# Are there arguments for the form of the Lagrangian in classical mechanics?

Consider some sort of entity/object with generalized coordinate $q$ and generalized velocity $\dot{q}$. Now suppose that we know the form of the object's Lagrangian when it's in free motion, not interacting with anything. Denote that by $\mathcal{L}_{free}$. Now we create conditions for the object to interact with its surroundings, and the Lagrangian becomes $\mathcal{L}$. In classical mechanics, and even in electrodynamics, we have that $\mathcal{L}=\mathcal{L}_{free}-U$, where $U$ is a function of the generalized position, velocity and time that's usually called the potential, and its form depends only on the type of interaction. Okay. But why?
Can we reach that conclusion without starting from the equations of motion (i.e. Newton's Second Law)? That we can always describe interactions by adding some sort of extra term to the Lagrangian of free motion?

UPDATE 1: I've fiddled around with the concept a bit, and started to think about lagrangian multipliers; describing interactions as some sort of constraint $f=0$, and finding stationary points for the free action, $$S_{free}=\int_{t_1}^{t_2}\mathcal{L}_{free}\,\mathrm{d}t$$ subject to the constraint $f=0$, which is equivalent to finding the stationary points of a new functional $$S=\int_{t_1}^{t_2}\mathcal{L}_{free}\,\mathrm{d}t-\lambda f=\int_{t_1}^{t_2}\mathcal{L}_{free}-U\,\mathrm{d}t$$ where $U\equiv -\lambda\frac{df}{dt}$, and so we define a new Lagrangian $\mathcal{L}=\mathcal{L}_{free}-U$.

But this still isn't satisfactory, given that I had to assume that the free action is still 'stationary' despite knowing nothing about how interactions may affect that original Lagrangian.

UPDATE 2: I see someone mentioned coupling constants and whatnot. I've elaborated on that and got no satisfactory results. Suppose we can modulate the strength of our interaction through some constant $\alpha$. Then we can say that our Lagrangian is a perturbation of the original, of the form $$\mathcal{L}=\mathcal{L}_{free}-\alpha U,$$ for $U\equiv-\frac{\partial\mathcal{L}}{\partial\alpha}$. If we then absorb the 'correct' value of our constant (i.e. the value that gets us the right equations of motion) into $U$, we get that $\mathcal{L}=\mathcal{L}_{free}-U$. The problem with this approach is that some interactions can't be modulated to the proper intensity, e.g. very quick isentropic processes in a gas or anything with hysteretic behavior, and therefore this model is at least flawed.​

• Generally, the interaction strength is specified by some coupling constant. We then want to recover the free Lagrangian when the coupling constant tends to zero. Therefore, an additive relationship is needed. – Virgo Aug 23 '16 at 2:25
• Comments to the post (v3): 1. Note that not all interactions/forces can be described by a (possibly velocity-dependent) generalized potential $U$, e.g. friction, cf. this Phys.SE post. 2. Related: physics.stackexchange.com/q/50075/2451 , physics.stackexchange.com/q/78138/2451 and links therein. – Qmechanic Sep 1 '16 at 1:47
• A quantum motivation due to Feynman is in Section 4 of hitoshi.berkeley.edu/221a/pathintegral.pdf – J.G. Sep 11 '16 at 21:01
• Are you reading page 8 of Landau's mechanics book by any chance? – John McVirgooo Sep 11 '16 at 21:19
• @J.G. I've seen that paper before. It's only showing it works, not arriving at that form from other reasonable arguments. – Gabriel Golfetti Sep 18 '16 at 22:35

• But what if we accept Hamilton's Principle before the knowledge of Newton's Laws? Can we somehow deduce that the lagrangian is of form $\mathcal{L}=\mathcal{L}_{free}-U$? – Gabriel Golfetti Aug 24 '16 at 18:56
• Sure, you could work backwards and find $L=T-V$ from assuming – Lewis Miller Aug 25 '16 at 14:58
• $H=T+V=constant$. But the conservation of energy, the principle of least action, the Lagrange and Hamiltonian functions would never have been discovered without Newton's laws of motion. – Lewis Miller Aug 25 '16 at 15:03