Timeline for Deriving the Poisson bracket relation of the Ashtekar variables
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| Jun 7, 2021 at 13:05 | answer | added | Jeremy Miller | timeline score: 1 | |
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| Oct 25, 2015 at 10:36 | history | edited | thenumbernine | CC BY-SA 3.0 |
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| Oct 24, 2015 at 0:23 | history | edited | thenumbernine | CC BY-SA 3.0 |
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| Oct 23, 2015 at 23:52 | history | edited | thenumbernine | CC BY-SA 3.0 |
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| Oct 23, 2015 at 20:08 | history | edited | thenumbernine | CC BY-SA 3.0 |
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| Oct 23, 2015 at 10:02 | comment | added | thenumbernine | Thanks much. I still don't think this is the correct conclusion, and I posted a counterexample to the assumption $e\delta e$ is symmetric and to the subsequently derived definition ${\partial\over\partial\gamma_{ab}} {e_c}^i = {1\over2} \delta_c^a e^{bi}$ | |
| Oct 23, 2015 at 9:51 | history | edited | thenumbernine | CC BY-SA 3.0 |
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| Oct 22, 2015 at 15:47 | comment | added | Alex Nelson | 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, 2015 at 12:51 | comment | added | thenumbernine | Even if you symmetrize the $cd$ on the deltas, it doesn't help. Without it we're showing a matrix equals a symmetrized matrix. With it we're showing a symmetrized matrix equals a symmetrized matrix. What you need to solve this is to show one matrix (not just the symmetric portion) equals another matrix (not just the symmetric portion). Just because $A^T + A = B^T + B$ doesn't mean $A = B$. And even with $A = B^T + B$, you can't get an exact solution for $B$. | |
| Oct 22, 2015 at 2:34 | comment | added | Alex Nelson | 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 22, 2015 at 0:42 | comment | added | thenumbernine | Yes the symmetry shows that $e \cdot e$ is symmetric, because this equals $\gamma$. But it says nothing about $\delta e \cdot e$. | |
| Oct 21, 2015 at 20:17 | comment | added | Alex Nelson | 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 21, 2015 at 11:50 | history | edited | thenumbernine | CC BY-SA 3.0 |
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| Oct 20, 2015 at 18:42 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
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| Oct 20, 2015 at 15:32 | answer | added | Alex Nelson | timeline score: 0 | |
| S Oct 20, 2015 at 15:24 | history | suggested | Alex Nelson | CC BY-SA 3.0 |
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| Oct 20, 2015 at 8:07 | history | edited | thenumbernine |
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| Oct 19, 2015 at 22:10 | history | asked | thenumbernine | CC BY-SA 3.0 |