Nonconservative lagrangian systems

[jdstokes]

Hi,

I'm in the process of self-teaching myself Lagrangian and Hamiltonian dynamics. In my readings, I've found that some velocity dependent forces (e.g. the Lorentz force) can be derived from a classical Lagrangian whereas others such as friction cannot.

Do there exist necessary and sufficient conditions to decide when velocity dependent forces are obtainable from a Lagrangian, in the same way that velocity independent forces can be obtained from a lagrangian iff they are the gradient of a potential function?

James

 

[pervect]

Hi,

I'm in the process of self-teaching myself Lagrangian and Hamiltonian dynamics. In my readings, I've found that some velocity dependent forces (e.g. the Lorentz force) can be derived from a classical Lagrangian whereas others such as friction cannot.

Do there exist necessary and sufficient conditions to decide when velocity dependent forces are obtainable from a Lagrangian, in the same way that velocity independent forces can be obtained from a lagrangian iff they are the gradient of a potential function?

James

If all the forces on a system can be derived from a potential, you can use Lagrange's equation.

If you have forces that can't be derived from a potential, one has the option of re-writing Lagrange's equation (Goldstein, Classical mechanics, pg 23-24)

$$\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q_j}} \right) - \frac{\partial L}{\partial q_j} = Q_j$$

where $Q_j$ is a generalized force that can include forces not derivable from a potential. In many cases, $Q_j$ can be specified via a dissipation function $\mathcal{F}$ so that $$Q_j = \frac{\partial \mathcal{F}}{\dot{q_j}}$$

One then needs to specify L and $\mathcal{F}$ to get the equations of motion.

 

[jdstokes]

Hi pervect,

Thanks for the reply.

I think we need to distinguish between deriving forces from the Lagrangian and adding them directly to the RHS of the E-L equations.

For the Lorentz force, we start with the Lagrangian $L = (1/2)m \dot{\mathbf{r}} \cdot \dot{\mathbf{r}} + e V(\mathbf{r},t) + e \dot{\mathbf{r}}\cdot \mathbf{A}(\mathbf{r},t)$ and then apply the standard, unmodified E-L equation $\frac{d}{dt}\frac{\partial L}{\partial \dot{\mathbf{r}}} - \frac{\partial L}{\partial \mathbf{r}}$ to obtain the velocity-dependent Lorentz force $m\ddot{\mathbf{r}} = e[\mathbf{E} + \dot{\mathbf{r}}\times \mathbf{B}]$. So we can say that these equations of motion are obtainable directly from a Lagrangian.

If we consider linear frictional forces, on the other hand, we cannot obtain these directly from a Lagrangian, we must add them in later to the RHS of the E-L equations as you describe.

My question: what determines whether a velocity-dependent force is directly obtainable from a Lagrangian?

Thanks

James

 

[StatusX]

The second to last line in the derivation of Lagrange's equations is:

$$\frac{d}{dt}\left(\frac{\partial T}{\partial \dot q_i} \right) - \frac{\partial T}{\partial q_i} = Q_i$$

where Q_i is the generalized force. Then if there is a function V(q_1,...,q_n,t) with:

$$\frac{\partial V}{\partial q_i} = - Q_i$$

Then you get lagrange's equations (note this holds in one coordinate system iff it holds in all of them). But more generally, as long as you can find a function $V(q_1,...,q_n,\dot q_1,...,\dot q_n,t)$ with:

$$Q_i = \frac{d}{dt}\left(\frac{\partial V}{\partial \dot q_i} \right) - \frac{\partial V}{\partial q_i}$$

Then you can just take it as your V when you define L=T-V and you get the correct equations. It can be shown that the lorentz force is of this form with:

$$V = q \phi - q \vec v \cdot \vec A$$