# Why does overall action need to have an extremum?

Quoting from Landau's and Lifshitz' Mechanics :

The integral ${\int\limits_{t_1}^{t_2}}L(q, \dot{q},t)\,dt$ for the entire path must have an extremum, but not necessarily a minimum. This, however, is of no importance as regards the derivation of equations of motions, since only the extremum condition is used.

Why is it so that the entire path must have an extremum, but not a minimum?

Also, what do they mean by saying that only the extremum condition is used? After all, when we consider motion, we usually work with infinitesimal displacements, and they have said that in such displacements we should use a minimum condition (since the extremum applies only to entire path).

• Have you derived the Euler-Lagrange equations for the "Lagrangian" $-L$? – Phoenix87 Oct 9 '15 at 14:13
• The action principle requires only $\delta S/\delta q=0$, it says nothing about $\delta^2S/\delta q^2$. – Ryan Unger Oct 9 '15 at 14:15
• The book says about it though. Also this should be applicable in every case. – Qwedfsf Oct 9 '15 at 14:27
• Possible duplicate: physics.stackexchange.com/q/122486/2451 – Qmechanic Oct 10 '15 at 17:20

Why is it so that the entire path must have an extremum, but not a minimum?

I mean, this isn't an exact proof, but: there is at least one path which satisfies the Euler-Lagrange equations since they are second-order in time and therefore admit two boundary conditions, which are generally sufficient to fix the start and end points. Exceptions would only really involve cases where your laws of motion forbid any path from going from point A to point B and you've asked for such a path anyway, like if you tried to use a space that is not connected (topologically speaking) and then put the points in disconnected parts.

Also, what do they mean by saying that only the extremum condition is used?

More precisely: let $\mathcal P$ denote the smooth functions from $\mathbb R$ ("time") to $\mathbb R^n$ ("particle coordinates"), which we'll call "paths": mathematicians would probably say $\mathcal P = C^\infty(\mathbb R, \mathbb R^n)$ or so. (We might occasionally need a weaker constraint like twice-differentiable paths or so.)

An action principle is a function $S_T :: \mathcal P \to \mathbb R$ mapping paths to real numbers, which we call the action of the path. The $T$ here is an optional time domain, useful for some action principles which can be represented with a Lagrangian $L :: (\mathbb R^n, \mathbb R^n, \mathbb R)\to\mathbb R$ as:$$S_T[p] = S_T\big[t \mapsto p(t)\big] = \int_T d\tau~L\big(p(\tau),~p'(\tau),~\tau\big).$$Notice that the Lagrangian does not know that its arguments are time-dependent or derivatives of each other or any such thing; the Lagrangian just sees some $(\vec a, \vec b, c)$ and maps it to some number. This helps you keep things straight when you start to wonder why, say, there is a total time derivative but only a partial spatial derivative: the sequence goes "do partial derivatives of $L(\vec a, \vec b, c),$ then substitute in $\vec a = p(t)$ and $\vec b = p'(t),$ then take this time derivative."

We introduce a path-perturbation $p(t) \mapsto p(t) + \epsilon~q(t)$ with $q[T] = (\vec 0, \vec 0)$ and then look for the path $p$ whose action is an extremum relative to path perturbations,$$S\big[t\mapsto p(t) + \epsilon~q(t)\big] = S[p] + O(\epsilon^2).$$We are therefore trying to study the paths for which the term linear in $\epsilon$ vanishes, and this gives the Euler-Lagrange equations for the action principle, $(\nabla_a L)_{a=p,b=p'c=t} - \frac{d}{dt}\big[(\nabla_b L)_{a=p,b=p',c=t}\big]= 0.$

We are therefore only using the extremum condition that $\delta S = 0$, not the minimum condition that as well for this path the next term is $\epsilon^2 F[q]$ for some strictly positive $F.$ The principle is often called "least action" but that is a misnomer because the action could also be a maximum or a saddle point or whatever, as long as its "derivative" (response to small path-perturbations) vanishes.

That's all they mean by saying "only the extremum condition is used."

[Of course, we then take this abstract mathematical idea (extremal inputs to an action principle) and imbue it with physical significance. The Lagrangian is kinetic minus potential energy, the extremal path is the path the system actually takes from the given start to end point, the Euler-Lagrange equations are therefore "equations of motion" for the system through its phase space. But they are making a statement at the mathematics-level, not at the physics-level.]

• How does your first proof prove that it must have an extremum? I'm not very acknowledged in higher mathematics. – Qwedfsf Oct 9 '15 at 15:28
• Sorry, I kept the first part brief because the mechanism was sort of detailed in the second half. So, if we path-perturb $a\mapsto a+\epsilon~\alpha,\;b\mapsto b+\epsilon~\beta$, we find $L(a,b,c)\mapsto L(a,b,c)+\epsilon(L_a~\alpha+L_b~\beta)+O(\epsilon^2);$ we substitute $a,b,c\mapsto~p,p',t$ and then integrate by parts to get the Euler-Lagrange equations, so in all cases $S[p+\epsilon~q]=S[p]+\epsilon~\int_T~dt~q~(L_a-\dot L_b)+O(\epsilon^2).$ So "extremal path" means the $\epsilon$-term vanishes if the ODE is solved, which does happen. – CR Drost Oct 9 '15 at 15:57
• In general the Euler-Lagrange equations define a "two point boundary-value problem" in a higher-dimensional space, and I don't know the exact details of the existence/uniqueness of their solution: that is a better question for Math.SE. What I'm trying to tell you is that if an action principle has a Lagrangian underlying it, then we can ask whether it has an extremal path based on whether the Euler-Lagrange equations are satisfied, which is much easier: in many cases we know that solutions must exist to such problems; in general the reasoning "two free parameters, two constraints" suffices. – CR Drost Oct 9 '15 at 16:01
• How would vanishing ${\epsilon}$ imply a function extremum? – Qwedfsf Oct 10 '15 at 13:48
• @Qwedfsf: That's either a really basic question or a really deep question. Assuming really deep, I think they're just using "extremum" to mean "critical point" and not "minimum or maximum." – CR Drost Oct 10 '15 at 15:58