I understand the lagrangian formulation of classical mechanics, to a degree. I can derive the Euler-Lagrange equations from the "least" action principle, and equivalently can determine the equations of motion from a given lagrangian. I can handle lagrangian exercises in textbooks with ease.

I don't quite grok it, though. If you erased it from my mind, I wouldn't be able to re-invent it. So I'm going back through it.

The way I understand the Euler-Lagrange equation is thus: In classical physics, by observation, there is a stationary quantity. This stationary quantity is called the "action", and it is the sum of energies over time (alternatively, a product of energy and time). Again by observation, energies can be calculated given the positions and velocities of all elements of the system. Call such a function $\mathcal{L}(x, \dot{x})$, treat it as a black box. Then we have $Action = \int dt \mathcal{L}(x, \dot{x})$. Making this stationary, we derive

$$\frac{\partial\mathcal{L}}{\partial x} = \frac{d}{dt}\frac{\partial\mathcal{L}}{\partial \dot{x}}$$

Informally, this says to me that the paths taken in actual physics are ones where small perturbations in a particle's position are canceled out precisely by the changes in velocity before and after that position (where the velocity changes are caused by the position perturbation). This has an almost economic elegance to it.

But it still doesn't completely make sense to me what a Lagrangian is. Its unit is energy, sure, but I also don't quite grok energy beyond the abstract. So I figured I'd play with a few simple Lagrangians, hoping to break the formulation and learn something from how the pieces fell. Take, for example, this trivial function:

$$\mathcal{L}(x, v) = x + v$$

It describes an unphysical world, surely. Energy is far from conserved. But I figured constructing a weird but simple lagrangian would give me insight as to the nature of the formulation. Let's derive the equations of motion:

$$\frac{\partial\mathcal{L}}{\partial x} = 1$$

$$\frac{\partial\mathcal{L}}{\partial v} = 1$$

$$\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial v} = 0$$

So the path taken is the path satisfying

$$1 = 0$$

...Huh. The laws of motion are unsatisfiable. I'm not sure how to take that.

What, precisely, went wrong here? I'm looking for a geometric or intuitive explanation -- the algebraic troubles I understand.

What sort of world was I trying to construct? Whence the contradiction?

More generally, when you are handed a Lagrangian, what does it really signify? I can integrate it to get the action (an abstract number assigned to a path) or I can plug it into the Euler-Lagrange equations to figure out motion, but what does it mean in its native form? How do I read a Lagrangian without twisting its arm?

  • 5
    $\begingroup$ well i know you didn't want to hear about your algebraic troubles, but adding a position and a velocity sure as hell won't get you an energy. $\endgroup$
    – wsc
    Jul 31, 2013 at 23:31
  • 3
    $\begingroup$ yeah, but a bad way to break things is to ignore physics. physics is dimensionally consistent. always. $\endgroup$
    – wsc
    Aug 1, 2013 at 0:56
  • 3
    $\begingroup$ That is an example of the well known fact that there are inconsistent Lagrangians. The easiest example is $L\propto x$. These action functionals don't have stationary points (or curves). It's like looking for the stationary point of $f(x)=x$ or $f(x)=e^x$. $\endgroup$ Aug 1, 2013 at 1:16
  • 1
    $\begingroup$ As far as we know there are classical and quantum (nonclassical) evolutions. Quantum evolutions don't make the action stationary, although the full action (almost) determine its evolution. If you wish it, you can ask as a separate question under what conditions a quantum evolution can be described by an inconsistent action. This system, if it exists, doesn't have a classical limit, as most of the systems do. $\endgroup$ Aug 1, 2013 at 1:43
  • 4
    $\begingroup$ @wsc: make his lagrangian $Ax + Bv$, where $A$ is in joules per meter and $B$ is in joule seconds per meter, and the same issue remains. $\endgroup$ Aug 1, 2013 at 3:03

2 Answers 2


First I want to remind you what is going on behind the scenes. You know where the particle is at some initial time $t_1$, and you know where the particle is at some final time $t_2$, and the question you are asking is, which path will get me from the initial position at the initial time to the final position at the final time in a way that minimizes the action. Mathematically you want the function $x(t)$ that satisfies the conditions $x(t_1)=x_1$ and $x(t_2)=t_2$ and that minimizes $S$.

You wrote two terms, let's consider them separately. The easy one to deal with is the $v$ term. That term is a total derivative:

$$S=\int_{t_1}^{t_2} dt \frac{dx}{dt} = x(t_2)-x(t_1) = x_2 - x_1$$

But we don't have any freedom to change $x_2$ or $x_1$, they are part of the known information. We don't vary the path at the end points. So this term is actually irrelevant, and it is why the Euler Lagrange equations give zero for this term. This is a crucial property of lagrangians: I can add a total derivative to a lagrangian without changing the equations of motion.

