What is a sufficient condition to determine if the action integral reaches an extremal through the trajectory described by $q(t)$? - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-06-20T03:24:58Z https://physics.stackexchange.com/feeds/question/160874 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/160874 3 What is a sufficient condition to determine if the action integral reaches an extremal through the trajectory described by $q(t)$? Stan Shunpike https://physics.stackexchange.com/users/66165 2015-01-22T20:13:48Z 2015-01-22T21:08:21Z <p>As I understand it, the Euler-Lagrange equation is a necessary but not a sufficient condition to determine if the action integral reaches an extremal through the trajectory described by $q(t)$. </p> <p>If this is so, what is a sufficient condition? </p> https://physics.stackexchange.com/questions/160874/-/160884#160884 1 Answer by Ryan Unger for What is a sufficient condition to determine if the action integral reaches an extremal through the trajectory described by $q(t)$? Ryan Unger https://physics.stackexchange.com/users/59215 2015-01-22T21:08:21Z 2015-01-22T21:08:21Z <p>Let me begin with an analogy to high school calculus. We know that a function $f(x)$ has a stationary point $x_0$ if at that point $$\left.\frac{df}{dx}\right|_{x_0}=0$$ There are three possibilities: minimum, maximum and saddle. This is called the <em>first derivative test</em>. To test for these conditions, we use the <em>second derivative test</em>. We check $$\left.\frac{d^2f}{dx^2}\right|_{x_0}$$ If this is positive, $f$ is concave up and thus $x_0$ is a minimum. If it is negative, $x_0$ is concave down and a maximum. If both first and second derivatives are zero at $x_0$, it is a point of inflection, more specifically, a saddle point (a point of inflection which has zero first derivative is a saddle).</p> <p>Now we come to variational calculus. Path integral quantum mechanics tells us that the action $$S[q]=\int L\,dt$$ is stationary along the classical path, a condition which we write as $$\delta S[q][h]=0$$ where the functional derivative of a functional $G[f]$ is defined as $$\delta G[f][h]=\left.\frac{d}{d\epsilon}G[f+\epsilon h]\right|_{\epsilon=0}$$ The action principle implies the Euler-Lagrange equations $$\frac{\partial L}{\partial q}-\frac{d}{dt}\frac{\partial L}{\partial\dot q}=0$$ But these equations are only necessary for extremal solutions, and saddle point solutions are possible.</p> <p>So suppose we solve the EL equations and are not sure if it is a minimum, maximum or saddle point. We perform the second derivative test for functionals as follows: The second functional derivative is $$\delta^2 G[f][h]=\left.\frac{d^2}{d\epsilon^2}G[f+\epsilon h]\right|_{\epsilon=0}$$ so we look at the integral $$\delta^2 S[q_\text{c}][h]$$ where $q_\text{c}$ is the curve which solves the EL equations. If it is positive, $q_\text{c}$ is a minimum, etc.</p> <p>You should take a look at this paper, "<a href="http://www.eftaylor.com/pub/Gray&amp;TaylorAJP.pdf" rel="nofollow">When action is not least</a>". It discusses the differences between minima, maxima and saddle points in great detail.</p>