*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.

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$^1$ Hat tip: Valter Moretti.

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