Least action principle : is $ \delta \frac{dx}{dt} = \frac{d \delta x}{dt} $ always true?

Question

(Just some recalls)

We have an action on which we want to apply Least action principle.

$$ S=\int_{t_i}^{t_f} L(q,\dot{q},t)dt$$

We assume that $t \mapsto q(t)$ is the function that will extremise the action, thus if we replace :

$$q(t) \rightarrow q(t)+\delta q(t)$$

with $ \delta q(t_i)=\delta q(t_f)=0$

We will have at first order in $\delta q$ the variation of the action that will be zero.

$$ \delta S = \int_{t_i}^{t_f} \frac{\partial L}{\partial q} \delta q + \frac{\partial L}{\partial \dot{q}} \delta \dot{q} = 0 $$

And we say that because : $ \delta \frac{dx}{dt} = \frac{d \delta x}{dt} $, we can integer by part (the boundary term vanish because of (*)): $$ \delta S = \int_{t_i}^{t_f} (\frac{\partial L}{\partial q} - \frac{d}{dt} \frac{\partial L}{\partial \dot{q}} )\delta q = 0 $$

---

My question :

It makes sense to me to say that, as :

$$q(t) \rightarrow q(t)+\delta q(t)$$

then by derivation I will have :

$$\dot{q}(t) \rightarrow \dot{q}(t)+\frac{d}{dt} \delta q(t)$$

We immediately find that $ \delta \frac{dx}{dt} = \frac{d \delta x}{dt} $, so it seems to be very general for any transformation.

But my teacher said it could'nt be the case for any transformation. I don't see why ??

What's more, just to understand :

In a Lagrangian, we consider $q(t)$ and $\dot{q}(t)$ as independant variables, because at a given point, if I have a value of a function, its derivative can take any values.

But the functions $q$ and $\dot{q}$ are not independant functions (they are linked by derivative operation).

So, we always have $\delta q(t)$ and $\delta \dot{q}(t)$ that are independant, and to proove that $ \delta \frac{dx}{dt} = \frac{d \delta x}{dt} $, we use the "trick" that, the functions associated are not independant.

In a sense, we "go" in the function space where functions are not independent and then we go back in "variable" spaces where they are independent.

Am I right ? And why did my teacher said that $ \delta \frac{dx}{dt} = \frac{d \delta x}{dt} $ is not always true, was him wrong?

Answer 1

The $\delta$ "symbol" can be viewed as an operator. When you are talking about variations, what you actially have is a family of curves $$ q_\epsilon(t), $$ where $q_{\epsilon=0}(t)$ is the physically realized curve. If $F[q]$ is a function or functional of the curve, then we write $$\delta F[q]=\frac{d}{d\epsilon}(F[q_\epsilon])|_{\epsilon=0},$$ so in particular, $\delta q(t)=\frac{d}{d\epsilon}q_\epsilon(t)$ at $\epsilon=0$.

Therefore, we have $$\delta \dot{q}(t)=\frac{d}{d\epsilon}(\dot{q}(\epsilon,t))=\frac{\partial^2}{\partial\epsilon\partial t}q(\epsilon,t)=\frac{\partial}{\partial t\partial\epsilon}q(\epsilon,t), $$ where the $\epsilon$-derivatives are evaluated at 0. We could switch to partials because $\epsilon$ and $t$ are completely independent variables.

So if the function $q(\epsilon,t)=q_\epsilon(t)$ is well-behaved enough to exchange the partials (we take this to be $C^\infty$ usually, so it is), then this identity stands.

One example I know where this is not true is in general relativity, for field-theoretic use of the variational principle. There we use covariant derivatives, and if the metric is the field to be varied, then the variational $\delta$ derivative is not exchangeable with the covariant derivative $\nabla$, because $\nabla$ itself depends on the metric.