How do I know if the Hamiltonian is constant?

[komodekork]

Lets say $$H = \frac{m}{2} (\dot{Q}^2 - \omega^2 Q^2 )$$
where Q is the generalized coordinate.
It doesn't explicitly depend on time, but the Q and the $$\dot{Q}$$ does.
If i differentiate it with respect to time it should be zero if it's constant, right?
So i guess my question is should i treat the Q's as constants or as functions depending on time when i differentiate?

 

[RedX]

Lets say $$H = \frac{m}{2} (\dot{Q}^2 - \omega^2 Q^2 )$$
where Q is the generalized coordinate.
It doesn't explicitly depend on time, but the Q and the $$\dot{Q}$$ does.
If i differentiate it with respect to time it should be zero if it's constant, right?
So i guess my question is should i treat the Q's as constants or as functions depending on time when i differentiate?

If you want to view it like that, then if the Hamiltonian is explicitly independent of time, then it just comes from equality of mixed partials.

$$\dot{H}=\frac{\partial H}{\partial q}\dot{q}+\frac{\partial H}{\partial p}\dot{p} = \frac{\partial H}{\partial q} \left(\frac{\partial H}{\partial p}\right)+ \frac{\partial H}{\partial p}\left(-\frac{\partial H}{\partial q}\right) =0$$

 

[komodekork]

If you want to view it like that, then if the Hamiltonian is explicitly independent of time, then it just comes from equality of mixed partials.

I don't understand what you mean, could you please explain?

 

[RedX]

I don't understand what you mean, could you please explain?

oops, you're right. it's not equality of mixed partials, but comes from Hamilton's equations.

so take H=.5(p²+q²)

Then dH/dp=p and dH/dq=q
But p^.=-q and q^.=p

So H^.=p(-q)+q(p)=0

 

[komodekork]

Hm...
I don't know if I see what you are getting at. I'm not sure if you are telling me that what you did is right or wrong.

Should it be $$\frac{d}{dt}(H) = m ( \dot{Q} \ddot{Q} - \omega^2 Q \dot{Q} )$$

or just

$$\frac{d}{dt}(H) = 0$$

 

[RedX]

Hm...
I don't know if I see what you are getting at. I'm not sure if you are telling me that what you did is right or wrong.

Should it be $$\frac{d}{dt}(H) = m ( \dot{Q} \ddot{Q} - \omega^2 Q \dot{Q} )$$

or just

$$\frac{d}{dt}(H) = 0$$

oops, sorry. First of all, your Hamiltonian doesn't represent an oscillator. If it does, then the second term should be + rather than -:
$$H = \frac{m}{2} (\dot{Q}^2 + \omega^2 Q^2 )$$

So as you say: $$\frac{d}{dt}(H) = m ( \dot{Q} \ddot{Q} + \omega^2 Q \dot{Q} )$$

Now from the equations of motion of an oscillator, $$\ddot{Q}=-\omega^2 Q$$. PLugging that in should get you zero.

 

[komodekork]

I don't know what my Hamilton represents. I may have done something wrong.

I all i got is this lagrangian $$L = \frac{m}{2} (\dot{q}^2 sin^2(\omega t) + \dot{q} q \omega sin(2\omega t) + \omega^2 q^2)$$
and this new coordinate $$Q = q sin(\omega t)$$

after this substitution i get $$L = \frac{m}{2} (\dot{Q}^2 + \omega^2 Q^2)$$

then i make the Hamiltonian $$H = p\dot{Q} - L$$

and get $$H = \frac{m}{2}(\dot{Q}^2 - \omega^2 Q^2)$$



Then the question is, is the Hamiltonian constant?

Thanks for helping me out btw.

 

[komodekork]

But forget all of that, just in a genreal case,
if $$H=H( \dot{q},q)$$ is it constant just because it's not $$H=H(p,q,t)$$?
Or because it cancels out?

 

[qbert]

you need to write your Hamiltonian in terms of p and q not in terms of q and q-dot,
and THEN if time doesn't appear explicitly it's conserved.

 

[RedX]

I don't know what my Hamilton represents. I may have done something wrong.

I all i got is this lagrangian $$L = \frac{m}{2} (\dot{q}^2 sin^2(\omega t) + \dot{q} q \omega sin(2\omega t) + \omega^2 q^2)$$
and this new coordinate $$Q = q sin(\omega t)$$

after this substitution i get $$L = \frac{m}{2} (\dot{Q}^2 + \omega^2 Q^2)$$

then i make the Hamiltonian $$H = p\dot{Q} - L$$

and get $$H = \frac{m}{2}(\dot{Q}^2 - \omega^2 Q^2)$$



Then the question is, is the Hamiltonian constant?

Thanks for helping me out btw.

Can't you just plug in $$Q = q sin(\omega t)$$ into your Hamiltonian $$H = \frac{m}{2}(\dot{Q}^2 - \omega^2 Q^2)$$, and see if the time-dependence vanishes?