Zero Hamiltonian and its energies

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
40 replies · 13K views
By the way if we have [tex]\bold H =0[/tex] can this H be split into the terms.

[tex]H^{(0)} (g_{00} , \pi _{00} )+H^{(3)}(g_ {ij} , \pi _ {ij} )=0[/tex]


so we quantizy the term... [tex]H^{(3)}(g_ {ij} , \pi _ {ij})\Phi = E_{n} \Phi[/tex]


and from this we get the energies..:-p :-p
 
Physics news on Phys.org
Karlisbad said:
By the way if we have [tex]\bold H =0[/tex] can this H be split into the terms.

[tex]H^{(0)} (g_{00} , \pi _{00} )+H^{(3)}(g_ {ij} , \pi _ {ij} )=0[/tex]


so we quantizy the term... [tex]H^{(3)}(g_ {ij} , \pi _ {ij})\Phi = E_{n} \Phi[/tex]


and from this we get the energies..:-p :-p
First of all in GR, you have four constraints, second the constraints themselves do only contain the [tex]g_{ij}, \pi_{ij}[/tex] :biggrin:. Moreover, these energies as you call them have no physical meaning whatsoever. :-p
 
By the way appart from constraint (i finally learned how to apply Dirac's method to them using Wikipedia :-p) Does the Einstein Lagrangian has an acceleration term?..i have some troubles to quantize Lagrangians with an acceleration term so:

[tex]L= A(v)^{2}+BV(x)+C(\dot V)^{2}[/tex]
 
Karlisbad said:
By the way appart from constraint (i finally learned how to apply Dirac's method to them using Wikipedia :-p)

Perhaps you should consult a better reference :-p

Karlisbad said:
Does the Einstein Lagrangian has an acceleration term?..

Yes, the Ricci scalar contains second derivatives with respect to time.
 
Last edited:
dextercioby said:
Usually, second order derivative terms in the HE action can be removed by a part integration. See Dirac (1975) for details.

Daniel.
I guess you need to demand in such cases that spacetime is asymptotically static; in either that [tex]\partial_t g_{\mu \nu} = 0[/tex] in the infinite past and infinite future (in some a priori chosen coordinate time t).
 
Last edited:
Of course, in fact Dirac didn't mention this assumption. He just used it to get a better looking Lagrangian and a faster way to derive the field eqns. Actually, a boundary term occurs which can be set to 0, iff your assumption is validated.

Daniel.
 
dextercioby said:
Of course, in fact Dirac didn't mention this assumption. He just used it to get a better looking Lagrangian and a faster way to derive the field eqns. Actually, a boundary term occurs which can be set to 0, iff your assumption is validated.

Daniel.

But that would kill off asymtotically de Sitter universes. I guess you do not want that.
 
And to get the Semi-classical Energies..could we make or use Bohr-Sommerfeld formula?..i mean:

[tex]\oint_{S}d^{4}x \pi _{ab}=\hbar (n+1/2)[/tex]

where S is an hyper-surface, and the "pi's" are the conjugate momenta to the metric g_ab, could we from this expression get the energies? :confused: :confused:
 
Careful said:
But that would kill off asymtotically de Sitter universes.


Which is weird because such solutions are solutions (excuse the tautology) of the field equations one derives by making such an assumption. :rolleyes: :biggrin:

Daniel.
 
dextercioby said:
Which is weird because such solutions are solutions (excuse the tautology) of the field equations one derives by making such an assumption. :rolleyes: :biggrin:

Daniel.

I was merely worried about the fact that de Sitter allows for no global timelike Killing field and that therefore any such construction involves a cosmological horizon associated to one worldline I guess :-p (since the metric gets degenerate in such coordinate systems). Therefore, it seems you cannot apply this trick to get out full de Sitter, that is all I meant (working in special coordinate systems can be pretty dangerous, at least that is how I understood your comment about Dirac's trick).