Does the Hamiltonian is always equal to the energy of the system??

Casco

I have this doubt since a few weeks ago. For the newtonian case we have that H=K+U, kinetical energy plus potential energy, but given that the definition of the Hamiltonian is H=[itex]\dot{q}[/itex]P-L, my question is, Does exist a system or a "type" of systems that their Hamiltonian is not H=K+U? A example of this could be a Hamiltonian with the form H=K+U+f(t).

Perhaps my question can be modeled like this, Does exist a theorem which proves that the hamiltonian for a given system is always of the form H=K+U??

fluidistic

I think that when there are dissipative forces -like friction force- in a system, the Hamiltonian of the system is not worth K+U.

andresordonez

I think the answer is no, but I'm not sure, you should check section 2.7 of Classical Mechanics 3rd Ed, Goldstein.

Try to improve your English writing and LaTeX skills, some of the words you wrote (like dude meaning doubt) I was able to guess just because my first language is Spanish.

Casco

Does your answer is for my first or second question?? And sorry about my writing, I have a few weeks writing in english (my first language is spanish too), I'll try to get better. About latex, I'm already looking for a tutorial.

juanrga

First, [itex]H=\dot{q}p-L[/itex] is a Legendre transformation for obtaining a Hamiltonian from a given Lagrangian. It can inverted [itex]L=\dot{q}p-H[/itex] for obtaining a Lagrangian from a given Hamiltonian.

Your question in the body message has not answer without first explaining what do you mean by K and U. Different people defines K and U in different ways, therefore the answer to your question varies.

The question in your title message is different and the answer is yes. Energy is defined as [itex]\left\langle E \right\rangle = Tr \{H \rho\}[/itex] and for classical mechanics this reduces to [itex]\left\langle E \right\rangle = E = H[/itex]

fluidistic

To me it means the kinetic and potential energy respectively.
I disagree with this. Let me quote a passage from the book "Introduction to Theoretical Mechanics" by Robert A.Becker, page 338:

juanrga

I was not asking what mean by the letters K and U, but what do you mean by kinetic energy and potential energy. As said

The definition of energy given is fundamental and the more general possible (I cannot imagine a physical system that does not verifies it).

As stated in previous post the equality E=H is only valid for some systems. Therefore I do not understand your disagreement.

andresordonez

Sorry, I wasn't clear at all with my answer. What I tried to say was that I think (I am not sure of this, you should check the reference I gave you) that the Hamilonian does not always represents the energy of the system. I'm not sure if that was what you were asking at all! But I'm sure it is somehow related to your question.

Regarding the zero level of the energy (the constant added in the Hamiltonian), as far as I know it doesn't really matter where you put it as long as you be consistent throughout your calculations. However, and again I'm not sure of this, in general relativity the absolute energy (instead of the relative energy) is important (See the discussion in I. J. K. Aitchison, Contemp. Phys., 26 (1985), 333.) I haven't read this discussion yet, but it is cited in Quantum Optics, Gerry when discussing the problem of relative vs. absolute energy.

fluidistic

Ah ok I understand now.



I do not see where in your post you say that this equality is valid only for certain system. Rather I see that the answer to the title question, namely "Does the Hamiltonian is always equal to the energy of the system??" is yes, while it isn't so. This is where I disagreed, though I may misunderstand your post once again, in which case I'm sorry.
Also I don't think the OP wants to go outside classical mechanics since he posted this question in the classical physics forums.

Casco

For me K would be [itex]\frac{1}{2}m\dot{x}^2[/itex] kinetical energy and U [itex]U(x)[/itex] potential energy.

I read Goldstein section 2.7 and got a good answer about my question.

It would be good to know for which systems this [itex]\left\langle E \right\rangle \neq H[/itex] is fulfilled. Goldstein says that this is for systems whose lagrangian is a explicit function of time. But it seems that he doesn't go deeper into the problem. Well, I don't know, these are just speculative comments of a undergraduated.

Casco

This doubt(about if the Hamiltonian is always equal to the energy E of the system) arose from one of my lectures on fluids last week, to be more accurate, last monday. My professor gave us a lecture about the ergodic hipothesis this states that,over long periods of time, the time spent by a particle in some region of the phase space of microstates with the same energy is proportional to the volume of this region, i.e., that all accessible microstates are equiprobable over a long period of time(as Wikipedia says). One of the requirements of the system is that its Hamiltonian always has to be equal to the energy.

The question now is, What happens if a system fulfills al the requirements except the one that the Hamiltonian must be equal to the energy E of the system? How do we aproach to that system?

That's the reason of my question, but I have seen that is nothing trivial at all. Of course all of this classically.

juanrga

Sorry my mistake . The response is no because

The Hamiltonian H equals system energy E only in some special cases.

juanrga

Ok, then the response to your other question «Does exist a system or a "type" of systems that their Hamiltonian is not H=K+U?» is yes. There are lots of systems for which its Hamiltonian is not given by H=K+U: General relativity, relativistic particle in an electromagnetic field...

juanrga

Precisely for microcanonical systems the equiprobability of microstates means that [itex]\rho \rightarrow 1/W [/itex] and substituting in the general expression

[tex]\left\langle E \right\rangle = Tr \{H \rho\} = H[/tex]