0
$\begingroup$

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 $$ \dfrac{\partial(\partial^{\gamma}A^{\mu})}{\partial(\partial_{\lambda}A_{\sigma})} = \eta^{\lambda \gamma} \eta^{\sigma \mu} $$

$\endgroup$
  • 1
    $\begingroup$ 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. Does that help? $\endgroup$ – Philip Jun 22 at 12:50
  • $\begingroup$ Ah thank you very much, that makes it simple to remember. Are the answers then $\eta_{\mu \nu}$ and $\eta_{\gamma \lambda} \eta_{\mu \sigma}$ ? $\endgroup$ – Mr Lolo Jun 22 at 13:07
  • 1
    $\begingroup$ Also remember that a "downstairs" index that is itself "downstairs" is equivalent to an upstairs index. For example: $$\frac{\partial}{\partial x^\mu} = \partial_\mu \quad \text{and} \quad \frac{\partial}{\partial x_\mu} = \partial^\mu.$$ You can check your answer for the second term in your question by setting $A_\mu = x_\mu$ and calculating it explicitly to see if it makes sense :) $\endgroup$ – Philip Jun 22 at 13:11
  • $\begingroup$ I have tried setting $A_{\mu} = x_{\mu}$ to obtain a new result. Any chance you could have a look if I ended up in the right place ? $\endgroup$ – Mr Lolo Jun 22 at 14:03
  • $\begingroup$ Seems largely to be going in the right direction. I was not suggesting that you solve the second equation by analogy though, apologies if there was a misunderstanding. At any rate, your answer for the second equation is nearly correct, but the upstairs and downstairs indices need to be interchanged. $\endgroup$ – Philip Jun 22 at 14:11
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.