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Jan
28
comment The cosmological constant as a Lagrange multiplier?
One might be interested in perusing Appendix X.7 in Zee's Einstein Gravity in a Nutshell which cites arXiv:gr-qc/0505104 and arXiv:0711.3170.
Jan
2
answered Is causality a total order?
Nov
19
answered Can a neutron decay to the gravitons?
Nov
3
revised Simple QFT simulation - how to do it
Improved TeX formatting
Nov
3
suggested approved edit on Simple QFT simulation - how to do it
Oct
27
revised Deriving the Poisson bracket relation of the Ashtekar variables
added 706 characters in body
Oct
22
comment Deriving the Poisson bracket relation of the Ashtekar variables
But look, the only reason you have the symmetry problem is because you arbitrarily introduced it into Eq (3). If you instead rewrite it as $\gamma_{ab} = {e_{a}}^{i}{e_{b}}^{j}\delta_{ij}$ without explicitly symmetrizing the RHS, you're golden. (Your reasoning and counter-example is quite excellent, though.)
Oct
22
comment Deriving the Poisson bracket relation of the Ashtekar variables
no, look at Eq (4), it should read $\delta^{a}_{(c}\delta^{b}_{d)} = \partial\gamma_{cd}/\partial\gamma_{ab}$, which fixes your problem.
Oct
21
comment Deriving the Poisson bracket relation of the Ashtekar variables
You didn't symmetrize the LHS of Eq (4), which leads to the discrepency. If you match the downstairs indices, you end up with (10) and hence (13).
Oct
20
comment Deriving the Poisson bracket relation of the Ashtekar variables
@thenumbernine It's just the product rule: $\delta_{jk}({e_{c}}^{j} (\partial {e_{d}}^{k}/\partial_{ab}) + {e_{d}}^{k} (\partial {e_{c}}^{j}/\partial_{ab}))$, then set it equal to $\delta^{(a}_{c}\delta^{b)}_{d}$, and you're done.
Oct
20
answered Deriving the Poisson bracket relation of the Ashtekar variables
Oct
20
revised Deriving the Poisson bracket relation of the Ashtekar variables
Improved the TeX
Oct
20
suggested approved edit on Deriving the Poisson bracket relation of the Ashtekar variables
Oct
14
revised Lagrangian isn't unique
Improved the TeX formatting
Oct
14
suggested approved edit on Lagrangian isn't unique
Oct
11
answered What is meant by a “c-number”?
Sep
26
comment Is “approximative reduction” general knowledge to physicists?
"Approximation reducation" a term used by philosophers, not real physicists...
Sep
8
comment What happens when you apply the path integral to the Einstein-Hilbert action?
@user1825464 "This unboundedness [of the Einstein-Hilbert action] is caused by the conformal mode of the metric, whose kinetic term enters the kinetic term of (both the Lorentzian and Euclidean action) with the “wrong” sign." Page 12 of arXiv:1203.3591.
Aug
6
comment Absolute time as a common basis of quantum gravity?
@Moonraker I think this preprint may be pertinent for what you are trying to get at.
Aug
6
comment Absolute time as a common basis of quantum gravity?
@ACuriousMind I think what Moonraker is trying to describe is the Hamiltonian density for dust, which does have that intuition.