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How do Hamilton's equations deal with non-conservative forces?

Physics Asked on June 2, 2021

I have searched everywhere I know to look but I cannot find out how Hamilton’s equations deal with non-conservative forces. In my understanding, Lagrangian mechanics deals with this as follows: the Euler-Lagrange equations no longer have a zero on the right, they have a term $$Sigma F_q$$ that is the sum of all the non-conservative forces encountered in the direction of the coordinate q.

begin{equation}
frac{d}{dt} frac{partial L}{partial dot{q}} – frac{partial L}{partial q}=Sigma F_q(t)
end{equation}

The only document I have been able to find about how hamiltonian mechanics deals with non-conservative forces has been: https://doi.org/10.1007/BF00692025

It requires you to buy it and I felt like just the plain equations and a little context for an example like a box sliding down a hill against friction would be enough.

2 Answers

Lorentz force is an example of non-conservative conservative force that is discussed in pretty much any theoretical mechanics textbook. The Lagrangian is: $$L = frac{m}{2}dot{mathbf{r}}cdotdot{mathbf{r}} + q mathbf{A}(mathbf{r})cdotdot{mathbf{r}} - qphi(mathbf{r}),$$ from which the momentum, the Hamiltonian, and the Hamilton equations follow as usual.

Answered by Roger Vadim on June 2, 2021

  1. If a variational formulation of Lagrange equations exists (e.g. because of the existence of a generalized velocity-dependent potential $U(q,dot{q},t)$ for the forces in the problem), then we can in principle derive a Hamiltonian formulation via a Legendre transformation in the standard manner.

  2. If no variational Lagrangian formulation exists for Lagrange equations $$begin{align}frac{d}{dt}frac{partial L}{partial dot{q}^j}-frac{partial L}{partial q^j}~=~&Q_j, cr j~in~& {1,ldots, n}, end{align}tag{L}$$ but the Lagrangian $L$ is regular/non-degenerate, then we may still derive a Hamiltonian $H$ via a Legendre transformation. However Hamilton's equations get modified $$begin{align} dot{q}^j~=~frac{partial H}{partial p_j}, & qquad dot{p}_j+frac{partial H}{partial q^j}~=~Q_j, cr j~in~&{1,ldots, n}. end{align}tag{H}$$ So there is no conventional Hamiltonian formulation. Nevertheless, several unconventional approaches exist, cf. this related Phys.SE post.

Answered by Qmechanic on June 2, 2021

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