# Why are there only derivatives to the first order in the Lagrangian?

Why is the Lagrangian a function of the position and velocity (possibly also of time) and why are dependences on higher order derivatives (acceleration, jerk,...) excluded?

Is there a good reason for this or is it simply "because it works".

• The question is a lot more specific than the title, so I want to point out that a successful explanation of what's asserted in the question doesn't imply the truth of the assertion in the title. Higher derivatives are sometimes important in physics. Engineers designing camshafts work very hard to minimize the "jerk" $j=d^3x/dt^3$, because high jerk damages the cam follower. The Lorentz-Dirac radiative reaction force on a charged particle is proportional to jerk. The common thread is that only a changing force can cause mechanical or electromagnetic waves to be radiated. – user4552 Jul 5 '13 at 2:00
• For the derivation of higher-order Euler-Lagrange (EL) equations for higher-order Lagrangians, see this Phys.SE post. – Qmechanic Apr 25 '19 at 16:23
• It is kind of implied in the question, but I would like to mention it nontheless - you consider only derivatives with respect to time. However, higher-order derivatives with respect to the space variable $x$ are no problem in classical mechanics. For example, the Lagrangian that gives the Korteweg-de Vries equation has $\partial^2_x \phi$ dependence. See en.wikipedia.org/wiki/Korteweg-de_Vries_equation for more details. – FusRoDah Sep 14 '19 at 17:07

I reproduce a blog post I wrote some time ago:

We tend to not use higher derivative theories. It turns out that there is a very good reason for this, but that reason is rarely discussed in textbooks. We will take, for concreteness, $L\left(q,\dot q, \ddot q\right)$, a Lagrangian which depends on the 2nd derivative in an essential manner. Inessential dependences are terms such as $q\ddot q$ which may be partially integrated to give ${\dot q}^2$. Mathematically, this is expressed through the necessity of being able to invert the expression $$P_2 = \frac{\partial L\left(q,\dot q, \ddot q\right)}{\partial \ddot q},$$ and get a closed form for $\ddot q \left(q, \dot q, P_2 \right)$. Note that usually we also require a similar statement for $\dot q \left(q, p\right)$, and failure in this respect is a sign of having a constrained system, possibly with gauge degrees of freedom.

In any case, the non-degeneracy leads to the Euler-Lagrange equations in the usual manner: $$\frac{\partial L}{\partial q} - \frac{d}{dt}\frac{\partial L}{\partial \dot q} + \frac{d^2}{dt^2}\frac{\partial L}{\partial \ddot q} = 0.$$ This is then fourth order in $t$, and so require four initial conditions, such as $q$, $\dot q$, $\ddot q$, $q^{(3)}$. This is twice as many as usual, and so we can get a new pair of conjugate variables when we move into a Hamiltonian formalism. We follow the steps of Ostrogradski, and choose our canonical variables as $Q_1 = q$, $Q_2 = \dot q$, which leads to \begin{align} P_1 &= \frac{\partial L}{\partial \dot q} - \frac{d}{dt}\frac{\partial L}{\partial \ddot q}, \ P_2 &= \frac{\partial L}{\partial \ddot q}. \end{align} Note that the non-degeneracy allows $\ddot q$ to be expressed in terms of $Q_1$, $Q_2$ and $P_2$ through the second equation, and the first one is only necessary to define $q^{(3)}$.

We can then proceed in the usual fashion, and find the Hamiltonian through a Legendre transform: \begin{align} H &= \sum_i P_i \dot{Q}_i - L \ &= P_1 Q_2 + P_2 \ddot{q}\left(Q_1, Q_2, P_2\right) - L\left(Q_1, Q_2,\ddot{q}\right). \end{align} Again, as usual, we can take time derivative of the Hamiltonian to find that it is time independent if the Lagrangian does not depend on time explicitly, and thus can be identified as the energy of the system.

However, we now have a problem: $H$ has only a linear dependence on $P_1$, and so can be arbitrarily negative. In an interacting system this means that we can excite positive energy modes by transferring energy from the negative energy modes, and in doing so we would increase the entropy — there would simply be more particles, and so a need to put them somewhere. Thus such a system could never reach equilibrium, exploding instantly in an orgy of particle creation. This problem is in fact completely general, and applies to even higher derivatives in a similar fashion.

