Is Landau&Lifshitz's argument for the classical Lagrangian's symmetries too strict?

I realize that this paragraph has raised more questions on stackexchange, but I wanted to ask this question nevertheless since I want to discuss it in terms of a counter-example.

I’ve already completed a course in classical mechanics following the first part of Goldstein (that is, elementary Lagrangian and Hamiltonian mechanics), but I wanted to see how Landau & Lifshitz tackle the subject. So far, they have only postulated the least action principle, the existence of some “Lagragian” function $$L(q,\dot{q},t)$$, and the existence of inertial frames that are both homogenous and isotropic. In terms of proofs, the following has been proven:

• The familiar form of the Euler-Lagrange equations

• The fact that Lagrangians aren't unique; rather, adding the time derivative of a function of generalized coordinates and time ($$\frac{d}{dt}f(q,t)$$) to the Lagrangian does not change the motion.

At this point, authors I'm familiar with tend to postulate the form of the Lagrangian $$L = T - V$$; L&L however attempt first to derive the form of the free-particle Lagrangian from these principles.

The argument is that due to homogeneity and isotropy of space and time, the Lagrangian must not depend on the position and the direction of velocity of the particle. Starting from here, they derive the familiar $$L = \frac{1}{2}mv^2$$. They continue to use the condition that the Lagrangian is homogenous and isotropic w.r.t. space and time throughout chapter 2 to derive all sorts of interesting ideas.

But why does the homogeneity and isotropy of space and time imply a homogenous and isotropic Lanrangian? As far as I know, the structure of space and time only implies that the equations of motion are homogenous and isotropic:

$$\frac{d}{dt}\frac{\partial L}{\partial \dot{q_j}} - \frac{\partial L}{\partial q_j} = 0$$

But demanding isotropy and homogeneity here implies very little about the structure of the Lagrangian. In fact, there are many free particle Lagrangians explicitly not isotropic, such as $$L = \frac{1}{2}mv^2 + 4x^3v_x.$$ I'm not saying it's a particularly useful Lagrangian, but it exists :-)

So my question becomes: why do L&L place to much emphasis on obtaining a Lagrangian that is homogenous and isotropic? Is there perhaps a way to show that there exists some Lagrangian (not unique) that is homogenous and isotropic? Perhaps from the condition that the equations of motion are homogenous and isotropic? I could argue for myself that a H&I Lagrangian is useful from there, I reckon.

Bonus question: I always thought that Lagrangians were postulated, not derived? L&L's exposition isn't a full derivation (they throw in the potential energy ad hoc) but it still makes me wonder whether it wouldn't be cleaner to just postulate $$L = T - V + \frac{d}{dt}f(q,t)$$ "because it works" and take it from there.

• I am having a bit of trouble with your counterexample, could you spell it out in more detail? – lalala Jul 16 at 9:13
• Sure! According to my second bullet point (this can be lroved from the action principle or the Euler-Lagrange equations), the time derivative of any function of coordinates and time can be freely added to the Lagrangian. In my counter-example, I add the time derivative of the function $x^4$ to the Lagrangian. This Lagrangian is then not homogenous (it depends explicitly on an arbitrary choice of origin) but once you substititute it into the Euler-Lagrange equations we regain $ma=0$. This is Newton’s second law for a free particle, as was expected in the problem. – Heatherfield Jul 16 at 9:31
• L&L started from least action principle and they choose one representative of the whole family of lagrangians that leads to the same action. In my view, your question makes sense only if there exists homogeneous and isotropic EoM derivable from least action principle such that no lagrangian that leads to those EoM can be put in explicitly homogeneous and isotropic form. Otherwise, they simply choose the one that is the simplest to work with and if you want other lagrangian then you are free to transform it. – Umaxo Jul 16 at 9:50
• @Umaxo, right, that makes a lot of sense. What confused me specifically is the fact that L&L write these paragraphs as if the homogeinity/isotropy of the Lagrangian is some fundamental fact that must always be adhered to. Jan’s answer below nicely explains why this is not so (not to say that L&L are wrong, it’s mostly a misinterpretation on my side). – Heatherfield Jul 16 at 10:12
• Possible duplicates: physics.stackexchange.com/q/23098/2451 and links therein. – Qmechanic Jul 16 at 10:28

You are correct, a valid Lagrangian for a free particle does not have to be position or rotation independent. L&L imply that themselves when they show $$df(q,t)/dt$$ can be added with no consequence and your example gives a concrete demonstration of this.
The L&L argumentation when they seek Lagrangian for a free particle is more of an heuristic method than a solid argument. I think what they do is based on symmetry of the situation, they rule out functions that shouldn't be necessary to include in the Lagrangian and find that $$v^2$$ is the simplest function that cannot be ruled out by any such simple argument.