# Lagrange equation and a force derivable from a generalized potential

I was reading the solution of this exercise and I have a doubt:

A point particle moves in space under the influence of a force derivable from a generalized potential of the form $$U(r,v) = V(r)+\sigma\cdot L$$

where $r$ is the radius vector from a fixed point, $L$ is the angular momentum about that point, and $\sigma$ is a fixed vector in space.

I need to find the components of the force on the particle in Cartesians coordinates, on the basis of Lagrange equations with a generalized potential.

This exercise is from the book "Goldstein - Classical Mechanics". I have the solution, but I really don't understand a step. If I convert $r$ to Cartesian coordenates, I have $r = \sqrt{x^2+y^2+z^2}$, and if I put the expression into the Lagrange equation: $$Q_{j} = \frac{d}{dt}\frac{\partial }{\partial \dot q_{j}}\left(V\left(\sqrt{x^2+y^2+z^2}\right)+\sigma\cdot L\right) - \frac{\partial }{\partial q_{j}}\left(V\left(\sqrt{x^2+y^2+z^2}\right)+\sigma\cdot L\right)$$

$$= \frac{d}{dt}\frac{\partial}{\partial \dot v_{j}}(\sigma\cdot[(x\hat{i}+y\hat{j}+z\hat{k})\times(p_{x}\hat{i}+p_{y}\hat{j}+p_{z}\hat{k})])- \frac{\partial }{\partial x_{j}}\left(V\left(\sqrt{x^2+y^2+z^2}\right)+\sigma\cdot L\right)$$

What happened with $V\left(\sqrt{x^2+y^2+z^2}\right)$? and why $\partial\dot q_{j} = \partial\dot v_{j}$?

If anyone can give me an explanation I would appreciate it :)

• $\uparrow$ Which problem in Goldstein? – Qmechanic Mar 26 '15 at 0:42
• What do you mean by "What happened with $V(...)$?" – Kyle Kanos Mar 26 '15 at 0:46
• @Qmechanic exercise 15 from the page 33, Chapter "Survey of the Elementary Principles" – Sebastián Molina Mar 26 '15 at 0:46
• @KyleKanos because $V(...)$ has disappeared, or not? D: – Sebastián Molina Mar 26 '15 at 0:48
• Consider this question: What is $\partial_x f(y)$ (and $y$ not a function of $x$)? – Kyle Kanos Mar 26 '15 at 0:50

What happened with $V\left(\sqrt{x^2+y^2+z^2}\right)$?
You mean, why does V(r) disappear from the $\frac{\partial }{\partial \dot q_j}$ term, right? It's because V(r) is a function only of $q_j$ not $\dot q_j$. Those variables are treated as independent and so $\frac{\partial V}{\partial \dot q_j}=0$.
and why $\partial\dot q_{j} = \partial\dot v_{j}$?
Looks like you have an extra dot... Should be: $$\dot q_{j} = v_{j}$$ If I'm interpretting your notation correctly... I.e., the $q_j$ are the coordinates (which are just the cartesian coordinates. And the $\dot q_j$ are the velocities...