# For virtual displacement in the Lagrangian, why is $\delta \dot{x_i} = \delta \frac{dx_i}{dt} = \frac{d}{dt}\delta x_i \equiv 0$?

I am having trouble understanding why $$\delta \dot{x_i} = \delta \frac{dx_i}{dt} = \frac{d}{dt}\delta x_i \equiv 0.\tag{7.132}$$

you can see my explanation leading up to it below.

Book

The book: Classical Dynamics of Particles and Systems

Author: Stephen T. Thornton

Chapter: 6-7 (little bit of both)

My explanation/question

The brachistochrone problem uses Euler's equation to find the optimal path in which a particle under the influence of gravity, will reach a point $$(x_1,y_1)$$ in the shortest time.

And the way its done is that the $$y(x)$$ is varied until we reach the optimal path.

The way this path is represented as: $$y(x,\alpha) = y(x) + \alpha \eta(x)$$

where $$\alpha \eta(x)$$ is the variation and the $$y(x)$$ is the optimal path.

clearly $$\frac{\partial y}{\partial \alpha} = \eta(x)$$

I am guessing for Eulers equation, we never know what the function $$\eta(x)$$ is but just that its a there "may" be there theoretically.

Now going on to virtual displacement

$$\delta y = \frac{\partial y}{\partial \alpha} d\alpha = \eta(x) d\alpha$$

This represents a change in in $$y(x)$$ by $$\eta(x) d\alpha$$ amount where $$d\alpha \approx 0$$.

I understand that virtual displacement is instantaneous infinitesimal change, said that $$dt = 0$$

But this is where I am not sure.

I believe that $$\delta y$$ should be a function of $$x$$ as its a constant $$\times$$ $$\eta(x)$$ (see picture)

suppose that is true (now going to Lagrangian and I believe that our independent $$x$$ becomes a $$t$$)

then in my book:

"write the Lagrangian in terms of rectangular coordinates $$L = L(x_i,\dot{x_i})$$. The change in $$L$$ caused by the infinitesimal displacement $$\delta \mathbf{r} = \sum_i\delta x_i \mathbf{e}_i$$ is $$\delta L = \sum_{i}\frac{\partial L}{\partial x_i} \delta x_i +\sum_{i}\frac{\partial L}{\partial \dot{x_i}} \delta \dot{x_i} = 0\tag{7.131}$$ We consider only a varied displacement, so that the $$\delta x_i$$ are not explicit or implicit function of the time. Thus $$\delta \dot{x_i} = \delta \frac{dx_i}{dt} = \frac{d}{dt}\delta x_i \equiv 0"\tag{7.132}$$

But isn't $$x_i$$ the particles position in the $$i^{th}$$ coordinate dependent on time? meaning that $$x_i = x_i(t)$$

And thus $$x_i(t,\alpha) = x_i(t) + \alpha \eta(t)$$

and thus $$\delta x_i = \eta(t) d\alpha$$ which is time dependent?(or is it?) making $$\frac{d}{dt}\delta x_i$$ not necessarily equal to $$0$$

• Are you saying that your book puts $\delta\dot{y}_i=0$ for the Brachistochrone problem? i.e. $\eta'(x) = 0$ Commented May 12, 2022 at 11:18
• It would help if you could state the problem you are trying to solve, and what the Lagrangian is, to help see what the issue is. Commented May 12, 2022 at 11:26
• @Robbie the book does not continue with the Brachistochrone problem to talk about $\delta\dot{y_i}$, as the problem is in the beginning of chapter 6 and on the last page he talks about $\delta$ notation. But according to a later on part of the book (chapter 7) with a different problem, we should $\delta\dot{y_i}$ should also $= 0$ then? Commented May 12, 2022 at 11:27
• here is the online version of the book: eacpe.org/app/wp-content/uploads/2016/11/… (chapter 6, page: 225 of pdf) is the Brachistochrone (chapter 6, page: 238 of pdf) is the$\delta$ notation (chapter 7, page: 276 of pdf) is my problem Commented May 12, 2022 at 11:30
• Related post by OP: physics.stackexchange.com/q/708099/2451 Commented May 12, 2022 at 11:31

Generally speaking, you must consider a displacement with $$\delta x_i$$ to be $$t$$-dependent, when applying the principle of least action.
The section where $$\delta\dot{x}_i = 0$$ is about symmetries of the Lagrangian. We are choosing to consider what happens if we make a change of coordinates $$\delta x_i$$ which do not depend on time, and observing that if the Lagrangian is invariant (up to a boundary term) then we can conclude that momentum is conserved.
For example, if $$L =\frac 1 2 m \dot{x}^2 - mgx$$, then making a change of coordinates $$x \mapsto x + a$$, where $$a$$ is constant, we get $$L\mapsto \frac 1 2 m \dot{x}^2 - mgx - mga = L - mga$$, i.e. the Lagrangian changes by a constant. This is not going to affect dynamics, so we can conclude that momentum in the $$x$$-direction is conserved.
To solve the Brachistochrone problem, you must consider general $$\delta x_i$$. You could investigate symmetries by changing coordinates if you like, but this would be a different exercise.
• I understand the explanation of $L -mga$ an it makes sense. Now, for this specific case and setting $\delta x_i$ is a constant Eg: lets say $\delta x_i = a$ then obviously $\frac{d}{dt} a = 0$. How does $\delta x_i$ represent on the graph? This feels like a the graph representing the + $\delta x_i$ to the standard $x_i$ (showing the variation) will be a constant higher. But then at the end points, the variation will not vanish at the boundary conditions or does this not matter? Commented May 12, 2022 at 13:44