Rigorous Way to Obtain Classical Lagrangian

Read from the book Classical Mechanics written by Goldstein, Poole and Safko. Found it interesting. Thought I would like to share here.

Constraints

If the condition of the constraint can be expressed as the coordinates of the particles and time only $$f(\vec{r}_1,\vec{r}_2,...,t)=0,$$ then the constraint is said to be holonomic. An example would be the rigid body, in which the constraints are $$|\vec{r}_i - \vec{r}_j|^2 - c^2_{ij}=0,$$ for some constant $c_{ij}$.

Suppose there are $N$ particles in a system. Without constraints, there will be $3N$ degree of freedom. The $3N$ are represented by $3N$ independent coordinates. If there are $k$ equations of contraint, the degree of freedom will be reduced by $k$. Then, there will be $3N-k$ independent coordinates. Let $q_1,q_2,...,q_{3N-k}$ be the set of independent coordinates. The set of position vectors can be expressed in terms of these independent coordinates $$\begin{align} \vec{r}_1 &= \vec{r}_1(q_1,q_2,...,q_{3N-k},t)\\ & .\\ & .\\ & .\\ \vec{r}_N &= \vec{r}_N(q_1,q_2,...,q_{3N-k},t) \end{align}$$

Lagrange's Equations

Let $\delta \vec{r}_i$ be an infinitesimal change of virtual displacement ("virtual" because there is not really a change in the actual displacement of the system). If the system is in equilibrium, there is no work by the forces acted on each particle $$\sum_i\vec{F}_i\cdot\delta\vec{r}_i.$$ Express the force $\vec{F}_i$ in terms of the applied force, $\vec{F}^{\text{(app)}}_i$, and the force of constraint, $\vec{f}_i$, $$\vec{F}_i=\vec{F}^{\text{(app)}}_i + \vec{f}_i.$$ The virtual work is $$\sum_i\vec{F}^{\text{(app)}}_i\cdot\delta\vec{r}_i+\sum_i\vec{f}_i\cdot\delta\vec{r}_i=0.$$ The equation of motion is $$\vec{F}_i=\dot{\vec{p}}_i \implies \vec{F}_i-\dot{\vec{p}}_i=0.$$ If a force $-\dot{\vec{p}}_i$ is also acted on each particle, the system will be in equilibrium and we will have $$\sum_i(\vec{F}_i-\dot{\vec{p}}_i)\cdot\delta\vec{r}_i=0. \tag{1}$$ Decompose the force into applied force and force of constraint, we have $$\sum_i(\vec{F}^{\text{(app)}}_i-\dot{\vec{p}}_i)\cdot\delta\vec{r}_i+\sum_i\vec{f}_i\cdot\delta\vec{r}_i=0.$$ If the system has no virtual work by the forces of constraint, then $$\sum_i(\vec{F}^{\text{(app)}}_i-\dot{p}_i)\cdot\delta\vec{r}_i=0.$$ The displacement vectors $\vec{r}_i$ can be expressed in terms of generalized coordinates $$\vec{r}_i=\vec{r}_i(q_1,q_2,...,q_n,t)$$ The rate of change of the displacement is $$\vec{v}_i\equiv\frac{d\vec{r}_i}{dt}=\sum_k\frac{\partial\vec{r}_i}{\partial q_k}\dot{q}_k+\frac{\partial \vec{r}_i}{\partial t} \tag{2}$$ Similarly, the arbitrary virtual displacement $\delta\vec{r}_i$ is $$\delta\vec{r}_i=\sum_j\frac{\delta\vec{r}_i}{\delta q_j}\delta q_j,$$ in which the term $\frac{\partial\vec{r}_i}{\partial t}$ is vanished because by definition, the virtual displacement is not real and only displacement change is considered. Then the virtual work is $$\sum_i\vec{F}^{\text{(app)}}_i\cdot\delta\vec{r}_i=\sum_{i,j}\vec{F}^{\text{(app)}}_i\cdot\frac{\partial\vec{r}_i}{\partial q_j}\delta q_j$$ Define $$Q_j\equiv\sum_j\vec{F}^{\text{(app)}}_i\cdot\frac{\partial\vec{r}_i}{\partial q_j}$$ as the generalized force. Note that the generalized force does not neccessarily have the same dimension as force. The work by the rate of change of momentum in Eq.