In V.I. Arnold's Mathematical methods of classical and celestial mechanics, chapter IV, a definition of natural is provided.

It is stated that on a Riemannian manifold (i.e. a differentiable manifold M equipped with a positive-definite quadratic form or metric g), a Lagrangian system is natural if and only if $L=K−U$, where

  • for $v\in T_xM$, the kinetic energy K is a quadratic form defined as $K=\frac{1}{2}g(v,v)$ defined on all tangent bundles.

  • $U:M\rightarrow \mathbb{R}$ is a differentiable function corresponding to potential energy.

This definition raises the following concerns:

In relativistic mechanics, it appears that, since the manifold is not Riemannian (the metric is not positive-definite), no natural Lagrangian can be written: this seems to explain why the free particle Lagrangian writes as $L=-\gamma^{-1}mc^2$ and not $(\gamma-1)mc^2$.

But in classical mechanics, one always deals with Riemannian manifolds and yet, it is easy to find counterexamples for which $L\neq K-U$. For instance, the Lagrangian for a system undergoing viscous dissipation can be taken as

$$L(t,x,\dot{x})=e^{\frac{\gamma t}{m}}\left(\frac{1}{2}m\dot{x}^2-U(x)\right)$$

which obviously cannot be identified to $L=K-U$.

Here is my question based on the definition and above concerns. Suppose I know $K$ and $U$, but not $L$:

  1. From Arnold's standpoint, why is there no natural Lagrangian in my viscous force example (or more generally in examples like this), even though the manifold is Riemannian?

  2. Is there a general mathematical theorem accounting for how to build a Lagrangian on a manifold, regardless of whether it is Riemannian or not?

  • 3
    $\begingroup$ I don't really understand what this question is going for - "natural Lagrangians" on a manifold are simply a certain subset of all possible Lagrangian functions on that manifold. What sort of answer do you expect for "why" a system whose Lagrangian is not in that subset is not in that subset? I have a feeling you're interpreting "natural" as having some sort of implicit meaning like "This is a Lagrangian you should take" here, when it is just a definition of a specific kind of Lagrangian. $\endgroup$
    – ACuriousMind
    Sep 29, 2021 at 15:57
  • 1
    $\begingroup$ The calculus of variations is an entire branch of math, one can conceivably use any function of functions and their derivatives as a Lagrangian. Physics uses symmetry principles to restrict the form of the Lagrangian as much as possible to apply them to the real world. Arnold is not arguing that the use of a Riemannian manifold is why non-relativistic Mechanics looks the way it does, indeed he does not seem to (unlike Landau and Lifshitz) actually give a first principles argument for why the non-relativistic Lagrangian looks the way it does, he invokes Newton to justify it in Ch.'s 1, 2 and 3. $\endgroup$
    – bolbteppa
    Sep 29, 2021 at 16:45
  • 1
    $\begingroup$ What condition other than that $L-K$ is a function depending only on the $M$ part of $TM$ do you expect here? Either $L$ is of that form or it isn't, what difficulty could you have for any given $L$ in telling that so that you need a theorem with some other conditions? The (pseudo-)Riemannian aspect of the manifold is only here to define the kinetic term $K$ via the metric $g$. $\endgroup$
    – ACuriousMind
    Sep 29, 2021 at 17:19
  • 2
    $\begingroup$ Comment to the post (v6): 1. So by a Lagrangian formulation, you mean that there exists a stationary action principle (SAP)? Note that there are Lagrangian formulations that don't have a SAP. 2. What is the starting point/first principle if not the Lagrangian itself? E.g. in the viscous dissipation example, the viscous force should presumably also be given. 3. Related: physics.stackexchange.com/q/20298/2451 , physics.stackexchange.com/q/147341/2451 and links therein. $\endgroup$
    – Qmechanic
    Sep 30, 2021 at 7:21
  • 1
    $\begingroup$ The Lagrangian in the post can trivially be re-written as $$L = \frac{1}{2} m \dot{x}^2 + [e^{\gamma t/m} (\frac{1}{2} m\dot{x}^2 - U(x)) - \frac{1}{2} m \dot{x}^2] = T(\dot{x}) - U(x,\dot{x},t)$$ which is just expanding $e^{\gamma t/m} = 1 + ...$. From a fundamental perspective, adding velocity dependence (i.e. friction) to the non-relativistic potential is a model to deal with unaccounted for degrees of freedom e.g. a medium and how it interacts with a particle (See L&L Sec. 25). $\endgroup$
    – bolbteppa
    Sep 30, 2021 at 11:28

1 Answer 1


I am not particularly sure what is the essence of OP's question, but here are some remarks.

