therefore the metric, and then just re-express your system in terms of the new coordinates, while for active transformations you only transform whatever it is you have living on that background.

Often the difference isn't important, because the transformation you want to do is an isometry of your space anyway. For example, in flat Euclidean space, you can translate and rotate passively or actively all you want, it doesn't change your metric. On the other hand, you cannot actively scale things, because if you don't scale your metric along, you end up with a different system. In this case you can only perform your transformation passively.

Now, in GR, you promote your metric to a dynamical field, in a sense making it part of the 'foreground' rather than the background. But then there really isn't much of a distinction between passive and active transformations anymore, since the active ones also take along your metric, always.
 
FedEx said:
1) Diffeomorphic Invariance is the invariance of the theory under general coordinate transformations. For instance the Einstein-Hilbert action is diffeomorphic invariant.

Ok so far.

FedEx said:
2) Poincare Invariance : The invariance of the equations of physics under rotations and translations.

And boosts--i.e., Lorentz transformations (as distinct from spatial rotations and translations).

FedEx said:
Should Poincare be thought of as an Active Diffeomorphism?

I don't think so. It's a diffeomorphism (since it's a subset of general coordinate transformations, which are diffeomorphisms), but not an active one, because it doesn't change the underlying geometry, only the coordinates you use to describe it.
 
NanakiXIII said:
classically the difference between active and passive coordinate transformations is that for the passive ones you transform the background, and therefore the metric, and then just re-express your system in terms of the new coordinates, while for active transformations you only transform whatever it is you have living on that background.

This isn't my understanding of the standard usage of this terminology. (Although, to be fair, the standard usage of this terminology seems to be somewhat muddled; different references seem to use the terms in different ways.) My understanding is: an active diffeomorphism changes the actual geometry; a passive diffeomorphism only changes the coordinates used to describe the geometry.

For example: if I take a flat Euclidean plane and deform it into a bumpy, wiggly surface, that's an active diffeomorphism; I'm changing the actual geometry of the surface. But if I take a flat Euclidean plane and switch from Cartesian to polar coordinates, that's a passive diffeomorphism: the coordinate transformation from Cartesian to polar has to satisfy all the requirements of a diffeomorphism, but it doesn't change the underlying geometry of the plane.

NanakiXIII said:
Now, in GR, you promote your metric to a dynamical field, in a sense making it part of the 'foreground' rather than the background. But then there really isn't much of a distinction between passive and active transformations anymore, since the active ones also take along your metric, always.

There still is a distinction between passive and active in GR--at least, there is with the usage of the terms I gave above. For example: if I switch from Schwarzschild coordinates to Painleve coordinates in describing Schwarzschild spacetime, keeping the mass of the black hole constant, that's a passive diffeomorphism; I'm only changing the coordinates, not the underlying geometry. But if I take Schwarzschild spacetime and change the mass of the black hole, that's an active diffeomorphism, because I'm changing the actual geometry.

It's worth noting here as well that one of the requirements for a diffeomorphism is that the topology of the manifold does not change. That actually rules out a lot of conceivable active diffeomorphisms in GR. For example, you can't do an active diffeomorphism from Schwarzschild spacetime to Kerr spacetime, because the topologies of the underlying manifolds are different.
 
There is no mathematical difference between an "active" and "passive" transformation. It is something you will only find in physics books and math books will not spend time making the distinction. They are both diffeomorphisms and all they do is map you from a smooth manifold M to another smooth manifold N such that the smooth structure of M is preserved. There is no mention of the riemannian structure of the manifolds at all. A diffeomorphism need not preserve the metric tensor which is what I assume you mean by "geometry". If the diffeomorphism happens to be an isometry, THEN we can say the two manifolds have the same riemannian structure. One should not confuse diffeomorphisms with isometries, which are special diffeomorphisms: the former does not make any mention of the riemannian structure of the manifold whereas the latter does by definition.

If we are given a diffeomorphism \phi :M\rightarrow N then we speak of active transformations of tensorial quantites by simply speaking of the pushforward and pull back \phi _{*},\phi ^{*}. This active point of view makes no reference to a coordinate system. The passive transformation is simply to look at the coordinate transformation x^{\mu}\rightarrow y^{\mu} induced by \phi and talk about tensorial quantities transforming under the usual coordinate transformation rules. Again, there is no mention of riemannian geometry and mathematically there is no difference between the notion of passive and active diffeomorphisms.

By the way, the poincare group is the group of isometries of specifically (\mathbb{R}^{4},\eta _{ab}) and as Peter already mentioned, includes boosts, translations, and rotations. A general space - time need not admit any isometries and as such need not have any isometry group attatched to it.
 
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WannabeNewton said:
There is no mathematical difference between an "active" and "passive" transformation. It is something you will only find in physics books and math books will not spend time making the distinction.

Agreed, mathematically there is no difference, the difference is only in the physical interpretation of the math.

WannabeNewton said:
If the diffeomorphism happens to be an isometry, THEN we can say the two manifolds have the same riemannian structure.

Yes, but if both manifolds have a riemannian structure, you can still talk about it even if the diffeomorphism between them is not an isometry.
 
PeterDonis said:
This isn't my understanding of the standard usage of this terminology. (Although, to be fair, the standard usage of this terminology seems to be somewhat muddled; different references seem to use the terms in different ways.) My understanding is: an active diffeomorphism changes the actual geometry; a passive diffeomorphism only changes the coordinates used to describe the geometry.

For example: if I take a flat Euclidean plane and deform it into a bumpy, wiggly surface, that's an active diffeomorphism; I'm changing the actual geometry of the surface. But if I take a flat Euclidean plane and switch from Cartesian to polar coordinates, that's a passive diffeomorphism: the coordinate transformation from Cartesian to polar has to satisfy all the requirements of a diffeomorphism, but it doesn't change the underlying geometry of the plane.

