0
$\begingroup$

While we are deriving Lagrange's equation from D'Alembert's principle, when we argues as;

$$\delta \vec r_\alpha = \sum_i \frac{\partial \vec r_\alpha}{\partial q_i }\delta q_i + \frac{ \partial \vec r_\alpha}{\partial t } \delta t ,$$ but since a virtual displacement takes no time, $\delta t =0$

However, $\delta t$ is a virtual time translation, not a real one, similar to $\delta \vec r$ is a virtual displacement, not a real one.

However, in most of the applications, we are free to choose the time $t = t_0$ arbitrarily, but not all systems have such time translation, so in that case, shouldn't we not ignore that factor $\delta t$ also ?

Edit:

$\delta r_\alpha$ stands for the virtual displacement in the position of the particle $\alpha, and in general $\delta$ is used for a virtual change.

$\endgroup$
2
  • $\begingroup$ To avoid confusion, consider to provide references and definitions. $\endgroup$
    – Qmechanic
    Commented Feb 23, 2019 at 10:34
  • $\begingroup$ @Qmechanic see my edit; I hope this is enough. $\endgroup$
    – Our
    Commented Feb 23, 2019 at 10:36

1 Answer 1

1
$\begingroup$

FWIW,

  1. The concept of virtual displacements in classical mechanics is defined with frozen time $\delta t=0$ whether or not the system has explicit time dependence.

  2. We usually assume that D'Alembert's principle holds holds for all times $t$, not just for particular values of time $t$.

$\endgroup$
6
  • $\begingroup$ Correct me if I'm wrong, but we assume that the virtual displacement happens in a frozen real time $d t = 0$; not $\delta t = 0$. $\delta$ stands for a virtual change. $\endgroup$
    – Our
    Commented Feb 23, 2019 at 10:06
  • $\begingroup$ FWIW, $t$ is time, not the parameter for a virtual displacement. $\endgroup$
    – Qmechanic
    Commented Feb 23, 2019 at 10:36
  • $\begingroup$ I'm aware of that. $\endgroup$
    – Our
    Commented Feb 23, 2019 at 10:37
  • $\begingroup$ How do you define $dt$ vs. $\delta t$? $\endgroup$
    – Qmechanic
    Commented Feb 23, 2019 at 10:40
  • $\begingroup$ 1. $dt$ is the change in time that we can actually measure. 2. $\delta t$ is the time required for the system to cover $\delta r$ distance. $\endgroup$
    – Our
    Commented Feb 23, 2019 at 10:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.