classical mechanics - Why isn't the Euler-Lagrange equation trivial?

The Euler-Lagrange equation gives the equations of motion of a system with Lagrangian $L$. Let $q^\alpha$ represent the generalized coordinates of a configuration manifold, $t$ represent time. The Lagrangian is a function of the state of a particle, i.e. the particle's position $q^\alpha$ and velocity $\dot q^\alpha$. The Euler-Lagrange equation is

$$ \frac{d}{dt} \frac{\partial L}{\partial \dot q^\alpha } = \frac{\partial L}{\partial q^\alpha}$$

Why is this a law of physics and not a simple triviality for any function $L$ on the variables $q^\alpha$ and $\dot q^\alpha$? The following "proof" of the Lagrange Equation uses no physics, and seems to suggest that the Lagrange Equation is simply a mathematical fact that works for every function.

$$\begin{align} \frac{d}{dt} \frac{\partial L}{\partial \dot q^\alpha} & = \frac{\partial}{\partial \dot q^\alpha} \frac{dL}{dt} &\text{commutativity of derivatives} \\ \ \\ &= \frac{\partial \dot L}{\partial \dot q^\alpha} \\ \ \\ &= \frac{\partial L}{\partial q^\alpha} & \text{cancellation of dots} \end{align}$$

This can't be right, or else nobody would give a hoot about this equation and it would be totally useless to solve any problem. What is wrong with the logical reasoning above?

Answer

Ah, what a tricky mistake you've made there. The problem is that you've simply confused some notions in multivariable calculus. Don't feel bad though-- this is generally very poorly explained. Both steps 1 and 3 above are incorrect. Rest assured, the Euler-Lagrange equation is not trivial.

Let's first take a step back. The Lagrangian for a particle moving in one dimension in an external potential energy $V(q)$ is $$ L(q, \dot q) = \frac{1}{2}m \dot q^2 - V(q). $$ This is how most people write it. However, this is very confusing, because clearly $q$ and $\dot q$ are not independent variables. Once $q$ is specified for all times, $\dot q$ is also specified for all times.

A better way to write the above Lagrangian might be $$ L(a, b) = \frac{1}{2}m b^2 - V(a). $$ Here we've exposed the Lagrangian for what it really is: a function that takes in two numbers and outputs a real number. Likewise, we can clearly see that $$ \frac{\partial L}{\partial a} = -V'(a) \hspace{1cm} \frac{\partial L}{\partial b} = m b. $$ Usually, most people write this as $$ \frac{\partial L}{\partial q} = -V'(q) \hspace{1cm} \frac{\partial L}{\partial \dot q} = m \dot q. $$ However, $q$ and $\dot q$ must be understood as independent variables in order to do this correctly. Just as $a$ and $b$ were independent variables, $q$ and $\dot q$ are too when they're being put into the Lagrangian. In other words, we could put any two numbers into $L$; we just decided to put in $q$ and $\dot q$.

Furthermore, let's look at the total time derivative $\frac{d}{dt}$. How should we understand the following expression? $$ \frac{d}{dt} L(q(t), \dot q(t)) $$ Both $q$ and $\dot q$ are functions of time. Therefore, $L(q(t), \dot q(t))$ depends on time simply because $q(t)$ and $\dot q(t)$ do. Therefore, in order to evaluate the above expression, we need to use the chain rule in multivariable calculus. $$ \frac{d}{dt} L(q(t), \dot q(t)) = \frac{dq}{dt} \frac{\partial L}{\partial a}(q(t), \dot q(t)) + \frac{d \dot q}{dt} \frac{\partial L}{\partial b}(q(t), \dot q(t)) = \dot q(t) \frac{\partial L}{\partial a}(q(t), \dot q(t)) + \ddot q(t) \frac{\partial L}{\partial b}(q(t), \dot q(t)) $$

In the above expression, I once again used $a$ and $b$ in order to make my point clearer. We need to take partial derivatives of $L$ assuming $a$ and $b$ are independent variables. AFTER differentiating, we THEN evaluate $\partial L / \partial a$ and $\partial L / \partial b$ by plugging in $(q, \dot q)$ into the $(a,b)$ slots. This is just like how in single variable calculus, if you have $$ f(x) = x^2 $$ and you want to find $f'(3)$, you first differentiate $f(x)$ while keeping $x$ an unspecified variable, and THEN plug in $x = 3$.

In your first step, the derivatives DON'T commute because $t$ and $q$ are not independent. ($q$ depends on $t$.) Yes, partial derivatives commute, but ONLY if the variables are independent. In your third step, you can't "cancel the dots" because $L$ depends on two inputs. If $L$ only depended on $q$, then yes, you could "cancel the dots" (as this is equivalent to the chain rule in single variable calculus), but it doesn't, so you can't.

EDIT: You can see for yourself that the Euler-Lagrange equation is not identically $0$. If you take the Lagrangian $L(q, \dot q)$ I've written above and plug it into the Euler Lagrange equation, you get $$ m \ddot q(t) + V'(q(t)) = 0. $$ This is not the same as $0 = 0$. It is a condition that a path $q(t)$ would have to satisfy in order to extremize the action. If it was $0 = 0$, then all paths would extremize the action.

EDIT: As Arthur points out, this is also a good time to discuss the difference between $dL / dt$ and $\partial L / \partial t$. If we have a time dependent Lagrangian, $$ L(q, \dot q, t) $$ then $L$ can depend on $t$ explicitly, as opposed to just through $q$ and $\dot q$. So, for example, where as we might have the Lagrangian for a particle in a constant gravitational field $g$ is $$ L(a,b) = \frac{1}{2} mb^2 - m g a $$ if we let allow $L$ to depend on $t$ explicitly, we could have the gravitational field get stronger as time goes on: $$ L(a,b,t) = \frac{1}{2} mb^2 - m ( C t )a. $$ ($C$ is a constant such that $Ct$ has the same units as $g$.)

The quantity $$ \frac{\partial}{\partial t} L(a, b, t) $$ should be understood as differentiating the "$t$-slot" of $L$. In the above example, we would have $$ \frac{\partial}{\partial t} L(a,b,t) = - m C a. $$ The quantity $$ \frac{d}{d t} L(q(t), \dot q(t), t) $$ should be understood as the full time derivative of $L$ due to the fact that $q$ and $\dot q$ also depend on $t$. For the above example, \begin{align*} \frac{d}{d t} L(q(t), \dot q(t), t) &= \dot q(t) \frac{\partial L}{\partial a}(q(t), \dot q(t),t) + \ddot q(t) \frac{\partial L}{\partial b}(q(t), \dot q(t),t) + \frac{\partial L}{\partial t} (q(t), \dot q(t), t) \\ &= (\dot q) (-mC t ) + \ddot q(t) (m \dot q(t)) - mC q(t) \end{align*}