You're basically saying the same thing I was. If you describe a physical theory very heuristically as a set of 1) a manifold, 2) some structure on that manifold, e.g. the metric, and 3) the stuff that lives on your manifold, i.e.

(M, g, \psi)

then a passive transformation \phi brings you to a theory (\phi M, g, \psi) in your terminology or (M, \phi g, \phi \psi) in mine. An active transformation gets you (\phi M, \phi g, \psi) in your terminology or (M, g, \phi \psi) in mine. I believe they're equivalent. My terminology corresponds to the good old elementary mechanics way of using these terms (see for example the Wiki page on the matter.)


PeterDonis said:
There still is a distinction between passive and active in GR--at least, there is with the usage of the terms I gave above. For example: if I switch from Schwarzschild coordinates to Painleve coordinates in describing Schwarzschild spacetime, keeping the mass of the black hole constant, that's a passive diffeomorphism; I'm only changing the coordinates, not the underlying geometry. But if I take Schwarzschild spacetime and change the mass of the black hole, that's an active diffeomorphism, because I'm changing the actual geometry.

It's worth noting here as well that one of the requirements for a diffeomorphism is that the topology of the manifold does not change. That actually rules out a lot of conceivable active diffeomorphisms in GR. For example, you can't do an active diffeomorphism from Schwarzschild spacetime to Kerr spacetime, because the topologies of the underlying manifolds are different.

I don't see how you can change the mass of the black hole through a diffeomorphism on your manifold. It involves changing the matter content of your spacetime.
 
There's no standard definition of the terms. NanakiXIII is using eg. Giulini's convetion, while WannabeNewton is using something like Wald's (not exactly, since Wald says they are different, but effectively the same).
 
NanakiXIII said:
If you describe a physical theory very heuristically as a set of 1) a manifold, 2) some structure on that manifold, e.g. the metric, and 3) the stuff that lives on your manifold, i.e.

(M, g, \psi)

then a passive transformation \phi brings you to a theory (\phi M, g, \psi) in your terminology or (M, \phi g, \phi \psi) in mine.

I understand the way you've stated my terminology, and I agree it captures what I was getting at. But I don't understand why your version transforms \psi as well as g. A passive diffeomorphism doesn't change any of the stuff that lives on the manifold; it just changes the coordinates.

NanakiXIII said:
An active transformation gets you (\phi M, \phi g, \psi) in your terminology or (M, g, \phi \psi) in mine.

Here I don't understand either version. An active diffeomorphism changes the geometry; but physically, that means you have to change the "stuff living on the manifold" as well, at least in the context of GR, because the Einstein Field Equation links the two. So my version would have the transformation giving (\phi M, \phi g, \phi \psi).

(However, even here there is another potential issue: the "manifold" should include the topology, and as I said before, I don't think an active diffeomorphism can change the topology. So there is still something--a portion of M--that doesn't get changed, and your notation doesn't really capture that.)

I also don't understand why your version doesn't transform M or g.

NanakiXIII said:
I don't see how you can change the mass of the black hole through a diffeomorphism on your manifold. It involves changing the matter content of your spacetime.

See my comments above on an active diffeomorphism; since the matter content and the geometry are linked by the Einstein Field Equation, you can't change one without changing the other. As far as the geometry is concerned, you're just changing the M parameter in the line element; that's obviously a diffeomorphism. In the case of the maximally extended Schwarzschild spacetime, which is vacuum everywhere, there is no "matter content" to change; but in any real scenario, changing the M parameter in the line element amounts to changing the amount of matter that originally collapsed to form the black hole, so yes, it does mean changing the matter content of the spacetime.
 
  • #10
PeterDonis said:
Yes, but if both manifolds have a riemannian structure, you can still talk about it even if the diffeomorphism between them is not an isometry.
Sure but my point is that passive and active diffeomorphisms don't have anything to do, in general, with the geometry of the underlying manifold assuming by geometry one means the riemannian structure of the manifold. Say \phi :M\rightarrow N is a diffeomorphism and X:M\rightarrow TM is a vector field then the active point of view is to look at the point - wise pushforward \phi _{*}X_{p}, p\in M (this works nicely since \phi is a diffeomorphism). The passive point of view is to take the \phi - induced coordinates \left \{ x^{'\mu} \right \} and look at X^{'\mu} = \frac{\partial x'^{\mu}}{\partial x^{\nu}}X^{\nu}.
 
  • #11
WannabeNewton said:
Sure but my point is that passive and active diffeomorphisms don't have anything to do, in general, with the geometry of the underlying manifold assuming by geometry one means the riemannian structure of the manifold. Say \phi :M\rightarrow N is a diffeomorphism and X:M\rightarrow TM is a vector field then the active point of view is to look at the point - wise pushforward \phi _{*}X_{p}, p\in M (this works nicely since \phi is a diffeomorphism). The passive point of view is to take the \phi - induced coordinates \left \{ x^{'\mu} \right \} and look at X^{'\mu} = \frac{\partial x'^{\mu}}{\partial x^{\nu}}X^{\nu}.

Those are the Wald definitions (which are in the tradition of mathematics), let's abbreviate them WNA and WNP

NanakiXIII said:
This is very handwaving, but here is how I see it: classically the difference between active and passive coordinate transformations is that for the passive ones you transform the background, and therefore the metric, and then just re-express your system in terms of the new coordinates, while for active transformations you only transform whatever it is you have living on that background.

These are Giulini's definitions (which are on the tradition of Anderson and MTW). http://arxiv.org/abs/gr-qc/0603087 (actually he cal's them general covariance and general invariance). Let's call them GA and GP.

