- #1
quasar_4
- 290
- 0
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?).
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?).