Euler-Lagrange equation even if the action isn't stationary?

[Happiness]

It seems like I could get the Euler-Lagrange equation for any function that allows symmetry of second derivatives even when the action is not stationary.

Suppose $L=L(q_1, q_2, ... , q_n, \dot{q_1}, \dot{q_2}, ... , \dot{q_n}, t)$, where all the $q_i$'s and $\dot{q_i}$'s are functions of $t$. $(\dot{x}$ represents $\frac{dx}{dt}$.$)$

$\frac{d}{dt}\frac{\partial L}{\partial\dot{q}}=\Sigma_i\Big(\frac{dq_i}{dt}\frac{\partial}{\partial q_i}\frac{\partial L}{\partial\dot{q}}+\frac{d\dot{q_i}}{dt}\frac{\partial}{\partial\dot{q_i}}\frac{\partial L}{\partial\dot{q}}\Big)+\frac{\partial}{\partial t}\frac{\partial L}{\partial\dot{q}}$

$=\Sigma_i\Big(\frac{dq_i}{dt}\frac{\partial}{\partial\dot{q}}\frac{\partial L}{\partial q_i}+\frac{d\dot{q_i}}{dt}\frac{\partial}{\partial\dot{q}}\frac{\partial L}{\partial\dot{q_i}}\Big)+\frac{\partial}{\partial\dot{q}}\frac{\partial L}{\partial t}$

$=\frac{\partial}{\partial\dot{q}}\frac{dL}{dt}$

$=\frac{\partial\dot{L}}{\partial\dot{q}}$

$=\frac{\partial L}{\partial q}$

Therefore, $\frac{d}{dt}\frac{\partial L}{\partial\dot{q}}-\frac{\partial L}{\partial q}=0$.

What's wrong?

EDIT: I think I found the mistake. $\frac{\partial\dot{L}}{\partial\dot{q}}=\frac{\partial L}{\partial q}$ when $L=L(q, t)$ but not so when $L=L(q, \dot{q}, t)$.

 

[vanhees71]

No, your derivation was correct. The reason is that indeed concerning the partial derivatives used here you consider the $q$, the $\dot{q}$, and $t$ as independent variables, and the Euler-Lagrange equation are valid thus also if $L$ is explicitly time dependent.

 

[Happiness]

No, your derivation was correct. The reason is that indeed concerning the partial derivatives used here you consider the $q$, the $\dot{q}$, and $t$ as independent variables, and the Euler-Lagrange equation are valid thus also if $L$ is explicitly time dependent.

Consider $L=q\dot{q}$.

$\frac{\partial L}{\partial q}=\dot{q}$

whereas $\dot{L}=q\ddot{q}+\dot{q}^2$

$\frac{\partial \dot{L}}{\partial \dot{q}}=2\dot{q}\neq\frac{\partial L}{\partial q}$

 

[vanhees71]

Well, maybe I didn't follow your derivation carfully enough, but nevertheless the Euler-Lagrange equations are also correct for explicitly time-dependent Lagrangians. The reason is simply that in the Hamilton principle $\delta t=0$ by definition, i.e., you define the equations of motion as the stationary points of the action functional
$$A[q]=\int \mathrm{d} t L(q,\dot{q},t) \; \Rightarrow \; \delta A[q]=\int \mathrm{d} t \left (\delta q \frac{\partial L}{\partial q} + \delta \dot{q} \frac{\partial L}{\partial \dot{q}} \right ).$$
Now since $\delta t=0$ by definition you have
$$\delta \dot{q}=\frac{\mathrm{d}}{\mathrm{d} t} \delta q$$
and thus via integration by parts
$$\delta A[q]=\int \mathrm{d} t \delta q \left (\frac{\partial L}{\partial q} - \frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{q}} \right ).$$
This must vanish for all $\delta q$ in order to make $\delta A$ stationary, which leads to the Euler-Lagrange equations
$$\frac{\partial L}{\partial q} - \frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{q}}=0.$$
I didn't need the total time derivative of $L$ at all.

Sorry for the confusion.