Deriving Commutation of Variation & Derivative Operators in EL Equation

[hideelo]

I am trying to do go over the derivations for the principle of least action, and there seems to be an implicit assumption that I can't seem to justify. For the simple case of particles it is the following equality

δ(dq/dt) = d(δq)/dt

Where q is some coordinate, and δf is the first variation in f. In general, this can be seen more broadly, given a scalar field ψ

δ(∂ψ/∂x) = ∂(δψ)/∂x

Where x is any independent variable (i.e. x,y,z,t or any other coordinate system)

How are these equalities justified?

 

[stevendaryl]

I am trying to do go over the derivations for the principle of least action, and there seems to be an implicit assumption that I can't seem to justify. For the simple case of particles it is the following equality

δ(dq/dt) = d(δq)/dt

Where q is some coordinate, and δf is the first variation in f. In general, this can be seen more broadly, given a scalar field ψ

δ(∂ψ/∂x) = ∂(δψ)/∂x

Where x is any independent variable (i.e. x,y,z,t or any other coordinate system)

How are these equalities justified?

Well, if you have two functions $q(t)$ and $q'(t)$ that are related by:

$q'(t) = q(t) + \delta q(t)$

Then clearly:

$\dfrac{d q'}{dt} = \dfrac{dq}{dt} + \dfrac{d \delta q}{dt}$

So if $\delta \dfrac{dq}{dt}$ is defined to be $\dfrac{dq'}{dt} - \dfrac{dq}{dt}$, then the conclusion follows.

I suppose there are different ways to think about it, but I don't think of $\delta$ as being an operator. $\delta q$ is just a function of $t$, that's assumed to be small.

 

[vanhees71]

Another important point is that time (and also the endpoints of the action integral) are not varied in Hamilton's principle.