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The Euler-Lagrange equation is typically written in this form :

\begin{equation} \frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial \dot{q}}\right)-\frac{\partial}{\partial q}\mathcal{L}=0 \end{equation}

Since the Lagrangian is a functional form of $t$, $q$, $\dot{q}$, why do we use a full derivative for the time, instead of a partial derivative ?

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    $\begingroup$ If you read a proof that derives this equation from a suitable optimisation problem, you can see that the total derivative appears quite naturally. An answer to this question will most likely just copy a proof for you and point out where the total derivative happens. It is better if you read through a proof yourself and ask us to explain the parts you don't get. $\endgroup$
    – Arthur
    Mar 28, 2021 at 17:55
  • $\begingroup$ I tend to believe that the reason for the total derivative is that my question is wrong : the Lagrangian does not depend on time ? $\endgroup$ Mar 28, 2021 at 17:57
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    $\begingroup$ The Lagrangian may depend on time, and for a given solution path, the position $q$ and velocity $\dot q$ definitely depend on time in most cases. So the total derivative makes sense there and is also correct. $\endgroup$
    – Arthur
    Mar 28, 2021 at 18:00
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    $\begingroup$ A partial derivative would only track the EXPLICIT time dependence - often a Lagrangian will only depend on time implicitly, through its dependence on $q$ and $\dot q$ $\endgroup$
    – WW1
    Mar 28, 2021 at 18:03
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    $\begingroup$ Related Phys.SE question: physics.stackexchange.com/q/321668/2451 $\endgroup$
    – Qmechanic
    Mar 29, 2021 at 16:41

1 Answer 1

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@Arthur's advice is to revisit the standard proof of the ELE, in which an integration by parts with vanishing boundary terms gives $\int\frac{\partial L}{\partial\dot{q}}\delta\dot{q}dt=-\int\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}}\right)\delta qdt$. So the real question is why this IBP has the form$$\int_a^bu\frac{dv}{dt}dt=[uv]_a^b-\int_a^b\frac{du}{dt}vdt.$$By the FTA, this reduces to asking why the product rule should be $\frac{d(uv)}{dt}=u\frac{dv}{dt}+\frac{du}{dt}v$, instead of something similar in which a partial time derivative appears at least once.

In the case at hand $q$ is a function of time alone, since the action is a single integral. Then $\frac{\partial L}{\partial\dot{q}}$, being a function of $q,\,\dot{q}$, reduces to a function of time only.

Your question concerns classical mechanics, so we should contrast it with an ELE from classical field theory. The role of time is replaced with multiple variables $x^\mu$; the role of $q$ is replaced with a field, say $\phi$. An action of the form $S=\int\mathcal{L}(\phi,\,\partial\phi/\partial x^\mu)d^nx$ has ELE$$\frac{\partial}{\partial x^\mu}\frac{\partial\mathcal{L}}{\partial(\partial\phi/\partial x^\mu)}-\frac{\partial\mathcal{L}}{\partial\phi}=0$$(with implicit summation over $\mu$). You'll notice the generalization of $\frac{d}{dt}$, namely $\frac{\partial}{\partial x^\mu}$, is a partial derivative.

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