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Tensor contraction question

  1. Mar 1, 2010 #1
    Not sure if this is the right place for this but...

    I am a bit confused about taking derivatives of tensors. Let's say I have some tensors, R, S and T, and an expression like

    [tex] R^{abc} \nabla_a S_{bcd} T^{d}[/tex].

    Do I contract on the d index inside, to get an expression like

    [tex] R^{abc} \nabla_a U_{bc}[/tex] where U is a new tensor,

    then contract with the R tensor on the outside, e.g.

    [tex] \nabla_a V^{a}[/tex] where V is yet another tensor? Or, can I not contract at all with things that are on two different sides of a differential operator?

    I am also confused as to why suddenly the derivatives of scalars don't vanish... I guess the idea is that a derivative should raise the index of a tensor, and since a scalar is a rank 0 tensor, one index should go to 1. But why? I don't intuitively have a good feel for it (i.e., how does the derivative of a scalar act as a map that takes a vector as its argument?).
  2. jcsd
  3. Mar 2, 2010 #2

    Ben Niehoff

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    Science Advisor
    Gold Member

    Contraction follows the same rules with respect to differentiation as multiplication does. That's why it's written to resemble multiplication. So you can't do most of what you're describing; it would violate the product rule for the derivative operator.

    The gradient of a scalar belongs to the dual vector space because to get the rate of change in a direction [itex]\hat u[/itex] you act on the vector via

    [tex]\nabla f \cdot \hat u = df(\hat u) = u^a \partial_a f[/tex]
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