14
$\begingroup$

I'm doing some self reading on Lagrangian Mechanics and Special Relavivity. The following are two statements that seem to be taken as absolute fundamentals and yet I'm unable to reconcile one with the other.

  1. Principle of Least action states that the particle's trajectory under the influence of a potential is determined by minimising the action that is $\delta S =0$.
  2. The universal speed limit of information propagation is $c$.

Question: Consider a potential that exists throughout space (like gravity perhaps). Now, I give a particle $x,p$ and set it motion. As per principle of least action the motion is now fully deterministic since the Lagrangian $L$ is known. We arrive at this claim by assuming that the particle is able to "calculate" the trajectory such that $\delta S = 0$ using Principle of Least Action.

This is where my confusion is. The first principle is a global statement whereas the second principle is a local statement.

What is the modern day physics answer to this?

I hope I could get my questions across! Look forward to your valuable opinions!

$\endgroup$
3
  • 9
    $\begingroup$ the Euler-Lagrange equations tell you though that the 'global' property of minimising action corresponds to locally solving the 'local' Euler lagrange equations though? $\endgroup$
    – Joshua Lin
    May 1 at 0:54
  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/38348/2451 , physics.stackexchange.com/q/563624/2451 and links therein. $\endgroup$
    – Qmechanic
    May 1 at 3:43
  • 3
    $\begingroup$ Perhaps the problem is with the idea that a particle is somehow picking or calculating a trajectory (implying particles have free will!), rather than the trajectory being determined by the "shape" of space. $\endgroup$
    – jamesqf
    May 2 at 4:06
23
$\begingroup$

The key to this is that the Lagrangian cannot be just any old function. It has to be a function such that, when the action is stationary, its solution describes the kinematics of the system.

Thus, if we assert that there is a universal speed limit in the real world, then that says something about the Lagrangians which can be used. They must be chosen such that the motion of the particle perscribed by solving the Lagrangian Mechanics problem is such that the path of any particle is the same, regardless of the state of the universe outside of its light-cone.

These are not mutually exclusive. From one perspective, you have a collection of every particle, and every particle obeys some rules. From another perspective, you have a global system whose time evolution ensures said rules for each particle are obeyed.

Perhaps obvious: Physicists who use Lagrangian Mechanics to explore relativistic systems do indeed use Lagrangians that have this property.

$\endgroup$
2
  • 2
    $\begingroup$ As you state: "The path of any particle is the same, regardless of the state of the universe outside of its light-cone." As I understand it: the concern expressed in the question is along the following lines: It looks as if a particle must have instantaneous knowledge of the state of the potential over the entire extent of its trajectory. That is how I interpret the question: To satisfy motion law: does a particle in fact need instantaneous knowledge extending over the entire future trajectory? As opposed to: is particle motion in fact continuous negotiation of the local potential. $\endgroup$
    – Cleonis
    May 1 at 15:58
  • 2
    $\begingroup$ @Cleonis To construct a Lagrangian where they are the same, you simply ensure that the cross-terms between two distant particles is always multiplied by 0. Alternative phrasing would be to make sure they can always be separated. As a somewhat abusrd example, it's trivial to show that you can find the stationary paths for L=a^2+0.0*a*b + b^2 by solving for the stationary path of a, the stationary path of b, and then combining them into one universal state. $\endgroup$
    – Cort Ammon
    May 1 at 16:24
13
$\begingroup$

About global statement and local statement:

There is a lemma in variational calculus that was first stated by Jacob Bernoulli.

When Johann Bernoulli had presented the Brachistochrone problem to the mathematicians of the time Jacob Bernoulli was among the few who was able to find the solution independently. The treatment by Jacob Bernoulli is in the Acta Eruditorum, May 1697, pp. 211-217

Jacob opens his treatment with an observation concerning the fact that the curve that is sought is a minimum curve.

