# The Hamiltonian

1. Mar 8, 2006

### stunner5000pt

This is a question, not a homework problem, as i am currently studying for my test on classical mechanics
suppose $$H = \sum_{i} \dot{q_{i}}(p,q,t) p_{i} - L(p,q,t)$$
also i can prove that
$$dH = \sum_{i} (\dot{q_{i}}dp_{i} - \dot{p_{i}} dq_{i}) - \frac{\partial L}{\partial t} dt$$

suppose $$\frac{\partial L}{\partial t} = \frac{\partial H}{\partial t} = 0$$ then dH/dt = 0 i.e. H = E

now if i sub in to the equation above i get
$$dH = \sum_{i} (\dot{q_{i}}dp_{i} - \dot{p_{i}} dq_{i})$$

how would i transform the above dH into dH/dt formally?

Do i simply integrate by parts to get H and then differentiate wrt t??

Last edited: Mar 8, 2006
2. Mar 9, 2006

### dextercioby

Well, just use that

$$dH=\frac{dH}{dt} dt$$

which is an equality between 2 differential 1-forms which is valid iff H=H(t) is the curve it's being evaluated on.

Daniel.

3. Mar 9, 2006

### stunner5000pt

oh alright

what about going from this point
$$dH = \sum_{i} (\dot{q_{i}}dp_{i} - \dot{p_{i}} dq_{i}) - \frac{\partial L}{\partial t} dt$$

to go on and say that
$$\frac{dH}{dt} = \frac{\partial L}{\partial t}$$
what happens to the summed terms??

q dot doesnt depend on p and p dot doesnt depend on q so
what we get is
$$\dot{q} p - \dot{p} q$$
is that zero??

4. Mar 9, 2006

### stunner5000pt

do i simply use the idea which you suggested, dexter, on the dp and dq terms? THat was we get dots over all and they cancel out

5. Mar 9, 2006

### topsquark

You are "dividing" by dt in the whole expression, so you get terms of the form $$\dot{q_{i}} \dot{p_{i}} - \dot{p_{i}} \dot{q_{i}}$$ in the summation for all i. So what DOES happen to the summed terms?

-Dan

6. Mar 9, 2006

### pmb_phy

Simply because dH/dt = 0 it doesn't mean that H = E. E.g. See "Classical Mechanics 3rd Ed.," Goldstein, Safko and Poole page 334-346.

Pete

7. Mar 9, 2006

### stunner5000pt

let me correct that by saying H is constant with respect to time, but not necessarily the energy. I wil check Goldstein's book tomorrow

since we 'dividing' by dt we get this right
$$\frac{dH}{dt} dt = \sum_{i} (\dot{q_{i}}\frac{dp_{i}}{dt} dt - \dot{p_{i}} \frac{dq_{i}}{dt} dt) - \frac{\partial L}{\partial t} dt$$

and once we integrate we go $$\dot{p} \dot{q} - \dot{q} \dot{p}$$
so tahts why the summed terms are zero

8. Mar 9, 2006

### Physics Monkey

Hi stunner,

Indeed, if L contains no explicit time dependence then neither does H, and this means H is conserved since $$dH/dt = \partial H/\partial t$$ by virtue of the equations of motion. The question of rather H is equal to the energy has little to do with the question of time dependence and depends rather on the structure of the kinetic energy term.

9. Mar 9, 2006

### stunner5000pt

isnt hte kinetic energy term for all purpose always p^2/2m?? Perhaps i have not come across situations where it is otherwise. When would it be different?

10. Mar 9, 2006

### Physics Monkey

Very often the kinetic energy is of the form $$\frac{1}{2} m v^2$$, but your choice of generalized coordinates can change the formal structure of the Lagrangian. For example, consider a bead constrained to move on a rod that rotates in the plane at some frequency $$\omega$$. The kinetic energy is just given by $$T = \frac{1}{2} m \dot{r}^2 + \frac{1}{2} m r^2 \dot{\theta}^2$$. You can solve the constraint immediately by setting $$\dot{\theta} = \omega$$ and then leave $$\theta$$ out of your generalized coordinates. The Lagrangian then looks like $$L = T = \frac{1}{2} m \dot{r}^2 + \frac{1}{2} m r^2 \omega^2$$. If you construct the Hamiltonian you will find that it is $$H = \frac{p^2_r}{2 m} - \frac{1}{2} m r^2 \omega^2$$ where $$p_r = m \dot{r}$$ which is clearly not the total kinetic energy. Nevertheless, the Hamiltonian is conserved and the total energy is not. The key here was how you treat your constraints and generalized coordinates. If you had left $$\theta$$ as a generalized coordinate and used the Lagrange multiplier scheme to deal with the constraint, then your Hamiltonian would be the total energy. However, the Hamiltonian wouldn't be conserved because of the time dependent constraint (an important caveat to the general result I quoted above).

Last edited: Mar 9, 2006