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Hamilton's principle states that a dynamic system always follows a path such that its action integral is stationary (that is, maximum or minimum).

Why should the action integral be stationary? On what basis did Hamilton state this principle?

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    $\begingroup$ It should be noted that this is "Hamilton's principle", that is is not exactly the same as "Hamiltonian [classical] mechanics" (ie, where an actual Hamiltonian is involved) and that is as nothing specific about QM. $\endgroup$ – Cedric H. Nov 2 '10 at 20:40
  • $\begingroup$ In the Euler Lagrange equations. The neccesary condition L to be an extremal point is it satisfies the EL eqs. So Hamilton's principle is not actually a principle. You can think in QED, in QM,... but it is just because a mathematical reason. $\endgroup$ – Dog_69 Mar 22 '18 at 18:29
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The notes from week 1 of John Baez's course in Lagrangian mechanics (http://math.ucr.edu/home/baez/classical/#lagrangian) give some insight into the motivations for action principles.

The idea is that least action might be considered an extension of the principle of virtual work. When an object is in equilibrium, it takes zero work to make an arbitrary small displacement on it. I.E. the dot product of any small displacement vector and the force is zero (in this case because the force itself is zero).

When an object is accelerating, if we add in an "inertial force" equal to -m*a, then a small, arbitrary, time-dependent displacement from the objects true trajectory would again have zero dot product with F-ma, the true force and inertial force added. This gives

$(F-ma)\cdot \delta q(t) = 0$

From there, a few calculations found in the notes lead to the stationary action integral.

Baez discusses D'Alembert more than Hamilton, but either way it's an interesting look into the origins of the idea.

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There is also Feynman's approach, i.e. least action is true classically just because it is true quantum mechanically, and classical physics is best considered as an approximation to the underlying quantum approach. See http://www.worldscibooks.com/physics/5852.html or http://www.eftaylor.com/pub/call_action.html .

Basically, the whole thing is summarized in a nutshell in Richard P. Feynman, The Feynman Lectures on Physics (Addison–Wesley, Reading, MA, 1964), Vol. II, Chap. 19. (I think, please correct me if I'm wrong here). The fundamental idea is that the action integral defines the quantum mechanical amplitude for the position of the particle, and the amplitude is stable to interference effects (-->has nonzero probability of occurrence) only at extrema or saddle points of the action integral. The particle really does explore all alternative paths probabilistically.

You likely want to read Feynman's Lectures on Physics anyway, so you might as well start now. :-)

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    $\begingroup$ Feynman's Lectures on Physics are good, but best read after to have properly learnt the subject, in order to provide new/further insight, I feel. $\endgroup$ – Noldorin Jan 18 '11 at 22:31
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As you can see from the image below, you want the variation of the action integral to be a minimum, therefore $\displaystyle \frac{\delta S}{\delta q}$ must be $0$. Otherwise, you are not taking the true path between $q_{t_{1}}$ and $q_{t_{2}}$ but a slightly longer path. However, even following $\delta S=0$, as you know, you might end up with another extremum.

alt text

Following the link from j.c., you can find On a General Method on Dynamics, which probably answers your question regarding Hamilton's reasoning. I haven't read it but almost surely it is worthwhile.

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    $\begingroup$ This seems like a tautological answer as it is precisely Hamilton's principle which is used to arrive at the above picture in the first place. $\endgroup$ – Casimir Oct 18 '15 at 7:41
  • $\begingroup$ Maybe you were taught Hamilton's principle and arrived at that picture as an explanation, but the picture is perfectly general. It describes the variation of a function with fixed end points. $\endgroup$ – Robert Smith Oct 18 '15 at 16:44
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I generally tell the story that the action principle is another way of getting at the same differential equations -- so at the level of mechanics, the two are equivalent. However, when it comes to quantum field theory, the description in terms of path integrals over the exponentiated action is essential when considering instanton effects. So eventually one finds that the formulation in terms of actions is more fundamental, and more physically sound.

But still, people don't have a "feel" for action the way they have a feel for energy.

