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Conformal form of metric

  1. Feb 22, 2013 #1
    If you are given a metric gαβ and you are asked to find if it describes a flat space, is there any way to answer it without calculating the Riemman Tensor Rλμνσ?
    and how can I find for that given metric the coordinate transformation which brings it in conformal form?

    For example I'll give you the way I'm thinking on my problem.
    The metric I'm given is in :
    ds2= exp(ax+by) (-dx2+dy2)

    If I want to show that the space is flat, I would compute the Riemman tensor and so it vanishes... (is there a faster way to do it?).
    Then in order to find the transformation that would bring it in obvious conformal form, I would say that I want to write:
    ds2=exp(w(x',y')) (dx'2+dy'2)

    (translated to gαβ2 nαβ)
    Is that a correct approach?
    In the way that will it give me the x'(x,y) and y'(x,y) form?
     
  2. jcsd
  3. Feb 22, 2013 #2

    Ben Niehoff

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    You can't, in general, bring a metric to conformal form unless it is a 2-dimensional metric.

    Generally speaking, a metric is (locally) conformally flat if and only if the Weyl tensor vanishes vanishes. In 2d and 3d, however, the Weyl tensor vanishes identically, so they are exceptions. In 2d, every metric is (locally) conformally flat. In 3d, a metric is (locally) conformally flat if and only if the Cotton tensor vanishes.

    Also, in general, a metric is flat if and only if the Riemann tensor vanishes. So your options are either A) Get lucky and notice that there is a coordinate transformation that puts a metric in an obviously-flat form, or B) Calculate the Riemann tensor.

    But if you're working in 2 dimensions, it is really easy to calculate the Riemann tensor.
     
  4. Feb 22, 2013 #3

    bcrowell

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    Did you mean to switch the signature from -+ to ++? A change of coordinates can't change the signature.
     
  5. Feb 22, 2013 #4
    oops...yes it can't...that was my mistake...
    still =_= it's obviously in the simplest form
     
  6. Feb 23, 2013 #5
    In addition to what Ben Niehoff wisely said, I have to say that the Riemann tensor does not always tell you if the space is flat: there are coordinate systems in flat space where the Riemann tensor does not vanish. This is because it is not an invariant, as its Lorentz indices tell. On the other hand the curvature scalar (trace of the Ricci) is a scalar, so it really tells you about the curvature.

    There is another theorem that works in any dimension, called Bach's theorem. If the Bach tensor vanishes, then the space is conformaly flat. The Bach tensor is roughly the curl of the Ricci, you should check it.
     
  7. Feb 23, 2013 #6

    Nugatory

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    Could I ask you for an example?
     
  8. Feb 23, 2013 #7
    If you really meant this as stated you have an important confusion here. Curvature is coordinate-independent and the Riemann tensor always tells you if there is Riemann curvature.
     
  9. Feb 23, 2013 #8
    The Riemann tensor is a tensor...Which means that if it vanishes in a Coordinate System, it will vanish in any others after a coord.transformation...

    But still what is the most obvious conformal form for a metric?
    I've seen that it needs to be written in the form:
    ds2= Φ(x,y) {dx2+dy2}
    of course now I cannot change the sign, I had forgotten of that theorem...
    So the needed form should just be:
    ds2= Φ(x,y) {-dx2+dy2}

    -_- however, it's already in that form...with Φ(x,y)=exp(ax+by)
    if I say the transformation:
    u= ax
    v= by

    du2=a2dx2
    dv2=b2dy2

    so
    ds'2= exp{u+v} (- (du/a)2 + (dv/b)2 )

    is that of any better?
     
    Last edited: Feb 23, 2013
  10. Feb 23, 2013 #9

    Dale

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    I also think this is wrong. Please provide an example, or preferably a reference.
     
  11. Feb 23, 2013 #10

    WannabeNewton

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    There is no example unfortunately. If the Riemann tensor vanishes identically in one coordinate system it will vanish in all coordinate systems by virtue of it being a tensor.
     
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