# Can a Hamiltonian be formed from this Lagrangian?

1. Jun 17, 2010

### pellman

$$L=\frac{1}{2}m(\dot{q}_1-\dot{q}_2)^2-V(q_1,q_2)$$

Because if we put

$$p_1=\frac{\partial L}{\partial \dot{q}_1}$$
$$p_2=\frac{\partial L}{\partial \dot{q}_2}$$

we get

$$p_1=-p_2=m(\dot{q}_1-\dot{q}_2)$$

We can't invert to get $$\dot{q_1}$$ in terms of the two momenta. We can still write down a Hamiltonian of sorts since

$$H=p_1\dot{q}_1+p_2\dot{q}_2-L$$

$$=p_1(\dot{q}_1-\dot{q}_2)-L$$

$$=\frac{p_1^2}{m}-L$$

$$=\frac{p_1^2}{2m}+V(q_1,q_2)$$

or equivalently

$$=\frac{p_2^2}{2m}+V(q_1,q_2)$$

or

$$=\frac{p_1^2}{4m}+\frac{p_2^2}{4m}+V(q_1,q_2)$$

The main thing then is that we can't get an equation of motion which looks like

$$\dot{q_j}=\frac{\partial H}{\partial p_j}$$

What do we do with Lagrangians like this? Does the Hamiltonian method just fail?

2. Jun 17, 2010

### K^2

I think you'll find that the only potential that doesn't cause a problem is the one where $V(q_1,q_2)=V(q_1-q_2)$ Otherwise, your problem is not well-defined.

The q1 and q2 can increase/decrease simultaneously with zero generalized momentum. If that makes change in potential, you effectively have a mode with zero mass and finite force. That's bad.

That's really what you are seeing near the end. You really only have one degree of freedom: q1-q2. You tried to write the equation as if there are two. You ended up with a degeneracy. So naturally, the Hamiltonian cannot be used to distinguish two degenerate degrees of freedom. Define the new coordinate q=(q1-q2), and the problem resolves itself.

3. Jun 17, 2010

### Mute

Try changing your lagrangian's basis. Let $Q_1 = q_1 - q_2$, $Q_2 = q_1 + q_2$. Take derivatives with respect to these new variables.

$$\mathcal L = \frac{\dot{Q_1}^2}{2m} - V\left(\frac{Q_1+Q_2}{2},\frac{Q_2-Q_1}{2}\right)$$

4. Jun 17, 2010

### pellman

Quite right. Thanks, both.