Hamiltonian does NOT equal energy?

  • #1
nonequilibrium
1,437
2
Hello,

Just to be sure: is the following correct?

Imagine a long rod rotating at a constant angular speed (driven by a little motor). Now say there's a small ring on the rod that can move on the rod without friction. The ring is then held onto the rod by an ideal binding force (I don't know the correct term in English, a force that doesn't deliver virtual work). This ring, free to move along the rod, can be described using the Lagrangian formalism.

Now is it true that the binding forces are non-holonomic? And this is the reason why [tex]H \neq E[/tex]? It seems like it anyway, but I remember reading a thread on here a while ago looking for simple systems for which [tex]H \neq E[/tex] while all examples given were very mathematical and far-fetched in a physical sense, so this makes me a bit unsure about what I've written above, hence the double-check.
 

Answers and Replies

  • #2
nonequilibrium
1,437
2
Oops, important EDIT:

"Now is it true that the binding forces are non-holonomic?"

I meant non-SCLERONOMIC, i.e. rheonomic.
 
  • #3
facenian
436
25
Hello,

Just to be sure: is the following correct?

Now is it true that the binding forces are non-holonomic? And this is the reason why [tex]H \neq E[/tex]?

Yes,the constraint forces are rheonomic and this is whry [itex]H\neq E=T+V[/itex]
 

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