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Lagrangian and Hamiltonian EOM with dissipative force

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I am trying to write the Lagrangian and Hamiltonian for the forced Harmonic oscillator before quantizing it to get to the quantum picture. For EOM $$m\ddot{q}+\beta\dot{q}+kq=f(t),$$ I write the Lagrangian $$ L=\frac{1}{2}m\dot{q}^{2}-\frac{1}{2}kq^{2}+f(t)q$$ with Rayleigh dissipation function as $$ D=\frac{1}{2}\beta\dot{q}^{2}$$ to put in Lagrangian EOM $$0 = \frac{\mathrm{d}}{\mathrm{d}t} \left ( \frac {\partial L}{\partial \dot{q}_j} \right ) - \frac {\partial L}{\partial q_j} + \frac {\partial D}{\partial \dot{q}_j}. $$

On Legendre transform of $L$, I get $$H=\frac{1}{2m}{p}^{2}+\frac{1}{2}kq^{2}-f(t)q.$$

How do I include the dissipative term to get the correct EOM from the Hamiltonian's EOM?


This post imported from StackExchange Physics at 2015-07-29 19:11 (UTC), posted by SE-user user56199

asked Nov 18, 2014 in Theoretical Physics by user56199 (15 points) [ revision history ]
edited Jul 29, 2015 by Dilaton

1 Answer

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Problem: Given Newton's second law

$$\tag{1} m\ddot{q}^j~=~-\beta\dot{q}^j-\frac{\partial V(q,t)}{\partial q^j}, \qquad j~\in~\{1,\ldots, n\}, $$

for a non-relativistic point particle in $n$ dimensions, subjected to a friction force, and also subjected to various forces that have a total potential $V(q,t)$, which may depend explicitly on time.

I) Conventional approach: Following the terminology of this Phys.SE post, there is a weak formulation of Lagrange equations of second kind

$$\tag{2} \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{q}^j}\right)-\frac{\partial L}{\partial q^j}~=~Q_j, \qquad j~\in~\{1,\ldots, n\},$$

where $Q_j$ are the generalized forces that don't have generalized potentials. In our case (1), the Lagrangian in eq. (2) is $L=T-V$, with $T=\frac{1}{2}m\dot{q}^2$; and the force

$$\tag{3} Q_j~=~-\beta\dot{q}^j$$

is the friction force. It is shown in e.g. this Phys.SE post that the friction force (3) does not have a potential. As OP mentions, one may introduce the Rayleigh dissipative function, but this is not a genuine potential.

Conventionally, we demand that the Lagrangian is of the form $L=T-U$, where $T=\frac{1}{2}m\dot{q}^2$ is related to the LHS of EOM (1) (i.e. the kinematic side), while the potential $U$ is related to the RHS of EOM (1) (i.e. the dynamical side).

With these requirements, the EOM (1) does not have a strong formulation of Lagrange equations of second kind

$$\tag{4} \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{q}^j}\right)-\frac{\partial L}{\partial q^j}~=~0,\qquad j~\in~\{1,\ldots, n\}, $$

i.e. Euler-Lagrange equations. The Legendre transformation to the Hamiltonian formulation is traditionally only defined for a strong formulation (4). So there is no conventional Hamiltonian formulation of the EOM (1).

II) Unconventional approach$^1$: Define for later convenience the function

$$\tag{5} e(t)~:=~\exp(\frac{\beta t}{m}). $$

A possible strong formulation (4) of Lagrange equations of second kind is then given by the Lagrangian

$$\tag{6} L(q,\dot{q},t)~:=~e(t)L_0(q,\dot{q},t), \qquad L_0(q,\dot{q},t)~:=~\frac{m}{2}\dot{q}^2-V(q,t).$$

The corresponding Hamiltonian is

$$\tag{7} H(q,p,t)~:=~\frac{p^2}{2me(t)}+e(t)V(q,t).$$

The caveat is that the Hamiltonian (7) does not represent the traditional notion of total energy.

--

$^1$ Hat tip: Valter Moretti.

This post imported from StackExchange Physics at 2015-07-29 19:11 (UTC), posted by SE-user Qmechanic
answered Nov 18, 2014 by Qmechanic (2,790 points) [ no revision ]

For what stands "EOM"?

EOM = equation of motion

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