# Variables in lagrangian vs hamiltonian dynamics

1. Sep 20, 2014

### copernicus1

In the lagrangian formalism, we treat the position $q$ and the velocity $\dot q$ as dependent variables and talk about configuration space, which is just the space of positions. In the hamiltonian formalism we talk about canonical positions and momenta, and we consider them independent. Is the independence based on the additional parameters in momenta (namely the mass), or is it based on the fact that the canonical momentum is separate from the physical momentum?

Thanks!

Last edited by a moderator: Sep 20, 2014
2. Sep 20, 2014

### Einj

The choice of which parameters are independent from each other is completely up to you. In lagrangian mechanics one chooses $q$ and $\dot q$, so you have two independent variables. When you define the conjugate momentum you are going to have:
$$\frac{\partial L}{\partial \dot q}=p(q,\dot q).$$
Now, there is nothing wrong in inverting this relation to obtain $\dot q(q,p)$ and then re-write everything in terms of $q$ and $p$, you will still have two independent variables, just with a different meaning. This is nothing but a change of variables.

3. Sep 20, 2014

### copernicus1

I think you may have misread my question. My question was referring to the dependence of q and q-dot in the Lagrangian formalism versus the independence of q and p in the Hamiltonian. In Lagrangian dynamics, q and q-dot are not independent. I'm wondering about the difference between the two formalisms.

4. Sep 20, 2014

### Einj

Actually in the lagrangian formalism q and q-dot are treated as independent variables. When, for example, you take the variation of the action you derive first w.r.t. q and then w.r.t. q-dot, this can only be done if they are considered as independent, otherwise you would have to use the chain rule.

5. Sep 21, 2014

### voko

That is false. You would not have to use the chain rule, you merely might, which would lead you nowhere. Lagrange's insight was that instead of using the chain rule, one should use integration by parts, which eliminates q-dot (except he did not use the dot notation all) and results in Euler-Lagrange equations immediately.

Neither in Lagrangian nor in Hamiltonian formalism are the variables truly independent. As an example, take an oscillator, whose (reduced) Lagrangian is ${\dot q^2 \over 2} - k{q^2 \over 2}$, the E-L equation is $\ddot q + k q = 0$. This equation can be trivially converted to $\dot q^2 + k q^2 = h$, where the dependence between $q$ and $\dot q$ is manifest. Why this example proves that the variables in the Hamiltonian formalism are likewise not independent is left as an exercise.