Jacob Bernoulli lemma diagram

Lemma. Let ACEDB be the desired curve along which a heavy point falls from A to B in the shortest time, and let C and D be two points on it as close together as we like. Then the segment of arc CED is among all segments of arc with C and D as end points the segment that a heavy point falling from A traverses in the shortest time. Indeed, if another segment of arc CFD were traversed in a shorter time, then the point would move along ACFDB in a shorter time than along ACEDB, which is contrary to our supposition.



Jacob's lemma generalizes to any case where the curve that you are trying to find is characterized by satisfying a global extremum condition.

If the curve satisfies the extremum condition globally then every subsection of the trajectory satisfies the extremum condition too, down to infinitissimally short subsections.


It follows that if the curve satisfies an extremum condition then a differential equation must exist that has the sought after curve as its solution.

Back in the time of Jacob and Johann Bernoulli there was no systematic way of identifying the corresponding differential equation.

This problem was addressed by Euler and Lagrange, hence their systematic approach is called the 'Euler-Lagrange equation'.

I will come back to the Euler-Lagrange equation further down in this answer.



Comparison: numerical analysis

I recommend the following way to see the connection between variational calculus and differential calculus: implementation in numerical analysis.

There is a Java applet created by Todd Timberlake that offers a numerical implementation of Hamilton's stationary action


For differential calculus the simplest algorithm is Euler integration. Time is divided in small increments, the calculation proceeds increment by increment.

The numerical analysis implementation of variational calculus is as follows: time is divided in small increments. Then there is an iterative calculation: In each step the algorithm evaluates the middle point of a triplet of points, advancing one time increment each step. In each step the middle point is adjusted relative to its directly adjacent points, to make it satisfy the extremum condition locally:
$t_0$, $t_1$, $t_2$
$t_1$, $t_2$, $t_3$
$t_2$, $t_3$, $t_4$
And so on.

When the last point is reached the iteration start again with the first point

The simplest seed for the iterative calculation is a flat line. The iteration can start with any seed.

Over each iteration the trial trajectory shifts towards the true trajectory. Over a sufficiently large number of iterations the trial trajectory converges to the true trajectory to within any desired level of computational accuracy. So: the numerical analysis implementation arrives at satisfying the extremum condition globally by iteration of evaluating the extremum condition locally.

And of course: the smaller the time increments the higher the level of computational accuracy.


It is occasionally claimed: variational calculus is unique in that it is based on evaluating whether a global condition is satisfied. However, that claim is untenable; there is only one way to implement the numerical analysis: by division in small time increments, and the global evaluation is built up by way of iteration of local evaluation.



Is Hamilton's stationary action a form of knowing the future?

I surmise that the underlying thought of your question is: "Is a particle able to know the future in advance?"

Well, it's not the case that Hamilton's stationary action as a calculation strategy implies "knowing the future in advance".

Hamilton's stationary action is a form of extrapolation.

As we know: the simplest form of extrapolation is for motion along a straight line: The distance between point A and point B is X units of distance. How fast must you travel to make it to point B to arrive in Y units of time?

Next level up is to make the trajectory curvilinear: a projectile deflected by gravity. The distance between point A and point B is X units of distance. A projectile is fired at an angle to the horizontal of $\alpha$ degrees. At what velocity must the projectile be fired in order to make it to point B in Y units of time?

The trajectory involves deflection now, but we can still solve for the required velocity, and we do not find in in the least surprising that we can still still solve the problem.

(It so happens we know that with gravity the trajectory follows a parabola, but not knowing that in advance does not present a problem: solving for that parabola will be an organic part of the overal course of the solution.)

Either you get the velocity as initial information, and you use that to calculate the expected duration, or you get the expected duration as initial information, and you use that to calculate required velocity.

Hamilton's stationary action as a calculation strategy is no different from the above. You don't know the shape of the trajectory in advance, but you solve for that organically in the overall process of solving the problem.



the Euler-Lagrange equation

In this section I examine the strategy of the derivation of the Euler-Lagrange equation.

Mathematicians always push for more generalisation: the Euler-Lagrange equation is the general strategy for the general variational problem.

For transparency I will narrow down to application of the Euler-Lagrange equation in dynamics, more specifically classical dynamics.