So if we throw out that irrelevant term really the action you wrote down was

$$S=\int_{t_1}^{t_2} dt x(t)$$

You want the function that minimizes this integral.

But there is no function that will do that. For example, imagine a parabola in the x,t plane connecting $x_1$ and $x_2$. By changing the height of the parabola I can make the integral arbitrarily positive, negative, or zero. The key point though is that there is nothing that makes any of the parabolas special.

Explicitly for parabolas of the form $x(t)=3/2 a t^2$, and choosing the origin of time so that $t_1=-T$ and $t_2=T$, we have

$$S = \int_{-T}^{T} 3/2 a t^2 = a T^3$$

Then $dS/da=T^3$. In other words as I vary a I always get the same answer for the action. There is nothing here picking out a special parabola to minimize the action.

It's really exactly the same as trying to find the minimum of the function $f(u)=u$. In calculus you're told to take the derivative and set it equal to zero, that gives you the critical points. But if you do that here you get $f'(u)=1=0$, so what's up? The point is that there are no values of $u$ where $1=0$ so there is no minimum. It's the same problem here.

What this illustrates is that you really need the kinetic term $v^2$ in the lagrangian. The kinetic term penalizes paths that have unnecessary wild changes (the kinetic term punishes parabolas with large a in the example above). In the example above where $x_1=x_2=0$ adding the kinetic term would pick out one parabola that the particle actually follows. where you threw the particle in the air and watched it come back down (your potential $x$ is just the potential of the earth's gravitational field). The kinetic term is really the crucial thing. One thing you can try is to only consider parabolic paths in your action where you add a $v^2$ term, compute the action explcitly for those paths, and then just find the value of $a$ that minimizes the action. You don't have to use the Euler Lagrange equations at all here. (of course this only works since you know that the final path is a parabola, but it's proving a conceptual point)

By the way, to respond to some comments above the fact that the dimensions don't line up isn't actually that big a deal in this case, you can just put a parameter $\tau$ in front of the velocity with units of time and then work in units with $\tau=1$. The problem you are finding isn't affected by keeping $\tau$.

  • 3
    $\begingroup$ Here's another way to look at it. The kinetic term 1/2 mv^2 becomes ma in F=ma. Dropping the kinetic term is thus equivalent to setting m=0. That's a particle with zero mass, so if it experiences a net force you will get nonsense. You could salvage the situation by having a potential with a minimum (or maximum or saddle point) where there are no net forces. But your potential, V(x)=x, has no minimum. So the problem in Newton's laws is that you have a finite force on a zero mass particle. The problem in the Lagrangian formulation is that there isn't a minimizing path. The physics is the same. $\endgroup$
    – Andrew
    Aug 1, 2013 at 2:15
  • 1
    $\begingroup$ "This is a crucial property of lagrangians: I can add a total derivative to a lagrangian without changing the equations of motion." That really cool. Can you give an example of how that's useful in a real physical application? $\endgroup$
    – So8res
    Aug 1, 2013 at 12:39
  • $\begingroup$ (1) You need that fact to derive the Euler Lagrange equations at all, if you go through the derivation you end up throwing out a total derivative term. $\endgroup$
    – Andrew
    Aug 1, 2013 at 14:05
  • 1
    $\begingroup$ (pressed enter without shift, whoops) (2) The gem of the Lagrangian formulation is Noether's theorem. If the action is invariant under a transformation then there's a conserved current. However when you prove it you see the Lagrangian only has to be invariant up to a total derivative for Noether's theorem to work. To prove energy conservation from time translation invariance you need to throw out a total derivative term. This is a huge subject though. $\endgroup$
    – Andrew
    Aug 1, 2013 at 14:17

When you get a nonsensical equation of motion like this, it means that the action has no local extrema over the space of all possible paths. (Like how the curve $y = x$ has no local minimum or maximum.) There's no path $x(t)$ such that varying the path produces only a second-order (or higher) variation in the action.

