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

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?

• 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. Nov 2, 2010 at 20:40
• 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. Mar 22, 2018 at 18:29

The notes from week 1 of John Baez's course in Lagrangian mechanics 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 $$\,-ma\,$$, 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.

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 Feynman's Thesis — A New Approach to Quantum Theory or A call to action, by Edwin F. Taylor.

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. :-)

• 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. Jan 18, 2011 at 22:31
• But why is the least stationary action principle valid in quantum theory? That is even harder question to answer. Explaining Newtonian mechanics as "it is a limit of quantum theory" is not very pedagogical. May 21, 2021 at 12:59
• Because of the stationary phase approximation. May 21, 2021 at 13:51

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.

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.

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.

• 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. Oct 18, 2015 at 7:41
• 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. Oct 18, 2015 at 16:44

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.

• 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. Nov 1, 2012 at 6:34
• Amazingly, there is now a least action principle operating only with the initial conditions! prl.aps.org/abstract/PRL/v110/i17/e174301 Apr 23, 2013 at 7:52
• 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. Jun 15, 2013 at 22:07
• > "it is implied that the equations have physical solutions. This is so in the Classical Mechanics but is wrong in the Classical Electrodynamics" Classical mechanics has non-physical solutions too. With the right system and initial conditions, one can break determinism (Norton's dome) and the (expected from thermodynamics) Gibbs entropy non-decreasing in time in isolated systems. It seems physical solutions are always chosen by physicists, our theories always allow also for unphysical ones. May 21, 2021 at 13:06

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: $\textbf{What path does the particle take?}$

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 $S$:

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.

• 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. Aug 27, 2018 at 5:25

It is possible in classical physics to derive the Euler-Lagrange equations from D'Alembert principle, without any reference to the notion of action. They come from Newton's laws with the additional assumption that the forces are conservative. In this case there is a Lagrangian, and the equation of movement (EOM) is the Euler-Lagrange equation.

Suppose that a function q(t) is a solution for the EOM in a certain interval. q can be expanded as a Taylor series, that is a power series: $$q(t) = \sum_j a_jt^j$$.

The action is: $$S(L) = \int_{t1}^{t2} Ldt$$ where L is the Lagrangian that corresponds to the EOM. Because the integral is in $$t$$, and we are taking the derivative with respect to the coeficients $$a_j$$, it can go inside the integral. For each $$a_j$$. $$\frac{\partial S}{\partial a_j} = \int_{t1}^{t2} \frac{\partial L}{\partial a_j}dt$$

L is a function of $$q$$ and $$\dot q$$, so applying the chain rule:

$$\frac{\partial L}{\partial a_j} = \frac{\partial L}{\partial q}\frac{\partial q}{\partial a_j} + \frac{\partial L}{\partial \dot q}\frac{\partial \dot q}{\partial a_j}$$

Integrating this differential between 2 instants of time: $$\int_{t1}^{t2}\frac{\partial L}{\partial a_j}dt = \int_{t1}^{t2}\frac{\partial L}{\partial q}\frac{\partial q}{\partial a_j}dt + \int_{t1}^{t2}\frac{\partial L}{\partial \dot q}\frac{\partial \dot q}{\partial a_j}dt$$

The last term can be separated using integral by parts, using that differentiating with respect to time: $$d\left (\frac{\partial q}{\partial a_j}\right ) = \frac{\partial \dot q}{\partial a_j} dt$$: $$\int_{t1}^{t2}\frac{\partial L}{\partial \dot q}\frac{\partial \dot q}{\partial a_j}dt = \frac{\partial L}{\partial \dot q}\frac{\partial q}{\partial a_j}\bigg|_{t1}^{t2} - \int_{t1}^{t2}\frac {\partial \frac {\partial L}{\partial \dot q}}{\partial t} \frac{\partial q}{\partial a_j}dt$$

So: $$\int_{t1}^{t2}\frac{\partial L}{\partial a_j}dt = \int_{t1}^{t2}\frac{\partial L}{\partial q}\frac{\partial q}{\partial a_j}dt - \int_{t1}^{t2}\frac {\partial \frac {\partial L}{\partial \dot q}}{\partial t} \frac{\partial q}{\partial a_j}dt + \frac{\partial L}{\partial \dot q}\frac{\partial q}{\partial a_j}\bigg|_{t1}^{t2}$$

Joining the integrals, we get between parentheses the Euler-Lagrange equation, that is the EOM itself! If q is solution by hypothesis, this integral must be zero.

$$\int_{t1}^{t2}\frac{\partial L}{\partial a_j}dt = \int_{t1}^{t2}\left (\frac{\partial L}{\partial q} - \frac {\partial \frac {\partial L}{\partial \dot q}}{\partial t} \right) \frac{\partial q}{\partial a_j}dt + \frac{\partial L}{\partial \dot q}\frac{\partial q}{\partial a_j}\bigg|_{t1}^{t2}$$

For the last term, the second order integral needs 2 boundary conditions. If $$q(t_1)$$ and $$q(t_2)$$ are known, they are fixed and $$\frac{\partial q}{\partial a_j}\bigg|_{t1} = \frac{\partial q}{\partial a_j}\bigg|_{t2} = 0 \implies$$ this term vanishes.

Now, we get to the conclusion that the derivative of the action with respect to all coeficients must be zero in the interval, what is the same as to say that the action must be stationary.

• Yeah, derive from classical. The thing is, d'Alembert's virtual work is a non-transparent way of stating the work-energy theorem. Which is why today's books present the work-energy theorem, not d'Alembert. We have that the work-energy theorem is another way of stating $F=ma$ Demonstration: work-energy theorem: $$\int_{s_0}^s F\ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2$$ make initial velocity zero. Take the derivative with respect to position: $$\frac{d(\tfrac{1}{2}mv^2)}{ds} = ma$$ Hamilton's stationary action works because it is mathematically equivalent to the work-energy theorem. Apr 26, 2021 at 6:09
• @Cleonis, D'Alembert principle is not a statement about work or energy as in work-energy theorem. It is a reformulation of Newtonian equations of motion. This is equivalent to those equations and thus the principle is more powerful than the work-energy theorem is. May 21, 2021 at 13:21
• @JánLalinský The work-energy theorem is mathematically equivalent to $F=ma$ Take the derivative with respect to position and you recover $F=ma$ $$\frac{\int F \ ds}{ds} = F$$ and $$\frac{d(\tfrac{1}{2}mv^2)}{ds} = \tfrac{1}{2}m\left( 2v\frac{dv}{ds} \right) = m\frac{ds}{dt}\frac{dv}{ds} = m\frac{dv}{dt} = ma$$ May 21, 2021 at 13:47
• @Cleonis your calculation gives the same equation as Newton's laws do only for linear motion. If the particle moves along a curved trajectory, $dE_k/ds$ only gives us the tangential component of force(in direction of velocity), not the total force. For example, you won't get centripetal component of force during simple circular motion this way. The work-energy theorem can give us only that part of force that changes kinetic energy, it can't give us the normal component, hence it cannot give us the all the equations that Newton's laws do. May 21, 2021 at 17:47
• @JánLalinský When the force vector and the velocity vector are not aligned you use the vector dot product. Check out the following OpenCoursesWare PDF: Module 28: the Kepler Problem: planetary mechanics The motion is described in terms of polar coordinates. From section 28.6 on the treatment demonstrates energy evaluation can be used to solve the Kepler problem. General solution: it includes eccentric orbits. May 21, 2021 at 19:26