I believe FAPP WNA=WNP=GP=MTW's general covariance, but GA = MTW's "no prior geometry" is different.

General covariance is not very meaningful since all theories are generally covariant (you can use any coordinates you want). The difference between SR and GR is that in SR matter does not act on the spacetime metric, but in GR matter tells spacetime how to curve.

There is one more principle in GR that is important, called the Principle of Equivalence. To show how everyone's terminology is different, Weinberg calls the Principle of Equivalence the "Principle of General Covariance", whereas he calls MTW's general covariance "general covariance".

So to summarize there are 3 things:

1) general covariance: ability to change coordinates, true for all theories
2) no prior geometry: true for GR, not true for SR (with a tiny exception)
3) principle of equivalence: also known as minimal coupling, and is sufficient to obtain covariant energy conservation in metric theories (not quite sure about the exact statement, so let me point to http://arxiv.org/abs/gr-qc/0505128, http://arxiv.org/abs/0805.1726)
 
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  • #12
WannabeNewton said:
the active point of view is to look at the point - wise pushforward \phi _{*}X_{p}, p\in M

Just to make sure I understand the notation: this means that \phi _{*}X_{p} is a map \phi _{*}X_{p} : N \rightarrow TN, correct? (That is, it's a vector field on N, just as X is a vector field on M.)

WannabeNewton said:
The passive point of view is to take the \phi - induced coordinates \left \{ x^{'\mu} \right \} and look at X^{'\mu} = \frac{\partial x'^{\mu}}{\partial x^{\nu}}X^{\nu}.

Again, to make sure I understand the notation, here X^{'\mu} is a representation of a vector field on N in the coordinate basis \left \{ x^{'\mu} \right \}, and X^{\nu} is a representation of "the same" vector field on M in the coordinate basis \left \{ x^{\nu} \right \}, correct? ("The same" means that the diffeomorphism transforms one into the other.)
 
  • #13
The first one is \phi _{*p}:T_p(M)\rightarrow T_{\phi(p)}(N) (pushes forward vectors in the tangent space to M at p to the tangent space to N at the image of p under the diffeomorphism) and yeah the second one is exactly what you said.
 
  • #14
WannabeNewton said:
The first one is \phi _{*p}:T_p(M)\rightarrow T_{\phi(p)}(N) (pushes forward vectors in the tangent space to M at p to the tangent space to N at the image of p under the diffeomorphism)

Ah, I see. (The word "pointwise" should have clued me in.) So then the vector field on N that "corresponds" to X on M under the diffeomorphism would be a map N \rightarrow TN that takes the image \phi(p) of p under the diffeomorphism to the vector \phi _{*p} X(p), i.e., it uses the map between tangent spaces to find the "corresponding" vector at each point.
 
  • #15
WannabeNewton said:
Sure but my point is that passive and active diffeomorphisms don't have anything to do, in general, with the geometry of the underlying manifold assuming by geometry one means the riemannian structure of the manifold. Say \phi :M\rightarrow N is a diffeomorphism and X:M\rightarrow TM is a vector field then the active point of view is to look at the point - wise pushforward \phi _{*}X_{p}, p\in M (this works nicely since \phi is a diffeomorphism). The passive point of view is to take the \phi - induced coordinates \left \{ x^{'\mu} \right \} and look at X^{'\mu} = \frac{\partial x'^{\mu}}{\partial x^{\nu}}X^{\nu}.

Thats how I think about them too.

Thanks :)
 
  • #16
PeterDonis said:
i.e., it uses the map between tangent spaces to find the "corresponding" vector at each point.
Yeahp, and it works nicely because the map is a diffeomorphism otherwise there are more technical problems involved than one would like =D. As atyy noted, this is how Wald presents it and as you mentioned MTW has their way of saying it but unfortunately I don't have a copy of MTW (I'm far too weak to actually lift that thing) however I'm curious as to how they talk about the two notions of passive and active?
 
  • #17
atyy said:
So to summarize there are 3 things:

1) general covariance: ability to change coordinates, true for all theories
2) no prior geometry: true for GR, not true for SR (with a tiny exception)
3) principle of equivalence: also known as minimal coupling, and is sufficient to obtain covariant energy conservation in metric theories (not quite sure about the exact statement, so let me point to http://arxiv.org/abs/gr-qc/0505128, http://arxiv.org/abs/0805.1726)

I understand one and three.

Two has more content that one. Agreed.

However, I do not understand what you mean by "with a tiny exception"?
 
  • #18
WannabeNewton said:
Yeahp, and it works nicely because the map is a diffeomorphism otherwise there are more technical problems involved than one would like =D.

Yeah. Transforming mixed tensors would not be possible without diffeos. Right?

I think I have understood what is that was confusing me. I was confusing isometries with diffeomorphisms.

One should actually call Poincare Invariance as Poincare Isometries :p

And I presume the way one would go about finding them would be to find the Killing vector fields.

Hence as mentioned above, it is perfectly reasonable to ask for a theory to be Diffeomorphic Invariant; since that just asks for coordinate invariance. But a theory need not necessarily be poincare invariant.
 
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  • #19
FedEx said:
Yeah. Transforming mixed tensors would not be possible without diffeos. Right?
In general there is no operation of push - forward or pull - back for mixed tensor fields but yes in the case of a diffeomorphism it does work out nicely. If you want to work it out yourself, take a look at problem 11 - 6 in Lee's Smooth Manifolds book if you have access.
 
  • #20
FedEx said:
However, I do not understand what you mean by "with a tiny exception"?

To some extent, the physical content of general relativity can also be obtained by writing the theory in the form of a spin 2 field on flat spacetime. I don't know the exact limitations of this alternative formulation. It is mentioned eg. http://arxiv.org/abs/gr-qc/0411023 and http://arxiv.org/abs/1105.3735 (section 6.1).

