The Lagrangian formalism in general does not work for dissipative forces (fricti

Question

The Lagrangian formalism in general does not work for dissipative forces (friction, viscosity etc.) since such forces do not arise from a potential. Sometimes, however, we can find a clever formulation that achieves a Lagrangian description for some of these systems. Consider an action of the form L = 1/2mf(x)x.2 - U(x) where x is the coordinate of a particle and f(x),U(x) arbitrary functions of x. Derive the equation of motion for x and bring it to the form m = F(x,x.) to identify an effective force that may depend on x and x. Choose the function f(x) appropriately such that the force becomes - 1/2cx.2 - V'(x) with c a positive constant. This represents a particle feeling a conservative force -V'(x) as well as a resistance force -1/2cx.2 proportional to the square of its speed, like air resistance in the turbulent regime. What is the effective potential V(x) in terms of U(x) and c? With the above f(x) in the Lagrangian, find the canonical momentum p and the Hamiltonian H(x,p) for the particle. H is conserved, while in the presence of a resistance force there is no energy conservation. Does H represent the usual energy of a particle of mass m? Use the above constant of the motion to find the general solution for x(t) in the case U = 0 (only resistance force). Bonus question: does the above Lagrangian really describe a resistance force in all situations? (Hint: what happens to the force if x.

Explanation / Answer

Traditional Lagrangian and Hamiltonian mechanics cannot be used with nonconservative forces such as friction. A method is proposed that uses a Lagrangian containing derivatives of fractional order. A direct calculation gives an Euler-Lagrange equation of motion for nonconservative forces. Conjugate momenta are defined and Hamilton’s equations are derived using generalized classical mechanics with fractional and higher-order derivatives. The method is applied to the case of a classical frictional force proportional to velocity Hamiltonian, Lagrangian and Newtonian mechanics all predict the same behavior for systems, but they are of different difficulties, mathematically, for different systems. Some systems might be easy to solve Newtonianly, and a pain in the $&^% otherwise. Alternatively, the simplest forms of Lagrangian mechanics (and Hamiltonian, for that matter) only work on systems in which energy is conserved. Non-conservative forces, like aerodynamic drag, friction, damping in simple harmonic motion, are better dealt with using the Newtonian formulation. In many cases, the end result of a Newtonian analysis or a Lagrangian/Hamiltonian is a system of differential equations. And in most cases, that system will be the same (or equivalent*). They are just different methods for arriving at that system. Now, as to the nuts and bolts of the differences: In Newtonian mechanics, you are listing all the forces acting on a particle and using that build a differential equation that describes the behavior of that particle. Then you (attempt to) solve that equation and you have an analytic solution for the position (and therefore the velocity) of that particle. In Lagrangian mechanics, you are using what is called an action principle. For very subtle mathematical reasons, you know the action is minimized over the trajectory of the particle in a conservative force field. The Lagrangian action is L=T-U (T is kinetic energy, U is potential energy). Minimizing that action yields the Euler-Lagrange equations, a system of differential equations that describes the behavior of the particle. In Hamiltonian mechanics, you use the Lagrangian (L, from the previous paragraph) to calculate the Hamiltonian. In systems where energy is conserved, the Hamiltonian is just the total energy (all forms of kinetic, all forms of potential, etc.... all added up). Then you use Hamilton's equations to give you a system of first order differential equations for quantities that are called generalized momentum and generalized position. Quantum mechanics is formulated almost entirely in terms of the Hamiltonian. You'll also find other action principles (not necessarily the Lagrangian action) elsewhere in physics. For example, in optics, you are using an action principle (though you may not know it). *By "equivalent", I mean the following. A Newtonian analysis might give you an N-fold system of 2nd order differential equations whereas the Hamiltonian analysis will give you a 2N-fold system of 1st order differential equations. They will be equivalent if they yield the same solutions for the dependent variable and have the same number of arbitrary parameters (or necessary boundary conditions).

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