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World sheet EM tensor in complex coordinates

  1. Apr 23, 2007 #1
    In my Book (Becker, Becker, Schwarz) it is stated (eq 3.23) that the holomorphic component of the EM tensor is given by

    [tex]T_X(z)=T_{zz}=-2 : \partial _z X \cdot \partial _z X :[/tex]

    Now why is the expression for the (holonorphic, zz, component of the) energy momentum tensor in complex coordinates?
    Last edited: Apr 23, 2007
  2. jcsd
  3. Apr 24, 2007 #2
    I do not really understand the question, what is wrong with using complex coordinates?

  4. Apr 24, 2007 #3


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    See e.g. E. Kiritsis, hep-th/9709062, page 44.
    Essentially, you introduce new coordinates z=x+iy and \bar{z}=x-iy and then transform the components of the energy-momentum to new coordinates by the usual rules of tensor transformation.
    If your question is why complex coordinates are introduced at the first place, the answer is because the requirement of 2-dimensional conformal invariance is particularly easy to achieve with complex coordinates. See e.g. (6.1.6) and (6.1.7) in the reference above.
    Last edited: Apr 24, 2007
  5. Apr 24, 2007 #4
    Hmm, they were so short about it in the book that I think they presumed the relation should be intuitively clear or common knowledge or something. Well, I tried the transformation to complex coordinates:

    [tex] T(z)= T_{zz} = (\frac{\partial x^0}{\partial z})^2T_{00} + (\frac{\partial x^1}{\partial z})^2T_{11}[/tex]

    As the off-diagonal elements are zero. Now using [itex]x^0=\frac{1}{2}(z+/bar{z})[/tex] and [itex]x^1=\frac{i}{2}(z-/bar{z})[/tex] we get that

    [tex]T(z)=\frac{1}{4}(T_{00} -T_{11}[/tex]

    which is zero, as the terms are equal. This is the correct result, the ws em tensor should vanish, but leaves me blank as to how they got that expression...
  6. Apr 24, 2007 #5


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    The same way you did. :smile:
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