Simple four-vector partial derivatives

Question

I am a beginner in tensor calculus, and am finding it difficult finding the result to what I assume are basic identities. I am trying to compute the following :

$$ \partial_{\mu} x_{\nu} \quad and \quad \dfrac{\partial}{\partial(\partial_{\gamma}A_{\mu})}\partial_{\lambda}A_{\sigma} $$

I am assuming these will be simple results involving the metric tensor and the Kronecker delta, however I am lost as to wether I would end up with upper or lower indices.

-- My attempt so far --

Using Philip's comment : "Using the Einstein Summation Convention, the number of upper and lower indices in the left and right hand sides of the equation should be the same respectively."

$$\partial_{\mu} x_{\nu} = \eta_{\mu \nu}$$

With the identity $$ \eta^{ac} \eta_{cb} = \delta^{a}_{b}$$

We can construct $$ \partial_{\mu}x^{\nu} = \partial_{\mu}x_{\rho} \eta^{\rho \nu} \\ = \eta_{\mu \rho} \eta^{\rho \nu} \\ = \delta ^{\nu}_{\mu}$$

and $$\dfrac{\partial x_{\nu}}{\partial x_{\mu}} = \partial^{\mu}x_{\nu} = \eta^{\mu \rho} \partial_{\rho}x_{\nu} \\ = \eta^{\mu \rho} \eta_{\rho \nu} \\ = \delta_{\nu}^{\mu}$$

Now considering : $$\dfrac{\partial(\partial_{\gamma}A_{\mu})}{\partial(\partial_{\lambda}A_{\sigma})} $$

Keeping in mind a "downstairs" index that is itself "downstairs" is equivalent to an upstairs index as in Philips comment, andetting $A_{\mu} = x_{\mu}$ , we obtain :

$$\dfrac{\partial \eta_{\gamma \mu}}{\partial \eta_{\lambda \sigma}} = \delta^{\lambda}_{\gamma} \delta^{\sigma}_{\mu}$$

In analogy to the result for $\dfrac{\partial x_{\nu}}{\partial x_{\mu}}$

This suggests : $$ \dfrac{\partial(\partial_{\gamma}A_{\mu})}{\partial(\partial_{\lambda}A_{\sigma})} = \delta^{\lambda}_{\gamma} \delta^{\sigma}_{\mu} $$ and $$ \dfrac{\partial(\partial^{\gamma}A^{\mu})}{\partial(\partial_{\lambda}A_{\sigma})} = \eta^{\lambda \gamma} \eta^{\sigma \mu} $$

Answer 1

Possibly helpful comments-

1. $\partial _\mu x^\nu\equiv \frac{\partial x^\nu}{\partial x^\mu}=\delta^\mu_\nu$ should be no surprise-these are independent variables you're varying, and all it says is for example $\partial x/\partial y=0,\partial x/\partial x=1$.

2. $\partial_\mu x^\nu$ and $\partial_\mu x_\nu$ are different objects, because $x^\nu$ and $x_\nu$ are different objects, related by $x_\nu=\eta_{\nu\mu}x^\mu$ where the matrix coeffecients in this case are constant. Note that the 'independent variables' reasoning in 1. above fails here because $x_\nu$ and $x^\nu$ needn't be independent-they are related by linear transformation $\eta_{\mu\nu}$ which mix the $x^\mu$ among themselves.

3. The object is $\partial_\mu A_\nu$ is a rank 2 tensor, you might as well call it $T_{\mu\nu}$. Then, your second equation is of the form $\frac{\partial T^{\mu\nu}}{\partial T^{\alpha\beta}}$, and there must be corresponding kronecker deltas for each index.

4. There are subtleties with the point above, for example it assumes $T_{23}$ and $T_{32}$ must be treated as independent quantities. But this isn't true if $T$ is a symmetric tensor. You'll have additional factors of $1/2$ in that case, but it shouldn't be an important problem to worry about now.