In proving that the Lagrange equations hold in any coordinate system

[M. next]

I was checking the proof of this, when things came vague at one point.

It goes as follows, how to prove that Lagrange's equations hold in any coordinate system?

Answer:

Let q$_{a}$ = q$_{a}$(x$_{1}$,..., x$_{3N}$, t)

here the possibility of using a coordinate system that changes with time is included.

By chain rule: $\dot{q_{a}}$= $\partial$q$_{a}$/$\partial$x$^{A}$($\dot{x^{A}}$) + $\partial$q$_{a}$/$\partial$t

To be in a good coordinate system, we should be able to invert this relationship, call this equation (1):

$\dot{x_{a}}$= $\partial$x$_{a}$/$\partial$q$^{A}$($\dot{q^{A}}$) + $\partial$x$_{a}$/$\partial$t



Then the book said, now we can examine L(x$^{A}$, $\dot{x^{A}}$) when we substitute in x$^{A}$(q$_{a}$, t)

Using (1), we have

$\partial$L/$\partial$q$_{a}$ =

($\partial$L/$\partial$x$^{A}$)($\partial$x$^{A}$/$\partial$q$_{a}$) + $\partial$L/$\partial$$\dot{x^{A}}$($\partial$$^{2}$x$^{A}$/$\partial$q$_{a}$q$_{b}$*$\dot{q_{b}}$ + $\partial$$^{2}$x$^{A}$/$\partial$q$_{t}$q$_{a}$)

My question is I don't understand where the q$_{b}$ come from??

Thank you for your time!

 

[Valerie Park]

The q_b comes from the fact that the partial derivative of x^A with respect to q_a and q_b is a second-order partial derivative, which means that it depends on both q_a and q_b. This is because, in order for x^A to change, not only must q_a change (which is what the first partial derivative accounts for), but also q_b must change as well.