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Active/Passive Diffeomorphisms – clarification on Rovelli’s

  1. Feb 13, 2015 #1
    I am new here but tried to go through some of the posts on subject matter: I apologize if I am overlooking your input as I am sure you must have clarified already my naive doubts !

    I just completed a first reading of Carlo Rovelli's Quantum Gravity book (hardcover edition, 2004).

    I find the part starting at p. 63 with “Given a manifold M, an active diffeomorphism…” and ending at p. 64 with “Beware the formal similarity between (2.135) and (2.136)” a bit hard to follow and the formalism not clear. E.g. why the same symbol for the active and passive maps while domains are different? Why P(x) in the last line of p. 63 seems to me not defined? (too obvious? Is it the composition of T with the inverse of the coordinate map x?)

    I found in the attachment an explication (from A. Zee, Einstein Gravity in a Nutshell) which, while very similar to Rovelli’s, looks to me much more clear (beside the statement “Suppose also that the diffeomorphism moves the number T(P) to the point Q” (what does that mean?)).

    I would like your comment if you also find that part of Rovelli’s book confusing (or it is just me!) and if you can help me mapping the symbols between Rovelli's and Zen's texts.

    I am also checking what Thomas Thiemann writes on the subject in his truly magnificent Modern Canonical Quantum General Relativity (can post that later) but I would like first to get Rovelli’s introduction.

    Thank you in advance. Any help is appreciated!
     

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  3. Feb 13, 2015 #2
    Just wish to add that the p. 63 and p. 64 of Rovelli's book are given on-line HERE.
     
  4. Feb 14, 2015 #3
    The same symbol is presumably used because in the overlaps of coordinate patches, there is a duality between active and passive diffeomorphisms. See the discussion on P3 here for example.

    P(x) is the point on the manifold M with coordinates x.

    But yes.....the active/passive thing is a bit confusing!
     
  5. Feb 14, 2015 #4

    haushofer

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  6. Feb 14, 2015 #5

    Fredrik

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    Ouch, if this is more clear than Rovelli, then I think I'd rather not look at Rovelli.

    That statement doesn't mean anything, but the comment about the fluid makes it possible to guess what he was thinking. f(P)=Q is where the piece of the fluid that's currently at P will be a time Δt from now. T(P) is the temperature of the piece of fluid that's located at P right now. T(Q) is the temperature of the piece of fluid that's located at Q right now. ##\tilde T(Q)## is the temperature of the piece of fluid that will be at Q a time Δt from now. That's the piece that's currently at ##P##. So we have ##\tilde T(f(P))=T(P)## for all points ##P##. If ##T## and ##\tilde T## are coordinate systems, then ##\tilde T## is the coordinate system that assigns the old coordinates to the new points.
     
  7. Feb 15, 2015 #6

    Fredrik

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    A diffeomorphism is a smooth bijection ##\phi:M\to N##, where M and N are smooth manifolds. If ##f## is a real-valued function on ##N##, then ##f\circ\phi## is a real-valued function on ##M##, called the pullback of ##f##. Rovelli uses a diffeomorphism ##\phi:M\to M## and a scalar field (=real-valued function) ##T## to define a new scalar field ##\tilde T## by ##\tilde T(P)=T(\phi(P))##. So his ##\tilde T## is just the pullback of ##T##.

    I find his comments about "passive" diffeomorphisms (in the 2003 draft version, page 45 in the pdf) pretty strange. I would start with a coordinate system ##x:U\to\mathbb R^n ## and a smooth injective map ##\phi:x(U)\to\mathbb R^n##. Then we can use that map to define a new coordinate system ##x'=\phi\circ x##. But Rovelli says that the relationship between the new coordinate system ##x'## and the old one ##x## is
    $$x(P)=\phi(x'(P)).$$ So for some reason he's defining the new coordinate system by ##x'=\phi^{-1}\circ x##. I assume that he's also using the definitions ##t=T\circ x^{-1}## and ##t'=T\circ x'^{-1}##. With these definitions, we can write
    $$t'(x'(P))=(T\circ x'^{-1})(x'(P)) =T(P) =(T\circ x^{-1})(x(P)) =t(x(P)) =t(\phi(x'(P))).$$ This is my interpretation of his ##t'(x')=t(\phi(x'))##, which strictly speaking doesn't make any sense.
     
