Is there any restriction on the Lagrangian of a system?

Question

I have learned the calculus of variations in my previous semester, and now we are studying classical mechanics. What I found is that there is lots of lack of rigor in Lagrangian mechanics in comparison to the calculus of variations.

For example, the least action principle or Hamilton's principle is stated as:

Every mechanical system is characterised by a definite function $\mathcal{L}(q_1,..,q_n,\dot{q}_1,...,\dot{q}_n,t) $ or briefly $\mathcal{L}(q,\dot{q},t)$, and the motion of the system in such that a certain condition is satisfied.

Let the system occupy, at the instants $t_1$ and $t_2$, positions defined by two sets of values of the co-ordinates,$q^{(1)}$ and $q^{(2)}$. Then the condition is that the system moves between these positions in such a way that the integral \begin{equation}\label{key} S = \int_{t_1}^{t_2}\mathcal{L}(q,\dot{q},t)dt \end{equation}

takes the least possible value. The function $\mathcal{L}$ is called the Lagrangian of the system concerned, and the integral is called the action.

But here you see they didn't make any restriction on $\mathcal{L}$, neither they say it should by continuous nor the functional $S$ should be differentiable, and without these restriction they actually derive the Euler-Lagrange equation.

So the question is: Is there any restriction on the action (functional) or on the Lagrangian? If not, are there examples where the principle is valid but the Euler-Lagrange equation is not? If yes, is it right to specify them in principle?

Answer 1

Generally speaking, people do not consider the generic class of Lagrangians, since they tend to be interested in physical systems, but yes, there are examples of "bad Lagrangians".

A Lagrangian should certainly be an integrable function, as well as $C^1$ (or at least weakly differentiable) in its variables, but more importantly, it should have an extremum. The classic example of a bad Lagrangian is

\begin{equation} L(q, \dot{q}) = q \end{equation}

The Euler-Lagrange equation is then $1 = 0$, which is not an ideal system. You can see the reason why by simply considering random curves $q(t)$ and considering the action

\begin{equation} S[q] = \int_{t_1}^{t_2} q(t) dt \end{equation}

By taking arbitrarily long detours in positive or negative values of $q$, you could make the action arbitrarily high or low, there is therefore no minimum or maximum (and from the Euler-Lagrange equation, we can see no saddle points either) action.

A less terrible outcome is if there is more than one extremum, even considering boundary conditions. The simplest example we can come up with is the Lagrangian

\begin{equation} L(q, \dot{q}) = 1 \end{equation}

The action will be the same, no matter what path you choose for the particle :

\begin{equation} S[q] = \int_{t_1}^{t_2} dt = t_2 - t_1 \end{equation}

Or, from the Euler-Lagrange perspective, this is simply $0 = 0$, which is true no matter the function we consider. This is related to the problem of symmetries (in this case, our Lagrangian is symmetric for any function $q \to f(q, \dot{q})$, $\dot{q} \to g(q, \dot{q})$), and is dealt with in the theory of Lagrangian with constraints. This doesn't spell doom for such a Lagrangian, although it does mean that some of the variables will be gauge variables, and be somewhat arbitrary.

Another possible issue is the one of boundary conditions. You may remember that, in computing the Euler-Lagrange equation, we use integration by parts, with

\begin{eqnarray} \int_{U} \frac{\partial L}{\partial \dot{q}} \delta \dot{q}\ dt &=& [\frac{\partial L}{\partial \dot{q}} \delta q]_{\partial U} - \int_U \frac{d}{dt} \left[ \frac{\partial L}{\partial \dot{q}} \right] \delta q\ dt \end{eqnarray}

We got rid of the first term under the assumption that the variation $\delta q$ vanished on the integration boundary $\partial U$, but for more complex Lagrangian systems (especially ones for field theories or more complex objects like strings, or if the underlying spacetime is particularly nasty), this isn't necessarily true, and we may have non-vanishing boundary terms, spoiling the Euler-Lagrange equation, meaning that if we want to use it, we will have to narrow somewhat the class of functions we consider, or take such boundary effects into account.

Answer 2

Often we impose regularity conditions on a Lagrangian formulation to simplify the calculations, and/or so that we can work within some mathematical framework, such as e.g.,

• differentiability,
• that constraints are holonomic,
• that the rank of the Hessian is maximal, or at least don't jump,
• other rank conditions, see e.g. this Phys.SE post,
• locality,
• etc.

However, it may be that Nature or the model under investigation don't respect such regularity conditions. Then it becomes a matter of

• What is the physical interpretation?
• Can the Lagrangian formulation be salvaged or improved?
• etc.

If Euler-Lagrange equations don't have solution, or infinitely many solutions, it doesn't have to be a failure per se. It could be telling us something about the system.