# Derivation of the Euler-Lagrange Equation and the Principle of Least Action

. Hello, everyone

I have some trouble with the calculus in the derivation of the Euler-Lagrange equation for the principle of least action. Specifically:

$$\newcommand{\lagr}{\mathcal{L}}$$

$$\delta S = \delta \int_{t_1}^{t_2}\lagr(q,\dot{q},t)dt=0$$

$$\delta S = \int_{t_1}^{t_2} (\frac{\partial{\lagr}}{\partial{q}}\delta q + \frac{\partial{\lagr}}{\partial{\dot{q}}} \delta \dot{q})dt=0$$ (Eq. 2)

$$\delta \dot{q}= \frac{d}{dt} \delta q$$

Integrating the second term by parts we get

$$I_2 = \int_{t_1}^{t_2} (\frac{\partial{\lagr}}{\partial{\dot{q}}} \cdot \frac{d}{dt} \delta q) dt = [\frac {\partial{\lagr}} {\partial{\dot{q}}} \delta q]_{t_1}^{t_2} - \int_{t_1}^{t_2} \delta q \cdot \frac{d}{dt}(\frac{\partial{\lagr}}{\partial{\dot{q}}})dt$$

Substituting this into eq. 2

$$\delta S = [\frac{\partial{\lagr}}{\partial{\dot{q}}}\delta q]_{t_1}^{t_2} + \int_{t_1}^{t_2} (\frac{\partial{\lagr}}{\partial{q}} - \frac{d}{dt} \frac{\partial{\lagr}}{\partial{\dot{q}}}) \delta q dt=0$$

The parts I am having trouble with are:

1. During the integration by parts (at $$I_2$$) why can't we just take the $$\frac{d}{dt}$$ term outside the integral and end up with $$\frac{\partial{\lagr}}{\partial{\dot{q}}}\delta q$$ for the whole integral ? Is it because d/dt only applies to delta q?

2. If the boundary conditions dictate $$\delta q$$ must be equal to zero for $$t_1$$ and $$t_2$$, why do we say $$\delta q$$ can take on any value, and therefore the integrand $$(\frac{\partial{\lagr}}{\partial{q}} - \frac{d}{dt} \frac{\partial{\lagr}}{\partial{\dot{q}}})$$ must equal to zero to satisfy the equation?

3. What does $$\delta$$ actually describe? Is it kind of a total derivative?

Thank you for your help

## 1 Answer

More or less formally, what happens when deriving the Euler-Lagrange equations is what follows.

We consider the "space" of all the smooth paths $$q:[0,1]\to\mathbb{R}^3$$ (I take [0,1] for simplicity) and look at the Lagrangian as a functional on this space. Given a path $$q$$, we "deform it" by a path $$\delta q:[0,1]\to\mathbb{R}^3$$ with $$q(0)=q(1)=0$$ (so that the endpoints don't change; you can do this in greater generality but I won't do it here), i.e. we consider the path $$q+\epsilon\delta q$$ with small $$\epsilon\in\mathbb{R}$$. Now we find the extrema of the action by "differentiating in direction $$\delta q$$" \begin{align} 0={}&\left.\frac{d}{d\epsilon}\right|_{\epsilon=0}\int_0^1L(q+\epsilon\delta q,\dot q+\epsilon\delta\dot q)dt\\ ={}&\int_0^1\left.\frac{d}{d\epsilon}\right|_{\epsilon=0}L(q+\epsilon\delta q,\dot q+\epsilon\delta\dot q)dt\\ ={}&\int_0^1\left(\frac{\partial L}{\partial q}\delta q + \frac{\partial L}{\partial\dot q}\delta\dot q\right) dt\\ ={}&\int_0^1\left(\frac{\partial L}{\partial q}\delta q - \frac{d}{dt}\frac{\partial L}{\partial\dot q}\delta q\right) dt\\ ={}&\int_0^1\left(\frac{\partial L}{\partial q} - \frac{d}{dt}\frac{\partial L}{\partial\dot q}\right)\delta q\,dt\ , \end{align} where in the second to last line we used integration by parts. This is true if, and only if (and I'll leave it to you to prove) $$\frac{\partial L}{\partial q} - \frac{d}{dt}\frac{\partial L}{\partial\dot q} = 0\ .$$ I hope this helps to clarify the derivation of the Euler-Lagrange equations!