Least action principle is a lie

Ok, cool. Now that we have space to define things on, let's study it. A fundamental assumption from classical mechanics is that all systems are goverened by something called Lagragian. If you are like me and ever been remotely interested in classical mechanics, you probably know the drill.

Problem statement: the system evolves from state \(\vb{q_i} \in Q\) to \(\vb{q_f} \in Q\) within time interval \([t_i, t_f]\). We want to figure out how. Which path, i.e. a function \(\gamma: t \in [t_i, t_f] \to Q\) does it take?

Answer: to figure out the path, we exploit something that is called stationary action principle (more commonly, minimum action principle, I'll elaborate on that later). We define a functional called action \(S(\gamma) = \int\limits_{t_i}^{t_f} L(\gamma(t), \gamma'(t)) dt\) and claim that the system chooses a path, such that \(S\) is stationary to first order. This is something called Hamilton's principle and \(L\) is called Lagrangian.

Finding stationary points of a functional is quite tricky. To actually do that, we solve Euler-Lagrange (I'll just call them EL for brevity) equations. It's pretty much variational calculus 101, you can find the derivation here if you're interested. They look like:

\[\pdv{L}{\vb{q}} - \dv{}{t} \pdv{L}{\vb{v}} = 0\]

Ok, so far so good. If you see it for the first time it sort of blows your mind so you are curious what's next. After that, 99% of explanations are considering a classical particle in a potential well: \(L(\vb{q}, \vb{v}) = \frac{1}{2} m \vb{v} \cdot \vb{v} - V(\vb{q})\). Then they plug in the Euler-Lagrange equation and get Newthon's law of motion: \(\vb{F} = - \nabla V\). Yeah, awesome. The problem with those explanations is they stop there. So what? What's up with that weird thing which has units of energy but not quite? What is a Lagrangian? What it isn't? I was pretty comfy solving potential wells without having to minimize a functional, thank you very much.

A common viewpoint is a dogmatic one: Lagrangian is whatever function that corresponds to the actual physics observed in a system. For all practical purposes indeed you just guess a function, plug it into EL equations and if you got your desired laws of motion – you win. However, what if we want to reverse the process? As in, given a Lagrangian, can we come up with a universe where it would describe the physics? To do that, you of course need to figure out what a Lagrangian can be, that is, a definition. That turns out to be really not easy! The closest insights you get on this page are

Lagrangian is a function of the generalized coordinates, their time derivatives, and time, and contains the information about the dynamics of the system

and

the position coordinates and velocity components are all independent variables

I'm lucky to know Russian, but Russian wikipedia page is not any more helpful. You probably have a gut feeling that it's gotta at least be a smooth function or something like that, but it's not really mentioned anywhere. Okay, for now you decide that it must be any reasonble function of \(\vb{q}\), \(\vb{v}\) and \(t\). Actually, forget time dependency for now, it's tricky enough as it is.

So now that you established what's the object you are trying to study, you go through more examples. Next one is typically double pendulum or sliding pendulum. But yet again, Lagrangian is postulated to be the difference of kinetic and potential energy, which is not very convincing.

Next example is typically electromagnetism: \(L(\vb{q}, \vb{v}) = \frac{1}{2} m \vb{v} \cdot \vb{v} -e \phi(\vb{q}) + e \vb{v} \cdot \vb{A}\). Whoa guys, I'm still not done with simple systems! EM fields are scary, things spiral around and radiate, come on. Also EM fields are defined in three dimensions (if not 4), which would make the whole problem six-dimensional. It's a bit too much to begin with!

Ok fine, you think. I'll come up with my own Lagrangians and just see what happens. Let's assume that configuration space \(Q\) is 1D for now, so \(q\) and \(v\) are just real numbers.

1. If you have to come up with any real-valued function what would your first thought be? I bet it's zero identity, so \(L(q, v) = 0\). Nice, let's see what does that mean.

   Well, if you think about it, action is always zero, so any path would be stationary. This should probably ring an alarm but let's see what happens if we try to plug that in Hamilton's approach…

1. Okay, another attempt. \(L(q, v) = v^2\) is boring$, it's like typical example, but with no potential energy. Hmm… what if we instead we make the Lagrangian equal to potential energy only? \(L(q, v) = - q^2\).

So, if we plug \(L\) into EL equation \(\dv{}{t} \pdv{L}{v} = \pdv{L}{q}\), we get… \(0 = -2 q\).

