Q: Interpreting the Expanded Euler Lagrange Equations in terms of Newton's 2nd Law We have the classic Euler-Lagrange equations
$$
\frac{\partial L}{\partial q^k} - \frac{d}{dt} \frac{\partial L}{\partial \dot{q}^k} = 0
$$
We can expand this equation further with the chain rule:
$$
\frac{\partial L}{\partial q^k} - \frac{\partial^2 L}{\partial q^j \partial \dot{q}^k} \dot{q}^j - \frac{\partial^2 L}{\partial \dot{q}^j \partial \dot{q}^k} \ddot{q}^j = 0
$$
where I have assumed $\frac{\partial L}{\partial t} = 0$ for simplicity. Rearranging slighty, we obtain:
$$
\frac{\partial L}{\partial q^k} - \frac{\partial^2 L}{\partial q^j \partial \dot{q}^k} \dot{q}^j = \frac{\partial^2 L}{\partial \dot{q}^j \partial \dot{q}^k} \ddot{q}^j
$$
which is starting to look like Newton's 2nd law. Indeed, if we identify $F_k = \frac{\partial L}{\partial q^k}$ as a generalized force and $M_{kj} = \frac{\partial^2 L}{\partial \dot{q}^j \partial \dot{q}^k}$ as some kind of generalized inertia, we have
$$
F_k - \frac{\partial^2 L}{\partial q^j \partial \dot{q}^k} \dot{q}^j = M_{kj} \ddot{q}^j
$$
There is one term that remains unaccounted for, however: $\frac{\partial^2 L}{\partial q^j \partial \dot{q}^k}$. How should this term be interpreted? I note that it would vanish in inertial coordinates with the usual potential and kinetic terms, so I'm tempted to say that it represents inertial forces. But that can't be the full story, since, for example, in polar coordinates inertial forces like the Coriolis force show up in the $F_k = \frac{\partial L}{\partial q^k}$ term.
Is there a standard interpretation of the terms that arise in this expanded form of the Euler-Lagrange equation?

A: Newton equations
starting with the position vector $~\vec R$
\begin{align*}
 & \vec R=\begin{bmatrix}
            x \\
            y \\
            z \\
          \end{bmatrix}=
\begin{bmatrix}
  x(q_i) \\
  y(q_i) \\
  z(q_i) \\
\end{bmatrix}
\end{align*}
where $~q_i~$ are the generalized coordinates
Kinematic
\begin{align*}
&\frac{d R_i}{dt}=v_i= J_{ik}\,\dot{q}^k\quad~,
J_{ik}=\frac{\partial R_i}{\partial q^k}\quad,
J_{ki}=\frac{\partial R_k}{\partial q^i}\\
&\dot{v}_i=J_{ik}\,(\ddot{q})^k+\dot J_{ik}\,(\dot{q})^k=
J_{ik}\,(\ddot{q})^k+\left(\frac{\partial  J_{ik}\,\dot{q}^k}{\partial q^l}\right)\,\dot{q}_l\\
\end{align*}
Dynamic
\begin{align*}
&m\,\dot{v}_i=F_a^i+F_c^i\\
&m\,J_{ni}\,\dot{v}_i=J_{ni}\,F_a^i\quad\text{where}~J_{ni}\,F_c^i=0\\
&\text{substitue }~\dot{v}_i~\Rightarrow\\
&\underbrace{m\,J_{ni}\,J_{ik}}_{M_{nk}}\,(\ddot{q})^k=
\underbrace{J_{ni}\,F_a^i}_{F_n}-\underbrace{\,m\,J_{ni}
\left[\left(\frac{\partial  J_{ik}\,\dot{q}^k}{\partial q^l}\right)\,\dot{q}_l\right]}_{f_{n}}
\end{align*}
hence
\begin{align*}
&f_{n}=\frac{\partial^2 L}{\partial q^j\,\partial\dot{q}^n}\,\dot{q}^j
\end{align*}

*

*$~F_a~$ applied forces; conservative and non conservative

*$~F_c~$ constraint forces

*$~F_n~$ generalized forces

*$~f_n~$ generalized pseudo forces

*$~M_{nk}~$ "mass"  matrix


A: Well, we have that the Euler-Lagrange equation with ($E_k - E_p$) as input expresses $F=ma$.
Let me refer to the above as Hamilton's Euler-Lagrange equation. As we know: Hamilton's action is ($E_k - E_p$).
The expanded version must still be expressing $F=ma$, since Hamilton's Euler-Lagrange equation is expressing $F=ma$.

I think the expanded version is not doing you any favors.

Recapitulating: potential energy is defined as the negative of work done. Work done is the integral of force with respect to the position coordinate:
$$ \int F \ ds  $$
Differentiation of work done with respect to the position coordinate recovers the force:
$$ \frac{d ( \int F \ ds)}{ds} = F   \tag{1}  $$
Differentiation of kinetic energy with respect to the position coordinate recovers $ma$:
$$ \frac{d(\tfrac{1}{2}mv^2)}{ds} = \tfrac{1}{2}m\left( 2v\frac{dv}{ds} \right) = m\frac{ds}{dt}\frac{dv}{ds} = m\frac{dv}{dt} = ma \tag{2}  $$
That is: Hamilton's Euler-Lagrange equation takes kinetic energy as input, and then differentiates that kinetic energy with respect to the position coordinate:
(The chaining of differentiation with respect to velocity 'v' and then differentiation with respect to time 't' has the same effect as differentiation with respect to the position coordinate)
$$ \frac{d}{dt} \frac{d(\tfrac{1}{2}mv^2)}{dv} = \frac{d}{dt}mv = ma \tag{3} $$

Hamilton's stationary action can be derived from F=ma, the derivation is a two-stage process.

*

*Derivation of the work-energy theorem from $F=ma$

*Demonstration that in cases where the work-energy theorem holds good Hamilton's stationary action will hold good also.


The work-energy theorem:
$$ \int F \ ds = \tfrac{1}{2}mv^2 $$
I like to think of the work-energy theorem as the Rosetta Stone of Hamilton's stationary action. The work-energy theorem and Hamilton's stationary action have the following in common: both are expressed in terms of kinetic energy and potential energy, and from both $F=ma$ can be recovered.

So far I haven't mentioned generalized coordinates in this answer. The concept of generalized coordinates extends to force; there is a concept of generalized force Quoting: "A generalized force does not necessarily have the dimensions of force. However, the product $ Q_i q_i$ must have the dimensions of work."