  1. Natural Lagrangians as defined by Arnold are simply a particular class of Lagrangians. There is nothing particularly significant about them aside from them being very common examples of Lagrangians in mechanics.
  2. The Riemannian metrics that appear in Lagrangians are defined on the configuration space and not on physical space and do not have anything to do with physical distances. In particular if the Lagrangian has dimensions of energy, then the "distance" that one can calculate from the Riemannian metric has units of $M^{1/2}L$, which is not distance.
  3. In fact, the aforementioned Riemannian metrics have more to do with mass than distance. They can be seen as abstractions of mass, since the usual kinetic energy formula $\frac{1}{2}mv^2$ becomes $\frac{1}{2}m(v,v)$. It is simply useful to have a general abstract concept of mass and kinetic energy on a general abstract configuration space.
  4. Regarding OP's "From Arnold's standpoint, why is there no natural Lagrangian in my viscous force example (or more generally in examples like this), even though the manifold is Riemannian?" The thing is, if one's viscous force is defined by a Lagrangian not of the form $L=K-U$, then one's manifold is not Riemannian, or at least does not have a Riemannian metric that is meaningful with respect to the properties of the dynamical system itself. It is useful to separate $\mathbb R$ the physical space from $\mathbb R$ the configuration space. As the physical space, it has a preferred Riemannian metric that determines physical distances. As the configuration space it does not have any preferred Riemannian metric whose kinetic energy would enter the Lagrangian.
  5. Regarding OP's "Suppose I know $K$ and $U$, but not $L$": I believe this is probably at the core of OP's misunderstanding. It does not really make sense to talk about knowing $K$ and $U$ but not $L$. One can either know the Lagrangian, or the equations of motion directly to know the dynamical system. If one only knows $K$ and $U$ then they can assemble the Lagrangian $L=K-U$ but this will be of standard form and describes the motion under the potential $U$.
  6. "Is there a general mathematical theorem accounting for how to build a Lagrangian on a manifold, regardless of whether it is Riemannian or not?": A Lagrangian for what purpose? If one knows the equations of motion then there are procedures to construct Lagrangians, which I have detailed in my answer here. These Lagrangians will not be in natural form however, although for second order ordinary differential equations, they can always be put into standard form by integrating by parts. However if further multipliers are needed (as in the viscous force example), then it still won't be in standard form.

Finally, regarding OP's comment

I would like to know if there is a theorem giving general conditions under which you can write the Lagrangian as $K-U$"

A natural generalization of Newton's equation $\mathbf F(\mathbf r,\dot{\mathbf r},t)=m\ddot{\mathbf r}$ to a Riemannian manifold $(M,m)$ is $$ m_{ij}(q)\ddot q^j+\Gamma_{i,jk}(q)\dot q^j\dot q^k=F_i(q,\dot q, t), $$where $\Gamma_{i,jk}$ are the Christoffel symbols (of the first kind) of the metric $m_{ij}$ and the $F_i(q,\dot q,t)$ are force functions. The left hand side always appears as the Euler-Lagrange equations of the kinetic energy function $$ K=\frac{1}{2}m_{ij}(q)\dot q^i\dot q^j, $$although iirc the Euler-Lagrange equations are actually the negative of this.

Therefore we are interested in whether the force functions $F_i(q,\dot q,t)$ have a generalized potential or not.

This can be checked by applying the so-called Helmholtz operator to $F_i$, which in this case is $$ H^{(0)}_{ij}(F)=\frac{\partial F_i}{\partial q^j}-\frac{\partial F_j}{\partial q^i}+\frac{d}{dt}\frac{\partial F_j}{\partial \dot q^i} \\ H^{(1)}_{ij}(F)=\frac{\partial F_i}{\partial\dot q^j}+\frac{\partial F_j}{\partial\dot q^i}. $$

If these give zero, then we can construct a generalized potential by $$ U(q,\dot q,t)=-\int_0^1 F_i(sq,s\dot q,t)q^ids. $$

If the Helmholtz operators do not give zero, then it might still be possible to find a Lagrangian for the EoMs by using multipliers, but that will absolutely not be of the form $K-U$ (think of the viscous force example).

  • $\begingroup$ Thanks a lot, it clarifies and answers my questions. $\endgroup$
    – Seb
    Oct 5, 2021 at 8:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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