Err, that's looks bad. Does it only work if \(q = 0\)? Well, that's not too surprising I guess since we are solving a boundary value problem, and generally existence and uniqueness are tricky for them. Anyway, \(q(t) \equiv 0\), fine! Let's look at this solution then. First thing we notice is that it corresponds to a non-moving harmonic oscillator. Ok, that's a reasonable solution and makes physical sense. Now, if we look closely, the Lagrangian is actually taking its maximum value when \(q = 0\). Taking literally any other path than staying at \(0\) would decrease the value of Lagrangian, and hence the total action. So we actually ended up at a global maximum of this functional. Which is consistent with Hamilton's principle, since it needn't be a local/global minima, but a stationary point. I just found it interesting that such a simple example contradicts the 'minimum action' formulation yet it's so common. But maybe, this is because the who situation is kind of artifical, unphysical and not interesting? Or one could argue that we could have inverted the sign, defined \(L(q, v) = q^2\) and it would be minimum. So let's explore more.

One mystery is that we know that the harmonic oscillator admits a solution that… well, oscillates. And our lagrangian is just like harmonic oscillator except for kinetic term. So what happened to oscillating solutions here?

To investigate that and try to get some intuition, let's go back to the following 'standard' Lagrangian for harmonic oscillator \(L_{m, k}(q, v) = \frac{1}{2} m v \cdot v - \frac{k}{2} q \cdot q\). The Lagrangian of our interest is just like that but with stiffness fixed at \(k=1\) and mass equal to zero! So for \(L_{m, k}\), the E-L equation boils down to just \(m \ddot{q} = -k q\). We are ruling out \(q \equiv 0\) solution for now, so this must a periodic motion with frequency \(\omega = \sqrt{\frac{k}{m}}\) and the law of motion \(x(t) = A sin (\omega t + \phi)\)

What happens to a particle on a spring if we reduce its mass while keeping the stiffness fixed? The frequency goes up. And we can't take the limit \(m \to 0\), since \(\sin \frac{1}{m}\) is discontinuous at \(0\). So I guess intuitivel that's how we lose the 'oscillating' solution.

1. Now let's get back to the \(q \equiv 0\) solution for \(L_{m, k}\). Now that Lagrangian has got a velocity-dependent term in it, it's not that easy to classify what kind of stationary point in it with respect to the action. What we can tell for sure though, it definitely will not be a minimum!

To see that intuitively, just notice that the potential part of action proportional to \(q^2\) gets its contribution from the path's amplitude, whereas the kinetic part proportional to \(v^2\) gets it from the amplitude of path's time derivative. So, given a path \(\gamma(t)\), within the interval \([t_i, t_f]\), we know that \(\int\limits_{t_i}^{t_f} L_{m, k}(q, v) \ge -\frac{k}{2} \left( \max\limits_{t \in [t_i, t_f]} \abs{\gamma(t)} \right)^2\). The interesting thing is we can always increase the impact of kinetic term while keeping the function's maximum same by wiggling \(\gamma(t)\) it more frequently. With enough wiggling, you can make the action strictly positive. To be more specific: suppose we are considering the time interval \([0, T]\) and the variation \(v(t) = \sin 2 \pi n \frac{t}{T}\).

More rigorously: for any natural \(n\), \(z(t)\) is compatible with the boundary conditions \(z(0) = 0, z(T) = 0\). Now, if you compute the action:

, we get: \(\frac{4 \, \pi^{3} m n^{3} - \pi T^{2} k n + {\left(2 \, {\left(4 \, \pi^{2} m n^{2} + T^{2} k\right)} \cos\left(\pi n\right)^{3} - {\left(4 \, \pi^{2} m n^{2} + T^{2} k\right)} \cos\left(\pi n\right)\right)} \sin\left(\pi n\right)}{4 \, \pi T n}\), so you can see that for any fixed \(m, k, T\), we can choose \(n\) ('wiggling') big enough to make the action positive. Once again, that means that in this (perfectly physical!) case, the action is not a minimum anymore. I find that pretty interesting.

Yet again, one could argue that the oscillator that stands still is an extreme and boring case. Perhaps what we normally think of as the trajectories of a pendulum (i.e. periodically swinging) are doing the least action thing? So let's make things simpler and just take \(L(q, v) = \frac{1}{2} v^2 - \frac{1}{2} q^2\).

TODO link https://en.wikipedia.org/wiki/Harmonic_oscillator#Simple_harmonic_oscillator TODO spoiler should be greyish?

