[NOTES/CM-02003] From Newton's EOM to Euler Lagrange EOM

Start with Newton's Second Law

Let $\vec{F}_\alpha$ be total force acting on $\alpha^ {th}$ particle. The equations of motion are
\begin{eqnarray}
\label{Eq01}
\frac{d}{dt}\vec{p}_\alpha  &=& \vec{F}_{\alpha}\\
\label{Eq02}
(\vec{F}_\alpha-\frac{d}{dt} \vec{p}_\alpha)&=&0,
\end{eqnarray}
where $\vec{p}_\alpha$ is the momentum of $\alpha^{th}$ particle. Let $\delta\vec{r}_\alpha$ be small displacement in the position \(\vec{r}_\alpha\). Then we have \begin{equation} \label{Eq04} \sum_{\alpha=1}^{N}(\vec{F}_{\alpha}-\frac{d}{dt}\vec{p}_\alpha)\cdot\delta \vec{r}_\alpha=0 \end{equation}

Separate out the forces  due to constraint

We write total force $\vec{F_\alpha}$ as \begin{equation} \label{Eq05} \vec{F}_{\alpha}=\vec{F_\alpha}^{(e)}+\vec{f}_\alpha \end{equation} where $\vec{F}_{\alpha}^{(e)}$ is total  force on $\alpha^ {th}$ particle excluding  the force due to constraint $\vec{f}_\alpha$.

Eliminating forces of constraints
Eliminating forces of constraints Since the system is in equilibrium under forces of constraint, therefore virtual work principle gives \begin{equation} \label{Eq06} \sum \vec{f}_{\alpha}\cdot \delta\vec{r}_\alpha=0. \end{equation} \noindent {\sf Note that this important fact eliminates the forces of constraints.}}\\ Eqs.\eqref{Eq04} - \eqref{Eq06} give \begin{equation} \label{Eq07} \sum_{\alpha=1}^{N}(\vec{F}_{\alpha}^{(e)}-\frac{d}{dt}\vec{p_{\alpha}})\cdot \delta \vec{r}_\alpha=0 \end{equation} where $\delta\vec{r}_\alpha$ is virtual displacement of the system. This result \eqref{Eq07} is known  D' Alembert's principle.

 Express 'total work done' in terms of generalised coordinates

Next few steps below are straightforward differential calculus manipulations.

The first term in \eqref{Eq07}, \( \sum_{\alpha=1}^{N}(\vec{F}_{\alpha}^{(e)}\cdot \delta \vec{r}_\alpha)\) is the work done due to external forces. The Cartesian components of position vectors can be expressed as functions of the generalized coordinates. \begin{equation}\label{Eq08} \vec{r}_\alpha=\vec{r}_\alpha(q_1,q_2\dots q_N,t) \end{equation}

 The strategy now is

• to express $\delta\vec{r}_\alpha$ in terms of $\delta{q_k}$ using \begin{equation}\label{Eq09} \delta\vec{r}_\alpha= \sum_j\frac{\partial\vec{r}_\alpha}{\partial{q_j}}\delta{q_j}. \end{equation} and
• {\it to leave $\vec{v}_\alpha$, coming from \(\dd{t}\vec{p}_\alpha\) in D' Alembert's principle, alone.} No attempt will be made to express it in terms of ${\dot{q_k}},q_k$. This is because we shall try to get final answer in terms of $T=\frac{1}{2}{m_\alpha}\vec{v}_\alpha^{\,2}$.

Using \eqref{Eq09}, we now to rewrite the second term, \(\sum_\alpha\dd[\vec{p}_\alpha]{t}\cdot\delta{\vec{r}}_\alpha\), in D' Alembert's principle, as
\begin{eqnarray}\label{Eq12A}
\sum_\alpha\dd[\vec{p}_\alpha]{t}\cdot\delta{\vec{r}}_\alpha=
\sum_{j,\alpha}\Big\{\Big[\dd{t}(m_\alpha\vec{v}_\alpha)\Big]
\Big[\frac{\partial\vec{r}_\alpha}{\partial{q_j}}\Big]\Big\}\delta{q_j}
\end{eqnarray}
Remember that the coordinates $q_{k}$ in \(\vec{r}_\alpha\), see \eqref{Eq08}, change with  time during the motion. Hence \(\pp[\vec{r}_\alpha]{q_j}\) in \eqref{Eq09}, carry implicit time dependence. By adding and subtracting a suitable term, we can rewrite \eqref{Eq12A} as

\begin{eqnarray}\label{EQ12}
\sum_\alpha\dd[\vec{p}_\alpha]{t}\cdot\delta{\vec{r}}_\alpha
 = \sum_{j,\alpha}\Big\{\frac{d}{dt}
\Big[(m_\alpha\vec{v}_\alpha)
\Big(\frac{\partial\vec{r}_\alpha}{\partial{q_j}}\Big)\Big]-{m_\alpha}{v_\alpha}
\Big[\frac{d}{dt}\frac{\partial\vec{r}_\alpha}{\partial q_j}\Big]\Big\}\delta
q_j
\end{eqnarray}

 Bring in kinetic energy

Proof of \eqref{EQ12x} is easy:
We take up \eqref{EQ12x} first. Exchanging the order of differentiation w.r.t. \(t\) and \(\vec{r}_\alpha\) gives

\begin{eqnarray}\label{Eq16}
m_\alpha\vec{v}_\alpha\frac{d}{dt}\Big(\frac{\partial\vec{r}_\alpha}{\partial{
q_j }}\Big)
&=&m_\alpha\vec{v}_\alpha\cdot\pp{q_j}\Big(\dd[\vec{r_\alpha}]{t}\Big)=
m_\alpha\vec{v}_\alpha\cdot\frac{\partial\vec{v}_\alpha}{\partial{q_j}}
\nonumber\\
&=&\pp{q_j}\Big(\frac{1}{2}m_\alpha\vec{v}_\alpha^2\Big)
=\pp[T]{q_j} \label{Eq18}
\end{eqnarray} 

