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Diffeomorphic Invariance implies Poincare Invariance?

  1. Mar 14, 2013 #1
    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!
     
  2. jcsd
  3. Mar 14, 2013 #2
    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.
     
  4. Mar 14, 2013 #3

    PeterDonis

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    Ok so far.

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

    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.
     
  5. Mar 14, 2013 #4

    PeterDonis

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    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.

    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.
     
  6. Mar 14, 2013 #5

    WannabeNewton

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    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 [itex]M[/itex] to another smooth manifold [itex]N[/itex] such that the smooth structure of [itex]M[/itex] 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 [itex]\phi :M\rightarrow N[/itex] then we speak of active transformations of tensorial quantites by simply speaking of the pushforward and pull back [itex]\phi _{*},\phi ^{*}[/itex]. This active point of view makes no reference to a coordinate system. The passive transformation is simply to look at the coordinate transformation [itex]x^{\mu}\rightarrow y^{\mu}[/itex] induced by [itex]\phi[/itex] 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 [itex](\mathbb{R}^{4},\eta _{ab})[/itex] 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.
     
    Last edited: Mar 14, 2013
  7. Mar 14, 2013 #6

    PeterDonis

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

    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.
     
  8. Mar 14, 2013 #7
    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.

    [tex](M, g, \psi)[/tex]

    then a passive transformation [itex]\phi[/itex] brings you to a theory [itex](\phi M, g, \psi)[/itex] in your terminology or [itex](M, \phi g, \phi \psi)[/itex] in mine. An active transformation gets you [itex](\phi M, \phi g, \psi)[/itex] in your terminology or [itex](M, g, \phi \psi)[/itex] 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.)


    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.
     
  9. Mar 14, 2013 #8

    atyy

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    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).
     
  10. Mar 14, 2013 #9

    PeterDonis

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    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 [itex]\psi[/itex] as well as [itex]g[/itex]. A passive diffeomorphism doesn't change any of the stuff that lives on the manifold; it just changes the coordinates.

    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 [itex](\phi M, \phi g, \phi \psi)[/itex].

    (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 [itex]M[/itex]--that doesn't get changed, and your notation doesn't really capture that.)

    I also don't understand why your version doesn't transform [itex]M[/itex] or [itex]g[/itex].

    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.
     
  11. Mar 14, 2013 #10

    WannabeNewton

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    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 [itex]\phi :M\rightarrow N[/itex] is a diffeomorphism and [itex]X:M\rightarrow TM[/itex] is a vector field then the active point of view is to look at the point - wise pushforward [itex]\phi _{*}X_{p}, p\in M[/itex] (this works nicely since [itex]\phi[/itex] is a diffeomorphism). The passive point of view is to take the [itex]\phi[/itex] - induced coordinates [itex]\left \{ x^{'\mu} \right \}[/itex] and look at [itex]X^{'\mu} = \frac{\partial x'^{\mu}}{\partial x^{\nu}}X^{\nu}[/itex].
     
  12. Mar 14, 2013 #11

    atyy

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

    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)
     
    Last edited: Mar 14, 2013
  13. Mar 14, 2013 #12

    PeterDonis

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

    Again, to make sure I understand the notation, here [itex]X^{'\mu}[/itex] is a representation of a vector field on [itex]N[/itex] in the coordinate basis [itex]\left \{ x^{'\mu} \right \}[/itex], and [itex]X^{\nu}[/itex] is a representation of "the same" vector field on [itex]M[/itex] in the coordinate basis [itex]\left \{ x^{\nu} \right \}[/itex], correct? ("The same" means that the diffeomorphism transforms one into the other.)
     
  14. Mar 14, 2013 #13

    WannabeNewton

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    The first one is [itex]\phi _{*p}:T_p(M)\rightarrow T_{\phi(p)}(N)[/itex] (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.
     
  15. Mar 14, 2013 #14

    PeterDonis

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    Ah, I see. (The word "pointwise" should have clued me in.) So then the vector field on [itex]N[/itex] that "corresponds" to [itex]X[/itex] on [itex]M[/itex] under the diffeomorphism would be a map [itex]N \rightarrow TN[/itex] that takes the image [itex]\phi(p)[/itex] of [itex]p[/itex] under the diffeomorphism to the vector [itex]\phi _{*p} X(p)[/itex], i.e., it uses the map between tangent spaces to find the "corresponding" vector at each point.
     
  16. Mar 14, 2013 #15
    Thats how I think about them too.

    Thanks :)
     
  17. Mar 14, 2013 #16

    WannabeNewton

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    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?
     
  18. Mar 14, 2013 #17
    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"?
     
  19. Mar 14, 2013 #18
    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.
     
    Last edited: Mar 14, 2013
  20. Mar 14, 2013 #19

    WannabeNewton

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    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.
     
  21. Mar 15, 2013 #20

    atyy

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    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
     
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