    Last edited: Feb 15, 2015
  8. Feb 16, 2015 #7
    Yes, it must be so. Thank you for the link to Luca Lusanna’s paper and his explication of that duality: “…Let us recall that there is another, non-geometrical - so-called dual – way of looking at the active diffeomorphisms, which, incidentally, is more or less the way in which Einstein himself formulated the original Hole Argument. This duality is based on the circumstance that in each region of M4 covered by two or more charts there is a one-to-one correspondence between an active diffeomorpshism and a specific coordinate transformation (or passive diffeomorphism)…”

    With respect to that duality, it could also be useful to read the text from Thomas Thiemann, here.

    Sorry for having mislead you (as if this matter was not confusing enough!). Of course, I agree. What I meant was that P looked as the inverse coordinate map ##x^{-1}: \mathbb R^d\rightarrow\mathbb M##. Also, that means, in Rovelli’s notation (note there is a typo in the note 20 at p. 64), that ##t(x)=(T \circ x^{-1})(x)##.

    I am happy I am not the only one to be confused on the active/passive diffeomorphism stuff! Thank you for the link which is very helpful! I will keep it as a resource to understand, after I clear up my mind on the basic formalism, how it extends to the metric and other structures, which I understand was your main point.

    Yes, I see what you mean. Thank you for having basically re-worked all Zen’s explication! I also understand that statement as you describe. The fluid example he gives is also similar to the Rovelli’s “wind and temperature” notions when Rovelli introduces the concepts at p. 62 of the book.

    What I meant by “more clear” when comparing to Rovelli's (hoping I am not misleading myself!) is that I found more satisfactory the Zen’s notation when comparing his equations 29 and 30:

    ##\tilde T(Q)=T(P)=T(f^{-1}(Q))##

    ##\hat T’(x’)=\hat T(x)=\hat T(F^{-1}(x’))##

    First, I liked he used different symbols for the different maps ##f: \mathbb M\rightarrow\mathbb M ## and ##F: \mathbb R^d\rightarrow\mathbb R^d## as the domains are different. Second, I like making the intuitive correspondences between points when working in the manifold and points when working in ##R^d## i.e. point P corresponding to x, transformed point Q corresponding to transformed coordinate x’.

    I would have liked Rovelli following a similar notation which is, btw, recovered from Zen's when you put ##t=\hat T##, ##t’=\hat T’##, ##\phi=f^{-1}=F^{-1}##. For some reason, I arrive in this case to follow better the formalism from Zen than the one from Rovelli, go figure out why ... :-)

    Thank you for having clarified this passage. I now also read it that way :-)
     
  9. Feb 16, 2015 #8

    atyy

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    The definition of these things depend on who's saying it. Wald's appendix, for example, defines the terms so that they are not different (except to a mathematician). If one is using the terms to distinguish between GR and other theories like SR, then one of them is simply invariance under coordinate changes which is a property of SR also, since it is just a theory on a manifold. The other idea is "no prior geometry", to use MTW's clear phrase.

    Defined to be the same: Wald https://www.amazon.com/General-Relativity-Robert-M-Wald/dp/0226870332 (Appendix C.1)

    Defined to be the same: Giulini http://arxiv.org/abs/gr-qc/0603087v1 (Eq 5)

    Defined to be different: Giulini http://arxiv.org/abs/gr-qc/0603087v1 (Eq 6)
     
