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(1) For formulating a lagrangian for a system of particles compared to one free particle, we start with the kinetic energy and formulate a potential energy term that is in terms of each of the radius vectors of the particles in the system that is meant to encapsulate the interaction between the particles. If a particle $q_a$ is in a constant external field, (let's say the Earth), then $U=mgh$ by the Euler-Lagrange equations. But, for a system of particles, why is the potential energy of the system the sum of the individual gravitational potential energies? This potential energy takes into account the influence of the external field on each of the individual particles but not the particles on each other within the system.

(2) We have by definition $F_{\text{ath particle}}=-\frac{\partial L}{\partial r_a}.$ But, this is equal to $-\frac{\partial U}{\partial r_a}=-m_ag.$ This is the force on a particle due to gravity. Then, how do I get the net force on the particle?

Qmechanic
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Arpit Mittal
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2 Answers2

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Lagrangian theories do often contain interactions. It all depend on the model.

OP's question is quite broad but at its core it seems OP is essentially observing an interesting and useful feature of the Lagrangian point mechanics (which goes hand in hand with d'Alembert principle, cf. my Phys.SE answer here), namely that we only have to take the applied forces into account, not necessarily all forces. If a force produces no virtual work, it can be omitted. This is one of the advantages of Lagrangian mechanics (as opposed to Newtonian mechanics).

The definition of an applied force depends on the model, see e.g. this related Phys.SE post.

Once the acceleration of the $i$th particle is determined from Lagrange equations, one can then e.g. use Newton's 2nd law to find the total force on $i$th the particle.

Qmechanic
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I think your question can be solved by thinking about how you choose to define your Lagrangian. Whether or not to include 2-body interactions, 3-body interactions, etc. in a potential is a choice made by the physicist who is creating that model of the system. You could easily add a term to your Lagrangian to account for the potential between any two balls, but it would greatly complicate the math and you know intuitively that the gravitational forces between two balls are extremely small.

In fields like nuclear physics, it becomes important to include 2- and 3-body interaction terms in the potential to describe the dynamics of a nucleus, for example. But for a macroscopic system, you don't need to include these terms. It doesn't mean the potential isn't there, it just means you're choosing to ignore it for simplicity.

klippo
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