Kinetic term in the Hamiltonian

[Silviu]

Homework Statement

Consider a point mass m attached to a string of slowly increasing length $l(t)$. Them motion is confined to a plane. Find L and H. Is H conserved? Is H equal to the total energy? Is the total energy conserved? Assume $|\dot{l}/l|<<\omega$

Homework Equations

The Attempt at a Solution

So I have $$L=\frac{1}{2}m(\dot{l}^2+l^2\dot{\theta}^2)+mglcos\theta$$. For H, for some reason the solution in the book doesn't calculate the $p_l$ but only $p_\theta$ and I am not sure why. The way I proceeded is: $$p_\theta=ml^2\dot{\theta}$$ and $$p_l=m\dot{l}$$ so $$H=\frac{1}{2}\frac{p_\theta^2}{ml^2}+\frac{1}{2}\frac{p_l^2}{m}-mglcos\theta$$. Is this correct by now? Now for the conservation I calculate $$\frac{dH}{dt}$$ which is not zero, as I will have terms such as $\dot{l}$ which are not 0 (is $\dot{p_\theta}=0$?). Now by the form of $H$ it is clear that $H=E_{tot}$ so the total energy is not conserved either. Is my reasoning correct?

 

[Orodruin]

$l$ is not a dynamic variable, it has an external constraint and hence there is no corresponding canonical momentum.

 

[Silviu]

$l$ is not a dynamic variable, it has an external constraint and hence there is no corresponding canonical momentum.

Thank you for the reply. Could you please give me a bit more details about it (I am new to hamiltonian mechanics)? What do you mean by the fact that it's not a dynamic variable? And what would be the external constrain? And when does a variable have a canonical momentum? I thought that each term in the kinetic part has a corresponding canonical momentum.

 

[Orodruin]

Thank you for the reply. Could you please give me a bit more details about it (I am new to hamiltonian mechanics)? What do you mean by the fact that it's not a dynamic variable? And what would be the external constrain? And when does a variable have a canonical momentum? I thought that each term in the kinetic part has a corresponding canonical momentum.

The length is already a fixed function of time, this is your constraint and thus not a degree of freedom.

 

[Silviu]

The length is already a fixed function of time, this is your constraint and thus not a degree of freedom.

I am not sure I understand. I agree it is a fixed function of time and it should be an external force acting on it (so that it compensate the tension in the string for example). But can't you say the same thing about the angle. It is also a fixed function of time and it is due to an external force (i.e. gravity). Why is l different than $\theta$?

 

[Orodruin]

It is also a fixed function of time and it is due to an external force (i.e. gravity).

No, you are not correct here. The function $l(t)$ is set from the beginning and the constraining force is whatever it needs to be to accomplish this. The dynamics appear only in the angle $\theta$, which is not externally constrained. The difference lies in whether or not you allow your system to have displacements. You can have different initial conditions on $\theta$, leading to different solutions, but $l(t)$ is being fixed to some function.

 

[Silviu]

No, you are not correct here. The function $l(t)$ is set from the beginning and the constraining force is whatever it needs to be to accomplish this. The dynamics appear only in the angle $\theta$, which is not externally constrained. The difference lies in whether or not you allow your system to have displacements. You can have different initial conditions on $\theta$, leading to different solutions, but $l(t)$ is being fixed to some function.

Oh ok, I got it now. However, isn't this change in l associated with a kinetic energy? And a kinetic energy has a momentum associated with it? I understand that both mathematically and physically $\dot{l}$ and $\dot{\theta}$ are different but why can't you call $\frac{\partial L}{\partial l}$ a momentum?

 

[Orodruin]

However, isn't this change in l associated with a kinetic energy?

Yes, but you already have that term in the Lagrangian. The term does not disappear.

And a kinetic energy has a momentum associated with it?

No it does not. A degree of freedom represented by a generalised coordinate has a canonical momentum associated with it.

but why can't you call $\frac{\partial L}{\partial l}$ a momentum?

You can call it whatever you want, it will not be standard nomenclature though. Canonical momenta are related to the degrees of freedom of the system.