How Does Loop Quantum Gravity Address These Key Questions?

  • Context: Graduate 
  • Thread starter Thread starter Karlisbad
  • Start date Start date
  • Tags Tags
    Introduction Lqg
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
1 reply · 3K views
Karlisbad
Messages
127
Reaction score
0
MNy questions about the QG are vast..i have some math knowledge about GR and only a bit about Regge calculus and Canonical Quantization...

1) if we have the Hamiltonian constraint [tex]\bold H =0[/tex] then you must have energies (eigenvalues) are all zero¡¡¡ then how does LQG overcome this apparent "paradox"

2) DOes space-time quantization arises from the Boundary conditions of the Wave function of the universe? (as it happened with the quantization of momentum [tex]p=n\hbar/L[/tex] in usual QM for a box of width L

3) HOw the Hell do you solve [tex]\bold H \Phi =0[/tex]?? i suppose that some functional derivatives will appear so it makes it a harder task to recover or obtain the Wave function of Universe...:frown: :frown:

4) GR with Torsion is hard for me to understan (Einstein-Cartan theory) so could we suppose that all spins are "polarized" in the same directions as an approximation to avoid spin-effects?? ( i don't know what has spin to do with GR..could someone explain)

5) By the way if we make a Wick rotation so the propagator becomes just "some kind of partition function" does the approach:

[tex]\sum_{n}e^{-E(n)/\hbar}\sim\int_{V}D[\Phi]e^{iS_{E-H}/\hbar}[/tex]

makes sense?..(we consider if the partition function is still the "trace" of the operator [tex]exp{-H/\hbar}[/tex] where H would be the Hamiltonian constraint.
 
Physics news on Phys.org
Karlisbad said:
MNy questions about the QG are vast..i have some math knowledge about GR and only a bit about Regge calculus and Canonical Quantization...

1) if we have the Hamiltonian constraint [tex]\bold H =0[/tex] then you must have energies (eigenvalues) are all zero¡¡¡ then how does LQG overcome this apparent "paradox"

I can answer this question. If you formulate your theory correctly, such that "time" is present inside your Hamiltonian constraint. You won't run into this paradox. Your hamiltonian constraint will encode the dynamics of the theory.

It's best to illustrate with a simple example. Toy model (the following will be included in my thesis, where it is explained clearer.)

[tex]S = \int(\frac{1}{2}m\dot{q}^2N^{-1}(\tau) - V(q)N(\tau)d\tau[/tex]

This system is invariant under time-reparametrization
[tex]\tau = f(\tau')[/tex]
[tex]N'(\tau') = N(\tau)f'(\tau')[/tex]

If we interpret [tex]N(\tau)[/tex] as another variable, we have two first class constraints:

[tex]p_N = 0[/tex]
[tex]H = p^2/2m + V = 0[/tex]

We run into trouble, but this can be remedied by adding a term into the action [tex]EN(\tau)[/tex] (cosmological term)

then we have correctly [tex]H = p^2/2m + V = E[/tex]
but the dynamics of the theory are lost. Lee Smolin took this as "the problem of time" in his lectures in PI. the part he explained a simple toy model where [tex]L = \sqrt{T/V}[/tex] or something like that.

If we interpret [tex]N(\tau)[/tex] as the velocity of a "hidden variable" the true time, that is [tex]N(\tau) = \dot{t}[/tex], we have only one first class constraint:

[tex]H = p_t + p^2/2m + V = 0[/tex]

when you quantize the theory, it becomes the time-independent Schrödinger equation.

[tex]p_t \rightarrow -i\hbar\frac{\partial}{\partial t}[/tex]
[tex]p \rightarrow -i\hbar\frac{\partial}{\partial q}[/tex]

whereas the previous constraint becomes only the time-dependent Schrödinger equation. You see if we interpret the variable wrongly we will run into paradox.

part of my research is try to implement this view on ADM's formalism. you will see it when I publish my thesis. In ADM there were efforts to locate the time variable inside the Wheeler-DeWitt operator, and DeWitt actually showed that there is a Lorentz signature in it showing "time" does exist inside. But in LQG, this nice feature is lost when we use Spin Networks.
 
Last edited: