# Why the least action principle is always (?) used in this particular form?

The least action principle in (mathematical) physics says the following. Given a system, e.g. collection of particles, whose motion satisfies a known system of differential equations (of second order). Then there exists a so called action functional $$S$$ on the space of paths of all the particles such that solutions of the above differential equations are precisely the critical paths of $$S$$. Moreover in all known to me cases it is assumed that $$S$$ can be chosen in the form $$S=\int L(x,\dot x,t)dt,\,\,\,(1)$$ where $$L$$ is called Lagrangian.

The least action principle is satisfied in this form for many problems of interest in physics, but not for arbitrary system of differential equations.

Question. Why it is important that $$S$$ has the form (1) for some Lagrangian? Are there situations of interest in (mathematical) physics where the action functional is not given by any Lagrangian?

Remark. If one takes the action functional in the form $$S_1:=\exp(S)=\exp(\int L(x,\dot x,t)dt)$$ then $$S_1$$ and $$S$$ have obviously the same critical paths.

• The action being the integral of a Lagrangian leads to the Euler-Lagrange equations as the equations of motion. – gmvh Sep 26 at 9:42
• Occasionally there are higher-order PDE of physical interest that arise as Euler-Lagrange equations of Lagrangians involving more than one derivative. For instance the biharmonic equation en.wikipedia.org/wiki/Biharmonic_equation is the Euler-Lagrange equation for the higher order Dirichlet energy $\int |\nabla^2 u|^2$ (or $\int |\Delta u|^2$). – Terry Tao Sep 27 at 0:30

In the form (1), if you compute the variation $$\delta S / \delta x(t) = E(t)$$, you find that $$E(t) = E(x(t),\dot{x}(t), \ddot{x}(t) ,t)$$ is a local/differential expression (the value of $$E(t)$$ does not depend on $$x(t')$$ or its derivatives at other times $$t'\ne t$$). This is no longer true if you use $$\exp(S)$$ instead of $$S$$. There is no dispute that $$S$$ and $$\exp(S)$$ have the same critical points (N.B.: first order variations cannot distinguish between critical points of different kinds like maxima, minima, or saddle points). But if like your $$\delta S / \delta x(t)$$ to be local (and some people do), then you are stuck with local action functionals, namely those in the form (1).
UPDATE: The proof that locality of $$E(t)$$ implies locality of $$S$$ is straightforward, essentially an application of the fundamental theorem of calculus. Morally, $$E(x(t), \dot{x}(t), \ddot{x}(t), t)$$ is the gradient of $$S$$ with respect to $$x$$. Conversely, $$S$$ is the primitive/anti-derivative of $$E(t)$$, and any two such primitives must differ by a constant. One primitive can be constructed by preserving locality: $$S = \int \left(\int_0^1 x(t) E(s x(t), s\dot{x}(t), s\ddot{x}(t), t) ds \right) dt,$$ where the expression in parentheses is known as the Vainberg-Tonti Lagrangian (Google the keywords for references). So all other primitives must differ by a constant. There may be some funny ways to express a constant which may not appear local in the way we have been discussing, but such non-locality can be dismissed as trivial. This discussion has obvious generalizations to more dependent and independent variables, as well as higher differential orders.
• @IgorKhavkine: is it obvious that the Vainberg-Tonti Lagrangian does not depend on $\ddot{x}$? Or is it just accepted as okay? – Willie Wong Sep 27 at 14:35