Commutation of first variation and derivative operators in formulation Euler Lagrange equation

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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 cant 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 independant variable (i.e. x,y,z,t or any other coordinate system)

How are these equalities justified?
 

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stevendaryl
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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 cant 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 independant variable (i.e. x,y,z,t or any other coordinate system)

How are these equalities justified?
Well, if you have two functions [itex]q(t)[/itex] and [itex]q'(t)[/itex] that are related by:

[itex]q'(t) = q(t) + \delta q(t)[/itex]

Then clearly:

[itex] \dfrac{d q'}{dt} = \dfrac{dq}{dt} + \dfrac{d \delta q}{dt}[/itex]

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

I suppose there are different ways to think about it, but I don't think of [itex]\delta[/itex] as being an operator. [itex]\delta q[/itex] is just a function of [itex]t[/itex], that's assumed to be small.
 
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vanhees71
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Another important point is that time (and also the endpoints of the action integral) are not varied in Hamilton's principle.
 

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