    Last edited: Feb 16, 2015
  10. Feb 16, 2015 #9

    stevendaryl

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    The distinction between active and passive transformations SEEMS clear enough. If [itex]\mathcal{M}[/itex] is an N-dimensional manifold (space, or spacetime), then we can define a coordinate system [itex]\mathcal{C}[/itex] to be a function from [itex]\mathcal{M}[/itex] to [itex]R^N[/itex]. That is, if [itex]\mathcal{P}[/itex] is a point, then [itex]\mathcal{C}(\mathcal{P}) = \langle x^0, x^1 ...\rangle[/itex] are its coordinates in coordinate system [itex]\mathcal{C}[/itex]. Then an active transformation is a function [itex]m : \mathcal{M} \rightarrow \mathcal{M}[/itex], which moves around points on your manifold, and a passive transformation is a function [itex]\mathcal{T} : R^N \rightarrow R^N[/itex], which just changes the labels given to points: If [itex]\vec{x} = \langle x^0, x^1, ...\rangle[/itex] are the original coordinates for a point [itex]\mathcal{P}[/itex]. then [itex]\mathcal{T}(\vec{x})[/itex] are the new coordinates for the same point. Passive and active transformations are very different types of objects.

    However, it becomes murkier when you talk about a law of physics being invariant (or covariant) under passive or active transformations. You can always rewrite a theory so that it has the same form in any coordinate system. Basically, what it takes to rewrite a theory is that you have to figure out what are the fundamental, non-dynamic (meaning that they are fixed and have no associated equations of motion) geometric objects in the theory: Scalars, vectors, tensors. Then you rewrite the theory explicitly in terms of those objects. The descriptions of those objects change from one coordinate system to another, but the equations involving them can be written to make no reference to any specific coordinate system. In Newtonian physics, the fundamental objects are the scalar universal time and the (partial) metric tensor that gives the distance between two events that occur at the same time. Special Relativity has a fundamental, nondynamic object, the metric tensor.

    These theories are not invariant under active transformations, though (or at least, not all of them). If you distort the manifold [itex]\mathcal{M}[/itex] so that the metric tensor for SR is no longer constant everywhere, then SR no longer applies. In contrast, GR applies to an arbitrarily distorted manifold, so it is invariant under active transformations.

    The reason that the distinction between active and passive transformations becomes a little murky in GR is, roughly speaking, this: there is no way to distinguish points in the manifold EXCEPT through the values of the various fields (the metric tensor, the stress-energy tensor, the electromagnetic field, etc.) at that point. The first observation is that there is no meaningful distinction between:
    1. Changing the metric tensor everywhere (say, by doubling it everywhere)
    2. Changing the manifold (by stretching it in all directions by a factor of 2).
    So an active transformation (moving points around in the manifold) is the same as modifying the metric (together with other fields).

    To simplify the discussion, let's only consider the metric tensor. Let [itex]\mathcal{C}[/itex] and [itex]\tilde{\mathcal{C}}[/itex] be two different coordinate systems, and let [itex]g[/itex] and [itex]\tilde{g}[/itex] be two different metric tensors. Let me introduce the notation [itex]\mathcal{C}(g)_{\mu \nu}[/itex] to mean the components of [itex]g[/itex] in coordinate system [itex]\mathcal{C}[/itex] (and similarly for [itex]\tilde{\mathcal{C}}(g)_{\mu \nu}[/itex] etc.)

    Now suppose [itex]g[/itex] and [itex]\tilde{g}[/itex] are chosen so that:

    [itex]\mathcal{C}(\tilde{g})_{\mu \nu} = \tilde{\mathcal{C}}(g)_{\mu \nu}[/itex]

    Then there is no physically meaningful distinction between the following two situations:
    1. We've changed coordinate systems from [itex]\mathcal{C}[/itex] to [itex]\tilde{\mathcal{C}}[/itex], but the metric tensor has not changed.
    2. The metric tensor has changed from [itex]g[/itex] to [itex]\tilde{g}[/itex], but we're still using the same coordinate system, [itex]\mathcal{C}[/itex]
    The first sounds like a passive change, since we've only changed coordinate systems. The second sounds like a physical change, since we've actually changed the metric tensor (which is equivalent to changing the manifold). But as far as any observations and measurements are concerned, there is no difference between these two situations. So there is no difference (for GR) between an active and a passive transformation.
     
