# Lagrange's equation for system not having time translation

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. • To avoid confusion, consider to provide references and definitions. Commented Feb 23, 2019 at 10:34 • @Qmechanic see my edit; I hope this is enough. – Our Commented Feb 23, 2019 at 10:36 ## 1 Answer 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 $$t$$. • 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. – Our Commented Feb 23, 2019 at 10:06 • FWIW,$t$is time, not the parameter for a virtual displacement. Commented Feb 23, 2019 at 10:36 • I'm aware of that. – Our Commented Feb 23, 2019 at 10:37 • How do you define$dt$vs.$\delta t$? Commented Feb 23, 2019 at 10:40 • 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.
– Our
Commented Feb 23, 2019 at 10:44