# Is the principle of least action a boundary value or initial condition problem?

Here is a question that's been bothering me since I was a sophomore in university, and should have probably asked before graduating: In analytic (Lagrangian) mechanics, the derivation of the Euler-Lagrange equations from the principle of least action assumes that the start and end coordinates at the initial and final times are known. As a consequence, any variation on the physical path must vanish at its boundaries. This conveniently cancels out the contributions of the boundary terms after integration by parts, and setting the requirement for minimal action, we obtain the E.L. equations.

This is all nice and dandy, but our intention is finding the location of a particle at a time in the future, which we do not know a priori; after we derive any equations of motion for a system, we solve them by applying initial values instead of boundary conditions.

How are these two approaches consistent?

• Most treatments contain a footnote or remark that the method can be extended to include variation of the endpoints and do not elaborate (e.g. both Goldstein and Marion & Thorten). I'm afraid that I have never pursued the matter. – dmckee --- ex-moderator kitten Sep 26 '12 at 3:06
• Hi Benji, and welcome to Physics Stack Exchange! This is (and was) actually an excellent question :-) The only thing I'd suggest you remember for the future is not to put in a signature or greeting (including the "hope I'm not doing anything wrong" type); generally, we like to have questions be as concise as possible. Also, it often helps to phrase the title as a question, but it's not a requirement. – David Z Sep 26 '12 at 3:32
• – Qmechanic Sep 26 '12 at 5:38

I) Initial value problems and boundary value problems are two different classes of questions that we can ask about Nature.

Example: To be concrete:

1. an initial value problem could be to ask about the classical trajectory of a particle if the initial position $q_i$ and the initial velocity $v_i$ are given,

2. while a boundary value problem could be to ask about the classical trajectory of a particle if the initial position $q_i$ and the final position $q_f$ are given (i.e. Dirichlet boundary conditions).

II) For boundary value problems, there are no teleology, because we are not deriving a (100 percent certain deterministic) prediction about the final state, but instead we are merely stating that if the final state is such and such, then we can derive such and such.

III) First let us discuss the classical case. Typically the evolution equations (also known as the equations of motion(eom), e.g. Newton's 2nd law) are known, and in particular they do not depend on whether we want to pose an initial value question or a boundary value question.

Let us assume that the eom can be derived from an action principle. (So if we happen to have forgotten the eom, we could always rederive them by doing the following side-calculation: Vary the action with fixed (but arbitrary) boundary values to determine the eom. The specific fixed values at the boundary doesn't matter because we only want to be reminded about the eom; not to determine an actual solution, e.g. a trajectory.)

IV) Next let us consider either an initial value problem or a boundary value problem, that we would like to solve.

Firstly, if we have an initial value problem, we can solve the eom directly with the given initial conditions. (It seems that this is where OP might want to set up a boundary value problem, but that would precisely be the side-calculation mentioned in section III, and it has nothing to do with the initial value problem at hand.)

Secondly, if we have a boundary value problem, there are two possibilities:

1. We could solve the eom directly with the given boundary conditions.

2. We could set up a variational problem using the given boundary conditions.

V) Finally, let us briefly mention the quantum case. If we would try to formulate the path integral

$$\int Dq ~e^{\frac{i}{\hbar}S[q]}$$

as an initial value problem, we would face various problems:

1. The concept of a classical path would be ill-defined. This is related to that the concept of the functional derivative $$\frac{\delta S[q]}{\delta q(t)}$$ would be ill-defined, basically because we cannot apply the usual integration-by-part trick when the (final) boundary terms do not vanish.

2. To specify both the initial position $q_i$ and the initial velocity $v_i$ would violate the Heisenberg uncertainty principle.

