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When studying classical mechanics using the Euler-Lagrange equations for the first time, my initial impression was that the Lagrangian was something that needed to be determined through integration of the Euler-Lagrange equations$^1$. Of course, now I know it's something that's a given for a mechanical system, and we integrate it via the Euler-Lagrange equations to get the evolution of the coordinates of the mechanical system as a function of time.

But are there alternative uses for the Euler-Lagrange equations where the Lagrangian isn't known before hand?


To all the teachers out there, for god's sake, explain right at the start that the Lagrangian is a functions known before hand for a system.

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3 Answers 3

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I'm not completely certain what OP is asking. The Euler-Lagrange equations are by definitions derived from an action principle. If the action doesn't exist, it does not make sense to talk about Euler-Lagrange$^1$ equations.

Instead what OP might want to ask is the following interesting question:

If one is given a set of equations of motion for some physical system, could they (secretly) be Euler-Lagrange equations for some action? In other words, does there exists an action principle for the system?

This is already discussed in several posts on Phys.SE, for instance here.

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$^1$ There exists a notion of Lagrange's equation (without Euler!),

$$\tag{1} \frac{d}{dt} \left(\frac{\partial T}{\partial \dot{q}^j} \right)-\frac{\partial T}{\partial q^j}= Q_j,$$

which doesn't require an action principle. (The generalized force $Q_j$ might not have a generalized potential $U$.) See e.g. Goldstein, Classical Mechanics, Chap. 1. Equation (1) is not called Euler-Lagrange equation unless an action principle exists. Equation (1) may be derived from D'Alembert's principle, which in turn may be derived from Newton's laws under additional assumptions, such as, e.g. no sliding friction. See also this Phys.SE post.

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The Euler-Lagrange equations can be derived from the principle of least work. –  Larry Harson Mar 16 '13 at 21:45
    
The Euler-Lagrange equation were originally derived without an extremum principle at all. You can see a modern version of that derivation in Goldstein. –  dmckee Mar 16 '13 at 22:07
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D'Alembert's principle may be used to derive Lagrange's equations (without Euler!) $\frac{d}{dt} \left(\frac{\partial T}{\partial \dot{q}^j} \right)-\frac{\partial T}{\partial q^j}= Q_j$ as done in Goldstein, Classical Mechanics, Chap. 1. However it is still not clear that an action principle exists. This only follows if the generalized force $Q_j$ has a generalized potential $U$. –  Qmechanic Mar 16 '13 at 22:08

Let me give a sketch of the flow of logic in classical mechanics.

Assume Newton's laws are true. One can, by using the D'Alembertian principle, and on holonomic systems, find that we can rewrite Newton's second laws as 'Lagrange's equations' $\frac{d}{dt} \frac{\partial L}{\partial \dot{q}_j} - \frac{\partial L}{\partial q_j} = 0$, where we have defined $L = T - V$, called the Lagrangian. (This construction of $L$ is of course not unique).

Now let's abandon Newton's 2nd law as your starting point. Let's say that instead we write the Lagrangian $L = T - V$ of the system. Also let's take the so-called Hamilton's principle as an axiom: the motion of the system from time $t_1$ to $t_2$ is such that the line integral (called the action) $S = \int_{t_1}^{t_2} L dt$ has a stationary value for the actual path of the motion. It turns out that the stationary value for $S$ is given by paths which satisfy the Euler-Lagrange equations, $\frac{d}{dt} \frac{\partial L}{\partial \dot{q}_j} - \frac{\partial L}{\partial q_j} = 0$. Hence we have recovered Newton's 2nd law.

So what do we have here? We see that Newton's 2nd law $\iff$ Lagrangian with Hamilton's principle. In other words, Newton's 2nd law and the Lagrangian description of mehchanics are equivalent (at least for holonomic systems, though that can be generalized).

Now which starting point do we take to be more sacred? Newtonian or Lagrangian? When we first learned classical mechanics the Newtonian approach was the holy path, but for various reasons (aesthetics, generalization to field theories / QM, explicit symmetry dependence etc.) we prefer the Lagrangian approach.

That is to say, we postulate that every system has a (or a family of equivalent) Lagrangian(s) that describe(s) it, and taking Hamilton's principle as our axiom, apply the Euler-Lagrange equations to $L$, thereby obtaining the equations of motions.

Now the question is, what $L$ works? Earlier I wrote $L = T - V$, but actually, this is a matter of guesswork, because ultimately, physics is an empirical science so whatever theories we come up with had better corroborate with real-world measurements. It turns out that $L = T - V$ is a good guiding principle, but there are some cases in which it is not so clear. For example, in writing a Lagrangian for the electromagnetic field or for GR, we are 'guided' by the known EOMs to write $L$ down. Then, once we find an $L$ that works, we take it to be the holy grail from which everything is derived from (including the EOMs), discarding the fact that we used the EOMs to 'derive' the Lagrangian.

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Your description of the flow of logic helped me a lot in understanding the Lagrange formalism. Thanks. –  abenthy May 9 '13 at 21:50

First of all I would like to point out that Lagrangian is not always given. Usually, you are given a system. Then you have to construct Lagrangian for that particular system, and then play around with it. Of course, Lagrangians of simple systems are well-known, so many teachers focus on applications rather than derivations of Lagrangian.

Regarding your question. Yes, sometimes you can apply Euler-Lagrange formalism even in cases when you don't know/don't want to calculate Lagrangian. For example (unfortunately, cannot provide you with the link at the moment) in gauge filed theories you can manipulate Lagrangians without calculating them up to some point. Also, Hamiltonian, which are in some sense related to Lagrangian mechanics may be used in order to prove some phase space relations. In this case you absolutely don't need to know its explicit form, as these relations are general.

Don't worry if you are not familiar with some terminology I have used. I just wanted to give you some examples. Don't worry, you will learn everything at it's high time.

To conclude: explicit form of Lagrangian/Hamiltonian is not required for general relations, working for any system. Obviously, there are not many of them, but there are some. Usually we work with a particular system and use Lagrangian formalism to make our analysis easier (in fact, for many systems this approach is crucial).

Hope this answered your question.

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