# Lagrangian and Hamiltonian EOM with dissipative force

+ 3 like - 0 dislike
154 views

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

edited Jul 29, 2015

+ 5 like - 0 dislike

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 (2,840 points)

For what stands "EOM"?

EOM = equation of motion

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysi$\varnothing$sOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.