# When and why is $\frac{d}{dt}\delta q^{i}=\delta \frac{dq^{i}}{dt}$ true? [duplicate]

Apparently my question is different from Lagrangian Mechanics - Commutativity Rule $$\frac{d}{dt}δq=δ\frac{dq}{dt}$$. I hadn't noticed because the answer given in the comments to this question was satisfactory for my immediate purposes. My question pertains to time-independent variation, whereas the other question pertains to time-dependent variations.

By time-independent variation I mean an active variation of the generalized position coordinates which does not change with time for a given configuration point. That is, one that may be written without an explicit time argument. Position as a function of time $$\left\{q\right\}=\left\{q^{i}\left[t\right]\right\}$$ is assumed to be some sufficiently smooth trajectory in configuration space representing the actual physical system being modeled.

At the point in the discussion where the question occurred to me, the only variations considered are active rigid transformations. These are assumed "infinitesimal" in the case of rotations. No further definitions for $$\delta{q^i}$$ and $$\delta{\dot{q}^i}$$ were given.

See: The Theoretical Minimum: What You Need to Know to Start Doing Physics, by Susskind and Harbovsky.

Apparently, in the case of time-independent variations, $$\delta{\dot{q}^i}$$ has to be given independently of $$\delta{q^i},$$ or be defined using the chain-rule, treating the argument $$\left\{q\right\}$$ in $$\delta q^{i}=f^{i}\left[\left\{q\right\}\right]\varepsilon$$ as a function of time.

In the case where $$\delta{q^i}$$ and $$\dot{\delta{q^i}}$$ are given independently the commutativity has to be part of the definition.

Following Susskind, a variation of the form

$$\delta q^{i}=f^{i}\left[\left\{q\right\}\right]\varepsilon,$$

such that the consequent variation of the Lagrangian is $$\delta{L}=0,$$ is said to be a symmetry. Using Einstein summation, the variation of the Lagrangian may be written

$$\delta{L}=\delta L=\frac{\partial L}{\partial q^{i}}\delta{q^{i}}+\frac{\partial L}{\partial\dot{q}^{i}}\delta\dot{q}^{i}.$$

Assuming the trajectory of the configuration point has a stationary action integral, and therefore satisfies the Euler-Lagrange differential equation, we may write

$$\delta L=\dot{p_{i}}\delta q^{i}+p_{i}\delta\dot{q}^{i}.$$

If we can justify its application in this context, the basic calculus rule for the derivative of the product gives

$$\delta L=\frac{d}{dt}\left[p_{i}\delta q^{i}\right].$$ Apparently, this applicability depends upon the validity of

$$\frac{d}{dt}\delta q^{i}=\delta \frac{dq^{i}}{dt}.$$

Within the context of discussing the rectilinear motion of a single particle and variations which are purely translational or purely rotational, the equation

$$\delta \dot{q}^{i}=f^{i}\left[\left\{\dot{q}\right\}\right]\varepsilon$$

was asserted to follow from the corresponding variation of the position coordinates. The claim is fairly obvious for those circumstances. But it is less obvious for more general variations such as those applied to a path when seeking a stationary action integral. I believe the product rule is applicable to all "well-behaved" variations, but I'm having difficulty formulating a satisfactory argument (informal or formal proof) of that fact.

I am assuming that the operators $$f^i$$ are point functions of the coordinate system and are time-independent. I believe they will always be affine transformations.

So, my question is: in the context of time-independent variations of the configuration of a system, when and why is it true that

$$\frac{d}{dt}\delta q^{i}=\delta \frac{dq^{i}}{dt}?$$

## marked as duplicate by AccidentalFourierTransform, Qmechanic♦ classical-mechanics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 18 '18 at 2:09

• Well, from what I know it is a general fact that $\delta\frac{dq^i}{dt}=\dot{q^i_2}-\dot{q^i_1}=\frac{d}{dt}(q^i_2-q^i_1)=\frac{d}{dq}\delta q^i$. I have never encountered some kind of variation that the above rule doesn't apply. – G K Sep 17 '18 at 22:36
• – AccidentalFourierTransform Sep 17 '18 at 23:06
• What exactly is your definition of time-independent variation? Only implicit time dependence? What variables do $\delta{\dot{q}^i}$ and $\delta q^i$ depend on? Consider to make the question more precise. Which pages in Susskind & Harbovsky? – Qmechanic Oct 20 '18 at 20:41
• The new question (v3) seems still to be a duplicate. – Qmechanic Oct 21 '18 at 3:55
• @Qmechanic They are only the same question if other assumptions are made. I'm not sure what all of those assumptions are, but one of them is that only one trajectory is under consideration. If a congruence of world-lines were considered, then variation as a function of time would be path-dependent. – Steven Thomas Hatton Oct 21 '18 at 19:20