Historically, the first relativistic theory of gravity was Nordstrom's, formulated as a field on flat spacetime. Einstein and Fokker then reformulated it as a theory of curved spacetime. It is not phenomenologically viable because it gets the perihelion of mercury wrong, but it was an important precursor to GR. http://arxiv.org/abs/gr-qc/0405030

Similarly, Newtonian gravity has its usual formulation as well as a formulation as curved spacetime. http://arxiv.org/abs/gr-qc/0506065
 
  • #21
I think the OP is already satisfied, but for completeness let me answer this.

PeterDonis said:
I understand the way you've stated my terminology, and I agree it captures what I was getting at. But I don't understand why your version transforms \psi as well as g. A passive diffeomorphism doesn't change any of the stuff that lives on the manifold; it just changes the coordinates.

You can either transform the manifold and not touch the geometry or the content, or you can transform the content and take along the metric in such a way that your physical system doesn't change. For example, if my system lives on \mathbb{R} with standard geometry (i.e. \forall x, y \in \mathbb{R}: d(x,y) = |x-y|), I can stretch my manifold by a factor 4

\phi(x) = 4x

or I can move around my matter content and take along the metric. Suppose my content consists of two point charges at points x and y and a potential V(|x-y|). I can affect the same change of coordinates by taking my charges and putting them closer together:

\phi(x,y) = (\frac{x}{4},\frac{y}{4})

However, this is not an isometry of the metric, so this is an active transformation that actually changes the physics (in this case the potential energy between the particles), unless we also take along the metric:

\phi(d(x,y)) = 4 d(x,y)

Again, I'm aware this is highly heuristic and it is meant to be.

PeterDonis said:
Here I don't understand either version. An active diffeomorphism changes the geometry; but physically, that means you have to change the "stuff living on the manifold" as well, at least in the context of GR, because the Einstein Field Equation links the two. So my version would have the transformation giving (\phi M, \phi g, \phi \psi).

That's because, in GR, you have no prior geometry, so stated in the notation I'm using it's more accurate to say GR looks like

(M, \emptyset, \Psi); \Psi = (g, \psi)

This reflects your statement that the EFE links geometry and content. I just think of it as the geometry becoming part of the content.

Doubling back, my idea of an active transformation is taking the stuff in your universe and moving it around, just the content. Like I mentioned in the above example, that's fine if your transformation is an isometry, but if it's not, you're changing your system in a nontrivial way. Using the same example:

\phi(x,y) = (\frac{x}{4},\frac{y}{4})

This is an active transformation, just acting on the content. It changes the physics because scaling is not an isometry. You end up with your particles at position (\frac{x}{4},\frac{y}{4}) and with a distance \frac{|x-y|}{4} between them. I can of course obtain the same thing by doing

\phi(x) = 4x; \phi(d(x,y)) = \frac{1}{4} d(x,y)

The first map stretches the manifold out underneath the particles so that they end up at the desired positions and the second map makes the metric reflect this stretching in the geometry.

In GR it makes less sense to make this distinction, because you cannot transform the content and geometry separately: every active transformation necessarily also takes along the metric, getting rid of the problem altogether. Starting with (M, \emptyset, \Psi), under an active transformation we get (M, \emptyset, \phi\Psi) and under a passive transformation we get (\phi M, \emptyset, \Psi) = (M, \emptyset, \phi\Psi).

I still think we're saying more or less the same thing.


PeterDonis said:
As far as the geometry is concerned, you're just changing the M parameter in the line element; that's obviously a diffeomorphism.

It seems to me this is a mapping in the parameter space you additionally put into your system. What kind of diffeomorphism \phi : M \to M (M the manifold now, not the mass, which I'll call m) changes this parameter m? Suppose in my simple example above, I choose my metric to be d(x,y) = |mx-y|. There is no change of coordinates I can perform that will change m into anything else. I can perform an arbitrary active transformation, changing x into a and y into b, and I will get |mx-y| \to |ma-b| \neq |m'x-y|, in general.

Of course I can change m using some mapping I define, but I don't see how you can generally write that as a diffeomorphism on your space-time manifold.
 
  • #22
WannabeNewton said:
I don't have a copy of MTW (I'm far too weak to actually lift that thing) however I'm curious as to how they talk about the two notions of passive and active?

I don't remember them talking about it much, if at all. They have a chapter on differential topology but IIRC it's mostly about affine parametrization and the meaning of vectors and tensors (mainly to show how those concepts can *have* a meaning in the absence of a metric) and differential forms. When I have a chance I'll hook a crane to my copy and haul it out to check. :wink:
 
  • #23
NanakiXIII said:
I still think we're saying more or less the same thing.

Your post clarifies what you meant by your notation, but I probably need to read some more references to get a better feel for it. I think some of the links that atyy posted will help.

NanakiXIII said:
What kind of diffeomorphism \phi : M \to M (M the manifold now, not the mass, which I'll call m) changes this parameter m?

It's the same sort of thing as the scaling transformation you gave above; m basically sets the scale of the r coordinate: more precisely, it tells you how the metric scales with the r coordinate. So a transformation that gives \phi (m) = 2m and \phi (r) = 2r will make the metric at \phi (p), the image of the point p under the diffeomorphism, be the same as the metric at p. I think you would call this an active transformation, not a passive one.
 
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  • #24
atyy said:
To some extent, the physical content of general relativity can also be obtained by writing the theory in the form of a spin 2 field on flat spacetime. I don't know the exact limitations of this alternative formulation.

I believe the key limitation is topology: since the field is on flat spacetime, this version of the theory can only describe spacetimes with the same global topology as flat spacetime. If you don't need to deal with the global topology, though, AFAIK the spin-2 field version is exactly equivalent to the curved spacetime version of GR; they have the same Lagrangian and the same field equation.
 
