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.

I would greatly appreciate any help you can offer.


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.

enter image description here

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)

enter image description here

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 $0$

  • $\begingroup$ Are you saying that your book puts $\delta\dot{y}_i=0$ for the Brachistochrone problem? i.e. $\eta'(x) = 0$ $\endgroup$
    – Robbie
    Commented May 12, 2022 at 11:18
  • $\begingroup$ 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. $\endgroup$
    – Robbie
    Commented May 12, 2022 at 11:26
  • $\begingroup$ @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? $\endgroup$
    – Reuben
    Commented May 12, 2022 at 11:27
  • $\begingroup$ 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 $\endgroup$
    – Reuben
    Commented May 12, 2022 at 11:30
  • $\begingroup$ Related post by OP: physics.stackexchange.com/q/708099/2451 $\endgroup$
    – Qmechanic
    Commented May 12, 2022 at 11:31

1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$
    – Reuben
    Commented May 12, 2022 at 13:44
  • $\begingroup$ @Reuben that's correct, it will have different end-points. That doesn't matter, because you're not varying the action, you're just changing co-ordinates. $\endgroup$
    – Robbie
    Commented May 12, 2022 at 13:46

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