In fact, you can see this explicitly for your Lagrangian $\mathcal{L} = \alpha x + \beta \dot{x}$ (constants inserted to make the units work): start with an arbitrary path $x(t)$, construct the path

$$y(t) = x(t) + \epsilon\sin\biggl(\pi\frac{t - t_1}{t_2 - t_1}\biggr)$$

(for example), and computing the action, you get

$$\begin{align}S[y] &= \int_{t_1}^{t_2} \left[\alpha x(t) + \alpha\epsilon\sin\biggl(\pi\frac{t - t_1}{t_2 - t_1}\biggr) + \beta \dot{x}(t) + \frac{\beta\pi\epsilon}{t_2 - t_1}\cos\biggl(\pi\frac{t - t_1}{t_2 - t_1}\biggr)\right]\mathrm{d}t \\ &= \int_{t_1}^{t_2} \left[\alpha x(t) + \beta \dot{x}(t)\right]\mathrm{d}t + \int_{t_1}^{t_2} \alpha\epsilon\sin\biggl(\pi\frac{t - t_1}{t_2 - t_1}\biggr)\mathrm{d}t + \int_{t_1}^{t_2} \frac{\beta\pi\epsilon}{t_2 - t_1}\cos\biggl(\pi\frac{t - t_1}{t_2 - t_1}\biggr)\mathrm{d}t \\ &= S[x] + \frac{2\alpha\epsilon(t_2 - t_1)}{\pi} + 0 \end{align}$$

which is strictly greater than $S[x]$ if $\alpha\epsilon > 0$ and strictly less if $\alpha\epsilon < 0$. In other words, given any path $x(t)$ with action $S[x]$, you can make another, arbitrarily close path with a larger (or smaller) action. That means, by definition, there is no extremal path.

That, in turn, means that this Lagrangian is meaningless, to put it bluntly. Remember, the whole validity of the stationary action principle depends on that local extremum of the action existing in the first place. Without that, the derivation of the Euler-Lagrange equations doesn't even really work. It's kind of like how the validity of division in algebra depends on the denominator not being zero - if you break that underlying assumption, your derivation loses logical consistency. And similarly, when you try to use action extremization on an action which has no extremum, you just get illogical nonsense.

  • $\begingroup$ Thanks. I get this sort of stuff, but I'm still confused. For example, division by zero is not just illogical nonsense. A lot can be said about division by zero. Your result diverges to positive or negative infinity. Following this "logical impossibility" can lead you to infinitesimals. I get that this lagrangian is nonphysical. I know it's nonsense. I'm wondering what kind of nonsense it is. Declaring that energy is now x + v clearly breaks everything, but it's not clear to me that it breaks things so hard that we have to stop talking about them. $\endgroup$
    – So8res
    Aug 1, 2013 at 2:00
  • 2
    $\begingroup$ @So8res When people say things like "x divided by zero diverges to infinity" they are really talking (shorthand) about a limiting process like "$x/\epsilon\to \infty$ as $\epsilon\to 0$," which is a well defined process at every step. So to connect this to your example you need to define your action as some kind of limit of a well defined action and examine what happens in the limit. That may be interesting, but it is different from just asking "what does $x/0$ mean?" or "what does this inconsistent action mean?" $\endgroup$
    – Michael
    Aug 1, 2013 at 2:28
  • 2
    $\begingroup$ @So8res Diverging to positive or negative infinity, or infinitesimals, is not division by zero though, it's a limit. I'm talking about actual division by zero, the kind that is often used in false proofs that 1=2 or the like. That's much like what you're doing here: you've constructed a false proof that 1=0, by assuming the validity of the Euler-Lagrange equation in a situation where it doesn't actually work. (Also: what kinds of nonsense are there?) $\endgroup$
    – David Z
    Aug 1, 2013 at 2:31
  • $\begingroup$ I think you miss the point. I know that the above Lagrangian is broken, but I don't buy that it's completely meaningless. For example, you could look at what happens as you smear the stationary point into a wider and wider interval. Or you could look at $\mathcal{L} = \alpha x^n + \beta v$ starting with $n = 2$ and observe what happens as $n$ goes to 1. Or you could play with similar limits using $v$. I was hoping people would point me towards illustrative interpretations. Of course the Lagrangian is broken, but many seem to think that unbroken means undiscussable. $\endgroup$
    – So8res
    Aug 1, 2013 at 12:39
  • 1
    $\begingroup$ You are taking a singular limit. In this case you are taking the $m\rightarrow 0$ limit of a particle experiencing a net force, which causes infinite acceleration. I agree that it is good to take a limit if you find something blows up. However you might find the limit doesn't converge, which is what happens here. You have to be careful with $V(x)=x^n$ btw. For $n=2$ it will work bc theres a minimum with 0 force. But for $n$ non-integer you get complex roots for negative $n$. You need the absolute value. But then you need to check the potential is smooth at the origin, or you get infinite force $\endgroup$
    – Andrew
    Aug 1, 2013 at 14:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.