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Does the Hamiltonian is always equal to the energy of the system? 
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#1
Feb412, 12:33 PM

P: 81

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]PL, 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?? 


#2
Feb412, 12:40 PM

PF Gold
P: 3,188




#3
Feb412, 03:57 PM

P: 68

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. 


#4
Feb412, 04:14 PM

P: 81

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



#5
Feb412, 04:18 PM

P: 476

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] 


#6
Feb412, 04:55 PM

PF Gold
P: 3,188




#7
Feb412, 05:08 PM

P: 476

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


#8
Feb412, 05:12 PM

P: 68

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. 


#9
Feb412, 05:27 PM

PF Gold
P: 3,188

Also I don't think the OP wants to go outside classical mechanics since he posted this question in the classical physics forums. 


#10
Feb412, 05:32 PM

P: 81

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. 


#11
Feb412, 08:34 PM

P: 81

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. 


#12
Feb512, 06:46 AM

P: 476




#13
Feb512, 06:53 AM

P: 476




#14
Feb512, 06:59 AM

P: 476

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


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