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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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    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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