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In the four-gradient notation, if $\mathcal{L(\phi, \partial_\mu\phi)}$ is a Lagrangian density, what is it $\partial_\mu \mathcal{L}$? Is it a vector $(\frac{1}{c}\frac{\partial \mathcal{L}}{\partial t}, \nabla \mathcal{L})$ or is it a total derivative with respect to the arguments of the Lagrangian. I'm having problems with this, can you provide a full expression of $\partial_\mu \mathcal{L}$?

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  • $\begingroup$ By the chain rule$$\partial_\mu\mathcal{L}=\frac{\partial\mathcal{L}}{\partial\phi}\partial_\mu\phi+\frac{\partial\mathcal{L}}{\partial\partial_\nu\phi}\partial_\mu\partial_\nu\phi\sim\partial_\nu\left(\frac{\partial\mathcal{L}}{\partial\partial_\nu\phi}\partial_\mu\phi\right),$$with $\sim$ holding on-shell. $\endgroup$
    – J.G.
    Commented Oct 14, 2023 at 16:25
  • $\begingroup$ I don't understand though, for example on Wikipedia I read that $\partial_{\mu}$ is a $4$ component vector, so from where the total derivative comes from? $\endgroup$
    – Salmon
    Commented Oct 14, 2023 at 17:10
  • $\begingroup$ $\partial_\nu X_\mu{}^\nu$ is a very different beast from $\partial_\nu J^\nu$. $\endgroup$
    – J.G.
    Commented Oct 14, 2023 at 17:28

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The notation $\partial_\mu \mathcal{L}$ means that you are computing the derivative of the lagrangian density $\mathcal{L}$ (which is a scalar) with respect to a certain coordinate $x^{\mu}$, where $\mu$ runs from $0$ to $3$, i.e., $$ \partial_\mu \mathcal{L} \equiv \dfrac{\partial \mathcal{L}}{\partial x^\mu}\ , $$ independently of whether you want to explicitly write it in terms of the fields that compose the lagrangian density by the chain rule or not.

Now, given a 4-vector with components $A^\mu$, it is also common to use this symbol to refer to the whole 4-vector itself, and you may see expressions like $$ A^\mu = (A^0, A^1, A^2, A^3) \quad\text{or}\quad A^\mu = (A^0, \mathbf{A})\ . $$ In this sense, you can make an statement like ''$\partial_\mu \mathcal{L}$ is a 4-vector'', whose components are $$ (\partial_0 \mathcal{L}, \partial_1 \mathcal{L}, \partial_2 \mathcal{L}, \partial_3 \mathcal{L}) = (\partial_t \mathcal{L}, \nabla \mathcal{L})\ . $$

Additionaly, notice that not everything that has an index need to be a vector (or 4-vector). In particular, $\partial_\mu \mathcal{L}$ is a vector (a covariant vector) because it transforms in the right way under a coordinate transformation as $$ \dfrac{\partial \mathcal{L}}{\partial x'^\mu} = \dfrac{\partial x^\nu}{\partial x'^\mu} \dfrac{\partial \mathcal{L}}{\partial x^\nu} $$

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