For dynamics the Euler-Lagrange equation expresses the derivative with respect to position of the Action. In dynamics one might expect to evaluate derivative with respect to time, but the EL-equation expresses derivative with respect to position.

The potential energy is a function of the trajectory, and the kinetic energy is a function of the trajectory. So you have the freedom to take the derivative with respect to position instead of the derivative with respect to time.


The starting point is Hamilton's stationary action.

The first step is to express the derivative of Hamilton's action with respect to the variation.

In the course of the derivation of the EL-equation it is demonstrated that the derivative with respect to position and the derivative with respect to variation are the same, since the variation is variation of position.

So, while the starting point was derivative with respect to variation, the derivation then transforms that in to derivative with respect to position.



On how to derive Hamilton's stationary action

For any curve that satisfies a global extremum condition we have that a differential equation exists that has the sought after curve as its solution.

We can take that in reverse, and transform a problem in differential calculus to a problem in variational calculus.

We start with the Work-Energy theorem.

$$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \qquad (1) $$

I noticed that this theorem is often referred to as Work-Energy principle. It is a theorem in the following sense: if we take the derivative with respect to position we recover $F=ma$ immediately.

Simplify to initial position coordinate zero and initial velocity zero

$$ \int F \ ds = \tfrac{1}{2}mv^2 \qquad (2) $$

Derivative with respect to position:

$$ \frac{\int F \ ds}{ds} = \frac{\tfrac{1}{2}mv^2}{ds} \qquad (2) $$

$$ F = ma \qquad (3) $$

From here on I will refer to the negative of the integral of force over distance as the 'potential energy'.

From the work-energy theorem it follows:

$$ \Delta(E_k) = \Delta(-E_p) \qquad (4) $$

Coming from the work-energy theorem (4) is fundamentally limited in scope as compared to the Principle of conservation of energy. The principle is unconditional, but the work-energy theorem is applicable only if there is a defined integral of the force over spatial coordinate.

(4) is valid down to infinitisimal increments.

$$ d(E_k) = d(-E_p) \qquad (5) $$

The derivative with respect to position:

$$ \frac{d(E_k)}{ds} = \frac{d(-E_p)}{ds} \quad (6) $$

About integration:
Integration is a linear operation. If you take a curve, and you multiply that curve with a factor $a$ then the integral of that curve will be larger by that factor $a$.

$$ \int ax \ dx = a \int x \ dx $$

We take the property that is expressed in (6) and state it as integrals of the respective energies:

$$ \frac{d(\int E_k \ dt)}{ds} = \frac{d(\int -E_p \ dt)}{ds} \quad (7) $$

The property expressed by (6) propagates to (7) because integration is a linear operation.

We know from the derivation of the Euler-Lagrange equation that the derivative with respect to position and the derivative with respect to variation are the same, hence (7) is mathematically equivalent to stating Hamilton's stationary action.

$\endgroup$
2
  • 1
    $\begingroup$ "If the curve satisfies the extremum condition globally then every subsection of the trajectory satisfies the extremum condition too, down to infinitissimally short subsections." However, the converse does not hold. A geodesic is locally the shortest distance curve, but it can globally fail to be the shortest curve. For instance, given two non-antipodal points on Earth's equator, there are two paths along the equator between them, and they are both locally minimal, but only one is globally minimal. $\endgroup$ May 2 at 1:02
  • 1
    $\begingroup$ @Acccumulation I deleted an earlier version of this comment. You raise an interesting point. When I wrote my answer I was careful to switch to 'extremum' immediately. As soon as the discussion moves to generalize the following becomes acute: variation can identify an extremum, but cannot by itself identify whether the extremum is a minimum or a maximum. The case of the non-antipodal points is an instance of that. As you point out: the longest path has the following property: for small variation the derivative of pathlength with respect to the variation is zero. $\endgroup$
    – Cleonis
    May 2 at 6:09
8
$\begingroup$

You shouldn't think of the principle of least action as something a particle "does." It is one of several mathematical representations of the laws of motion.