  11. Feb 21, 2015 #10
    Thank you attyy and stevendaryl for your clarifications.

    I think stevendaryl has nailed it down well. Adding an extra step to the argumentation, I think it is precisely realizing that you can always find a transformation such that [itex]\mathcal{C}(\tilde{g})_{\mu \nu} = \tilde{\mathcal{C}}(g)_{\mu \nu}[/itex] that brought Einstein to his argumentation (aka “hole argument”) that you end up to his equations being non-deterministic. He had then either to accept they were not generally covariant, in the sense that they did not have the same form when you change, by the diffeomorphism, both the dynamical as well the non dynamical structures (see Giulini eq. 5, thank you attyy), or there was no (physical) meaning associated to the spacetime points. Rovelli makes clear Einstein went for the latter (see p. 68 and 69 of the book or p. 48 and 49 of the pdf draft). So I understand that to recover a physical meaning to the spacetime points you need to associate them to a physical quantity, be it the gravitational field or another field.
     
    Last edited: Feb 21, 2015
  12. Feb 22, 2015 #11

    haushofer

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    I would say that in GR you can only use the metric to interpret points as physical events, not "any other field" :)
     
  13. Feb 22, 2015 #12

    haushofer

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    One thing which is still puzzling for me concerns the Lie derivative. In some texts this is done using the active interpretation only, e.g. Zee's section 9.6 ( or https://www.physicsforums.com/threads/passive-vs-active-transformations-and-variations.648747/ ). In other texts a Lie derivative is interpreted as a combination of an active and a passive transformation, see e.g. page 12 of the notes http://arxiv.org/pdf/gr-qc/0605010.pdf ("Lie dragging").

    How can this be? Is the "passive dual" of the active transformation used or vice versa in these definitions (such that in the combination of active and passive one of them, say the passive, is reinterpreted as an active one)?
     
    Last edited: Feb 22, 2015
  14. Feb 22, 2015 #13

    Fredrik

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    The definition of the Lie derivative of a vector field is
    $$(\mathcal L_XY)_p=\lim_{t\to 0}\frac{((\phi_{-t})_*X)_p- X_p}{t}.$$ The ##i##th coordinate of the first term in the numerator in the coordinate system ##x## is
    $$ \big((\phi_{-t})_* X\big)_p(x^i)= \big( (\phi_{-t})_ * X_{\phi_t(p)}\big)(x^i) =X_{\phi_t(p)}\big((\phi_{-t})^* x^i\big) =X_{\phi_t(p)}\big(x^i\circ\phi_{-t}\big).$$ The right-hand side is the ##i##th coordinate of ##X_{\phi_t(p)}## in the pullback coordinate system ##(\phi_{-t})^*x=x\circ\phi_{-t}##. This is the coordinate system that the authors of the pdf denote by ##\bar x##. So ##\phi_{-t}## induces the coordinate change ##x\mapsto\bar x##.
     
  15. Feb 22, 2015 #14

    Fredrik

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    Some thoughts about the active/passive terminology:

    Let ##R## be an invertible linear operator on ##\mathbb R^3##. ##R## induces the following transformation of the coordinate triple of an element of ##\mathbb R^3##.
    $$x'_i=(Rx)_i=R_{ij}x_j.$$ But it also induces a second transformation, defined by ##e'_i=Re_i## and ##x_ie_i=x'_ie'_i##.
    \begin{align}
    &e'_i=Re_i =(Re_i)_j e_j =R_{ji} e_j\\
    &x_j e_j =x'_i e'_i =x'_i R_{ji} e_j\\
    &x_j =x'_i R_{ji}\\
    &(R^{-1})_{kj} x_j = (R^{-1})_{kj} R_{ji} x'_i =(R^{-1}R)_{ki} x'_i =\delta_{ki} x'_i =x'_k\\
    & x'_i = (R^{-1})_{ij} x_j.
    \end{align} The first transformation is defined in a very straightforward way, and the second transformation is defined in a more convoluted way. Also, it seems to be more intuitive to think of the former as a physical transformation of the specimen (the object on which we do measurements) and the latter as the same physical transformation of the laboratory, than vice versa. (Think of the walls of the laboratory as defining a preferred orthonormal basis). So it seems natural to refer to these two transformations respectively as the active and passive transformations of the coordinate triples that are induced by ##R##.

