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In classical mechanics the dynamics on a manifold $M$ are characterised by the minimisation of a functional $$ \min_{q \in C^\infty(\mathbb{R},M)} \int_{\mathbb{R}}L(q(t),\dot{q}(t))dt, $$ where $L:TM\to\mathbb{R}$ is a smooth function, the "Lagrange function". In the case of a particle of mass $m$ constrained to a submanifold $M\subseteq \mathbb{R^n}$ moving in a potential $V: M \to \mathbb{R}$ the Lagrangian is given by $L(x,v):=\frac{1}{2}m\|v\|^2 - V(x),$ so the functional to minimise (the "action") is $$ C^\infty(\mathbb{R},M)\ni q \mapsto\int_{\mathbb{R}}\left( \frac{1}{2}m\|\dot{q}(t)\|^2 - V(q(t)) \right)dt \in \mathbb{R}.$$ This is all standard in physics and many "derivations" of this from variants of Newton's second law on $M$ are given in the physics literature. What I've never seen attempted is an explanation of this: What is the intuition behind formulating the dynamics in terms of a variational principle? Are there any heuristics on the interpretation of the action?

It seems that the action is some sort of "currency": Every path in $M$ comes at a cost and nature somehow chooses that path that costs the least amount. But how to interpret this amount? Why should the currency take the form:

$$\text{"kinetic energy} - \text{potential energy"} ?$$

EDIT 1: I'm not looking for a "historical explanation" on how this principle was discovered. I'm more looking for a direct interpretation of the action if this makes sense.

EDIT 2: As pointed out by arsmath the dynamics needn't be characterised by a minimum of the action, it can also happen that merely the derivative of the action vanishes and we have a stationary point (or a maximum!).

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    $\begingroup$ I think Fermat's principle that light follows the path of least time came first, and then various authors produced increasingly refined versions to be applicable to more situations, ending with "Hamilton's principle". $\endgroup$ – Robert Furber Apr 5 at 10:43
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    $\begingroup$ I think you mean Newton's 2nd law rather than 3rd law. The 2nd law tells you how motion of a particle is related to the potential it sees, which is what you're discussing. The 3rd law is actio=reactio - it ties together what different objects in a system experience when they interact with each other. (And the 3rd law breaks down once the objects interact, e.g., via magnetic forces) $\endgroup$ – Michael Engelhardt Apr 5 at 13:34
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    $\begingroup$ @Nemo I don't understand why you deleted your answer? $\endgroup$ – Jannik Pitt Apr 6 at 20:51
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    $\begingroup$ I don't think this is detailed enough for an answer, but here is my rough intuition as a physicist. The quantity given by the Lagrangian function evaluated on the (classical) trajectory of a system $\dot{S}(t) := L(q(t), dq(t)/dt,t)$ has units of energy, but it is perhaps best thought as a measure of the mechanical activity of the system at time $t$. A ball sitting on top of a hill is intuitively less active that the same ball rolling down the hill, so it makes sense that kinetic energy contributes positively and potential energy contributes negatively to the activity. $\endgroup$ – pregunton Apr 7 at 9:51
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    $\begingroup$ The trajectory's action $S = \int_{t_i}^{t_f} \dot{S}(t) dt$ is the time integral of activity; thus a trajectory with minimal action is roughly one where the system maximizes the amount of time being inactive and minimizes the time being active given the appropriate constraints (e.g. in a pendulum, the bob spends the most time near the points with the highest potential energy). Of course this isn't applicable to every system, but it may help to understand what is happening in some simple examples. $\endgroup$ – pregunton Apr 7 at 9:52
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It's not the case that the action is always minimized in physics -- the result is purely that it's a stationary point of the action. This has come up several times at Physics Stack Exchange:

This raises the question of "Why a stationary point?", which Carlo's answer explains, but the examples above undermine the "currency" implication.

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    $\begingroup$ Yes of course you're right, I knew that but somehow didn't think of these cases when writing the question. I'll edit it. $\endgroup$ – Jannik Pitt Apr 5 at 12:31
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Even though the historical order is the other way around, it is helpful to start from wave/quantum mechanics and arrive at classical mechanics in the limit that the wave length of the particle goes to zero. Mathematically, that limit is the stationary phase approximation, meaning that classical trajectories are perpendicular to surfaces of constant phase. The phase is the action = integral of Lagrangian $L$, as first realized by Dirac, and as follows immediately by calculating the phase $\phi$ accumulated in a time $T$, $$\phi=\int_0^T(p\dot{q}-H)\,dt=\int_0^T L(q,\dot{q})\,dt.$$ From this equation you see that the Lagrangian is "kinetic minus potential energy" because it is the difference of $p\dot{q}$ = twice the kinetic energy and the sum of kinetic and potential energy (the Hamiltonian $H$). In this way stationary phase amounts to stationarity of the action.