So in the spirit of Lagrangian formalism, we pose the following problem: the particle starts at \(q = q_0\) and ends at \(q = q_0\) (yes, same point) after time \(T = 2 \pi\) (spoiler: it better do because the frequency is \(1\) here). We want to know the equations of motion.

Now, you know it's gotta be some kind of periodic motion. However the question TODO TODO suppose you were solving the IVP: given \(q_0\) and \(v_0\), where does the particle end up at \(T = 2 \pi\)? So we know the solution is of form \(q(t) = A \cos(t + \phi)\), since \(q(0) = q_0\) and \(q'(0) = v_0\), we get: \(q_0 = A \cos \phi\) \(v_0 = -A \sin \phi\)

\(q(t) = q_0 \cos t + v_0 \sin t\)

Now if we consider \(q\) at a fixed time \(T\) as a function of \(v_0\), we get:

\(q_T(v_0) = q_0 \cos T + v_0 \sin T\), which is a linear function. So in general you can invert it and consider \(v_0\) as a function of \(q_T\), which would uniquely solve our BVP. However there is one special case: when \(\sin T = 0\), i.e. \(T = \pi k, k \in \bZ\). In that case \(q_T\) ends up being constant function \(q_T = q_0\), and can't be inverted! TODO ???? That means that the EL solution can't be global minimum??? TODO check that!

Now let's fix \(q_0\) and vary \(v_0\), so consider the function $E(v₀) = $ <TODO position of that particle after \(T\). Of course it will depend continously on \(v_0\), but it also won't be injective!

TODO demonstrate that?… TODO compute action to make sure TODO for pi, action is 0 to TODO just come up with smth negative and positive… that would prove the saddle point thing TODO positive – same argument as with zero solution? A bit more handwavy? TODO very negative – just stay at \(q_0\) and don't move

Let's try out some simple variations satisfying boundary conditions. First things I can think of are something periodic: \(z_1(t) = \sin^2(t)\) and something quadratic: \(z_2(t) = \pi^2 - (t - \pi)^2\).

As a result, we get: \(A(q) = 0\), \(A(q_1) = -243.73\dots \varepsilon^2\), \(A(q_2) = 0.78\dots \varepsilon^2\).

Few interesting observations:

1. \(A(q) = 0\), which means the action is independent of initial conditions. TODO that might be trivial consequence though??
2. \(A(q_1) > 0\) and \(A(q_2) < 0\), which implies \(A(q)\) is a stationary point.

In this case, you can't argue the motion is unphysical or trivial, but yet again, the solutions are stationary points!

So what does that mean? I've managed to do some fruitful googling.

TODO here the guy claims about small T in lagrangian approach. other than that it's only a saddle point TODO wonder still if the minimum possible velocity is the global minimum?

it even includes the harmonic oscillator as one of the answers; and the claim is that you've got to consider TODO blah blah

TODO maybe that's a demonstration Lagrangian approach doesn't make sense on large time scales, only locally? cause it's not clear which solution to choose

from a quick glance, conjugate points are something like opposing point on a sphere – one could intuitively expect that BVP would not give unique solutions for these.

'sufficiently long trajectories of solutions of Euler-Lagrange equations that are trajectories of the dynamical system cease to be the minimum'

TODO move questions about what's so special about stationary closer to feynman and baez

I'm not really criticising, I'm just hoping someone would come up with some intuition why would the action want to be stationary in a similar fashion.

TODO The functional S[x]=∫baL(t,x,x˙)dt, x(a)=A, x(b)=B must satisfy the following conditions in order to have a weak minimum for x=x(t):

The curve x(t) satisfies the Euler-Lagrange equation, namely it is an extremal, ∂x˙∂x˙L|x(t)>0, The interval [a,b] contains no points conjugate to a. The definition of conjugate points is in p.114.

TODO why can't a particle jump up high? TODO from the physical point of view why does it want to oscillate more often??

TODO consider if we start and ed in the rightmost position within a period timespan. Why this path is not stationary? TODO basically, because we could shift everything in the middle in both directions and decrease/increase the lagrangian?

Right. So we found out that not having terms dependent on \(v\) (which is kind of like limiting mass to \(0\)) can get us in trouble. Still unclear what's wrong with such a Lagrangian mathematically, as a function though! TODO do a better analysis of second variation?

TODO chap6.pdf – good remark 4 on page 7. considers harmonic oscillator and slight variation to reason the stationary function is not a maximum look at prob 6.6. – minimum of saddle?