    Note that the latter is the inverse of the former, as one would expect from the intuitive view: If we rotate the specimen clockwise by 45 degrees, it should affect experiments in the same way as if we rotate the entire laboratory counterclockwise by 45 degrees. Also note that the above is making it clear what's being transformed and by what. It's unambiguous to talk about the passive transformation of the coordinate triple induced by ##R##.

    There is however plenty of ambiguity when one merely speaks of an "active transformation". An active transformation of what? Is the thing that's sloppily called an active transformation really the thing that should be called an active transformation? If yes, then is there something that induces it? Does that thing also induce the inverse transformation in some convoluted way?

    These GR texts seem to think of diffeomorphisms between subsets of manifolds as active transformations and diffeomorphisms between subsets of ##\mathbb R^n## as passive transformations. This would make the active/passive terminology completely pointless. Just say what the domain of the transformation is, and there's no need to label it "active" or "passive".

    At the moment, I don't see a way to really make sense of the terminology in the context of differential geometry.
     
  16. Feb 22, 2015 #15

    martinbn

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    I don't even see why one would want to try.
     
  17. Feb 22, 2015 #16
    Yes, you are right. If that was not the case, then the general covariance (or active diffeomorphism invariance or background independence) would have been broken and I would end up with an a priori structure defined by that particular field.
     
  18. Feb 22, 2015 #17

    stevendaryl

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    I think that the difference is exactly the domain, and yes, it is sort-of trivial and pointless. Except...

    The only way we have to refer to specific points in a manifold is by coordinates. So the only way to specify a function: [itex]m : \mathcal{M} \rightarrow \mathcal{M}[/itex] is by fixing a coordinate system [itex]\mathcal{C} : \mathcal{M} \rightarrow R^N[/itex] and then defining a function [itex]\mathcal{T}: R^N \rightarrow R^N[/itex]. Then the same function [itex]\mathcal{T}[/itex] can denote a coordinate transformation:

    [itex]\tilde{\mathcal{C}}(\mathcal{P}) = \mathcal{T} (\mathcal{C}(\mathcal{P}))[/itex]

    or a diffeomorphism:

    [itex]m(\mathcal{P}) = \mathcal{C}^{-1} (\mathcal{T}(\mathcal{C}(\mathcal{P}))[/itex]
     
  19. Feb 24, 2015 #18

    Fredrik

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    A follow-up to my previous post...

    I've been thinking that if ##\phi:M\to N## is a diffeomorphism, then the pushforward maps ##\phi_*:T_pM\to T_{\phi(p)}N## (one for each ##p\in M##) are linear. So the pushforward map at ##p## induces two different transformations of the n-tuple of components of a tangent vector ##v\in T_pM##, corresponding to the two induced transformations discussed in my previous post. The "active" transformation of the n-tuple of components would be the one given by ##v^i{}'=(\phi_*v)^i##, and the "passive" transformation would be the one implied by ##v^i e_i =v^i{}' (\phi_* e_i)##.

    There is however a complication. What does ##(\phi_*v)^i## even mean? Yes, the ##i##th component of ##\phi_*v##, but in what coordinate system? If x is the coordinate system we use at p, then a natural choice for the coordinate system at ##\phi(p)## is ##x'=x\circ\phi^{-1}##. (This is the pullback of x with respect to ##\phi^{-1}##, i.e. we have ##x'=(\phi^{-1})^*x##). But the ##i##th component of ##\phi_*v## in this coordinate system is
    $$(\phi_*v)^i{}' =(\phi_*v)(x'^i)=v(\phi^* x'^i) =v(x'^i\circ\phi)=v(x^i\circ\phi^{-1}\circ\phi)=v(x^i)=v^i.$$ If we use this coordinate system, the "active" transformation is just the identity map. So this can't be what anyone means by an active transformation.

