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As a guide to deeper understanding: In classical mechanics we have that theory of motion can be expressed in terms of energy exchange. As we know, newtonian gravity is a conservative force, hence we can define a potential energy. As we know, it is customary to take infinite radial distance (to the source of gravity) as the zero point of potential energy. At ...

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As pointed out in an earlier answer: in the history of mathematical physics there have been multiple instances of putting calculus of variations to use. Those instances aren't necessarily connected, and the specific Lagrangian may not contain energy terms. My overall aim in this answer is to discuss the relation between differential calculus and calculus of ...

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user318039's answer is exactly right: An action principle only has to reproduce the given equations of motion (EOMs) as its Euler-Lagrange (EL) equations. The Lagrangian does not have to have dimension of energy. For instance, if one scales a Lagrangian with a dimensionfull (non-zero) constant, the EOMs will stay the same. See also e.g. this & this Phys....

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Historically, the Lagrangian approach may have been first introduced for mechanics, but the concept of the Lagrangian has been generalized across many different areas of physics. A key property across these formulations is that the Lagrangian is a function such that the stationary trajectories of the action give the equations of motion. In mechanics, we know ...

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In the case of Hamilton's stationary action there is no significance in the non-uniqueness you refer to. The point where the rubber meets the road is the constraint that as an object is moving along, subject to acceleration due to a potential gradient, the rate of change of kinetic energy must match the rate of change of potential energy. Inserting the ...

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It is wellknown that given a set of EOMs, the action $S$ is not necessarily unique, cf. e.g. this Phys.SE post. OP points out that the Euler-Lagrange (EL) equations are not affected if we add a boundary term, cf. e.g. this Phys.SE post. However, the caveat is that the boundary conditions (BCs) [which are necessary to impose in order to make the variational ...

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First, the Lagrangian is not a conserved quantity, so $\frac{1}{2} m x \ddot{x} + V$ is not conserved. For example, for a particle in a constant gravitational potential with $V = m g x$, taking the solution $x(t) = - \frac{1}{2} g t^2$ would yield for this quantity $\frac{1}{4} m g^2 t ^2 - \frac{1}{2} m g^2 t^2 = -\frac{1}{4} m g^2 t^2$, which is ...

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