  • #25
PeterDonis said:
It's the same sort of thing as the scaling transformation you gave above; m basically sets the scale of the r coordinate: more precisely, it tells you how the metric scales with the r coordinate. So a transformation that gives \phi (m) = 2m and \phi (r) = 2r will make the metric at \phi (p), the image of the point p under the diffeomorphism, be the same as the metric at p. I think you would call this an active transformation, not a passive one.

I'm not sure this is true, since the angular part of the Schwarzschild metric just depends on r, not on M/r.
 
  • #26
NanakiXIII said:
I'm not sure this is true, since the angular part of the Schwarzschild metric just depends on r, not on M/r.

Hm, good point, I need to think about this some more.
 
  • #27
PeterDonis said:
I believe the key limitation is topology: since the field is on flat spacetime, this version of the theory can only describe spacetimes with the same global topology as flat spacetime. If you don't need to deal with the global topology, though, AFAIK the spin-2 field version is exactly equivalent to the curved spacetime version of GR; they have the same Lagrangian and the same field equation.

I've seen this widely said, but can't find a good reference - do you know one?

I would especially like to know if the flat spacetime picture still holds with a cosmological constant.

http://arxiv.org/abs/hep-th/0007220 (Eq 1.8) seem to get a cosmological constant starting from flat spacetime, but I don't understand this work well.
 
  • #28
atyy said:
I've seen this widely said, but can't find a good reference - do you know one?

I originally learned about this from the Feynman Lectures on Gravitation; I don't know if that's still in print. MTW also discusses it, and I believe Wald does too. And IIRC Weinberg's 1972 GR text basically uses this approach as its primary approach (not surprising since Weinberg is a field theorist).

atyy said:
I would especially like to know if the flat spacetime picture still holds with a cosmological constant.

I'll take a look at the paper, but one general note is that the "spin-2 field on a background metric" trick can be done with background metrics other than the flat Minkowski metric. For example, you can model gravitational waves as a spin-2 field perturbation on a curved background metric.

In the case of trying to recover GR with a cosmological constant in this way, you would use de Sitter spacetime instead of Minkowski spacetime as the background. More precisely, you would use de Sitter for a positive cosmological constant and Anti-de Sitter for a negative cosmological constant.

In a way this is "cheating", since you're putting in at least some curvature "by hand" instead of having it "pop out" from the spin-2 field theory. But you can still recover *any* spacetime with a cosmological constant by this method, not just de Sitter (or Anti-de Sitter), at least locally (globally there is still the topological limitation, but now the restriction is to spacetimes with the same topology as dS or AdS).

It's been a while, but I'm pretty sure I've seen review papers on arxiv that go into this. I'll see if I can find some.
 
  • #29
PeterDonis said:
I'll take a look at the paper, but one general note is that the "spin-2 field on a background metric" trick can be done with background metrics other than the flat Minkowski metric. For example, you can model gravitational waves as a spin-2 field perturbation on a curved background metric.
Just for someone who wants to read up on this, section 7.5 in Wald goes into this.
 
  • #30
PeterDonis said:
I originally learned about this from the Feynman Lectures on Gravitation; I don't know if that's still in print. MTW also discusses it, and I believe Wald does too. And IIRC Weinberg's 1972 GR text basically uses this approach as its primary approach (not surprising since Weinberg is a field theorist).

Feynman talking about this says he doesn't know about cosmological solutions. Weinberg says it's good if there's harmonic coordinates, but it's more general than that - unfortunately he doesn't say what the more general conditions are.

PeterDonis said:
I'll take a look at the paper, but one general note is that the "spin-2 field on a background metric" trick can be done with background metrics other than the flat Minkowski metric. For example, you can model gravitational waves as a spin-2 field perturbation on a curved background metric.

In the case of trying to recover GR with a cosmological constant in this way, you would use de Sitter spacetime instead of Minkowski spacetime as the background. More precisely, you would use de Sitter for a positive cosmological constant and Anti-de Sitter for a negative cosmological constant.

In a way this is "cheating", since you're putting in at least some curvature "by hand" instead of having it "pop out" from the spin-2 field theory. But you can still recover *any* spacetime with a cosmological constant by this method, not just de Sitter (or Anti-de Sitter), at least locally (globally there is still the topological limitation, but now the restriction is to spacetimes with the same topology as dS or AdS).

Yes, I'm aware of this. It does indeed feel like "cheating" to me, since I don't see how one could have guessed the vacuum background without the traditional formulation.

PeterDonis said:
It's been a while, but I'm pretty sure I've seen review papers on arxiv that go into this. I'll see if I can find some.

Thanks, I'd like to know whether I'm doomed to be a cheater;)
 
  • #31
NanakiXIII said:
However, this is not an isometry of the metric, so this is an active transformation that actually changes the physics (in this case the potential energy between the particles), unless we also take along the metric:

NanakiXIII said:
In GR it makes less sense to make this distinction, because you cannot transform the content and geometry separately: every active transformation necessarily also takes along the metric, getting rid of the problem altogether.
I still think we're saying more or less the same thing.

The entire mechanism goes through only cause the metric is taken along as well. Which as someone mentioned is equivalent to saying that there is no background. The metric along with mass etc form the "foreground". Great. :-D

Now in this context, can one say that there might be solutions to EFE, say two black holes having different masses, i.e. two inequivalent physical situations, which are related by a diffeomorphism?

And if this is true, that would mean two different observers would see different curvature. But we know that co ordinate transformations cannot change the curvature. Does this imply that Active != Passive? And how would this affect the statement "Active and Passive are the same mathematically"

Sorry for all the far fetched (confused) implications.
 
