What is the Proper Way to Take Derivatives of Tensors?

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SUMMARY

The proper method for taking derivatives of tensors involves understanding the contraction of indices and the application of the product rule for derivatives. In the expression R^{abc} \nabla_a S_{bcd} T^{d}, one must contract the d index to form a new tensor U_{bc}, followed by contracting with the tensor R. It is essential to recognize that derivatives of scalars do not vanish because they act as maps that take vectors as arguments, thus raising the index of the tensor. The gradient of a scalar is a member of the dual vector space, which is crucial for understanding the behavior of derivatives in tensor calculus.

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  • Understanding of tensor notation and operations
  • Familiarity with differential operators and their properties
  • Knowledge of the product rule in calculus
  • Basic concepts of dual vector spaces
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  • Study the properties of differential operators in tensor calculus
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This discussion is beneficial for mathematicians, physicists, and engineers who work with tensor calculus, particularly those involved in fields such as general relativity, continuum mechanics, or differential geometry.

quasar_4
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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

R^{abc} \nabla_a S_{bcd} T^{d}.

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

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

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

\nabla_a V^{a} 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?).
 
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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 \hat u you act on the vector via

\nabla f \cdot \hat u = df(\hat u) = u^a \partial_a f
 

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