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I have trouble figuring out derivatives in tensor notation in SR. I haven't been able to find a simple recipe for writing down a solution. For example what would be the solution to the following derivative?

$$\frac{\partial{A}^{\mu}}{\partial{A}_\nu}$$

Is it 0? Is it $\eta{^\mu_\nu}$? Is it $\delta^\mu_\nu$?

If there are several steps involved I'd appreciate if you could show and explain all of them. Also what if there are several several quantities like in

$$\frac{\partial({\partial^\rho{A^\sigma})}}{\partial({\partial_\mu}{A_\nu)}}$$

Qmechanic
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subgr0ph
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  • Related: http://physics.stackexchange.com/q/3005/2451 and links therein. – Qmechanic Jan 30 '17 at 12:25
  • Ah, I see. And if I have upper indices like in my examples? Would I lower them first with the metric, then take the derivative and then replace the index with the delta? So I'd get $\eta^{\mu\nu}$ as a solution for the first one and $\eta^{\rho\mu}\eta^{\sigma\nu}$ for the second one? – subgr0ph Jan 30 '17 at 12:43

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You can begin by looking at the transformation properties of the Kronecker Delta which may clear things up slightly

$$ \delta'^a_b = \frac{\partial x'^a}{\partial x^i} \frac{\partial x^j}{\partial x'^b} \delta^i_j = 0 \qquad \forall i \neq j $$ but for $i=j$ we have

$$\delta'^a_b = \frac{\partial x'^a}{\partial x^i} \frac{\partial x^i}{\partial x'^b}$$ which is the equivalent to

$$ \delta'^a_b = \frac{\partial x'^a}{\partial x'^b} $$

Which looks very much like the first partial derivative you've presented above.

Are you able to generalise from here?

Rumplestillskin
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  • Yes, thank you. That did indeed help. I'm assuming I'd find more information in Differential Geometry textbooks? – subgr0ph Jan 30 '17 at 13:01
  • @subgr0ph yes for sure! A book that I loved was called tensors, relativity and cosmology by Dilarsson & Dilarsson. Another classic is Introducing Einstein's Relativity by d'Inverno. They have great introductions to tensors and tensor calculus. Good luck!! – Rumplestillskin Jan 30 '17 at 13:03