On covariance of the Lagrange equations

[wrobel]

Here is a sketch of deduction of the Lagrange equations by means of the covariance argument. I believe that it is a suitable substitute for the archaic terminology that is employed in most textbooks.

Assume we have $\nu$ particles with masses $m_1,\ldots,m_\nu$ and with position vectors $$\boldsymbol r_i=(x^{3(i-1)+1},x^{3(i-1)+2},x^{3(i-1)+3})\in\mathbb{R}^3,\quad i=1,\ldots,\nu.$$
Thus the position of the system is characterized by a vector $x=(x^1,\ldots,x^m)^T\in\mathbb{R}^m,\quad m=3\nu.$
Assume also that the system is restricted by ideal holonomic constraints so that $x=x(t,y)$ where $y=(y^1,\ldots,y^r)$ are the local coordinates (generalized coordinates) on a smooth $r-$dimensional manifold $Y$ and $y\mapsto x(t,y)$ is a smooth embedding of $Y$ into $\mathbb{R}^m$ for each $t$.
The Lagrange-D'Alembert principle is
$$\sum_i(m_i\boldsymbol{\ddot r_i}-\boldsymbol F_i,\delta \boldsymbol r_i)=0.\qquad (1)$$
Here $\boldsymbol F_i$ are the given forces.
Formula (1) can be rewritten as follows
$$([\mathscr T]_j-f_j)\frac{\partial x^j}{\partial y^s}=0,\qquad (2) $$
where $f=(f_1,\ldots,f_m)=(\boldsymbol F_1,\ldots,\boldsymbol F_\nu),$
$$\mathscr T=\frac{1}{2}\dot x^TG\dot x,\quad G=\mathrm{diag}(m_1,m_1,m_1,\ldots,m_\nu,m_{\nu},m_{\nu});$$
and
$$[\mathscr T]_j=\frac{d}{dt}\frac{\partial \mathscr T}{\partial \dot x^j}-\frac{\partial \mathscr T}{\partial x^j}.$$
Now employ the covariance:
$$[\mathscr T]_j\Big|_{x=x(t,y)}\frac{\partial x^j}{\partial y^s}=[T]_s,\qquad (3)$$
where $T(t,y,\dot y)=\mathscr T\mid_{x=x(t,y)}.$

Formula (3) is very general it holds for any smooth $\mathscr T=\mathscr T(t,x,\dot x).$
It is like to accomplish the pullback of an 1-form from the manifold $\mathbb{R}^m$ to the manifold $Y$.

Thus (2) takes the form
$$[T]_s=Q_s,\quad Q_s=f_j\frac{\partial x^j}{\partial y^s}.$$