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Passive vs active transformations and variations

  1. Nov 1, 2012 #1


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    In a desparate attempt, not for the first time, to understand some issues on coordinate transformations in GR
    from a passive and active point of view, I opened this thread. I already read references of Wald, Rovelli and others,
    (actually, I think I saw almost all articles and notes which try to explain this) and still don't get it, so I try to find my understanding here, hopefully with some help from you ;)

    I will use scalar fields to keep things as simple as possible, and use the following notation:

    \phi : scalar field
    x^{\mu} : coordinates (in the text I simply call them x)
    M : manifold
    p,q : points on the manifold
    O, O': charts (frames of reference)

    A scalar field is characterized by the fact that under a coordinate transformation

    x^{\mu} \rightarrow x^{'\mu} \ \ \ \ (1)

    one has

    \phi '(x') = \phi(x) \ \ \ \ (2)

    With other words: the functional change of the field, [itex]\phi'[/itex], is such that this [itex]\phi'[/itex] with the new
    coordinate x' as argument gives the same numerical value as [itex]\phi[/itex] at the old argument x.

    Now I want to look at the variation

    \delta \phi (x) \equiv \phi '(x) - \phi(x) \ \ \ \ \ (3)

    from both a passive and active interpretation of (1). First the active one, because that one I seem to understand.
    The active interpretation of (1) means that we stay in one chart O, and that x and x'
    refer to the different points p and q on the manifold M respectively. Then (3) has the interpretation of looking at the functional
    change of the field, while staying in one point. Namely, first we perform (2) to go from
    [itex]\phi(x)[/itex] to [itex]\phi'(x')[/itex] (which gives us the transformed field evaluated in the point q!), and then reset the argument x' to the old value x. So (3) is evaluated in the point p. If [itex]\phi[/itex] would be the metric tensor instead, this is the way to find isometries: we look at the functional change of the metric. That active view makes perfect sense to me, and this is the way to derive the form of the Lie derivativeon fields if we write the coordinate transformation infinitesimally as

    x^{\mu} \rightarrow x^{'\mu} = x^{\mu} - \xi^{\mu}(x) \ \ \ \ (4)

    giving that via a Taylor expansion (3) becomes

    \delta \phi (x) = - \xi^{\mu}\partial_{\mu} \phi(x)

    But now the passive interpretation of the coordinate transformation, which should be easy (dark music is coming up now). Then (1) means we go from chart O to O'. Both x and x' label the same point p on the manifold M. But now we do the same thing as above. We perform the coordinate transformation (1) to go from [itex]\phi(x)[/itex] to [itex]\phi'(x')[/itex], and then reset the argument x' in [itex]\phi'(x')[/itex] back to x. This gives us [itex]\phi'(x)[/itex]. But now in the chart O' the coordinate x indicates a different point q (namely, the point p was labeled by x'!), meaning that now in (3) we compare field values at different points! Or: both the arguments x at the RHS of (3) refer to different points, which of course is quite confusing. How is this to be reconciled with the passive point of view? What is the meaning of (3) from a passive point of view? And to what point does the argument x refer to on the LHS of (3)?

    Can someone make this point (:D) clear? I'm sorry if people are fed up with this discussion, but I'd really like to understand this thoroughly.

    In the end I would like to understand the transformation of the metric

    \delta g_{\mu\nu}(x) \equiv g_{\mu\nu}'(x) - g_{\mu\nu}(x)

    and I can hardly believe that such an expression only makes sense in the active point of view (where it gives the Lie-derivative of the metric if one applies (4)).
    Last edited: Nov 1, 2012
  2. jcsd
  3. Nov 1, 2012 #2
    Got into a little bit of this on the quantum board. Here's the way I made sense of it: an active transformation remaps the point ##x## to ##x'##, having different components but on the same set of basis vectors. A passive transformation remaps the basis vectors, so that ##x## is describes with different components and different basis vectors. If you're clever, you may have realized that these descriptions slightly different--it's not really an apples to apples comparison for one to talk about a new point and another to talk about the same point.

    The way to resolve this problem is to consider a passive transformation to also remap the point ##x##, but to an ##x'## with the same components but new basis vectors. See, both active and passive transformations will agree on the expression for ##x##--either with the old components and old basis vectors, or with new components and new basis vectors.

    Thinking about passive transformations as remapping points like active ones may seem unintuitive, but it makes the ultimate equivalence between active and passive transformations a bit easier to grasp. It should be clear that any ##x'## can be expressed in terms of (new components + old basis vectors) or (old components + new basis vectors) compared to ##x##.

    The reason passive and active transformations are usually presented the way they are is to talk about new coordinates only. ##x'## only has new coordinates in the active transformation, and ##x## only has new coordinates in the passive one. But this is really a misleading way of talking about the transformations. It is, however, important to note that these new coordinates should be the same if the active and passive transformations are equivalent--again, this reinforces that the distinction is conceptual more than practical.
  4. Nov 2, 2012 #3


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    Hi Muphrid,

    this seems a bit strange to me; if I look at the formal definitions of coordinate transformations (in the passive sense), this is a map from R4 to R4, not from the manifold to the manifold. So no points are moved.
  5. Nov 2, 2012 #4


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    Well, let's look at a part of my problem. Is my interpretation of [itex]\phi'(x)[/itex] in the passive view right? So:

    We perform the coordinate transformation to go from [itex]\phi(x)[/itex] to [itex]\phi'(x')[/itex], and then reset the argument x' in [itex]\phi'(x')[/itex] back to x. This gives us [itex]\phi'(x)[/itex]. But now in the chart O' the coordinate x indicates a different point q (namely, the point p was labeled by x'!), meaning that now in (3) we compare field values at different points.