(1) can be express as $$\sum_i\dot{\vec{p}}_i\cdot\delta\vec{r}_i=\sum_im_i\ddot{\vec{r}}_i\cdot\delta\vec{r}_i=\sum_{i,j}m_i\ddot{\vec{r}}_i\cdot\frac{\partial\vec{r}_i}{\partial q_j}\delta q_j.$$ Using product rule, we have $$\sum_im_i\ddot{\vec{r}}_i\cdot\frac{\partial\vec{r}_i}{\partial q_j} = \sum_i\left[\frac{d}{dt}\left(m_i\dot{\vec{r}}_i\cdot\frac{\partial \vec{r}_i}{\partial q_j}\right)-m_i\dot{\vec{r}}_i\cdot\frac{d}{dt}\left(\frac{\partial\vec{r}_i}{\partial q_j}\right)\right] \tag{3}$$ By Eq.(2), the second term can be written as $$\begin{align} \frac{d}{dt}\left(\frac{\partial \vec{r}_i}{\partial q_j}\right) &= \sum_k\frac{\partial^2\vec{r}_i}{\partial q_j\partial q_k}\dot{q}_k + \frac{\partial^2\vec{r}_i}{\partial q_j\partial t}\\ &= \frac{\partial \vec{v}_j}{\partial q_j} \end{align}$$ Also by Eq.(2), since $\vec{r}_i$ does not depend on $\dot{q}_j$ and $q_j$ are a set of independent coordinates, $$\begin{align} \frac{\partial \vec{v}_i}{\partial \dot{q}_j} &= \frac{\partial }{\partial \dot{q}_j} \left( \sum_k\frac{\partial \vec{r}_i}{\partial q_j}\left(\frac{d q_k}{dt}\right) + \frac{\partial \vec{r}_i}{\partial t} \right)\\ &= \sum_k\frac{\partial \vec{r}_i}{\partial q_j}\left(\frac{\partial \dot{q}_k}{\partial \dot{q}_j}\right) + \frac{\partial}{\partial t}\frac{\partial \vec{r}_i}{\partial \dot{q_j}}\\ &= \sum_k\frac{\partial \vec{r}_i}{\partial q_j}\delta_{kj} + 0\\ &= \frac{\partial \vec{r}_i}{\partial q_j} \end{align}$$ Substitute these into Eq.(3), $$\sum_im_i\ddot{r}_i\cdot\frac{\partial \vec{r}_i}{\partial q_j}=\sum_i\left[\frac{d}{dt}\left(m_i\vec{v}_i\cdot\frac{\partial \vec{v}_i}{\partial \dot{q}_j}\right)-m_i\vec{v}_i\cdot\frac{\partial \vec{v}_i}{\partial q_j}\right].$$ So, $$\sum_j\dot{p}_j\cdot\delta\vec{r}_j=\sum_j\left[\frac{d}{dt}\frac{\partial}{\partial\dot{q}_j}\left(\sum_i\frac{1}{2}m_iv_i^2\right)-\frac{\partial}{\partial q_j}\left(\sum_i\frac{1}{2}m_iv_i^2\right)\right]\delta q_j$$ Since $\sum_i\frac{1}{2}m_iv_i^2$ is the system kinetic energy , $T$, Eq.(1) becomes $$\sum_j\left[\left[\frac{d}{dt}\left(\frac{\partial T}{\partial \dot{q}_j}\right)-\frac{\partial T}{\partial q_j}\right]-Q_j\right]\delta q_j=0.$$ If the constraints are holonomic, then a set of independent coordinates $(q_1,1_2,...,q_n)$ can be found. Since the independent coordinates are independent, $q_i$ does not depend on $q_k$, where $i \neq k$. Then the equation holds only if for each term $$\frac{d}{dt}\left(\frac{\partial T}{\partial \dot{q}_j}\right)-\frac{\partial T}{\partial q_j}=Q_j.$$ If a scalar potential function $V$ can be defined for the forces $$\vec{F}^{\text{(app)}}_i=-\nabla_iV,$$ the generalized forces can be written as $$\begin{align} Q_j &= \sum_i\vec{F}^{\text{(app)}}_i\cdot\frac{\partial\vec{r}_i}{\partial q_j}\\ &= -\sum_i\nabla_iV\cdot\frac{\partial \vec{r}_i}{\partial q_j}\\ &= - \frac{\partial V}{\partial q_i}. \end{align}$$ Hence, the equation becomes $$\frac{d}{dt}\left(\frac{\partial T}{\partial \dot{q}_j}\right)-\frac{\partial(T-V)}{\partial q_i}=0.$$ As the potential $V$ does not depend on the generalized velocities, $\frac{\partial V}{\partial \dot{q}_j}=0$, $$\frac{d}{dt}\left(\frac{\partial(T-V)}{\partial\dot{q}_j}\right)-\frac{\partial(T-V)}{\partial q_j}=0.$$ Define the Lagrangian $L$ to be $$L=T-V,$$ we have $$\bbox[5px,border:2px solid #666] { \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_j}\right)-\frac{\partial L}{\partial q_j}=0 }$$