  • #32
No, if \left \{ T^{(i)} \right \}_{i} are the collection of tensor fields on the space - time M that represent physically measurable quantities of interest (which we for simplicity take to be a finite collection) and similarly we have \left \{ T'^{(i)} \right \}_{i} for the space - time N and N,M are not related by a diffeomorphism, then (M,T^{(i)}) will be physically distinguishable from (N,T'^{(i)}) when experiments are carried out and measurable quantities are obtained with regards to the tensor fields.
 
  • #33
WannabeNewton said:
No, if \left \{ T^{(i)} \right \}_{i} are the collection of tensor fields on the space - time M that represent physically measurable quantities of interest (which we for simplicity take to be a finite collection) and similarly we have \left \{ T'^{(i)} \right \}_{i} for the space - time N and N,M are not related by a diffeomorphism, then (M,T^{(i)}) will be physically distinguishable from (N,T'^{(i)}) when experiments are carried out and measurable quantities are obtained with regards to the tensor fields.

True. There are ofcourse different physically inequivalent situations which are not related by a diffeo.

But I was looking for : Two descriptions related by a diffeo can they correspond to different physically inequivalent situations? They should, shouldn't they? Since, active diffeos will change the metric(since an arbitrary diffeo need not be an isometry, as you rightly mentioned in the earlier posts) which might correspond to a different mass distribution.
 
  • #34
WannabeNewton said:
No, if \left \{ T^{(i)} \right \}_{i} are the collection of tensor fields on the space - time M that represent physically measurable quantities of interest (which we for simplicity take to be a finite collection) and similarly we have \left \{ T'^{(i)} \right \}_{i} for the space - time N and N,M are not related by a diffeomorphism, then (M,T^{(i)}) will be physically distinguishable from (N,T'^{(i)}) when experiments are carried out and measurable quantities are obtained with regards to the tensor fields.

Just to inject some more handwaving here :wink:, I still think there's a "gap", so to speak, when we're talking about active diffeomorphisms. The above seems to imply that no transformation that changes physically measurable quantities can be a diffeomorphism; but is that true? It seems like an active diffeomorphism that is not an isometry could change physically measurable quantities.

For example, consider a simple case: we have a 2-sphere with a given radius, and we have an atlas of coordinate charts on the sphere--for concreteness, say the atlas consists of the stereographic projections from the North and South poles, along with an appropriate metric.

A passive diffeomorphism would be changing coordinate charts to, say, two "latitude-longitude" type charts that overlap in such a way as to cover the entire sphere (with an appropriate change in the metric).

An active diffeomorphism that was an isometry would be rotating the sphere without changing coordinate charts; the coordinate labels of points in each chart would change, but the metric would not change, and neither would any other tensor field at any point.

An active diffeomorphism that was *not* an isometry would be changing the sphere's radius, again without changing coordinate charts: the metric would have to change because the curvature changes (and probably any other tensor field would have to change too).

So the two spheres with two different radii would seem to be physically distinct manifolds related by a diffeomorphism.
 
  • #35
FedEx said:
True. There are ofcourse different physically inequivalent situations which are not related by a diffeo.

But I was looking for : Two descriptions related by a diffeo can they correspond to different physically inequivalent situations? They should, shouldn't they? Since, active diffeos will change the metric(since an arbitrary diffeo need not be an isometry, as you rightly mentioned in the earlier posts) which might correspond to a different mass distribution.

They change the metric, but they also change anything that lives in your manifold that might care about the metric to compensate.

PeterDonis said:
Just to inject some more handwaving here :wink:, I still think there's a "gap", so to speak, when we're talking about active diffeomorphisms. The above seems to imply that no transformation that changes physically measurable quantities can be a diffeomorphism; but is that true? It seems like an active diffeomorphism that is not an isometry could change physically measurable quantities.

For example, consider a simple case: we have a 2-sphere with a given radius, and we have an atlas of coordinate charts on the sphere--for concreteness, say the atlas consists of the stereographic projections from the North and South poles, along with an appropriate metric.

A passive diffeomorphism would be changing coordinate charts to, say, two "latitude-longitude" type charts that overlap in such a way as to cover the entire sphere (with an appropriate change in the metric).

An active diffeomorphism that was an isometry would be rotating the sphere without changing coordinate charts; the coordinate labels of points in each chart would change, but the metric would not change, and neither would any other tensor field at any point.

An active diffeomorphism that was *not* an isometry would be changing the sphere's radius, again without changing coordinate charts: the metric would have to change because the curvature changes (and probably any other tensor field would have to change too).

So the two spheres with two different radii would seem to be physically distinct manifolds related by a diffeomorphism.

There's no such thing as a diffeomorphism that changes the radius of the sphere, at least not in any meaningful sense. There are two ways to look at the diffeomorphism, either they

1) act on your manifold only, in which case the "radius" of your sphere is not defined at all, because you need to impose a geometry first;

2) act on your geometry and your other content, in which case the fact that you're changing the radius is unimportant because you're also scaling along anything that might care about this radius.
 
  • #36
Take the following example: Consider an observer S and another observer S' who set up respective coordinates x^{\mu },x'^{\mu} on some open subset U\subseteq M where (M,g_{ab}) is a space - time. For each p\in U, the observers use their respective coordinate bases \left \{ \frac{\partial }{\partial x^{i}}|p \right \}_{i},\left \{ \frac{\partial }{\partial x'^{i}}|p \right \}_{i} for T_{p}(M). Say we have a rank 2 tensor T\in T_p(M)\otimes T_p(M) representing a physically measurable quantity. In this passive point of view, what general covariance tells us is that if S makes a measurement, using the measuring apparatus used to define his local coordinates, of the physical quantity represented by T and S' does the same then the values T^{\mu \nu } obtained in measurement by S will relate to the values T'^{\mu \nu } measured by S' by T'^{\mu \nu } = \frac{\partial x'^{\mu }}{\partial x^{\alpha}}\frac{\partial x'^{\nu}}{\partial x^{\beta}}T^{\alpha\beta}. If in particular, we instead talk about the value of a scalar field at a point then such observers, S,S' will measure the same exact value for the value of the scalar field at that point.

