How to get lagragean when hamiltonian is given

[vrinda mukund]

hai everyone,

the question is " the hamiltonian of a particle is H = [(p*p)/2m + pq] where q is the generalised coordinate and p is the corresponding canonical momentum. the lagragean is ....?

i know that H = p(dq/dt) - L. but the answer should not contain p. how can i solve it? answer is (m/2)(dq/dt - q)2

 

[G01]

Remember Hamilton's equations, specifically:

$$\dot{q}=\frac{\partial H}{\partial p}$$

Use this to get an expression for $p$ in terms of $\dot{q}$.

 

[vanhees71]

You've given the answer yourself, already: You calculate the Lagrangian,

$$L=\dot{q} p -H$$

and express $p$ via $\dot{q}$, where

$$\dot{q}=\frac{\partial H}{\partial p}.$$

In your case you get

$$\dot{q}=\frac{p}{m}+q \; \Rightarrow \; p=m(\dot{q}-q).$$

and

$$L=\dot{q} p-\frac{p^2}{2m}-p q<br /> =m \dot{q} (\dot{q}-q) - \frac{m}{2} (\dot{q}-q)^2 - m q(\dot{q}-q).$$

A bit simplified this gives

$$L=\frac{m}{2} (\dot{q}-q)^2.$$

 

[vrinda mukund]

thank u sir!

 

[jewbinson]

An important point here is that the Lagrangian is always a function of q, dq/dt, and time, whereas the Hamiltonian is always a function of q, p and time. So to get from the Hamiltonian to the Lagrangian we must get rid of p, and to get from the Lagrangian to the Hamiltonian we must get rid of dq/dt