E=MC^2 Mass and Energy, synonymous?

[VantagePoint72]

But how would you take into account the potentially non-conservative external forces doing work on the system in the Lagrangian if you were to include them in the system? There are ways of taking into account non-conservative forces in the Euler-Lagrange equations themselves using virtual work but how would you incorporate them into the Lagrangian itself?

What makes you so sure the forces would be non-conservative? It's just a cyclic process of the hoop doing work on the (whatever) and the (whatever) doing work on the hoop. Admittedly, I can't, off hand, imagine what (whatever) needs to look like in order to drive the hoop at a constant angular frequency, but it looks to me like it would be a non-dissipative process (in the idealization, naturally).

Besides, non-conservative forces are generally just a short-cut. A full (and I mean full) Lagrangian of the bead+hoop+environment would not have them: friction is nothing more than the transfer of kinetic energy of a macroscopic object to a bunch of microscopic particles. However, rather than writing down a Lagrangian for every single particle in the system, we just take a short cut and say that friction causes some of the kinetic energy to be lost to thermal energy. But thermal energy is just kinetic energy by a different name! Of course, to write down this full Lagrangian, you have to go well beyond classical mechanics, which is why I think it's fair to criticize the Hamiltonian-as-energy definition—at least, it's fair to do so in classical physics.

So, to sum up, I don't think it's immediately clear that a Lagrangian for the full, isolated system would require dissipative forces. Even if it did, that's just because we're only using an approximate description of the system in which the composition of matter is neglected. So, as I said, I think interpreting your example as proof that the Hamiltonian is not a good definition of the total energy is a mistake.

 

[WannabeNewton]

I should have been clear in that I wasn't referring to the hoop scenario necessarily but rather a general scenario.

 

[VantagePoint72]

I see. While, from the very beginning I think I was clear that I criticizing your particular example, even going as far as saying that nonetheless "I think I agree with the point". With respect to the question of whether the Hamiltonian is suitable as the definition of energy, I am arguing that the bead and hoop example do not prove anything either way because they don't fulfill the conditions of the proposed definition (one of which is that the system be isolated). Do you agree?

 

[WannabeNewton]

I see. While, from the very beginning I think I was clear that I criticizing your particular example, even going as far as saying that nonetheless "I think I agree with the point". With respect to the question of whether the Hamiltonian is suitable as the definition of energy, I am arguing that the bead and hoop example do not prove anything either way because they don't fulfill the conditions of the proposed definition (one of which is that the system be isolated). Do you agree?

Oh sure I wasn't using it as an example refuting the universality of the Hamiltonian within classical systems. I was just giving an example of a situation where you cannot naively assume the Hamiltonian and the total mechanical energy of the assumed system are the same which is what I interpreted HomogenousCow as asking an example of. I think my above post regarding the ADM energy-momentum if anything provides somewhat more of a support that the Hamiltonian notion of energy even carries over to entire space-times (albeit restricted to asymptotically flat ones).

My overall point, examples aside, was that one cannot take the Hamiltonian as the end all be all, unequivocal "definition" of energy.

 

[HomogenousCow]

Is the hamiltonian always equal to total energy when we have a closed system where the particles only interact with each other through the field?

 

[WannabeNewton]

There are explicit conditions that the Lagrangian must satisfy in order for the Hamiltonian to equal to the total energy for a given system. See Goldstein edition 3 page 339.

 

[HomogenousCow]

What makes you so sure the forces would be non-conservative? It's just a cyclic process of the hoop doing work on the (whatever) and the (whatever) doing work on the hoop. Admittedly, I can't, off hand, imagine what (whatever) needs to look like in order to drive the hoop at a constant angular frequency, but it looks to me like it would be a non-dissipative process (in the idealization, naturally).

Besides, non-conservative forces are generally just a short-cut. A full (and I mean full) Lagrangian of the bead+hoop+environment would not have them: friction is nothing more than the transfer of kinetic energy of a macroscopic object to a bunch of microscopic particles. However, rather than writing down a Lagrangian for every single particle in the system, we just take a short cut and say that friction causes some of the kinetic energy to be lost to thermal energy. But thermal energy is just kinetic energy by a different name! Of course, to write down this full Lagrangian, you have to go well beyond classical mechanics, which is why I think it's fair to criticize the Hamiltonian-as-energy definition—at least, it's fair to do so in classical physics.

So, to sum up, I don't think it's immediately clear that a Lagrangian for the full, isolated system would require dissipative forces. Even if it did, that's just because we're only using an approximate description of the system in which the composition of matter is neglected. So, as I said, I think interpreting your example as proof that the Hamiltonian is not a good definition of the total energy is a mistake.

This is what I was thinking of, in a full relativistic theory all the interactions should be mediated by the field.

 

[Dale]

Our Lagrangian is then $L = \frac{1}{2}mR^{2}\dot{\varphi}^{2} + \frac{1}{2}m(R\sin\varphi)^{2}\Omega^{2} + mgR\cos\varphi$. Notice that $\frac{\partial L}{\partial t} = 0$ therefore the Hamiltonian $H = \sum p_{i}\dot{q_{i}} - L = \text{const.}$ and is given by $H = \frac{1}{2}mR^{2}\dot{\varphi}^{2}- \frac{1}{2}m(R\sin\varphi)^{2}\Omega^{2} - mgR\cos\varphi$ whereas the total mechanical energy is given by $E = \frac{1}{2}mR^{2}\dot{\varphi}^{2}+ \frac{1}{2}m(R\sin\varphi)^{2}\Omega^{2} - mgR\cos\varphi$ so the total energy is not equal to the Hamiltonian in this case. The reason $E$ is not conserved is because whatever is keeping the hoop rotating at a prescribed constant speed must be doing work on the system.

So you have basically gone over all of this with other posters, but I thought I would put in my thoughts anyway.

In this case, it seems to me that H is the correct expression for the total energy of the system. You are absolutely correct that it is not equal to the mechanical energy of the bead. The difference between the mechanical energy of the bead and the total energy is some other unspecified energy that must be present in order for the system to have a time-invariant Lagrangian. (It cannot be a non-conservative force or the Lagrangian could not be time invariant since eventually the energy lost to a non-conservative force will be used up.)

This definition of energy is still not general since it only applies to systems described by a Lagrangian and whose Lagrangian is time-invariant. But I think that where it does apply it correctly defines energy.

One thing that I like about your example is that it is a counter-example to Popper's assertion that the posted definitions define forms of energy. In this Lagrangian the form of the additional energy is completely undefined. We don't know what form it is, but we nonetheless know that it is present. He could still take a philosophical stance that the definitions define the amount of energy but not what energy itself "actually really" is (which is what I believe Feynman was doing). In any case, the "forms" of energy argument is disproven.

 

[WannabeNewton]

Interestingly enough, you can give meaning to the term in $H$ that differs from the term in $E$ if you identify that term in $H$ as the "potential energy" of the centrifugal force on the bead in the frame co-rotating with the hoop. This allows you to interpret $H$ in the "conventional" way.

 

[Dale]

Interestingly enough, you can give meaning to the term in $H$ that differs from the term in $E$ if you identify that term in $H$ as the "potential energy" of the centrifugal force on the bead in the frame co-rotating with the hoop. This allows you to interpret $H$ in the "conventional" way.

That is interesting. I was aware that centrifugal forces etc. do come out naturally when you have an angular coordinate, I guess that it should not be surprising that they come out when you have an angular constraint that is not a coordinate.