Problem: Given Newton's second law
m¨qj = −β˙qj−∂V(q,t)∂qj,j ∈ {1,…,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
ddt(∂L∂˙qj)−∂L∂qj = Qj,j ∈ {1,…,n},
where Qj 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=12m˙q2; and the force
Qj = −β˙qj
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=12m˙q2 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
ddt(∂L∂˙qj)−∂L∂qj = 0,j ∈ {1,…,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 approach1: Define for later convenience the function
e(t) := exp(βtm).
A possible strong formulation (4) of Lagrange equations of second kind is then given by the Lagrangian
L(q,˙q,t) := e(t)L0(q,˙q,t),L0(q,˙q,t) := m2˙q2−V(q,t).
The corresponding Hamiltonian is
H(q,p,t) := p22me(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