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 $$\partial_\mu \mathcal{L}$$?

• 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.
– J.G.
Commented Oct 14, 2023 at 16:25
• 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? Commented Oct 14, 2023 at 17:10
• $\partial_\nu X_\mu{}^\nu$ is a very different beast from $\partial_\nu J^\nu$.
– J.G.
Commented Oct 14, 2023 at 17:28

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 $$\dfrac{\partial \mathcal{L}}{\partial x'^\mu} = \dfrac{\partial x^\nu}{\partial x'^\mu} \dfrac{\partial \mathcal{L}}{\partial x^\nu}$$