1
$\begingroup$

. 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

$\endgroup$
1
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.