    I think we will have to restrict the discussion to the special case M=N, and to situations where ##p## and ##\phi(p)## are both in the domain of some coordinate system ##x##. Then there's an even more natural choice for ##x'##. We simply take ##x'=x##. Now the active transformation changes ##v^i## to
    $$(\phi_*v)^i =(\phi_*v)(x^i)=v(\phi^*x^i)= v(x^i\circ\phi),$$ and the passive transformation is going to be the inverse of the active transformation, just like before.

    This shows how a diffeomorphism induces two linear maps on the tangent space at a point, and that it makes sense to call one of them "the active transformation" and the other "the passive transformation. But it doesn't shed any light on the terms "active diffeomorphism" and "passive diffeomorphism". I think that the only interpretation of those terms that makes sense is that an active diffeomorphism is from a manifold to another, while a passive diffeomorphism is from a subset of ##\mathbb R^n## to another. That subset is of course a manifold too, but it's not the one we're interested in.
     
    Last edited: Feb 24, 2015
  20. Feb 24, 2015 #19

    Ben Niehoff

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    I think the main complication here, and a big reason why I prefer to avoid talking about "active" vs "passive" diffeomorphisms, is that while there is certainly a sense in which two manifolds (or regions thereof) can be "the same, but with different coordinate systems", there is NO sense in which two coordinate systems can be "the same, but on different manifolds". What would that even mean?

    As Fredrik points out, the usual distinction between "active" and "passive" is that "active" transformations change the system itself, whereas "passive" transformations change how we describe the system. I think, however, that the distinction is only meaningful when there is something to change. For example, the physics of some experiment might be rotationally-invariant, but some given apparatus might not be (unless your apparatus is a perfect sphere floating in space...not sure what experiments you can do with that). Then an active transformation rotates the apparatus, whereas a passive transformation would be the experimenter choosing to look at the apparatus from a different angle.

    But if one is talking purely about differential manifolds (i.e. with no additional structure such as metric tensors, etc.), then diffeomorphisms do not actually transform anything. In fact, diffeomorphisms define what is "the same" in the category of differential manifolds: two manifolds are isomorphic as differential manifolds precisely when there exists a diffeomorphism between them.

    To get a notion of what a diffeomorphism actually does, you need to put something on your manifold that breaks diffeomorphism invariance. For example, if you have a 2-sphere, you could write the letter F on its surface. Now imagine the sphere is made of rubber sheet. An active diffeomorphism is any smooth deformation you can make to this rubber sheet, provided you do not a) cut it, or b) pinch any part of it into a cusp. You are even free to turn it inside-out, if you want. (One caveat: you ARE actually allowed to cut it, provided that you glue the cut back together exactly as it was before. This operation generates Dehn twists on the torus, for example).

    A passive diffeomorphism is where you leave the rubber sheet as-is, but label the points differently.

    Another object you can put on a manifold to break diffeomorphism invariance is a metric tensor. Gasp! you might say! Einstein gravity uses metric tensors, and it's supposed to be diffeomorphism invariant! Well, yes, the physics is invariant, but the specific apparatuses (i.e. manifolds with metric tensor that solve the EFE) are not.

    Looking back at our rubber sphere, the metric tensor is what allows you to distinguish between, say, a perfectly round sphere vs. an ellipsoid. These two shapes are the same as differential manifolds (they are diffeomorphic), but different as Riemannian manifolds.

    At this point, you may be wondering, "If a diffeomorphism, strictly speaking, is allowed to change my geometry (i.e. metric tensor), then what about all this stuff I've learned about pushforwards and pullbacks?". You would be correct that if I have a metric tensor on some manifold M, and put it through a diffeomorphism to a manifold N, then I can get some tensor out the other end by pushing it forward. But the catch is, there is no rule that says the resulting tensor needs to be the metric tensor on N. If I do impose this rule, then what I have is a kind of map that is more restrictive than a diffeomorphism: I would have an isometry, which is an isomorphism in the category of Riemannian manifolds.
     
  21. Feb 27, 2015 #20

    haushofer

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    Just want to say that right now my energy is devoted to finishing a PhD-thesis, but I'll take a close look as soon as I can to the responses here. Many thanks.
     
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