• Nice answer, +1. However, it leaves me somewhat unsatisfied. The final paragraph assumes that this is QFT, and $q$ is a field, but that's a very narrow context. We don't see higher-derivative theories describing any aspect of physics. And even in QFT, what about a Dirac sea picture, where all the negative energies are full? – user4552 Jul 4 '13 at 21:07
• Excellent answer... with respect to the "orgy of particle creation"... isn't that one aspect of the big bang? Can we have an unstable higher order theory that stabilizes itself into an effective theory of the usual first order form as it creates a sufficiently large pool of interacting particles? – CuriousOne Jan 1 '15 at 21:13
• Author's original blog post: tcm.phy.cam.ac.uk/~gz218/2010/01/… – Shital Shah Oct 20 '18 at 3:18
• Why can't the explanation simply be that stability is only achieved at extrema and saddle points of the potential , i.e., wherever its second derivative is zero. Any higher order terms are therefore irrelevant. Am I oversimplifying things? – Tfovid Jul 9 '19 at 6:58

Excellent question, and one that I've never really found a completely satisfactory answer for. But consider this: in elementary classical mechanics, one of the fundamental laws is Newton's second law, $\mathbf{F} = m\mathbf{a}$, which relates the force on an object to the object's acceleration. Now, most forces are exerted by one particular object on another particular object, and the value of the force depends only on the positions of the source and "target" objects. In conjunction with Newton's second law, this means that, in a classical system with $N$ objects, each one obeys an equation of the form

$$\ddot{\mathbf{x}}_i = \mathbf{f}(\{\mathbf{x}_j|j\in 1,\ldots,N\})$$

where $\mathbf{f}$ is some vector-valued function. The point of this equation is that, if you have the positions of all the objects, you can compute the accelerations of all the objects.

By taking the derivative of that equation, you get

$${\dddot{\mathbf{x}}}_i = \mathbf{f'}(\{\mathbf{x}_j\})\{\dot{\mathbf{x}}_j\}$$

(I'm getting quite loose with the notation here ;p) This allows you to compute the jerk (third derivative) using the positions and velocities. And you can repeat this procedure to get a formula (at least in some abstract sense) for any higher derivative. To put it in simple terms, since Newton's second law relates functions which are two orders of derivative apart, you only need the 0th and 1st derivatives, position and velocity, to "bootstrap" the process, after which you can compute any higher derivative you want, and from that any physical quantity. This is analogous to (and in fact closely related to) the fact that to solve a second-order differential equation, you only need two initial conditions, one for the value of the function and one for its derivative.

The story gets more complicated in other branches of physics, but still, if you look at most of them you will find that the fundamental evolution equation relates the value of some function to its first and second derivatives, but no higher. For example, in quantum mechanics you have the Schrodinger equation,

$$i\hbar\frac{\partial\Psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2} + U(x)\Psi$$

or in quantum field theory, the Klein-Gordon equation,

$$-\frac{\partial^2\phi}{\partial t^2} + \frac{\partial^2\phi}{\partial x^2} - m^2\phi = 0$$

and others, or Maxwell's equations (equivalently, the wave equation that can be derived from them) in classical electromagnetism. In each case, you can use a similar argument to at least motivate the fact that only position or its equivalent field and its first derivative are enough to specify the entire state of the system.

Of course, you might still wonder why the equations that describe the universe relate functions that are only two derivatives apart, rather than three or four. That part is a mystery, but one that falls in the realm of philosophy rather than physics.

• Nice answer +1. As to why we have only two derivatives, or more generally, why lagrangians have the form they do take a look at this paper by B. Roy Frieden. For a long time he has been trying to emphasize the role of information in physics. Only recently has the mainstream begun to appreciate this fact. Do not be turned off by the titles of some of Frieden's work. Grandiosity is a forgivable offence especially for workers in such unforgiving territory :) – user346 Jan 29 '11 at 10:35
• super answer. Re. physics.stackexchange.com/q/4102 . Are there any references for "To put it in simple terms, since Newton's second law relates functions which are two orders of derivative apart, you only need the 0th and 1st derivatives, position and velocity, to "bootstrap" the process," a book or paper? – user45664 Sep 12 '18 at 17:47

There are implications for causality when a equation of motion contains higher than second derivatives of the fields, EM radiation from charged bodies goes over the derivative of the acceleration

i don't know the details of WHY but this book should give more details: (Causality and Dispersion Relations) http://books.google.com/books?id=QDzHqxE4anEC&lpg=PP1&dq=causality%20dispersion%20relations&pg=PP1#v=onepage&q&f=false

• +1 I think this topic is highly underrated (at least in my experience) and extremely interesting. – Arturo don Juan Feb 28 '16 at 7:41

There are formulations involving higher order derivatives, however, you made a fair characterization.