The active point of view, instead of looking at coordinates and coordinate transformations on a given space - time, looks at the image of the space - time under the diffeomorphism and the pull back and pushforward, under this diffeomorphism, of tensor fields representing physically measurable quantities and again asserts general covariance.
 
  • #37
NanakiXIII said:
They change the metric, but they also change anything that lives in your manifold that might care about the metric to compensate.

The mass distribution for instance.

Hence the statement " Two descriptions related by a diffeo can correspond to different physically inequivalent situations " should hold good.
 
  • #38
FedEx said:
The mass distribution for instance.

Hence the statement " Two descriptions related by a diffeo can they correspond to different physically inequivalent situations? " should hold good.

They're not physically inequivalent. Suppose my mass distribution is two point particles. If I change my metric so that the particles are now twice as far apart, but I also act on the particles, moving them closer together, I end up with the same situation. See my examples in post #21.

Yes, the metric changes and yes, the mass distribution changes, but they change in such a way that the physics is identical.
 
  • #39
NanakiXIII said:
They're not physically inequivalent. Suppose my mass distribution is two point particles. If I change my metric so that the particles are now twice as far apart, but I also act on the particles, moving them closer together, I end up with the same situation. See my examples in post #21.

Yes, the metric changes and yes, the mass distribution changes, but they change in such a way that the physics is identical.

That helped. :-)

Thanks a lot.
 
  • #40
PeterDonis said:
So the two spheres with two different radii would seem to be physically distinct manifolds related by a diffeomorphism.
I don't disagree with you at all here but I think the issue is in the great ambiguity of the phrase "represents the same physical space - time" and what "physically the same" really means. I was thinking of it in terms of what I stated in my above post but I have not found like a standard definition for this notion of "physically equivalent".
 
  • #41
PeterDonis said:
Just to inject some more handwaving here :wink:, I still think there's a "gap", so to speak, when we're talking about active diffeomorphisms.

Yes, a diffeomorphism that is not an isometry will change physical quantities (Hawking & Ellis). IIRC, Wald's definition of "active diffeomorphism" is the same as Hawking & Ellis's "isometry".
 
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  • #42
Perhaps it would be useful to try and agree about a few concepts in order to avoid the different degrees of confusion that discussions about "diffeomorphism invariance" often stir up.

-First I think it is important to acknowledge that there is differences between what the different terms used mean in differential geometry texts like Lee's to mention a well known here good reference about smooth and Riemannian manifolds, versus what they convey in GR and physics textbooks. This comes from the fact that most differential geometry books center much more in Riemannian rather than pseudoRiemannian manifolds/dynamical spacetimes but not only from this. This can lead to confusion IMO.

-In purely mathematical terms a tensor equation like the EFE is completely indifferent to coordinate transformations/diffeomorphisms so it has diffeomorphism invariance (a.k.a. general covariance) by definition. The problem is that GR as a theory is a bit more than just the EFE, basically it adds that the context in which the EFE must be applied is that of a Lorentzian manifold, a dynamical spacetime, and this basic tenet of GR can make the general covariance ambiguous at the least because it is either trivial as a differential topology statement at the smooth manifold level, or not true when taken to the manifolds with metric differential geometry level of (global)isometry invariance.

-GR is concerned only with the local geometry of the manifold not with the global topology, so I always understood that when GR books talk about both both diffeomorphisms and isometries they refer to local diffeomorphisms and local isometries, but this distinction is never addressed in GR books. It is however perfectly explained in diff. geometry texts like Lee(WN I'm sure you know what I'm referring to since you mentioned Lee's books).
 
  • #43
NanakiXIII said:
There's no such thing as a diffeomorphism that changes the radius of the sphere, at least not in any meaningful sense.

This might be another issue of terminology. Let me give a concrete example. Suppose we start out with a sphere of radius 1, a coordinate chart ( \theta, \varphi ) on it, similar to the standard "latitude, longitude" coordinates, and a curve going from (0, 0) to (1, 0) in coordinate values. (We'll gloss over the fact that we need two such charts to cover the whole sphere; everything I say will apply to both charts and won't affect the transition map between them.) Then, using your ( M, g_{\alpha \beta}, \psi ) terminology, we have M = S^2, g_{\alpha \beta} = diag ( 1, sin^2 \theta ), and \psi = [ (0, 0), (1, 0) ].

Now we transform to (M', g'_{\alpha \beta}, \psi'), where M' = S^2, g'_{\alpha \beta} = diag ( 2, 2 sin^2 \theta ), and \psi = [ (0, 0), (1, 0) ]. We haven't changed M or \psi, but we have changed the metric to that of a sphere with radius 2. (This will also change the arc length of \psi to 2 instead of 1; if we treat the arc length as part of \psi, then \psi does change as a result of the transform. Or we could treat the arc length as a function of the metric and \psi.)

Is this transformation a diffeomorphism? I don't see why not. The transformation of the manifold itself, S^2, is just the identity, so the differentiable structure of the manifold itself is certainly preserved. The metric changes, but I don't see how that would have any impact on differentiability. You could say that the diffeomorphism is trivial because the manifold doesn't change at all, but that just means it's trivial; it doesn't mean it isn't a diffeomorphism. And it does change the metric, without changing anything else, so it does change things that are "physically measurable", like the arc length of the curve \psi.
 
