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Contravariant derivative tensor problem

  1. Aug 9, 2009 #1
    Tensor [tex]\left (
    \begin{array}{cc}
    1\\1
    \end{array}
    \right )[/tex] is [tex]T=T^\alpha_\beta \omega^\beta \otimes \vec e_\alpha[/tex]Why is contravariant derivative tensor [tex]\left (\begin{array}{cc}1\\1 \end{array}
    \right )[/tex]
    [tex]V^\alpha_{;\beta}\vec {e_\alpha}[/tex] - contravariant derivative
     
  2. jcsd
  3. Aug 10, 2009 #2

    Fredrik

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    I don't understand the question. What do you mean by contravariant derivative? Do you mean covariant? What do you mean by those matrices? Are you saying that the components of T in the basis you're considering are (1,1)?
     
  4. Aug 10, 2009 #3
    I made a mistake I meant covariant. T is general tensor. I don't understand why
    [tex]\frac{\partial \vec V}{\partial x^\beta}=V^\alpha_{;\beta} \vec e_{\alpha}[/tex]
    can be associated with a (1,1) tensor
     
  5. Aug 10, 2009 #4

    haushofer

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    A covariant derivative is per definition a derivative which transforms as a tensor. Loosely, the covariant derivative of a tensor T adds another covariant index, so if T is type (k,l) then the covariant derivative of T is type (k+1, l) (I always forget the convention, but here the k is the number of covariant indices).

    Here's a nice exercise: you know how the components of a vector V tranform:

    [tex]
    V^{\mu'}(x') = \frac{\partial x^{\mu'}}{\partial x^{\nu}}V^{\nu}(x)
    [/tex]

    You also know how a partial derivative transforms:

    [tex]
    \frac{\partial}{\partial x^{\lambda'}} = \frac{\partial x^{\alpha}}{\partial x^{\lambda '}}\frac{\partial}{\partial x^{\alpha}}
    [/tex]

    Now calculate the transformation the components of the partial derivative of a vector:

    [tex]
    \frac{\partial}{\partial x^{\lambda'}}V^{\mu'}(x') = \frac{\partial x^{\alpha}}{\partial x^{\lambda '}}\frac{\partial}{\partial x^{\alpha}} [\frac{\partial x^{\mu'}}{\partial x^{\nu}}V^{\nu}(x)]
    [/tex]

    If you write this out, you'll see that this doesn't transform as a (1,1) tensor as you would naively expect on basis of the index structure of the expression. However, in derivatives you compare tensors at different points on the manifold and you need a certain prescription to perform this comparison: the connection [itex] \Gamma [/itex].

    If you want to write things down with a basis, you can define the connection and the covariant derivative as

    [tex]
    \nabla_{\mu}e_{\nu} \equiv \Gamma^{\lambda}_{\mu\nu}e_{\lambda}
    [/tex]

    Here [itex]\nabla_{\mu} [/itex] means covariant derivation with respect to the basis vector

    [tex]e_{\mu} = \partial_{\mu}[/tex]

    This second-last formula states that the covariant derivative of the basis vector should be expressible in terms of your basis vectors. If you then write down a tensor in a certain coordinate basis, you can act on this tensor with the covariant derivative. For instance,

    [tex]
    \nabla_{\mu} (V) = \nabla_{\mu}(V^{\alpha} e_{\alpha}) = [(\nabla_{\mu}V^{\alpha})e_{\alpha} + V^{\alpha}\nabla_{\mu}e_{\alpha}]
    [/tex]

    Demanding that the covariant derivative acting on scalar functions gives the same result as a partial derivative (and some basic rules such that it obeys Leibnitz), you arrive at the same answer as in a component-only treatment.

    Hopefully this makes things a bit clear :)
     
  6. Aug 10, 2009 #5

    Fredrik

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    I like the approach to covariant derivatives that starts with a connection. I'll quote myself (and fix a couple of mistakes at the same time):

    The above only defines the action of [itex]\nabla_X[/itex] on vector fields, but note that condition (iii) above suggests a way to extend the definition to scalar fields. If we define

    [tex]\nabla_Xf=Xf[/tex]

    condition (iii) looks like the Leibnitz rule for derivatives:

    [tex]\nabla_X(fY)=(\nabla_Xf)Y+f\nabla_XY[/tex]

    So we choose to define [itex]\nabla_Xf[/itex] that way. Can we do something similar for covector fields? It turns out we can. Suppose that [itex]\omega[/itex] is a covector field. The closest thing to a Leibnitz rule we can get is this:

    [tex]\nabla_X(\omega(Y))=(\nabla_X\omega)(Y)+\omega(\nabla_XY)[/tex]

    so we choose to define [itex]\nabla_X\omega[/itex] by

    [tex](\nabla_X\omega)(Y)=\nabla_X(\omega(Y))-\omega(\nabla_XY)[/tex]

    for all Y. Note that this means that we define [itex]\nabla_X\omega[/itex] to be a covector field.

    The same idea can be used to find the appropriate definition of [itex]\nabla_X[/itex] acting on an arbitrary tensor field. You can probably figure it out on your own.

    Don't forget that the covariant derivative you're used to is the special case [itex]X=\partial_\mu[/itex].
     
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