# Simple four-vector partial derivatives

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

• 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? – Philip Jun 22 at 12:50
• Ah thank you very much, that makes it simple to remember. Are the answers then $\eta_{\mu \nu}$ and $\eta_{\gamma \lambda} \eta_{\mu \sigma}$ ? – Mr Lolo Jun 22 at 13:07
• 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 :) – Philip Jun 22 at 13:11
• 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 ? – Mr Lolo Jun 22 at 14:03
• 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. – Philip Jun 22 at 14:11

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.