  • #44
TrickyDicky said:
GR is concerned only with the local geometry of the manifold not with the global topology

I'm not sure I agree with this; I wouldn't say that GR is not concerned with global topology, just that solving the EFE, by itself, doesn't tell you the global topology, since you can have solutions that are locally identical but have different global topology. But I would agree that GR texts don't stress the fact that the EFE is local, so all the talk about invariance under coordinate transformations is also local.
 
  • #45
PeterDonis said:
This might be another issue of terminology.
It isn't Peter. You are correct and he/she isn't if what was typed is what was intended. Note first that closed balls in \mathbb{R}^{n} of all radii are diffeomorphic to one another therefore so are their manifold boundaries, which happen to be n-1 - spheres.
 
  • #46
I suppose while we hairsplitting, even two manifolds related by an isometry need not be physically equivalent - as NanakiXIII points out one has to move everything so that nothing moves - an isometry only moves the metric, and not matter, so it should be iso-everything:)
 
  • #47
WannabeNewton said:
I don't disagree with you at all here but I think the issue is in the great ambiguity of the phrase "represents the same physical space - time" and what "physically the same" really means. I was thinking of it in terms of what I stated in my above post but I have not found like a standard definition for this notion of "physically equivalent".

It means all physical observables are the same:)

As for what is physically observable, that's defined by the theory. If your theory doesn't say what is physically observable, it's not a theory of physics:)

Ok, joking aside, within general relativity, my guess is that it's something like all quantities which remain the same after arbitrary changes of coordinates.

But there may be more than that. For example, sticking to SR, if one formulates electromagnetism in terms of the scalar and vector potential, then one would also have to add, and under arbitrary changes of gauge. But one wouldn't have to specify the additional condition if electromegnetism had been formulated in terms of electric and magnetic fields.

I think the main problem in GR is that in pure gravity there are no local observables. So in a vacuum solution, one usually puts distinguishable test particles all over and the intersection of their worldlines are then events. But since in real GR, there are no test particles (ie. particles that don't contribute to spacetime curvature), then one has to add matter, something like the discussion in the introduction of http://arxiv.org/abs/gr-qc/9404053. Another interesting discussion is http://arxiv.org/abs/gr-qc/0110003.

Another question then is why do we ever introduce gauge descriptions - why can't we work with gauge invariant objects only? I think the answer is that in many cases, the gauge invariant objects are nonlocal - things like Wilson loops. So if we want to describe physics with local equations, we use a description with gauge redundancy.
 
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  • #48
WannabeNewton said:
It isn't Peter. You are correct and he/she isn't if what was typed is what was intended. Note first that closed balls in \mathbb{R}^{n} of all radii are diffeomorphic to one another therefore so are their manifold boundaries, which happen to be n-1 - spheres.

PeterDonis said:
This might be another issue of terminology. Let me give a concrete example. Suppose we start out with a sphere of radius 1, a coordinate chart ( \theta, \varphi ) on it, similar to the standard "latitude, longitude" coordinates, and a curve going from (0, 0) to (1, 0) in coordinate values. (We'll gloss over the fact that we need two such charts to cover the whole sphere; everything I say will apply to both charts and won't affect the transition map between them.) Then, using your ( M, g_{\alpha \beta}, \psi ) terminology, we have M = S^2, g_{\alpha \beta} = diag ( 1, sin^2 \theta ), and \psi = [ (0, 0), (1, 0) ].

Now we transform to (M', g'_{\alpha \beta}, \psi'), where M' = S^2, g'_{\alpha \beta} = diag ( 2, 2 sin^2 \theta ), and \psi = [ (0, 0), (1, 0) ]. We haven't changed M or \psi, but we have changed the metric to that of a sphere with radius 2. (This will also change the arc length of \psi to 2 instead of 1; if we treat the arc length as part of \psi, then \psi does change as a result of the transform. Or we could treat the arc length as a function of the metric and \psi.)

Is this transformation a diffeomorphism? I don't see why not. The transformation of the manifold itself, S^2, is just the identity, so the differentiable structure of the manifold itself is certainly preserved. The metric changes, but I don't see how that would have any impact on differentiability. You could say that the diffeomorphism is trivial because the manifold doesn't change at all, but that just means it's trivial; it doesn't mean it isn't a diffeomorphism. And it does change the metric, without changing anything else, so it does change things that are "physically measurable", like the arc length of the curve \psi.

I was under the impression a diffeomorphism is only defined to act on manifolds, not on their geometry. If that is a mistake, then what you're saying is correct.

However, it's not very meaningful to me. Of course you're going to change your system if you suddenly impose a different metric. But what is that kind of transformation supposed to signify? It's certainly not a change of coordinates, so I'm not comfortable calling it either an active or a passive coordinate transformation.

P.S.: I'd also like to point out that under your terminology, GR is not diffeomorphism invariant at all, so your definition does not seem to be the one used in the literature.

P.P.S.: Also, under my definition, spheres of all radii are also diffeomorphic to one another, precisely because in my terminology diffeomorphisms don't care about the geometry and hence they're all trivially identical.

P.P.P.S.: I had a look in Spivak and he clearly defines diffeomorphisms without making any mention of higher structure like geometry. His diffeomorphisms act on a manifold equipped with an atlas. It doesn't touch anything that lives on the manifold. So your transformation, Peter, consists of a (trivial) diffeomorphism on the manifold, but you added something separate to act on the metric.
 
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  • #49
Yes diffeomorphisms make no mention of geometry which is what I was trying to stress in many of my posts on the first page =D
 
  • #50
WannabeNewton said:
Yes diffeomorphisms make no mention of geometry which is what I was trying to stress in many of my posts on the first page =D

Then Peter's concrete example is not a diffeomorphism, but something more, since there is something specifically acting on the geometry.
 

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