Lagrange's Equation

The Cartesian equations of motion of our system take the form

$\begin{displaymath} m_j\,\ddot{x}_j = f_j, \end{displaymath}$

(600)

for $j=1,{\cal F}$ , where

are each equal to the mass of the first particle,

are each equal to the mass of the second particle, etc.Furthermore, the kinetic energy of the system can be written

$\begin{displaymath} K = \frac{1}{2}\sum_{j=1,{\cal F}} m_j\,\dot{x}_j^{\,2}. \end{displaymath}$

(601)

Now, since $x_j=x_j(q_1,q_2,\cdots, q_{\cal F},t)$ , we can write

$\begin{displaymath} \dot{x}_j= \sum_{i=1,{\cal F}} \frac{\partial x_j}{\partial q_i}\,\dot{q}_i + \frac{\partial x_j}{\partial t}, \end{displaymath}$

(602)

for $j=1,{\cal F}$ . Hence, it follows that $\dot{x}_j = \dot{x}_j(\dot{q}_1,\dot{q}_2,\cdots, \dot{q}_{\cal F},q_1,q_2,\cdots,q_{\cal F},t)$ . According to the above equation,

$\begin{displaymath} \frac{\partial \dot{x}_j}{\partial\dot{q}_i} = \frac{\partial x_j}{\partial q_i}, \end{displaymath}$

(603)

where we are treating the $\dot{q}_i$ and the

as independent variables.

Multiplying Equation (603) by $\dot{x}_j$ , and then differentiating with respect to time, we obtain

$\begin{displaymath} \frac{d}{dt}\!\left(\dot{x}_j\,\frac{\partial \dot{x}_j}{\pa... ...\frac{d}{dt}\!\left( \frac{\partial x_j}{\partial q_i}\right). \end{displaymath}$

(604)

Now,

$\begin{displaymath} \frac{d}{dt}\!\left(\frac{\partial x_j}{\partial q_i}\right)... ...,\dot{q}_k + \frac{\partial^2 x_j}{\partial q_i\,\partial t}. \end{displaymath}$

(605)

Furthermore,

$\begin{displaymath} \frac{1}{2} \,\frac{\partial\dot{x}_j^{\,2}}{\partial \dot{q}_i} = \dot{x}_j\,\frac{\partial \dot{x}_j}{\partial \dot{q}_i}, \end{displaymath}$

(606)

and

$\displaystyle \frac{1}{2}\,\frac{\partial \dot{x}_j^{\,2}}{\partial q_i} = \dot{x}_j\,\frac{\partial \dot{x}_j}{\partial q_i}$	$\textstyle =$	$\displaystyle \dot{x}_j\,\frac{\partial}{\partial q_i}\!\left(\sum_{k=1,{\cal F... ...partial x_j}{\partial q_k}\,\dot{q}_k + \frac{\partial x_j}{\partial t}\right)$
	$\textstyle =$	$\displaystyle \dot{x}_j\left(\sum_{k=1,{\cal F}}\frac{\partial^2 x_j}{\partial ... ...rtial q_k}\,\dot{q}_k + \frac{\partial^2 x_j}{\partial q_i\,\partial t}\right)$
	$\textstyle =$	$\displaystyle \dot{x}_j\,\frac{d}{dt}\!\left(\frac{\partial x_j}{\partial q_i}\right),$	(607)

where use has been made of Equation (605). Thus, it follows from Equations (604), (606), and (607) that

$\begin{displaymath} \frac{d}{dt}\!\left(\frac{1}{2}\,\frac{\partial \dot{x}_j^{\... ... + \frac{1}{2}\,\frac{\partial \dot{x}_j^{\,2}}{\partial q_i}. \end{displaymath}$

(608)

Let us take the above equation, multiply by

, and then sum over all

. We obtain

$\begin{displaymath} \frac{d}{dt}\!\left(\frac{\partial K}{\partial \dot{q}_i}\ri... ...\partial x_j}{\partial q_i} + \frac{\partial K}{\partial q_i}, \end{displaymath}$

(609)

where use has been made of Equations (600) and (601). Thus, it follows from Equation (597) that

$\begin{displaymath} \frac{d}{dt}\!\left(\frac{\partial K}{\partial \dot{q}_i}\right) = Q_i + \frac{\partial K}{\partial q_i}. \end{displaymath}$

(610)

Finally, making use of Equation (599), we get

$\begin{displaymath} \frac{d}{dt}\!\left(\frac{\partial K}{\partial \dot{q}_i}\ri... ...rac{\partial U}{\partial q_i}+\frac{\partial K}{\partial q_i}. \end{displaymath}$