Meaning of the Hamiltonian when it is not energy

[Happiness]

Suppose the initial radial position and radial velocity of the bead are $r_0>0$ and $0$ respectively. Then $E$ is negative. Is there any significance to the negative value of $E$? Note that $E$ is defined by (5.52) and given by (5.144) below.

 

[BvU]

Hi happy,
Well what do you think of the paragraph 'The main point here ... ' ?

 

[Happiness]

Hi happy,
Well what do you think of the paragraph 'The main point here ... ' ?

Hi BvU,

It suggests that $E$ is the energy of the system $+$ work done on the system $+$ a constant, right? And in this case, that constant is negative.

 

[BvU]

In this case L = T (there is no V). That is all the energy in the system. If the driving force to keep $\omega$ constant stops acting, then I suppose T will remain constant.
I think you have to solve the EL equations to find the constraint forces and that way you'll find their work input is $m\omega^2$. All the kinetic energy change comes from the driving force (torque).

 

[Happiness]

In this case L = T (there is no V). That is all the energy in the system. If the driving force to keep $\omega$ constant stops acting, then I suppose T will remain constant.
I think you have to solve the EL equations to find the constraint forces and that way you'll find their work input is $m\omega^2$. All the kinetic energy change comes from the driving force (torque).

I can't find the constraint force using the standard method of EL equations with Lagrange undetermined multipliers:

$$\frac{d}{dt}\frac{\partial L}{\partial\dot q_i}=\frac{\partial L}{\partial q_i}+F_i\frac{\partial\eta}{\partial q_i}$$

where $F_i$ is the generalized constraint force and $\eta=0$ is the constraint equation. In this case, $\eta=\dot\theta-\omega=0$. Since $\eta$ does not depend on $r$ or $\theta$, $\frac{\partial\eta}{\partial q_i}=0$ and the above equations reduce to the usual EL equations without $F$.

How do you solve for the constraint force?

 

[vanhees71]

I do not understand, why the textbook says, $E$ is not the energy of the bead. It depends on the definition of the book. For me energy is the conserved quantity defined via Noether's theorem as the generator of spatial translations, and that's the Hamiltonian. It's clear that the kinetic energy for itself is not conserved since you apply work via the external force to keep $\omega=\text{const}$, but why $E$ is not called the total energy of the bead here, is not clear to me.

 

[Happiness]

I do not understand, why the textbook says, $E$ is not the energy of the bead. It depends on the definition of the book. For me energy is the conserved quantity defined via Noether's theorem as the generator of spatial translations, and that's the Hamiltonian. It's clear that the kinetic energy for itself is not conserved since you apply work via the external force to keep $\omega=\text{const}$, but why $E$ is not called the total energy of the bead here, is not clear to me.

Your question is answered by the following remark.

 

[vanhees71]

Ok, then the book uses a different definition of energy than I do. Then it's of course consistent in itself.