Proof of \eqref{EQ11y} requires a trick:
Differentiating \eqref{Eq08} w.r.t. to time \(t\) we get

\begin{eqnarray}\label{Eq13}
\dd[\vec{r}_\alpha]{t}&=&\sum_{j}\frac{\partial\vec{r}_\alpha}{\partial
q_j}\dot{q_j}+\frac{\partial\vec{r}_\alpha}{\partial t}\Big|_{q_k},
\\\label{Eq14}
\text{or} ~~~~~\vec{v}_\alpha&=&\sum_j\frac{\partial\vec{r}_\alpha}{\partial
{q_j}}\dot{q_j}+\frac{\partial\vec{r}_\alpha}{\partial t}.
\end{eqnarray}

 Note that the velocities $\vec{v}_\alpha$ depend on $q,\dot{q}$ and $t$. Differentiating \eqref{Eq14} w.r.t $\dot{q_{k}}$ we get \begin{equation}\label{Eq15} \frac{\partial\vec{v}_\alpha}{\partial\dot{q_k}}=\frac{\partial\vec{r}_\alpha} {\partial{q_j}}. \end{equation}
We shall use this relation to eliminate $\frac{\partial\vec{r}_\alpha}{\partial{q_j}}$ in \eqref{EQ11y}.

 \begin{eqnarray}\nonumber
\sum_{\alpha}\frac{d}{dt}\left[({m_\alpha}{v_\alpha})\Big(\frac{\partial\vec{r}
_\alpha}{\partial{q_j}}\Big)\right]
&=& \sum_{\alpha}\dd{t}\left[({m_\alpha}{v_\alpha})
\frac{\partial\vec{v}_\alpha}{\partial{\dot{q_j}}}\right]
\\\nonumber
&=&\dd{t}\sum_{\alpha}\Big({m_\alpha}\vec{v}_\alpha
\frac{\partial \vec{v}_\alpha}{\partial\dot{q_j}}\Big)
\label{Eq19}\\\nonumber
&=&\dd{t}\sum_{\alpha}\frac{\partial}{\partial\dot{q_k}}
\left(\frac{1}{2}{m_\alpha}\vec{v}_\alpha^{\,2}\right)\\
 &=& \frac{d}{dt}\Big(\frac{\partial{T}}{\partial\dot{q_k}}\Big)
\label{Eq20}
\end{eqnarray}

where \(T\) is the kinetic energy of the system $$T={\sum_\alpha}\frac{1}{2}{m_\alpha}\vec{v}_\alpha^{\,2}$$
Thus \eqref{EQ12} can, now, be written as

\begin{eqnarray}\nonumber
\sum_\alpha\dd[\vec{p}_\alpha]{t}\cdot\delta{\vec{r}}_\alpha &=&\nonumber
\sum_{j,\alpha}\Big\{\Big[\dd{t}(m_\alpha\vec{v}_\alpha)\Big]
\Big[\frac{\partial\vec{r}_\alpha}{\partial{q_j}}\Big]\Big\}\delta{q_j}\\
&=&\sum_{j,\alpha}\Big\{
\frac{d}{dt} \Big[(m_\alpha\vec{v}_\alpha)
\Big(\frac{\partial\vec{r}_\alpha}{\partial{q_j}}\Big)\Big]-{m_\alpha}{v_\alpha}
\Big[\frac{d}{dt}\frac{\partial\vec{r}_\alpha}{\partial q_j}\Big]\Big\}\delta
q_j \nonumber\\
&=& \sum_j\left[\dd{t}\Big(\pp[T]{\dot{q}_j}\Big) -
\pp[T]{q_j}\right]\delta q_j
\end{eqnarray}

Thus \eqref{EQ12x} can be cast in the form
\begin{equation}\label{Eq21}\\
\sum\Big(\frac{d}{dt}\frac{\partial{T}}{\partial\dot{q_k}}-
\frac{\partial{T}}{\partial{q_k}}\Big)\delta{q_k}
=\sum_{k,\alpha}\Big({F_\alpha}^{(e)}\cdot\frac{
\partial\vec{r}_\alpha}{\partial{q_k}}\Big){\delta{q_k}}
\end{equation}

Define generalized force
The expression
\begin{equation}\label{EQ21A} \sum_\alpha{F_\alpha}^{(e)}\cdot\frac{ \partial\vec{r}_\alpha}{\partial{q_k}}\equiv Q_k \end{equation}
will be called generalized force. In terms of generalized force, \eqref{Eq21} becomes
\begin{equation}\label{EQ22} \sum_k\left[\Big(\frac{d}{dt}\frac{\partial{T}}{\partial\dot{q_k}}-\frac{ \partial{T}}{\partial{q_k}}\Big)-  Q_k\right]{\delta{q_k}}=0 \end{equation}

The  variations in generalized coordinates are independent 

Calculus manipulations being over, it is time to use the fact that the  generalised coordinates are independent

Since $\delta{q_k}$ are independent and arbitrary, the coefficient of each $\delta{q_k}$ in the above equation can be set equal to zero.
\begin{equation}\label{EQ23} \Big(\frac{d}{dt}\frac{\partial{T}}{\partial\dot{q_k}}-\frac{\partial{T}}{ \partial{q_k}}\Big)=Q_k \end{equation}