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  • $\begingroup$ (Out of what hypotheses does it “follow immediately”?) $\endgroup$ – Francois Ziegler Apr 6 at 11:11
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    $\begingroup$ well, once you know that the phase increment of a plane wave (wave number $k$ or $p$, frequency $\omega$ or $H$) is $d\phi=kdx-\omega dt$, the integral follows for an arbitrary wave front --- I understood the question of the OP as a request for "intuitive understanding", and I hold these notions of wave fronts to be intuitively obvious. $\endgroup$ – Carlo Beenakker Apr 6 at 11:25
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    $\begingroup$ @CarloBeenakker: I fear that this answer raises more questions than it answers. In particular, it moves the core of the OP's question from "why stationary points of the action" to "why stationary phase"? $\endgroup$ – Alex M. Apr 6 at 13:43
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    $\begingroup$ why stationary phase? because classical mechanics is the short-wavelength limit (= rapid phase variation limit) of quantum mechanics; since the OP is a student of quantum mechanics, I figured it might be helpful to appeal to the intuition of wave interference, also because the OP explicitly said they were not looking for a derivation but for a heuristic. $\endgroup$ – Carlo Beenakker Apr 6 at 14:30
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While the following may be less “physically” intuitive than the question intends, I liked it when I heard it. If you believe in the importance of Poisson brackets, it is natural to ask if they make the Lie algebra of some transformation group, or in other words, if they can be realized as Lie brackets of vector fields.

This question was answered in the 1960s by “prequantization” of symplectic manifolds, but in the case of $X=\smash{\mathbf R^2}$ with points $x=(p,q)$ and 2-form $\omega=dp\wedge dq$, already by Sophus Lie in (1890, p. 270): one considers $L=X\times\mathbf U(1)$ with points $ξ=(x,z)$, projection $ξ\mapsto x$ and connection (contact) 1-form $\varpi = p\,dq + dz/iz$. Then one checks that an automorphism $g\in\operatorname{Aut}(L,\varpi)$, i.e. a diffeomorphism such that $g^*\varpi = \varpi$, must have the form $$ g(x,z) = (s(x),ze^{iS(x)}) \tag1 $$ where $s\in\operatorname{Aut}(X,\omega)$ is a symplectomorphism and the function $S$ is determined up to an additive constant1) by $$ p\,dq-s^*(p\,dq)=dS. \tag2 $$ Likewise the Lie algebra $\operatorname{aut}(L,\varpi)$ is isomorphic to $(C^∞(X), \{\cdot,\cdot\})$: to any $\varpi$-preserving vector field $Z$ we can attach the function $H(x) = \varpi(Z(ξ))$ called its Hamiltonian, and conversely any $H ∈ C^∞(X)$ gives rise to the infinitesimal automorphism $$ Z(x,z)= (\operatorname{drag}H(x),iz\ell(x)) \tag3 $$ where $\operatorname{drag}H=\smash{\bigl(-\frac{\partial H}{\partial q},\frac{\partial H}{\partial p}\bigr)}$ is the symplectic gradient, and $\ell= p\frac{\partial H}{\partial p}-H$ is the Lagrangian for you. This is the only “derivation of it from something else” I ever liked. Opinions may vary :-)

Now paths $t\mapsto x$ in $X$ “lift” to paths in $L$ where $z$ spins around the “internal clock” $\mathbf U(1)$ so eloquently described in Feynman’s QED, evolving with $S=\int\ell\,dt$. “Wave functions” live in the complex plane it spans, and constructive interference of the waves corresponds to stationary $S$.

Added: At a more classical level, I can only recommend reading the 18-page Introduction of Souriau (1997, French version) where he explains from first Newtonian (or really D’Alembert) principles, not the Poincaré-Cartan 1-form $p\,dq - H\,dt=(p\dot q-H)dt=\ell\,dt$ but the Lagrange 2-form $dp\wedge dq - dH\wedge dt$ and generalizations which are (he argues) the only intrinsic objects worth talking about at non-quantum level. At (12.98) he explains why he abandoned the variational approach of a previous book — just as Lagrange also did 200 years earlier. Worth pondering?


1) That is to say, one has a central extension $1\longrightarrow\mathbf U(1)\longrightarrow\operatorname{Aut}(L,\varpi)\longrightarrow\operatorname{Aut}(X,\omega)\longrightarrow 1$ which “integrates” $0\longrightarrow\mathbf R\longrightarrow C^∞(X)\overset{\operatorname{drag}}{\longrightarrow}\operatorname{aut}(X,\omega)\longrightarrow 0$.

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    $\begingroup$ Indeed, there is nothing particularly fundamental about the combination (kinetic-potential) energy. This is what the Lagrangean happens to be if the kinetic energy happens to be quadratic in $p$. In general, e.g., for a relativistic particle, the Lagrangean doesn't have the (kinetic-potential) energy form. $\endgroup$ – Michael Engelhardt Apr 5 at 13:42
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In a way, the action is really best understood in a quantum-theoretical context, where it naturally emerges from Feynman's principle: essentially, one evaluates matrix elements of the time evolution operator $U(t',t)={\rm e}^{-i\hat{H}(t'-t)}$ with $\hat{H}=T(\hat{p})+V(\hat{q})$ by applying the Lie-Trotter formula, inserting complete sets of eigenstates of $\hat{q}$ and $\hat{p}$, and integrating over the eigenvalues of $\hat{p}$, which leaves one with a time-discretized path integral containing (a discretization of) $\int L(q,\dot{q})\,{\rm d} t$, and in the limit one obtains the usual Feynman path integral involving the action.

