I am taking quantum field theory this semester. There we use the Einstein Informalism. But I have never seen it before. I understood that if there are two letters on one letter, that this is a matrix, and if there is only one letter on it, that this is a vector. I am only confused about the derivations. There is one letter on top and one on the bottom. I suppose that this is as follows:
\begin{align} \partial_\mu L = \frac{\partial L}{\partial x^\mu}, \hspace{30pt} \partial^\mu L = \nabla L + \frac{\partial L}{\partial t} \end{align}
But now I have to derive the following: \begin{align} L = \sqrt{\partial_\mu\phi\partial^\mu \phi}, \hspace{30pt} \frac{\partial L}{\partial \phi} = ?, \hspace{30pt} \partial_\mu \frac{\partial L}{\partial(\partial_\mu \phi)} = ? \end{align} And I am very unsure. Is $L$ still dependent on $\phi$? Because of the derivations this should disappear completely. But if I should derive it to $\partial_\mu \phi$, do I have to consider that $\partial^\mu \phi$ is dependent on $\partial_\mu \phi$? Theoretically $\partial^\mu \phi = (\frac{\partial \phi}{\partial x^0}, \frac{\partial \phi}{\partial x^1}, \frac{\partial \phi}{\partial x^2}, \frac{\partial \phi}{\partial x^3})$. But how do I write this in the Einstein summation convention? I have no idea.
$\partial_\mu$ is just shorthand for the partial derivative with respect to the corresponding coordinate: $$ \partial_\mu:=\frac{\partial}{\partial x^\mu} $$ $\partial^\mu$ is just $\partial_\mu$ with an index raised; that is $$ \partial^\mu:=\eta^{\mu\nu}\partial_\nu $$ Where $\eta^{\mu\nu}$ is the inverse metric. Some authors write expressions exclusively in terms of $\partial_\mu$, writing the $\eta^{\mu\nu}$ factors explecitly. It might be instructive to do this in your situation.