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Let us remember that the equations of motion with initial conditions $q(0), (dq/dt)(0)$ were advanced first and the least action principle was formulated later, as a sequence. Although beautiful and elegant mathematically, the least action principle uses some future, "boundary" condition $q(t_2)$, which is unknown physically. There is no least action principle operating only with the initial conditions.

Moreover, it is implied that the equations have physical solutions. This is so in the Classical Mechanics but is wrong in the Classical Electrodynamics. So, even derived from formally correct "principle", the equations may be wrong on physical and mathematical level. In this respect, formulating the right physical equations is a more fundamental task for physicists than relying on some "principle" of obtaining equations "automatically". It is we physicists who are responsible for correctly formulating equations.

In CED, QED, and QFT one has to "repair on go" the wrong solutions just because the physics was guessed and initially implemented incorrectly.

P.S. I would like to show how in reality the system "chooses" its trajectory: if at $t = 0$ the particle has a momentum $p(t)$, then at the next time $t+dt$ it has the momentum $p(t) + F(t)\cdot dt$. This increment is quite local in time, it is determined by the present force value $F(t)$ so no future "boundary" condition can determine it. The trajectory is not "chosen" from virtual ones; it is "drawn" by the instant values of force, coordinate, and velocity.

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  • $\begingroup$ I like to think that both options are merely mathematical models and so none is more real. Neither the system chooses its trajectory nor the future determines the least action path. The non-locality of QM leads to similar doubts. $\endgroup$ – Eduardo Guerras Valera Nov 1 '12 at 6:34
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    $\begingroup$ Amazingly, there is now a least action principle operating only with the initial conditions! prl.aps.org/abstract/PRL/v110/i17/e174301 $\endgroup$ – wnoise Apr 23 '13 at 7:52
  • $\begingroup$ Here is a free arXiv version. Without reading the article in detail, it smells like a classical Keldysh formalism, cf. this and this Phys.SE posts. $\endgroup$ – Qmechanic Jun 15 '13 at 22:07
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Instead of specifying the initial position and momentum just like we have done in Newton's formalism, let’s reformulate our question as following:

If we choose to specify the initial and final positions: $\textbf{What path does the particle take?}$

enter image description here

Let's assert we can recover the Newton's formalism by the following formalism, so-called Lagrangian formalism or Hamiltonian principle.

To each path illstrated on above figure, we assign a number which we call the action

$$S[\vec{r}(t)] = \int_{t_1}^{t_2}dt \left(\dfrac{1}{2}m\dot{\vec{r}}^2-V(\vec{r})\right)$$

where this integrand is the difference between the kinetic energy and the potential energy.

$\textbf{Hamilton's principle claims}$: The true path taken by the particle is an extremum of S.

$\textbf{Proof:}$

1.Change the path slightly:

$$\vec{r}(t) \rightarrow \vec{r}(t) +\delta \vec{r}(t)$$

2.Keep the end points of the path fixed:

$$ \delta \vec{r}(t_1) = \delta \vec{r}(t_2) = 0 $$

3.Take the variation of the action $S$:

enter image description here

finally, you will get

$$ \delta S = \int_{t_1}^{t_2} \left[-m\ddot{\vec{r}} - \nabla V\right] \cdot \delta \vec{r} $$

The condition that the path we started with is an extremum of the action is

$$\delta S = 0$$

which should hold for all changes $\delta \vec{r}(t)$ that we make to the path.The only way this can happen is if the expression in $[\cdots]$ is zero. This means

$$ m\ddot{\vec{r}} = -\nabla V$$

Now we recognize this as $\textbf{Newton’s equations}$. Requiring that the action is extremized is equivalent to requiring that the path obeys Newton’s equations.

For more details you could read this pdf lecture.

Hope it helps.

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  • $\begingroup$ If we see a particle constrained to move on a sphere, we get to paths one is a maximum or a minimum. I feel a particle follows path of least action but the mathematical equation δS=0 does give us an ambiguous answer, but a certain part of this answer contains a path of least action in it. You can see Arfken and Weber. $\endgroup$ – Chetan Waghela Aug 27 '18 at 5:25

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