• Thanks Qmechanic! I posed my question in a strict classical context, so while I agree that the limitation of uncertainty is indeed problematic, it is technically not existent in the Lagrangian formalism. – Benji Remez Sep 27 '12 at 6:07
• So should your second point be interpreted as follows: I know that a particle is at a given point at an initial time, and will be in some, yet unknown, location at a later time, which I will be able to measure to arbitrary accuracy (because I deny uncertainty). Demanding minimal action, I find the differential equation that governs the physical path between the two locations - but now, technically, I can solve the equation with initial values instead, to "find" which final position would have yielded this initial momentum had the problem been solved under boundary conditions. Am I close? – Benji Remez Sep 27 '12 at 6:18
• I updated the answer. – Qmechanic Sep 27 '12 at 20:47

The usual derivation of the Euler-Lagrange equations forces us to assume that both the initial and final conditions are fixed. However, when one actually derives the equations, he sees that there are differential equations in time so from the knowledge of the initial state, including the velocities (or whatever derivatives are needed to specify the initial point of the phase space), one may derive the values at an infinitesimally later moment.

That's why the "teleological", acausal character of the principle of least action is just an illusion. The trajectory at time $t$ doesn't really depend on any "assumed" values of the fields at later times. This fact may not be obvious immediately, when the principle is formulated, but it is nevertheless true and easy to see via the mathematical derivation of what the principle implies.

• As I understand you, you mean that just because the equations are differential in time, this necessitates that the problem is an initial value problem? I doubt that's really what you meant. Can you refer me to a derivation where this distinction is manifest? – Benji Remez Sep 26 '12 at 5:01
• @Lubosh Motl: You wrote: "The trajectory at time $t$ doesn't really depend on any "assumed" values ... at later times." But yes, it does: if only $q(t_1)$ is known, then the differential equation determines a whole family of possible $q(t_2)$ and only final known data help to fix it and find the intermediary trajectory. – Vladimir Kalitvianski Sep 26 '12 at 14:51
• Benji, I don't understand your concern. The fact that the equations are differential means that one may define and solve initial value problems. That means that both initial coordinates and velocities (or momenta) must be given. Alternatively, one may specify initial and final conditions but only coordinates: momenta/velocities must be left free. Trying to impose both initial and final conditions for both coordinates and momenta would clearly mean an overdetermined problem that generically has no solutions. – Luboš Motl Sep 26 '12 at 16:39
• I fully understand that imposing both constrains creates an overdetermined problem. Clearly, only one set of conditions is needed - what is unclear to me is which one is used in the context of least action. Why is it derived under one assumption and solved under the other? – Benji Remez Sep 27 '12 at 6:02
• "The trajectory at time t doesn't really depend on any "assumed" values of the fields at later times." Feynman's point is that, although this is mind-bending, your statement is not true. The particle's path -does- depend on where you choose its endpoint to be. One example he gives is that of light diffraction; the path followed by the photon actually changes depending on where you choose to observe the endpoint of the path. Mind-bending, but this is how it works. Section 26-5 feynmanlectures.caltech.edu/I_26.html – MajorChipHazard Feb 28 at 21:53

The best way to understand this teleological property is the path integral--- the final boundary condition is required because the action is the amplitude phase for a path, and the stationary condition is saying that you are taking the path of stationary phase, so that the contributions add coherently.

Then the relation between this and the differential formulation is manifest, as Lubos Motl explains. The condition for an extremal path is enforced by a local differential equation. This is not mysterious, because the sum over all paths is not teleological at all, it becomes teleological when you consider that you seem to have taken a stationary path, but this is a consequence of the cancellation away from the stationary path.

This paper:

Feynman's "derivation" of Schrodinger's equation

mentions in an important way that anything that is true of the "whole path" must also be true of portions of the path, including infinitesimal portions. If the action was not extremal over some portion of the path, then that portion could be replaced with another sub-path for which it was, conflicting with the initial assumption that the full path was extremal.

So the least action principle applies to path portions as well, including infinitesimal portions. I have a feeling that, strictly speaking, that is how we wind up getting the eom.

Mathematically it is a boundary value problem by definition, but physically it is practiced as an initial condition problem. Remember the Newton equation: $$p(t+dt)=p(t)+F(t)dt$$ This means the next moment value $p(t+dt)$ is determined with the local values of $p(t)$ and $F(t)$, and no future data is involved. This equation was first discovered as a differential one with the initial (known) conditions, and only later an "integral derivation" under condition of known initial and final coordinates was discovered. Mathematically the boundary value problem is correct and possible, but not physically.