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Simple questions about weyl rescaling/Conformal transfo

  1. Jun 12, 2007 #1

    nrqed

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    This is a very very simple question and I am sure it will look dumb because I won't be using the correct terminology but here I go.


    Consider the points in a manifold. Now we assign coordinates to those points.
    ne thing that I find confusing about any type of transformation is whether
    a) it is the points themselves which are moved about (in which case the manifold is being deformed in some way) with the coordinates "following" the points so that each actual point remains at the same coordinate or
    b) it is only the coordinate chart that is being changed, without the points in the surface being actually moved (so the surface is not deformed in any way)
    c) or if the surface is being deformed without the chart being changed so that the points acquire new coordinates.


    I am not even totally sure if one makes a distinction between cases b) and c) in that in both cases the points change coordinates so maybe it's a "active vs passive" type of thing. It does seem that there is a physical difefrence but maybe mathematically it's really considered to be the same. I don't know.


    To have specific examples in mind, consider a Weyl rescaling. If I understand correctly (I may be wrong on that), this is different from a conformal mapping in that there is no actual change of coordinates [itex] x \rightarrow x'(x) [/itex] involved. Only the metric is rescaled.

    On the other hand, a conformal transformation does involve a tranformation of the coordinates [itex] x \rightarrow x'(x) [/itex] which leads to a rescaling of the metric also.

    So the question is: what is the difference between the two cases? Which of the three cases a, b or c described above does each correspond to?

    Here is what my gut feeling is but it's probably completely incorrect.
    First there is the question of what a mapping [itex] x \rightarrow x'(x) [/itex] actually means. I woudl think that it means that the actual points are assigned new coordinates. So this corresponds to a change of the map which does not involve deforming the manifold itself. Now, the physical "distance" (as given by [itex] g_{\mu \nu} dx^\mu dx^\nu [/itex]) between two actual points has not changed, so the metric has to change in order to give the same distance between two given points (which now have different coordinates). A confomal transformation is simply one such that the metric is rescaled. But the key point is that the manifold itself (if I understand correctly) is not changed. we are talking about a pure change of the coordinate grid.

    On the other hand, what would be a Weyl rescaling? It seems to me that since there is no transformation [itex] x \rightarrow x'(x) [/itex] , any actual point remains at the same coordinate. But the fact that the metric is rescaled implies that the distance between two given coordinates (and therefore two given actual points since they retain the same coordinate) is changed. That would imply that the manifold is actually being deformed here, with the coordinate grid "following" the points.


    Two questions: is that correct? And if so, what would be the "mathspeak" way of conveying what I am talking about (so that I coudl recognize it in a math book)?

    Thanks!
     
  2. jcsd
  3. Jun 13, 2007 #2
    A conformal transformation has nothing to do with a change in coordinates. If you have a metric [itex]g[/itex] and a space [itex]F^+(M)[/itex] of (smooth) positive functions on a manifold [itex]M[/itex] then a conformal transformation is simply a pointwise operation

    [tex]\phi^2\cdot g[/itex]

    where [itex]\phi\in F^+(M)[/itex]. This is obviously coordinate independent.
     
  4. Jun 14, 2007 #3

    nrqed

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    Maybe we are not talking about the same thing. I will quote an article by Belavin,Polyakov and Zamalodchikov:


    Under the condition that the stress-energy tensor is traceless the theory possesses not only scale symmetry but is also invariant with respect to the coordinate transformations

    [itex] \zeta^a \rightarrow \eta^a(\zeta) [/itex]

    having the property that the metric tensor transforms as

    [itex] g_{ab} \rightarrow \frac{\partial \zeta^{a'}}{\partial \eta^a} \frac{\partial \zeta^{b'}}{\partial \eta^b} g_{a' b'} = \rho(\zeta) g_{ab} [/itex]


    where [itex] \rho(\zeta)[/itex] is a certain function. Transformations of this type constitute the conformal group.




    Emphasis theirs!


    Thank you at least for replying!

    Patrick
     
  5. Jun 14, 2007 #4
    Their paper, or at least your interpretation of it, is incorrect. (There may be a subtlety related to their mention of a stress-energy tensor, but what I'm talking about here is a purely differential geometric construction.) To see why, note that because the coordinate transformation

    [tex]\zeta^a \rightarrow \eta^a(\zeta)[/tex]

    is essentially arbitrary, this implies that the sign of the quantity

    [tex]\frac{\partial \zeta^{c'}}{\partial \eta^a} \frac{\partial \zeta^{d'}}{\partial \eta^b}[/tex]

    is indeterminate. In particular, we could have

    [tex]\frac{\partial \zeta^{c'}}{\partial \eta^a} \frac{\partial \zeta^{d'}}{\partial \eta^b} \le 0[/itex]

    However, this contradicts the definition of a conformal transformation of the metric where the conformal factor is required to be non negative.
     
    Last edited: Jun 14, 2007
  6. Jun 14, 2007 #5

    nrqed

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    I did not interpret it, I copied verbatim what they wrote.

    Let me quote the book by Di Francesco, Mathieu and Sénéchal: (section 4.1)

    We denote by [itex] g_{\mu \nu} [/itex] the metric tensor in a space-time of dimension d. By definition, a conformal transformation of teh coordinates is an invertible mapping [itex] x \rightarrow x'[/itex] which leaves the metric invariant up to a scale: [itex] g^{'}_{\mu \nu} = \Lambda(x) f_{\mu \nu} (x) [/itex]

    It seems to *me* that you are describing a Weyl rescaling. How do *you* define a Weyl rescaling??

    I understand your objection about the positivity of the factor multiplying the metric. That's a good point. I assume that it's an extra condition that they assume implicitly although I don't know why it's never mentioned.

    regards

    Patrick
     
  7. Sep 19, 2007 #6

    nrqed

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    I am sorry to bump up this thread but I am still puzzled by this question and I feel that I can`t really understand some basic results of GR or string theory without havig a clear understanding of what weyl and conformal transformations are
    I hoep someone can help me with this.

    Thanks
     
  8. Sep 20, 2007 #7

    Chris Hillman

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    It should be in math books, being a fundamental issue concerning smooth manifolds, but I can't think of any textbook on the theory of manifolds which discusses this issue. But see "hole problem" in D'Inverno, Introducing Einstein's Relativity. Ask again if this doesn't help.

    Right.
     
  9. Sep 20, 2007 #8

    nrqed

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    Ok, Thank you. I will look at D'Inverno.

    My problem is that the basic idea of what is going on is buried under the technical jargon and I am never sure if it's the actual points in the manifold that are moved, if it's the coordinates that are changed without moving the points, etc. There seems to be an aversion in math books to explaining the basic ideas behind the formulae. To me it sounds as if a Weyl rescaling corresponds to keeping the same coordinates assigned to the same points but changing the distance between the coordinates, which implies that the surface has been deformed. A conformal transformation seems to be a change of coordinates without deformation of the actual manifold. I don't know if this is correct but whether it's correct or not, it seems to me that math books could explain the basic meaning of these transformations using simple descriptions like this.
    But cases b and c still seem quite different to me in the following sense: if one calculate the distance between two physical points, the value will be different after the transformation in case c), whereas in case b, the actual distance between two actual points will remain unchanged.

    But again, maybe I am going about thiswith the wrong way of thinking.


    Thanks again

    Patrick
     
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