I think a rule of thumb would be to start looking for the simplest Lagrangian you can think of. In the general case, a good Lagragian should obey homogeneity of space, time and isotropy of space which means that it can't contain explicitly the position, time and velocity $\vec{v}$, respectively. Then, the simplest allowed possibility is to have a Lagrangian with a velocity squared. Since we don't need to look for more conditions to be fulfilled, there is no need to add terms involving higher derivatives or combinations of other terms.

You can see this procedure at work (quite a few times, actually) in Landau & Lifshitz, The Classical Theory of Fields.

• If you are right, then my answer is wrong. Could you provide an example of a Lagrangian which involves higher derivatives? – gabgoh Jan 28 '11 at 22:25
• @gabgoh: Without searching too hard, I found this as an example arxiv.org/abs/1012.2969. – Robert Smith Jan 28 '11 at 22:45
• Thanks for the answer. Unfortunately the paper seems to involve involves Gauge theory, something which I'm not very familiar with. I was hoping for a simple example form classical mechanics which involves a higher order Lagrangian. Or perhaps I need to venture beyond mechanics to find a solution? – gabgoh Jan 28 '11 at 22:52
• @gabgoh: I don't think there is a mainstream formulation involving higher order Lagrangian from classical mechanics (although I could be wrong). Maybe the post written by @lurscher can give such example. – Robert Smith Jan 28 '11 at 23:03

Well, the usual physics in classical mechanics is formulated in terms of second-order differential equations. If you are familiar with the process of deriving Euler-Lagrange equations from the Lagrangian then it should be natural that the kinetic term must be proportional to $(\partial_t x)^2$ to reproduce that.

If you'd considered more general Lagrangians (which you are certainly free to) you would obtain arbitrarily complicated equations of motions but these wouldn't correspond to anything physical. Nevertheless, some of those equations might describe some mathematical objects (because Lagrangian formalism and calculus of variations isn't inherent only to physics but also to lots of other mathematical disciplines).

This question actually needs a 2 steps answer:

1. Why Lagrangian has only derivatives to the first order?:

Lagrangian has been defined in such a way, that problem to be solved would produce a second order derivative with respect to time when Euler-Lagrange equation is produced. It includes an implicit derivation of the momentum (notice time derivative after minus sign in $\frac{\partial L}{\partial q} - \frac{d}{dt}\frac{\partial L}{\partial \dot q}=0$) which in turn, is a first order derivative of position. It means that, acceleration is actually taken care when the full problem is setup. One can verify it by simply checking that for most cases Euler-Lagrange equation just turns to be $\frac{\partial L}{\partial q}-m \ddot q=0$ and if one defines$\frac{\partial L}{\partial q}=F$ it becomes Newton’s second law. Having said that, we need to move to the next step, which is,

1. Why jerk (or any bigger time derivative) is not necessary?:

This question has already been replied (including one by me) here Why $F=ma$ and not $F=m \dot a$. The short answer is: “… second order derivative is all one needs to differentiate natural states of motion from affected states of motion”.

If we assume, say, a second derivative in the Lagrangian, the Euler-Lagrange equations which minimize action

$$A[q] = \int_{x_1}^{x_2} L(x,q,q',q'')\, dx$$

would be

$$\frac{\partial L}{\partial q} - \frac{d}{dt}\frac{\partial L}{\partial q'} + \frac{d^2}{dt^2}\frac{\partial L}{\partial q''} = 0$$

This is a fourth order differential equation. However, this can't be the case as we already know that $q''=F/m$, i.e acceleration is determined by Force, which is "outside" the initial conditions. In a gravitational force field, for example, you know, a piori, the forces at every point in the system, and hence the acceleration at every point in the system is already known. A fourth order DE would lead to an internal inconsistency.

The deeper question to ask, I suppose, is why $F=mq''$, not $F=mq'''$ or $F=mq''''$. I won't pretend to know the answer to this, but I suspect there might be one.