# Is the Lagrangian “math” or “science”?

I've seen in class that we can get from Lagrangian to derive equations of motion (I know its used elsewhere in physics, but I haven't seen it yet). It's not clear to me whether the Lagrangian itself follows from the equations of motion, or whether it represents a fundamentally different approach - whether it's a different "model", or whether it's largely a mathematical observation (it may illuminate/integrate what came before, but for all intents and purposes it is equivalent).

I'm just curious about how these relationships are generally understood. I ask because it's not at all obvious to me that the principle of least action should be true. I'd also be curious whether the answer to this question would be the same across different fields of physics, though if the answer's "no" much detail would probably go over my head.

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–  Qmechanic Sep 27 '12 at 6:41

Short answer: Lagrangian mechanics only applies to a subset of classical mechanics problems, but when it does, it is mathematically equivalent to Newtonian mechanics (which I take to mean direct application of $\vec{F} = m\vec{a}$).

More detailed rambling: Your class and whatever text you're using ought to cover the equivalence of the Newtonian and Lagrangian (and Hamiltonian) approaches, so I'll just give the overview. We begin by expressing the state of a system by specifying the locations of all the relevant parts: $\vec{r}_1, \ldots, \vec{r}_M$. In $d$ dimensions there are $dM$ components to worry about, but let's just say $d = 3$. Now there will also be some number $C$ of constraints on the system (things like "the mass won't fall through the table" or "the pendulum mass is always a distance $l$ from the pivot"), so the system actually has $N = 3M - C$ degrees of freedom.

We should be able to come up with $N$ generalized coordinates, often written $q_k$, and express the system in terms of $N$ equations in those coordinates. Here's where the first of two important restrictions comes into play. We are only interested in holonomic constraints - those that depend only on positions but not on velocities or other derivatives. In this way we can express each $\vec{r}_i$ as a function of $q_1, \ldots, q_{N}$ and possibly $t$, with no $\dot{q}_i$'s appearing. (Bonus vocabulary lesson: the holonomic constraints are rheonomic or rheonomous if there is explicit time dependence; they are scleronomic or scleronomous otherwise.) The classical Lagrangian method doesn't really apply to nonholonomic constraints.

It's simple enough to calculate the kinetic energy $T$ as a function of the $q_k$'s and $\dot{q}_k$'s. The other important restriction is that the forces on the system are conservative - i.e., that they come from the gradient of a potential $U$. We need there to be such a $U$ expressible in terms of the $q_k$'s, otherwise we're stuck. (Actually, there are some methods involving "generalized potentials" that get around this in a few cases.)

If we have holonomic constraints and conservative forces, the calculus of variations tells us that the $N$ (possibly coupled) differential equations of motion for the system are $$\frac{\mathrm{d}}{\mathrm{d}t} \left(\frac{\partial L}{\partial\dot{q}_k}\right) - \frac{\partial L}{\partial q_k} = 0,$$ where $L = T - U$ is considered a function of the $2N + 1$ variables $q_1, \ldots, q_N, \dot{q}_1, \ldots, \dot{q}_N, t$. These give exactly the same motion as $\vec{F}_i = m \ddot{\vec{r}}_i$ applied to each of the $M$ original coordinates. In this sense, Lagrangian mechanics is just some mathematical trickery for easily getting equations of motion in certain cases. It's not new physics in any way.

Final note: Now this was all done long ago in a purely classical setting, considering systems with small numbers of mechanical parts. It's a very different way of thinking, emphasizing the global properties of the system rather than just the local properties at points of interest. As it turns out, this method, together with the related Hamiltonian approach, lend themselves to quantum mechanics quite nicely. Quantum field theory and its offshoots are all about constructing Lagrangians to derive equations of motion, and this applies to settings where $\vec{F} = m\vec{a}$ doesn't even make much sense anymore.

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Thanks, this is very informative. –  Albert Sep 28 '12 at 7:39

It is a mathematical observation that follows from Newton's laws. See for example Lanczos

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