    Is this true?
  6. Nov 2, 2012 #5
    Agreed, usually we just say the point ##x## is expressed in terms of different coordinates and basis vectors. But that doesn't stop that there exists a point with the same coordinates as ##x## originally had, but evaluated on the transformed basis vectors.

    You're probing at the problem, yes. In my opinion, it is confusing to define in the passive transformation that ##x'## and ##x## describe the same point--this is part of the lack of apples-to-apples comparison that I was talking about. You don't look at the same single point in an active transformation. Why would you look at the same single point in the passive one? To do so introduces a fundamental difference in how the transformations are handled.

    If you let ##x'## describe a different point even in the passive transformation, then this confusion goes away. See, if ##x## and ##x'## describe the same point (just in two different coordinate systems) then ##\phi'(x') = \phi(x)## doesn't actually have any meaning. ##x'## is really ##x## and ##\phi'## is still ##\phi##.

    Now, lemme qualify that for a second: it's possible you're still approaching this from a GR standpoint instead of where I'm coming from, which is a QFT standpoint (i.e. with a flat background), so that when I'm talking about basis vectors for positions it makes no sense. If that is the case, though, I must ask the question: what is a passive transformation then? I ask this out of genuine curiosity.
  7. Nov 2, 2012 #6


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    Yes, but in different coordinates. So physically: for different observers. I would say that is very well-defined and meaningful.

    A passive transformation as is understood in any GR book is just a relabeling of the events in spacetime. You don't shift anything on the spacetime manifold. Physically, you don't shift the event, but simply change the observer.
  8. Nov 2, 2012 #7
    I'm not so sure. An observer can use whatever coordinate system they want.

    How is a relabeling of events in spacetime distinguishable from remapping points?

    Let me be concrete with this. Let there be a scalar field ##\phi(x^0, x^1)##. A 1+1 spacetime is sufficient to get the point across. Now, let there be a transformation that maps ##x^0 \mapsto {x'}^0## and similarly for ##{x'}^1##, and we enforce that a new scalar field ##\phi'## is defined such that ##\phi'({x'}^0, {x'}^1) = \phi(x^0, x^1)##. Is this an active transformation or a passive one?

    I mean, I get that the picture or interpretation of the two transformations are very different, but mathematically, this could describe either one, could it not? If all you have are coordinates to work with, I'm not sure you can tell the two transformations apart.
  9. Nov 2, 2012 #8


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    Sure, if you keep the origin of the coordinate system as it is, like e.g. if you go from cartesian to spherical coordinates. But if you e.g. boost, then you go from one observer to another. In the Galilean case this boost reads

    x^i \rightarrow x^{'i} = x^i + v^i t

    With this you go from one observer to another (the origin of your coordinate system changes from x^i = 0 to x^i = - v^i t).

    Well, it's not, and that's why in the end there is a duality between the passive and active point of view. But I would like to understand the variation from both points.

    I would say that depends. If you stay in one chart, then x and x' are different points. If you go from one chart O to another one, O', then it's a passive transformation.

    Coordinates are functions from the spacetime manifold to [itex]R^2[/itex] in your example. A coordinate transformation is then defined as a map

    R^2 \rightarrow R^2 \ \ \ \ \ (5)
    Of course, shifting the point on the manifold can induce such a transformation, but eq.(5) as it stands is really a relabeling of the coordinates, not a function from the manifold to the manifold! An active transformation is often called a diffeomorphism in GR literature, which is a function [itex]M \rightarrow M [/itex].

    I agree, and that's the whole puzzle I have: how can I make sense of the variation in my OP from the passive point of view?
  10. Nov 2, 2012 #9
    Hmm. I guess, in the passive transformation, you can "convert" to an active one by considering the point with coordinates ##{x'}^\mu## but on the original coordinate chart ##O##, and then go about your business as with the active transformation?

    Alternatively, can you use a point with the original ##x^\mu## as coordinates, but on the chart ##O'##, in some way?

    I'm just throwing some things out here. I mean, I know that may feel inelegant, but the underlying math is going to work out. Knowing that, it seems to me best just to find enough of a picture of what's going on to be comfortable with it, even if that picture is a bit strange.
  11. Nov 2, 2012 #10


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    Well, that's my question: how does the underlying math exactly work? :P
  12. Nov 2, 2012 #11
    It works just fine? :P

    In all seriousness, this topic of passive transformations is one I'm still tackling myself. I've usually just gone along with passive-active equivalence and always used only active transformations.
  13. Nov 2, 2012 #12

    D H

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    Here's how I look at the difference: An active transformation pertains to two different events, but only one observer. A passive transformation pertains to two different observers, but only one event.

    Example: Suppose you launch a probe at 0.43c to Alpha Centauri. Ten years later you calculate that the probe should have just arrived at the target star. The transformation that brings the probe from here/now to there/ten years from now is an active transformation.

    Next suppose that upon reaching the target star, the probe sends out a very brief "arrived" signal. The aliens at Alpha Centauri will receive this signal almost immediately; you'll have to wait 4.3 years. Rectifying those two different descriptions of where/when for that same event is a passive transformation.
  14. Nov 2, 2012 #13


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    I completely agree. How do you now apply this to my variation [itex]\delta\phi[/itex]? (or the variation of the metric, whatever you want)
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