The Euler-Lagrange equation for particles is given by
$$ \frac{d}{dt}\frac{\partial L}{\partial \dot{q}} = \frac{\partial L}{\partial q},\tag{1}$$
and for fields it is
$$ \partial_{\mu} \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)} = \frac{\partial \mathcal{L}}{\partial \phi}.\tag{2} $$
Comparing the two equations, the first one has a total time derivative $\frac{d}{dt}$ but the other one appears to have partial derivatives $\partial_{\mu}$. These derivatives come from integration by parts in the derivation of the EL equation. I was wondering why the field version has partial derivatives and the particle version has total derivatives?
I have also seen for the specific example (in Quantum Field Theory for the gifted amateur) of 1 dimensional waves on a string, the corresponding Euler-Lagrange equation is
$$ \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \left(\frac{d\phi}{dt}\right)} + \frac{d}{dx} \frac{\partial \mathcal{L}}{\partial \left(\frac{d\phi}{dx}\right)} = \frac{\partial \mathcal{L}}{\partial \phi}, \tag{3}$$
which uses total derivatives, so I am a bit confused.
Best Answer
No, one of the partial derivative symbol $\partial_{\mu}$ in OP's equation (2) is not correct if it is supposed to mean partial derivatives. The correct Euler-Lagrange (EL) equations read $$ \tag{2'} 0~\approx~\frac{\delta S}{\delta \phi^{\alpha}} ~=~\frac{\partial {\cal L}}{\partial \phi^{\alpha}} - \sum_{\mu} \color{Red}{\frac{ d}{dx^{\mu}}} \frac{\partial {\cal L}}{\partial (\partial_{\mu}\phi^{\alpha})} + \ldots,$$ where the $\approx$ symbol means equality modulo eoms, and the ellipsis $\ldots$ denotes possible higher derivative terms. Here $$ \color{Red}{\frac{ d}{dx^{\mu}}}~=~ \frac{\partial }{\partial x^{\mu}} +\sum_{\alpha}(\partial_{\mu}\phi^{\alpha})\frac{\partial }{\partial \phi^{\alpha}} + \sum_{\alpha, \nu} (\partial_{\mu}\partial_{\nu}\phi^{\alpha})\frac{\partial }{\partial (\partial_{\nu}\phi^{\alpha})} + \ldots $$ is the $\color{Red}{\text{total spacetime derivative}}$ rather than a partial spacetime derivative. See also this and this related Phys.SE posts.
Let us mention for completeness that the other appearance of the partial derivative symbol $\partial_{\mu}$ in OP's equation (2) is correct. It may be replaced with a total spacetime derivative $\color{Red}{d_{\mu}}$, since $\partial_{\mu}\phi\equiv\color{Red}{d_{\mu}}\phi$ by definition, cf. OP's eq. (3).