# Derivation of Lagrange's eqs

## Homework Statement

So I'm deriving Lagrange's equations using Hamilton's principle which states that the motion of a dynamical system follows the path, consistent with any constraints, that minimise the time integral over the lagrangian $L = T-U$, where $T$ is the kinetic energy and $U$ is the potential energy.

We define the lagrangian as $L = L(q_j,\dot{q}_j,t)$. Now I want to let $q_j = q_j^{(0)}+\delta q_j$, where $\delta q_j$ is the variation of $q_j$. We also define
$$S=\int_{t_1}^{t_2}L(q_j,\dot{q}_j,t)dt$$

So according to Hamilton's principle we would now like to minimise $S$. At extremum we have $\delta S = 0$, i.e. the variation of S is zero.

Now, my problem:

My first experience with calculus of variations was to find Euler's equation. We considered then the functional
$$J = \int_{x1}^{x2}f(y,y',x)dx$$
and our goal was to find the function $y$ that minimise S. To do this we let $y(x,\alpha) = y(x)+\alpha\eta(x)$, and set $\frac{\partial J}{\partial\alpha}|_{\alpha=0}=0$, where $\eta$ is some arbitrary function. This would give us an equation in $y(x)$ since we evaluate at $\alpha=0$.

So, returning to the derivation of Lagrange's equations. I set to find $q_j$ that minimise $S$ in similar fashion as we did for $J$. But this time I do not have any $\alpha$ that I can put to zero. Should I instead take $\delta S|_{\delta q_j = 0}=0$? For unless I have understood it wrong, it is actually $q_j^{(0)}$ that we want to find, right? I mean it seems strange to find $q_j$ to be some curve $q_j^{(0)}$ added by some arbitrary variations.

At the same time, I am not sure whether evaluating $\delta q_j$ at makes sense. What increase my doubts is that neither in my text book (Marion and Thornton) nor at any lecture has it been emphasized that we evaluate with $\delta q_j = 0$.

Anyhow, I am thankful for answers.