There is a natural connection from this to the principle of least action by considering quantization as the inverse (in some sense) of tropicalization: Consider $M$ as a category with points as objects and paths as morphisms, and consider dynamics as coming from a functor from $M$ into a rig; where quantum theory assigns each path an amplitude $U\in\mathbb{C}$ such that concatenation of paths corresponds to multiplication of amplitudes and multiple paths between the same points interfere via addition, the classical theory assigns each path an action $S\propto\log U\in\mathbb{R}$ such that concatenation of paths corresponds to addition and multiple paths between the same points interfere via taking the minimum. There is a body of work along those lines by John Baez (cf. these lecture notes) and others.

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I think it is more transparent to answer your question in the Hamiltonian formulation of mechanics rather than the Lagrangian one. In the nondegenerate case, that is, where the Legendre transform or fibrewise derivative is a diffeomorphism, they coincide anyway. Let $(M,d\alpha)$ be an exact symplectic manifold. For $-\infty < t_0 < t_1 < +\infty$ and $x_0,x_1 \in M$ define $$C^\infty_{x_0,x_1}([t_0,t_1],M) := \{\gamma \in C^\infty([t_0,t_1],M) : \gamma(t_0) = x_0 \text{ and } \gamma(t_1) = x_1\}$$ as well as the Hamiltonian action functional $$\mathcal{A}^H \colon C^\infty_{x_0,x_1}([t_0,t_1],M) \to \mathbb{R}, \qquad \mathcal{A}^H(\gamma) := \int_{\gamma}\alpha - \int_{t_0}^{t_1}H \circ \gamma.$$ Then the critical points of $\mathcal{A}^H$ coincide with the integral curves of the Hamiltonian vector field $X_H$. As was already correctly noted in the comments, we are interested in critical points rather than minima, as they do not necessarily exist. Thus to answer your question:

The Principle of least action is an equivalent way to encode the dynamics of a Lagrangian or Hamiltonian system.

Note that the terminology "principle of least action" is a bit unfortunate.

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  • $\begingroup$ Thanks, to me this is the only direct answer to the question so far. $\endgroup$ – Christian Chapman Apr 23 at 0:11
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One could also mention the Special and General Relativity. The postulate is that the dynamic of a particle is a geodesic. More precisely it minimizes the quantity $$ A=\int mc\sqrt{ds^2} $$ where $$ ds^2 = (1+\frac{2V}{mc^2})c^2dt^2 - dx^2-dy^2-dz^2 $$ with $V$ is the potential. See for example the Schwarzschild metric ( https://en.wikipedia.org/wiki/Schwarzschild_metric), where the Newton potential appears and we neglected the modifications on the space which are of smaller order for large $c$. Then $$ A= \int mc\sqrt{c^2+\frac{2V}{m}-\frac{dx^2}{dt^2}-\frac{dy^2}{dt^2}-\frac{dz^2}{dt^2}}dt $$ and for large $c$ $$A=\int mc^2dt -\int \Big(\frac{1}{2}m(\frac{dx^2}{dt^2}+\frac{dy^2}{dt^2}+\frac{dz^2}{dt^2})-V\Big)dt+\mathcal{O}(\frac{1}{c^2})$$ Therefore, the action can be seen as the first order of the path length.

EDIT: True we start from a minimising principle (geodesic) to obtain another minimizing principle (with the Lagragian), however the first one is somehow more fundamental than the second one.

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If you're interested in quantum-mechanical motivations/generalizations of results in classical mechanics, we can give one in particular for the given action. The transition probability amplitude between two states over time is proportional to $\exp\frac{i\delta S}{\hbar}$, where $S$ turns out to be the usual action. In particular, the exponent is a definite integral; the interval of time over which we integrate $L$ to give $S$ is the initial-to-final-state time period. We call this $S$ a path integral. Verifying $L=\frac12m\dot{x}^2-V(x)$ is a fairly involved calculation, explored at the linked section. The article as a whole shows the path integral formulation is equivalent to the Schrödinger equation; the $SE\implies PI$ part gives the form of $L$.

In quantum field theory, more general path integrals arise, but they too give rise to Euler-Lagrange equations. The reason is that if each transition has an amplitude $\propto\exp\frac{i\delta S}{\hbar}$, interference between $\delta S\ne0$ states is destructive, which is why we observe $\delta S=0$ states. (Note this is a motivation for stationary actions, not minimized ones.) This is related to $\int_{\Bbb R}\exp ikxdx=2\pi\delta(k)$.

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