7
$\begingroup$

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 by 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.

$\endgroup$
  • $\begingroup$ To get the Hamiltonian you extend the Lagrangian to $ L\rightarrow L\left( \overrightarrow{q},\overrightarrow{\dot q}\right) +\overrightarrow{q}\cdot \overrightarrow{f}\left( t\right) $ $\endgroup$ – Eli Jul 23 at 6:09
5
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$
  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, then there is no conventional Hamiltonian formulation. However, several unconventional approaches exist, cf. this related Phys.SE post.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.