The way locality / the finiteness of the speed of light emerges from Lagrangian mechanics, is that the classical path you find by minimizing the action will always involve particles traveling at or below the speed of light (for "standard" Lagrangians). One way to prove this will happen within the Lagrangian framework is to consider massless particles, which will follow null geodesics, or paths which minimize the spacetime distance between two spacetime events, and null geodesics follow paths through spacetime which travel at the speed of light.

The modern explanation of the principle of least action, is in terms of the quantum mechanical path integral. Every possible way a particle can travel between two points has an associated probability amplitude, and the probability for the particle to move between the two points is the square of the absolute value of the sum of the amplitudes for every path. The contribution of typical paths will tend to cancel because they will have different phases. The path of least action is the path that contributes most to the sum (in the "classical limit") because nearby paths do not tend to cancel out. This corresponds to the classical path, which minimizes the action.

$\endgroup$
6
$\begingroup$

The two statements

  1. Principle of Least action states that the particle's trajectory under the influence of a potential is determined by minimising the action that is $\delta S =0$.
  2. The universal speed limit of information propagation is $c$.

are not in contradiction with each other. If you take the relativistic action of a massive particle to be $S = - m \int ds = - m \int \sqrt{1 - v^2} dt$, then particles move in straight lines which are slower than the speed of light. Because the actual particles themselves are always moving slower than $c$, the speed limit is never broken.

$\endgroup$
5
$\begingroup$

There are typically an infinite number of solutions to Lagrange equations, corresponding to one starting point and an infinity of end points. If the system starts out on one trajectory, its end point is determined. The system does not "know" in advance what its endpoint is, but the physicist who has advance knowledge of the beginning trajectory and the Lagrangian (which "knows" everything about what's between the starting point and all possible trajectories) can predict the rest of the trajectory and the endpoint.

$\endgroup$
2
$\begingroup$

Obviously, a particle doesn't go ahead in time to explore all possible trajectories, compute the actions for each possible path, and chooses the one for which the action is least. How could a particle do that? It's rather us who do the calculations and choose the one with the least action to arrive at the trajectory the particle will take. For a particle, we choose a bunch of trajectories with fixed endpoints in spacetime (in the time of Lagrange these included trajectories on which the particle moves faster than light, but in our time these must be excluded). It's us who choose the initial and final points in spacetime. And it's us who imagine all the intermediate trajectories (globality). The actual trajectory (locality) taken by the particle is the one for which the action is least. When no force is present the trajectory will be straight while if a force is present, the trajectory will be curved.

While it's us who have chosen the initial and final fixed points, the particle will actually move from an initial point (the same as we have chosen) to a final point (the same as we have chosen) on a trajectory with the least action if it has the right velocities on its trajectory. We impose the velocities which the particle has on its trajectory by choosing the initial and final (fixed) points and demanding that the action must be the least. But the particle doesn't know in advance what velocities it should have on the trajectory to make the action the least. The particle doesn't actually travel along all possible trajectories (including faster than light trajectories, which could be considered in the time of Lagrange, but which indeed have to be excluded in the modern view, though there is a subtlety lurking in the modern view which I'll explain in the last paragraph).

So, a particle doesn't know in advance what velocities it should have along the trajectory it takes. It's rather us who impose these velocities by choosing the fixed points and demanding for least action. If, conversely, we choose the fixed points to be on an actual path (so without varying the path) we'll find that the action is a minimum for this path.

In the realm of quantum field theory though, the problem becomes more complicated. In QFT, a particle actually moves on each trajectory between two fixed points in spacetime at once (which confuses the idea of a particle moving on a fixed trajectory; all these trajectories, and thus the particle, are said to be an excitation of the particle field). So not only in the imagination for calculating the one with least action. Even on trajectories for which their speed is higher than the speed of light (off-shell trajectories). And even on trajectories that go backward in time (in which case the anti-particle comes in view). But not to find out which path makes the action the least (the paths are just there). Each path has an associated probability amplitude, which goes to zero very fast if the trajectory diverges from the one for which the action is the least. The actual path we observe is the one for which the action is stationary. The wave function of the particle can be seen as the combined probability amplitudes of all paths.

$\endgroup$